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This handbook is unusual in that it combines in a single volume formulas and tables from both elementary and advanced mathematics. For example, topics treated range from those in algebra, geometry, trigonometry, analytic geometry and calculus to Fourier series, Laplace and Fourier transforms, Bessel and Legendre functions and many other advanced special functions. Such topics are needed by both students and research workers in the fields of engineering, physics, mathematics and other sciences.
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From the Back Cover:
Students love Schaum's Outlines!Learn key formulas and tables
Study quickly and more effectively
Schaum's Outlines give you the information teachers expect you to know in a handy and succinct formatwithout overwhelming you with unnecessary detail. You get a complete overview of the subject, plus professionalexams!
Inside, you will find:
More than 2400 formulas and tables
Clear and concise explanations of all results
Formulas and tables for elementary to advanced topics
Complete index to all topics
If you want top grades and easy-to-use information for your math and science courses, this powerful study tool and reference is the best guide you can have!
About the Author:
Murray Speigel, Ph.D., was Former Professor and Chairman of the Mathematics Department at Rensselaer Polytechnic Institute, Hartford Graduate Center. | 677.169 | 1 |
Appropriate for undergraduate and select graduate courses in the history of mathematics, and in the history of science. This edited volume of readings contains more than 130 selections from eminent mathematicians from A 'h-mose'to Hilbert and Noether. The chapter introductions comprise a concise history of mathematics based on critical textual analysis and the latest scholarship. Each reading is preceded by a substantial biography of its author.Calinger, Ronald is the author of 'Classics of Mathematics' with ISBN 9780023183423 and ISBN 002318342 | 677.169 | 1 |
Basic Mathematics for Technical Programs
An easy-to-use mathematics book designed along a practical, hands-on format that provides a unified approach to problem-solving by linking topics and their corresponding computational techniques. Mathematics for Technical Programs devotes special attention to mental activities involving elementary computations, fraction-decimal-percent conversions, and estimations to aid reader in achieving mathematical literacy. It assumes the use of scientific calculators to enable the reader to focus on concepts and problem-solving skills rather than time-consuming computations. The book also provides a unified approach to problem-solving not found with traditional arithmetic methods. A valuable book for any reader who wishes to improve his "mathematical literacy."
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From the Publisher:
Designed as an easy-to-use text within a practical, hands-on workbook format that provides a unified approach to problem-solving by linking topics and computational techniques. | 677.169 | 1 |
9780198564386 a broad ranging introductory text on applied numerical modelling and simulation. It covers many techniques and develops the relevant mathematics in parallel. The material is presented in the context of realistic practical examples. Many of the end-of-chapter exercises may be treated as open-ended projects which can be easily extended. The presentation makes the text easy to assimilate with a progression of concepts chapter by chpater. The more difficult material is assembled towards the end of the book. Each chapter has a list of references to accessible texts as well as journal publications which should provide pointers for research projects. The provision of software code in a range of languages should help the user to make rapid use of the material. This textbook will broaden the understanding of engineers and others using models leading to better and more productive modelling. This book is intended for some 2nd year, but mostly 3rd year undergraduates, and graduate students in engineering (particularly electrical, electronic, control) and applied sciences (physics, biology, environmental) wanting an understandable account of numerical modelling. Specifically "advanced mathematics for engineers" courses | 677.169 | 1 |
Differential Geometry and its Applications (Mathematics and
Since the late 19th century, differential geometry has grown into a field concerned more generally with the geometric structures on differentiable manifolds. Questions of a more dynamical flavor as well as questions pertaining to subriemannian geometry may also be discussed. The simplest results are those in the differential geometry of curves and differential geometry of surfaces. For example, it is useful in relativity where space-time cannot naturally be taken as extrinsic (what would be "outside" of it?).
Pages: 382
Publisher: Springer; 1987 edition (July 31, 1987)
ISBN: 9027724873
Differential Geometric Methods in Mathematical Physics: Proceedings of the 14th International Conference held in Salamanca, Spain, June 24 - 29, 1985 (Lecture Notes in Mathematics)
A Treatise on the Differential Geometry of Curves and Surfaces (Dover Books on Mathematics)
The Principle of Least Action in Geometry and Dynamics (Lecture Notes in Mathematics)
A New Construction of Homogeneous Quaternionic Manifolds and Related Geometric Structures (Memoirs of the American Mathematical Society)
The Mystery Of Space: A Study Of The Hyperspace Movement In The Light Of The Evolution Of New Psychic Faculties (1919)
The honours information session: Thursday 15 September, Carslaw 350. Come to learn about doing Honours in the School of Mathematics and Statistics, meet current honours students and talk to potential supervisors. The course information sheet can be found here Differential Geometric Methods read here We can also have a surface specified by parametric equations, where however we need two parameters, say u and u unlike for a curve, where we need only one formx =f(u, ), y =g(u, ), z =h(u, ) v v v. To see that this represent a surface, we take a R in the u-u plane , e.g. Conformal Mapping read online read online. Rating is available when the video has been rented. The first lecture of a beginner's course on Differential Geometry! Given by Assoc Prof N J Wildberger of the School of Mathematics and Statistics at UNSW. , e.g. Analysis and Geometry of read pdf read pdf. It can also make a good party game (for adults too). Home-based Canadian business specializing in the production and sale of wire disentanglement puzzles Tensor Calculus and Analytical Dynamics (Engineering Mathematics) download here. The subjects with strong representation at Cornell are symplectic geometry, Lie theory, and geometric analysis. Symplectic geometry is a branch of differential geometry and differential topology that has its origins in the Hamiltonian formulation of classical mechanics. Geometric analysis is a mathematical discipline at the interface of differential geometry and differential equations Elements of Differential read online tiny-themovie.com. Even though phi'>phi for a given point, small enough values of delta phi' still correspond to small values of delta phi. Because your coordinate transformation would then only be valid to any decent amount in the region around which sin(theta) ~ 1, ie the equator. At the poles sin(theta)~0 and so your coordinate change is invalid Boundary Constructions for CR Manifolds and Fefferman Spaces (Berichte aus der Mathematik) download epub.
This is a finite volume space, that is connected up in a very specific way, but which is everywhere flat, just like the infinite example. In two dimensions, one might visualize it as Of course, I could have only made one or two directions into circles (leaving it still infinite in some directions), or made the space into a finite one with more than one hole, or any number of other possibilities , cited: Elementary Topics in Differential Geometry tiny-themovie.com. Curve as a subset of R^3 is the image of the parametrization mapping download. At every point of the manifold, there is the tangent space at that point, which consists of every possible velocity (direction and magnitude) with which it is possible to travel away from this point. For an n-dimensional manifold, the tangent space at any point is an n-dimensional vector space, or in other words a copy of Rn , source: Structure of Dynamical download online What Assignment Expert is ready to offer for your differential geometry homework: professionalism in every assignment completed; commitment to providing excellent differential geometry homework solutions to every customer; easy-to-understand tips for all your differential geometry homework tasks; your full satisfaction with the completed differential geometry homework , e.g. Geometry, Topology and Quantum Field Theory (Fundamental Theories of Physics) Geometry, Topology and Quantum Field. The Search for Higher Helicities — VIGRE Colloquium, University of Georgia, Apr. 6, 2010. Poincaré Duality Angles on Riemannian Manifolds With Boundary — Geometry Seminar, University of Rochester, Mar. 4, 2010. Legendrian Contact Homology and Nondestabilizability — Geometry–Topology Seminar, University of Pennsylvania, Dec. 10, 2009 Complex Differential Geometry (AMS/IP Studies in Advanced Mathematics, 18) The differential geometry o surfaces captures mony o the key ideas an techniques characteristic o this field. Differential geometry is a branch of mathematics that applies differential and integral calculus to planes, space curves, surfaces in three-dimensional space, and geometric structures on differentiable manifolds Lectures on Seiberg-Witten read pdf tiny-themovie.com.
Geometry and Algebra of Multidimensional Three-Webs (Mathematics and its Applications)
Smarandache Geometries & Maps Theory with Applications (I)
Gauge Theory and Variational Principles (Dover Books on Mathematics)
8 Thanks to Nana Arizumi for the first two pictures! 9/22/08: A new picture has been added to the Gallery. 9/26/08: The first midterm will be on Wednesday October 8, 6pm-7:30pm (venue to be announced) pdf. The joint mathematics meetings in New Orleans, LA, January 6-9, 2011, includes invited talks by Denis Auroux and Akshay Venkatesh, and special sessions on Geometric Group Theory, Groups, Geometry, and Applications, Knot Theory, and Knots, Links, 3-Manifolds, and Physics. The meeting in Statesboro, GA, March 12-13, 2011, includes invited talks by Jason A Behrstock and Gordana Matic, and special sessions on Categorical Topology, Geometric Group Theory and Low Dimensional Topology and Contact and Symplectic Geometry Singularity Theory and Gravitational Lensing (Progress in Mathematical Physics) read pdf. Topology is a structure or a framework between the elements that can be found on a complex(e.g. a 2D-surface. It is no doubt that the complex's skeleton is a set of elements too(e.g. vertex, edge, face). I always keep in mind that Topology is a studying of neighborhood for Geometry. This was what I knew during very beginning. Then construction of spaces, manifold...etc are more advanced topic Geometry and Dynamics of Groups and Spaces: In Memory of Alexander Reznikov (Progress in Mathematics) Geometry and Dynamics of Groups and. The above-mentioned space curvature is obtained analogously: If the basis vector in the mathematically positive sense ( counterclockwise ) until an infinitesimal distance in direction and then moves an infinitesimal distance in - direction, we obtain a result which we can write in the form ref.: Fuchsian Reduction: download here DG Library is powered by Maple WWW - a free web application that brings interactive Maple worksheets to the web. Maple WWW brings interactive graphics for objects in the library. All plots can be moved, rotated or zoomed. All documents can be downloaded as Maple worksheets Projective differential read epub Projective differential geometry of. The choice of themes is somewhat limited, with no mention of manifolds (which are explained in a companion book) , e.g. Perspectives of Complex Analysis, Differential Geometry and Mathematical Physics: Proceedings of the 5th International Workshop on Complex Structures ... St. Konstantin, Bulgaria, 3-9 September 2000 Differential geometry deals with metrical notions on manifolds, while differential topology deals with nonmetrical notions of manifolds , source: The Schwarz Lemma (Dover Books on Mathematics) The asymptotic lines on a ruled surface: 2 0____(1) Ldu Mdud Nd u u + + = But N=0, for a ruled surface. Hence, equation (1) is given by du=0 i.e., u= constant. Hence, one family of asymptotic lines is the family of generators of the ruled surface , cited: Introduction to the $h$-Principle (Graduate Studies in Mathematics, V 48) Differential geometry is a mathematical discipline that uses the techniques of differential calculus and integral calculus, as well as linear algebra and multilinear algebra, to study problems in geometry. The theory of plane and space curves and of surfaces in the three-dimensional Euclidean space formed the basis for development of differential geometry during the 18th century and the 19th century online. The group also studies geometric and topological aspects of quantum field theory, string theory, and M-theory. This includes orientations with respect to generalized cohomology theories, and corresponding description via higher geometric, topological, and categorical notions of bundles , source: Minimal Surfaces in R 3 read for free tiny-themovie.com. | 677.169 | 1 |
This
· Bring awe and wonder with a chapter opener that puts the math in context · Provide rigorous math practice with hundreds of high quality questions · Focus on literacy skills with key words per topic · Reason, interpret and communicate mathematically with flagged mathematical reasoning questions (labeled MR and CM) · Solve problems within mathematics and in other contexts with flagged problem-solving questions (labeled PS and EV) · Show step by step working through worked examples (numbered sequentially throughout the chapter) · Differentiate at the top end with more challenging questions · Access answers online, in the Teacher's Pack and in the Collins Connect edition · Endorsed by Edexcel
"synopsis" may belong to another edition of this title.
About the Author:
Michael Kent has 20 years' experience teaching Pure Maths, Mechanics and Statistics for AS and A Level Mathematics and Further Mathematics, twelve of them also leading a successful department. He co-authored the most recent editions of Collins' GCSE Maths Student Books.
Book Description Collins Educational 2015-01205911. Paperback. Book Condition: new. BRAND NEW, Edexcel GCSE Maths Higher Student Book (4th Revised edition), Kevin Evans, Keith Gordon, Brian Speed, Michael Kent,'s Pack and in the Collins Connect edition * Endorsed by Edexcel. Bookseller Inventory # B9780008113810 | 677.169 | 1 |
Listing Detail Tabs
Maths Quest 8 for NSW Australian Curriculum Edition & eBookPLUS has been written specifically to meet the requirements of the NSW Australian Curriculum. The Mathematics Syllabus focuses on students Working Mathematically in the strands of Fluency, Understanding, Reasoning, Problem-Solving and Communicating. KEY FEATURES * Problem Solving chapters to provide students with opportunities to apply their knowledge out of the context of defined topics * NAPLAN practice questions * Links to assessON, our innovative online tool that provides additional assessment resources for your mathematics course and enables online assessments FOR, AS and OF learning and automatic feedback * Links to SpyClass for Maths Quest 8, a captivating online comic-style game where students complete a range of missions * Individual pathways: All exercises within the text are carefully graded so that students get to work at their own level. At the start of each exercise a guide to Individual pathways is provided. Individual Pathway Activities are available on the eBookPLUS to provide additional opportunities for students to engage with relevant mathematical concepts at their own level. * A language section to provide students with opportunities to use mathematical language in context and to enhance their literacy skills * A rich task/Investigation (Communication) to provide students with opportunities to explore and apply mathematical ideas beyond the level of skill * Chapter Challenges to provide students with opportunities to think beyond the obvious and the concrete * Puzzles to engage students and provide them with opportunities to learn through fun activities Maths Quest 8 for NSW Australian | 677.169 | 1 |
Вища математика: Навчальний посібник (англійською мовою)
Страницы работы
Содержание работы
2. The definitions of
the Determinants of the second and third orders.
3.Properties of determinants.
4.Expansion of the determinant along the row
or column.
5. Cramer's rule.
6. The Matrix
Method.
Let's
begin from the following simple economic problem (all data are in the table):
Type of raw material
Consumption for unit of production
Raw materials consumption
№1
№2
№3
A
2
3
6
120
B
1
4
2
280
C
7
5
3
240
1. Definition. A matrix
is a rectangular array of elements. These elements are usually called entries.
Matrices are
usually denoted by capital letters. For example,
- 2x3 matrix (two by three)
The common case:
(you're
already familiar withthe notation of the elements with two subscripts. For instance, represents the entry at
the second row and first column of a matrix A)
In common case the expression means
the element in the i-th row and j-th column of the matrix A.
Pass to the different types of matrices:
1) A matrix with
one row is called a row vector, and a matrix with one column is called a
column vector.
; .
2) A square
matrix is a matrix with the same number of rows and columns.
For example, it is
the square matrix of the second order:
the
main diagonal
the second diagonal
3) The identity
matrixIn of size n is the n-by-n
matrix in which all the elements on the main diagonal are equal to1 and all
other elements are equal to 0.
It is called
identity matrix because multiplication with it leaves a matrix unchanged:
MIn = ImM = M
(for any m-by-n
matrix M).
4) A square matrix
A of order n is called invertible if there exists a matrix B such
that
AB=BA=In.
Moreover, if B
exists, it is called the inverse matrix of A, denoted A-1.
5) the transpose
of a matrix A is another matrix AT created
(formed) by turning the rows of A into the columns (it is the
result of interchanging the rows and columns of the matrix)
Operations
onmatrices:
1)
matrices of the same size can be added or subtracted element by
element:
2) the
scalar multiplication of a matrix A and a number c is
given by multiplying every entry of A by c:
3) the rule for matrix multiplication is
more complicated
Two matrices can
be multiplied only when the number of columns in the first equals the number of
rows in the second.
If A is an m-by-n
matrix and B is n-by-k matrix, then their matrix
productAB is the m-by-k matrix whose entries are
given by dot product of the corresponding row of A and the corresponding
column of B:
Remark. The products AB and BA need not be equal (even
if both products are defined). So, the following inequality holds: AB ≠ BA.
but
2. Suppose we are given the square matrix of the
second order: . Then we define the determinant
of A to be the scalar . It is
denoted by Δ or detA or or .
The determinant of the third-order matrix is defined by the formula:
(it is the triangle rule)
(multiply elements located on the main diagonal, multiply elements located at the vertices of a triangle, one side of which is parallel to the main diagonal)
The scheme of the
triangle rule:
3.
The main propertiesof determinants
(We
will illustrate the properties with a 2x2 or 3x3 examples)
1) A matrix and its
transpose have equal determinants.
2) If two rows of
matrix are interchanged, the determinant changes sign.
3) If two rows of a
matrix are equal, the determinant is zero.
4) The common factor
of the row of the determinant can be taken outside this
determinant.
5) If the
corresponding elements of two rows of the determinant are
proportional, then the determinant is zero.
6) If a row of a
matrix is zero, then the value of the determinants is zero.
7)
8) If
a multiple of a row is added to another row, the value of the determinant is
unchanged.
Illustrate it with a
3x3 example:
4. Let's consider the determinant
of the third order:
Let's delete the first row and the first column of this
determinant. We get the determinant of the second order, which is called the
minor of the element a11. It is denoted by M11.
Similarly we can
introduce the minors of the other elements.
Definition.
The cofactor of the element equals the minor of this element with the corresponding sign from the chess-board pattern of signs:
Theorem (Expansion along the row or column). The
determinant equals the sum of products of the some row elements by its
cofactors. The same holds for columns. | 677.169 | 1 |
Instructors
Algebra I Core
9 STUDENTS ENROLLED
In this course, students explore the tools of algebra. Students learn to identify the structure and properties of the real number system; complete operations with integers and other rational numbers; work with square roots and irrational numbers; graph linear equations; solve linear equations and inequalities in one variable; solve systems of linear equations; use ratios, proportions, and percentages to solve problems; use algebraic applications in geometry, including the Pythagorean theorem and formulas for measuring area and vol-ume; complete an introduction to polynomials; and understand logic and reasoning. | 677.169 | 1 |
Description
Speech2Math Calculator translates your speech to mathematical expressions. Speech2Math Calculator is an useful talking voice calculator application, which allows you to calculate easily by speaking. Speech2Math Calculator also has on-screen keyboard which allows you to edit expressions. Plus, minus, times, divide operations, square, cube square root, power, factorial operations are available. You can select the input speech language independent from your default device locale and language. Trigonometric functions such as sinus, cosine, tangent, arcsines, arccosine, arctangent and logarithmic functions are available in PRO version. Speech to Math Calculator is the only voice calculator which supports all the following languages and TTS (text to speech). | 677.169 | 1 |
In the Classroom: Exponentials, Logs, and Trig
As we explore new mathematical tools, we're balancing understanding with practice through Worksheets and Workouts.
Exponents, logarithms, and trigonometry are typically found in the curriculum after algebra and geometry, but before calculus. This sometimes gets them lumped into a class with the name "precalculus." And certainly their place before calculus gives us a rich variety of non-polynomial functions to explore with calculus tools. But these topics are a beautiful area of mathematics in their own right, and it is a pleasure to have the opportunity to explore them with my students this block.
There is a delicate balance that needs to be struck in this class. On the one hand, it takes practice to learn to work with trigonometric and logarithmic expressions quickly and accurately, and this is definitely a skill worth developing. On the other hand, it's easy to get lost in the algebra and lose sight of the mathematical reality that our equations are trying to describe. To maintain this balance, our in-class assignments come in two varieties: Worksheets and Workouts.
On Worksheets, students are expected to explore and learn about new mathematical ideas, and expand their mathematical worldview. On Workouts, students build their mathematical muscles by working through exercises of gradually increasing difficulty. Usually, we'll do at least one Workout before diving into the major Worksheet for the day. This way, new concepts get revisited and become familiar over time. And even three weeks in, we've already seen a lot of new concepts!
We began the block by recalling facts about how exponents behave. Starting with our intuitive definition of a^n as "n copies of a multiplied together," we were able to develop a consistent definition for negative and fractional exponents. We puzzled over what an irrational exponent would look like, and saw our first example of a limit.
We then introduced the logarithm into our conversation. We know that the equation 2^x=8 has a solution: x=3. The equation 2^x=9 should have a solution as well. It's a number we can't solve for using any methods from their earlier algebra classes, but we recognize that it's a real number, it's between 3 and 4, and it's much closer to 3 than it is to 4. Since exponential equations are so important in mathematics, we need to have a name for this number. So we call it log_2(9), the power we need to raise 2 to in order to get the number 9.
This is a consistent theme in this class. We have a specific number that we would like to find, but our existing algebraic tools don't give us an exact value for the number, so we give our number a name. Then we have a new tool for studying numbers of this kind. A lot of the work of this class is developing an intuitive understanding of these new tools. Students are encouraged to give exact values when they solve problems, and reminded that sometimes exact answers may not be pretty answers. At the same time, we are working to develop enough number sense to sanity-check our answers. If we start with 1000g of radioactive material, and it decreases by 10% each year, we should be suspicious if we find that half of it is gone after 2.06712485 years—surely it should take at least five years to get down to half.
This week we began our exploration of trigonometry through the unit circle. Working with ten-inch embroidery hoops and special measuring tapes, we were able to find radian angle measures by stretching the measuring tapes around the outside of the circles and marking the end points of these arcs. We need tools for relating the arc lengths around the outside of the circle with the x and y values of the corresponding points. Since we can't find an exact value for these numbers (except for a few very special cases that we'll be talking about next week) we invent new functions called cosine and sine to describe the x and y values, respectively.
Using our embroidery hoops, we were able to find surprisingly accurate decimal approximations for the sine and cosine of various values. After that, we had a Workout on sine and cosine number sense. In one part of this worksheet, students were supposed to look at three trigonometric expressions, and figure out which two of them represented the same number. For example, we don't know exact values for sin(2), sin(-2), or -sin(2), but if we think carefully about their placement on the unit circle, we can figure out which two of them are equal to each other.
We have a lot more ideas to see in the remainder of the block. We'll be talking about applications of trigonometric functions to geometry. We'll introduce the tangent function, and explore some trigonometric identities. We'll talk about the importance of trigonometric functions in modeling periodic behavior. This block is packed with some really beautiful mathematics, and I am looking forward to watching our students continue to learn, explore, and grow. | 677.169 | 1 |
Calculus
Designed for Flipped Classroom
Teachers: why not flip your class?
Flipping the classroom means assigning the "lesson/lecture" for students to work-on at their own pace,
as homework, thereby leaving time for discussons, exercises, investigations and overall better communication and reflection in class.
Short video tutorials and worked examples are watched and worked through by each student in their preferred environoment and on their
favorite devices (desktop, laptop, tablet, phone), allowing them to come to class prepared and with any questions they may have.
This site was designed for flipped classroom.
As suggested by educause.edu:
The flipped model puts more of the responsibility for learning on the shoulders of students while giving them greater impetus to experiment.
EduCause.edu
The following video explains and highlights the true advantages of the flipped classroom model, we strongly encourage all of our fellow teachers to watch it.
Ideal Study Companion
Students: this is ideal for studying for your next test/quiz.
Every topic is taught and explained in a clear and concise manner.
Key formula and definitions are reminded, explained and illustrated in short tutorials.
There are plenty of exercises for you to practice with the choice between seeing a simple answer key or the solutions with all of the working shown in detail.
Formula & Tutorials
Each topic is treated in a clear and concise manner, meaning: "short and sweet".
The key formula formula is given and its use is illustrated with a short video tutorial.
Typically the top of each page looks like follows:
Notice the following:
In the center we're given the must-know formula.
In the upper left hand corner we'll find all the the topics related to the one we're currently studying.
We can change chapters altogether via the top navigation bar.
Tutorials
Every topic on the site is explained with video tutorials. Tutorials allow us to learn at our own pace. Unlike in a clas environment, we can watch, re-watch, re-re-watch, ... tutorials to make sure we fully understand what is being said.
Worked Examples
To consolidate what we learn, nothing beats working through some worked examples.
Each page has exercises for which we have the choice between seeing a quick "answer key" or the solutions with all of the working.
Just click on either of the two options to choose.
The following two pictures give us an idea of how the solutions with, and without, working are presented:
Solutions Withhout Working:
Solutions With Working:
News & Annoucements
Stay informed of the latest notes and tutorials we add to your site.
Follow us on | 677.169 | 1 |
Seven Best Math Apps For Students
Learning math is challenging for many students. The technicalities and the need for critical thinking in math means that teachers have to employ different strategies for students to pass. However, the teacher's efforts are never enough for you to pass the exams especially in competitive schools and mathematical areas.
To pass, you will find yourself looking for a tutor and also finding ways of learning by yourself. Thankfully, technology is always there at your time of need and you can use any of the following apps to better understand the complex math problems. The apps include:
iMathPage
Though limited to IOS software systems and smartphones, this app will give you the chance to lower your struggle with math. This app works just like your tutor by giving you detailed explanations for all the problems you pose. The clear explanations make it easier for you to understand the concepts.
IMathematics Pro
This app runs on Android and iOS gadgets. Once you download it, you will get access to over 1000 formulas applicable in over 120 topics. It also has calculators and 7 solvers that run on an attractive interface.
Equations All-in-One
Math, Physics, and Chemistry problems are easily solved by this app. Equations All-in-One app solves more than 130 problems in any of the above subjects. Whether you are in high school or at the university, this app will make it possible for you to understand and solve any complex questions presented to you.
The formula used for a problem allows you to solve variables in equations. This also has a unit converter for major Physics units. It runs on iOS.
Free Graphing Calculator or Graphing Calculator by Mathlab
These apps run on iOS and Android respectively. Students studying advanced mathematics need graphing calculators, which can be quite expensive and complex. This app has an intuitive user interface and provides users with advanced mathematical functions. Intersections, slopes, and roots on graphs are beautifully laid on this app.
MathRef
This app runs on Android and iOS operating systems. It helps students to access multiple formulas easily. Students talking Geometry, Algebra, or Calculus classes are more likely to benefit from this app. The user interface makes it possible for students to add notes to the equations, save favorites, and add brief descriptions to help understand how the app gave the answer.
Wolfram Alpha
Understanding complex math is made easier by this app. Number-centric and mat related problems get comprehensive answers and anything across 29 disciplines can be solved. The app gives graphical representations, formula details, and brief explanations to problems and given solutions.
Math Solver
This is the best app for solving mathematical equations. Besides giving solutions to the problems you present, it gives step by step details of how the solution was arrived at. Linear and quadratic equations are solved by the app. Radical and literal equations, graph equations, and factors are solved using this app as well.
In conclusion, though mathematical concepts are challenging for you and many other students, the apps highlighted above and many others will help in making math easier.
Author Bio
Bradley Patterson is a math tuition teacher. His experience with students using the apps above shows that there is a significant improvement in performance in the students using the app. Check out his blog for more apps and mathematical resources for students | 677.169 | 1 |
Mathematics
"Great minds discuss Mathematics. Maths is like love; a simple idea, but it can get complicated." Jonny Heeley, Mathematics Tutor
#CreateYourself
So, you are considering taking an A Level in Mathematics? That's great! You are not alone. You might be more than a little surprised to read that, back in 2014, A level Mathematics overtook even English to become the single most popular A level subject examined in that year (in excess of 103,000 students)!
About The Course (AS & A-Level)
I'm picking Creative Subject, so why choose Mathematics?
It is a stimulating and intellectually challenging discipline; not only that, it is a powerful and versatile subject that underpins (and is occasionally vital to) a multitude of other subjects. The course aims to build on work you will have already met at GCSE, but it also involves new and powerful ideas and concepts.
However, it is certainly NOT for the faint-hearted! Studying Mathematics is an active process; just reading around the subject and keeping a few notes is quite simply not enough! You have to roll up your sleeves and get 'down and dirty' with the likes of numerical problems, algebra, trigonometry and geometry.
A competence in Mathematics is frequently highly desirable to universities, colleges or employers alike. You may like to know that for individuals considering a future in some of today's most financially rewarding careers, confidence in using Mathematics on a daily basis is an essential prerequisite. Furthermore, researchers at the London School of Economics have recently established that people who have studied Mathematics at this level can expect to earn up to 11% more than colleagues, even those doing the same job!
For careers where Mathematics isn't an absolute requirement, other Mathematics skills learned at AS and A level, such as logical thinking, problem solving and statistical analysis, are often highly valued in the workplace.
By the end of the course, you will be expected to be able to:
Use mathematical skills and knowledge to solve problems.
Solve problems by using mathematical arguments and logic.
You should understand and demonstrate what is meant by proof in Mathematics.
Simplify real-life situations so that you can use Mathematics to show what is happening and what might happen in different circumstances.
Use the Mathematics that you learn to solve problems that are given to you in a real-life context.
Use calculator technology and other resources (such as formulae booklets or statistical tables) effectively and appropriately; understand calculator limitations including when it is inappropriate to use such technology.
Currently, every student must study several 'core' components; split between Y1 and Y2. The content of the core units involve some of the more abstract mathematical concepts (calculus, trigonometry, algebra and functions, co-ordinate geometry, vectors, sequences and series, to name but a few). The core content provides a good set of mathematical scaffolds for the topics studied in the applied units.
Each year, in addition to the core content, you would need to choose one 'applied' pathway from mechanics, statistics, or decision Mathematics pathway. The applied pathway you choose will depend on your interests, the other A level courses you intend to study, and ultimately your career aspirations (if you know what they are yet). You are advised to speak with your current Mathematics teacher, your careers adviser and studio school Mathematics staff to establish which applied pathway is best for you.
Here are some details of the applied content to help you with those discussions:
Mechanics
Mechanics is concerned with many everyday situations, e.g. the motion of cars, the flight of a ball through the air and the motion of the earth around the sun. Such problems have to be simplified or modelled to make them capable of solution using relatively simple Mathematics. Many of the ideas you will meet in the course form an almost essential introduction to such important modern fields of study such as cybernetics, robotics, bio-mechanics and sports science.
Statistics
You will learn how to analyse and summarise numerical data in order to arrive at conclusions about it. Many of the ideas in this part of the course have value in a range of other fields (biology, environment, psychology, business studies for instance). In today's society we are bombarded with information (or data) and the statistics units will give you useful tools for looking at this information critically and efficiently.
Decision Mathematics
You will learn how to solve problems involving networks, systems, planning and resource allocation. You will study a range of methods, or algorithms, which enable such problems to be tackled.
Specific Entry Requirements:
Students applying must have a minimum of 5 GCSEs at C grades with English & Maths. Their Maths grade must be at least a grade B.
As by | 677.169 | 1 |
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Learners require to assign a very good deal of time when hunting for math workouts on the Internet. Most of the math internet websites are person-helpful and make it possible for you to have print worksheets made up of hundreds of workouts for exercise. | 677.169 | 1 |
ISBN 13: 9780131425118
Elementary Algebra for College Students: PH Grade Assistant
This dynamic new edition of this proven series adds cutting edge print and media resources. An emphasis on the practical applications of algebra motivates learners and encourages them to see algebra as an important part of their daily lives. The reader-friendly writing style uses short, clear sentences and easy-to-understand language, and the outstanding pedagogical program makes the material easy to follow and comprehend. KEY TOPICS Chapter topics cover real numbers, solving linear equations and inequalities, formulas and applications of algebra, exponents and polynomials, factoring, rational expressions and equations, graphing linear equations, systems of linear equations, roots and radicals, and quadratic equations. For the study of Algebra.
"synopsis" may belong to another edition of this title.
From the Publisher:
Angel's text is one that students can read, understand, and enjoy. With short sentences, clear explanations, many detailed worked examples, and outstanding pedagogy. Practical applications of algebra throughout make the subject more appealing and relevant for students. Key pedagogical features include: Preview and Perspective at the beginning of each chapter; Helpful Hints; Group Activities/Challenge Problems, Writing Exercises, Real-Life Application Exercises, and Calculator Corners.
From the Back Cover:
Key Benefit: This is one book that readers can read, understand, and enjoy. It includes short sentences, clear explanations, many detailed worked examples, and outstanding pedagogy. Practical applications of algebra throughout make the subject more appealing and relevant for readers. KEY FEATURES: Includes: Preview Perspectives at the beginning of each chapter; also includes Helpful Hints; Group Activities/Challenge, Writing, and Real-Life Application problems; Calculator and Graphing Calculator Corners; Spiral approach to learning | 677.169 | 1 |
GCSE Maths. What will I learn?. number, data handling and shape There is no coursework in GCSE mathematics. Data Handling Scheme. Your final GCSE grade is assessed by external exams and coursework but we. You will also be required to purchase a GCSE Maths Work. One Hundred Years Of Solitude Theme Essay.gcse astronomy coursework b11. Essay On Increasing Old Age Homes. gcse maths coursework data handling. No coursework. Content: Maths is more than just. Handling Data the increasing exam demands of Maths at GCSE, promote Maths with the family and to. Free Tips on GCSE Statistics Coursework Data Handling Project) is part of GCSE Maths coursework done. GCSE Statistics coursework depends on. Interactive Worksheets to help your child in English, Maths and Science. ARE YOU A TUTOR OR A TEACHER?. maths and science spans across Year 1 to GCSE. GCSE Maths Statics Coursework Examples Help Online by our Maths coursework helpers and avail our best quality Maths coursework help online at. A data handling. | 677.169 | 1 |
Algebra b: Functions
Functions are basic building-block sentences of mathematical reasoning. A function relates input values in a domain to output values in a codomain, and these associations can be depicted using plots. While different disciplines use slightly different definitions of a function, an essential stipulation familiar to scientists and mathematicians from a variety of fields is that a function associates each input value with precisely one output value. | 677.169 | 1 |
Mrs. Inselmann's Math Page!!!
Welcome Parents and Students! This site is intended to inform students and parents about what is going on in my Mathematics classes.
Students and parents can link to the course syllabus, assignments, and calendars at any time.
You will always be up to date on what's due when and the date of the next test. You'll also find
a link to contact me, as well as links to sites that will help students
succeed in the mathematics classroom. I am so excited to be teaching at
Denton High School, and I know this year is going to be a success!
It's a new year and we are all starting fresh in 2008. Be sure to check with your student to make sure he/she is
staying on top of assignments. Remember, Algebra I and Geometry are required credits for graduation, and I am hoping to see a
great pass rate this year!
News and Events
I want to remind parents and students of the importance of homework in the math classroom.
All of my students have homework every single class day, and it needs to be completed before
the next class meeting. Parents, please check to make sure your teenager is staying
on top of homework assignments. | 677.169 | 1 |
This book explores the mathematical tools and procedures used in modeling based on the laws of conservation of mass, energy, momentum, and electrical charge. This edition expands the range of applied examples to reach a wider audience. The text proceeds from simple models of real-world problems at the algebraic and ordinary differential equations (ODE) levels to more sophisticated models requiring partial differential equations. The traditional solution methods are supplemented with Mathematica, which is used throughout the text to arrive at solutions for many of the problems presented. | 677.169 | 1 |
Interior Design Peter Papayanakis Cover Design Eugene Lo Cover Image The famous thrust stage of the Stratford Festival of Canada's Festival Theatre. Photo: Terry Manzo. Courtesy of the Stratford Festival Archives. Production Services Pre-Press Company Inc. Director, Asset Management Services Vicki Gould Photo/Permissions Researcher Daniela Glass Photo Shoot Coordinator Lynn McLeod Set-up Photos Dave Starrett Printer Transcontinental Printing Ltd. Every effort has been made to trace ownership of all copyrighted material and to secure permission from copyright holders. In the event of any question arising as to the use of any material, we will be pleased to make the necessary corrections in future printings.
ALL RIGHTS RESERVED. No part of this work covered by the copyright herein, except for any reproducible pages included in this work, may be reproduced, transcribed, or used in any form or by any means— graphic, electronic, or mechanical, including photocopying, recording, taping, Web distribution, or information storage and retrieval systems—without the written permission of the publisher. For permission to use material from this text or product, submit a request online at
Reviewers and Advisory Panel
Paul Alves Mathematics Department Head Stephen Lewis Secondary School Peel District School Board Mississauga, ON Terri Blackwell Mathematics Teacher Burlington Central High School Halton District School Board Burlington, ON Karen Bryan Program Resource Teacher, Numeracy & Literacy 7–12 Upper Canada District School Board Brockville, ON Angela Conetta Mathematics Teacher Chaminade College School Toronto Catholic District School Board Toronto, ON Justin De Weerdt Mathematics Department Head Huntsville High School Trillium Lakelands District School Board Huntsville, ON Robert Donato Secondary Mathematics Resource Teacher Toronto Catholic District School Board Toronto, ON Richard Gallant Secondary Curriculum Consultant Simcoe Muskoka Catholic District School Board Barrie, ON Jacqueline Hill K–12 Mathematics Facilitator Durham District School Board Whitby, ON Punitha Kandasamy Classroom Teacher (Secondary— Mathematics) Mississauga Secondary School Peel District School Board Mississauga, ON Halyna Kopach Mathematics Department Head Mary Ward Catholic Secondary School Toronto Catholic District School Board Toronto, ON Richard Long Mathematics Department Head Centennial Secondary School Hastings and Prince Edward District School Board Belleville, ON Frank Maggio Department Head of Mathematics Holy Trinity High School Halton Catholic District School Board Oakville, ON Peter Matijosaitis Mathematics Department Head Loretto Abbey Catholic Secondary School Toronto Catholic District School Board Toronto, ON Cheryl McQueen Mathematics Learning Coordinator, 7–12 Thames Valley District School Board London, ON Ian McTavish Librarian/Mathematics Teacher Huntsville High School Trillium Lakelands District School Board Huntsville, ON Grace Mlodzianowski Mathematics Department Head Cardinal Newman High School Toronto Catholic District School Board Toronto, ON Elizabeth Pattison Head of Mathematics Westlane Secondary School District School Board of Niagara Niagara Falls, ON Kathy Pilon Mathematics Program Leader St. John Catholic High School Catholic District School Board of Eastern Ontario Perth, ON Joshua Plant Mathematics Teacher Huntsville High School Trillium Lakelands District School Board Huntsville, ON Margaret Russo Mathematics Teacher Madonna Catholic Secondary School Toronto Catholic District School Board Toronto, ON Patricia Steele Mathematics Resource Teacher Simcoe County District School Board Midhurst, ON Scott Taylor Head of Mathematics, Business, and Computer Science Bell High School Ottawa-Carleton District School Board Nepean, ON Salvatore Trabona Mathematics Department Head Madonna Catholic Secondary School Toronto Catholic District School Board Toronto, ON Dave Wright Mathematics Teacher Woburn Collegiate Institute Toronto District School Board Toronto, ON Krista Zupan Resource Teacher for Student Success Durham Catholic District School Board Oshawa, ON
Identify a function as a special type of relation Recognize functions in various representations and use function notation Explore the properties of some basic functions and apply transformations to those functions Investigate the inverse of a linear function and its properties
? Anton needs a summer job. How
would you help him compare the two offers he has received?
NEL
1
1
Study
Getting Started
SKILLS AND CONCEPTS You Need
1. Simplify each expression.
Begin by completing a table like the one shown. Then list similarities and differences among the three types of relations.
Property Equation(s) Shape of graph Number of quadrants graph enters Descriptive features of graph Types of problems modelled by the relation Linear Relations Circles Quadratic Relations
2
Chapter 1
NEL
Getting Started
APPLYING What You Know
Fencing a Cornfield
Rebecca has 600 m of fencing for her cornfield. The creek that goes through her farmland will form one side of the rectangular boundary. Rebecca considers different widths to maximize the area enclosed.
?
A. B.
YOU WILL NEED
• graph paper
How are the length and area of the field related to its width?
What are the minimum and maximum values of the width of the field? What equations describe each? i) the relationship between the length and width of the field ii) the relationship between the area and width of the field Copy and complete this table of values for widths that go from the least to the greatest possible values in intervals of 50 m.
Width (m) Length (m) Area (m2)
C.
D. E.
Graph the data you wrote in the first two columns. Use width as the independent variable. Describe the graph. What type of relationship is this? Now graph the data you wrote in the first and third columns. Use width as the independent variable again. Describe the graph. What type of relationship is this? How could you have used the table of values to determine the types of relationships you reported in parts D and E? How could you have used the equations from part B to determine the types of relationships you reported in parts D and E?
F. G.
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Introduction to Functions
3
i) 8 ii) 10 iii) 13 Use your plot to predict what shoe size corresponds to each height.5 12 10 11 Height (cm) 156 161 179 178 173 177 165 182 177 192 157 163 183 168 180
?
A. Describe the relationship shown in the scatter plot. B-11. C. i) 153 cm ii) 173 cm iii) 177 cm Draw a line of good fit on your plot.1
YOU WILL NEED
Relations and Functions
• graphing calculator or
graph paper
GOAL
Recognize functions in various representations.5 8.5 12 11 10.
INVESTIGATE the Math
Ang recorded the heights and shoe sizes of students in his class.5 10 9 9 10 11 8 8 11 8 7. Write the equation of your line.5 8 12 13 13 7. see Technical Appendix.
Shoe Size 10 11. How are your results different from those in part B?
NEL
Tech
Support
For help drawing a line of best fit on a graphing calculator. and use it to determine the heights corresponding to the shoe sizes in part B. D.
4
Chapter 1
. using shoe size as the independent variable.1.5 10 11 11 10 Height (cm) 158 175 173 164 167 170 172 160 174 175 166 153 171 181 171 170 Shoe Size 8 7. Use your plot to predict the height of a person with each shoe size.5 8.
Can you predict a person's height from his or her shoe size?
Plot the data. B.
Tuesday). F. or second elements—to write the range. Stefanie. Amit} Range 5 {Tuesday. in a set. Saturday}
NEL
Introduction to Functions
5
.1.
Describe the domain and range of the relationship between shoe size and height in Ang's class. Explain why the relation plotted in part A is not a function. K. Tuesday). Is the relation drawn in part D a function? Explain. (student's name. (Amit. state the domain and range and then determine whether or not the relations are functions. Which of the relations in parts A and D could be used to predict a single height for a given shoe size? Explain.1
E. or first elements. in each ordered pair.
APPLY the Math
EXAMPLE
1
Representing functions in different ways
The ages and soccer practice days of four students are listed. 8. J. or elements. I listed the day for practice—the dependent variable. Thursday. Thursday). Saturday)} Domain 5 {Craig. G. For example. a) students and the day for soccer practice b) ages and the day for soccer practice
Jenny's Solution: Using Set Notation
a) {(Craig.
(Magda.
For each of the given relations. H.
I wrote the relation as a set of ordered pairs.
domain the set of all values of the independent variable of a relation range the set of all values of the dependent variable of a relation relation a set of ordered pairs. the set of the first five even numbers is {2. 4. (Stefani. Magda.
How did the numbers in the table of values show that the relation was not a function? How did the graph of the linear function you drew in part D differ from the graph of the relation you plotted in part A? Explain why it is easier to use the linear function than the scatter plot of the actual data to predict height.
Student Craig Magda Stefani Amit Age 15 16 15 17 Soccer Practice Day Tuesday Tuesday Thursday Saturday
Communication
Tip
Use braces to list the values. values of the independent variable are paired with values of the dependent variable function a relation where each value of the independent variable corresponds with only one value of the dependent variable
Reflecting
I. 6. I wrote the domain by listing the students' names—the independent variable. day for practice). 10}.
The diagram is called a mapping diagram. but another practiced on Thursday. so the relation between students and their soccer practice day is a function. (17. Amit} Range 5 {Tuesday. days. 17} Range 5 {Tuesday. maps to two different values of the dependent variable. Saturday} The value 15 of the independent variable. This is not a function. 16. This relation is not a function. Then I drew arrows to match the students with their practice days.
Each element of the domain has only one corresponding element in the range.
Domain 5 {Craig. age. Thursday. so this relation is not a function. Saturday)} Domain 5 {15. Thursday. Stefanie. since it maps the elements of the domain onto the elements of the range. The first elements appear only once in the list of ordered pairs.
I noticed that one 15-year-old practiced on Tuesday.
b) Age
15 16 17
Practice day
Sat Thurs Tues
Domain 5 {15. Saturday}
The elements in the left oval are the values of the independent variable and make up the domain. so the relation is a function. 17} Range 5 {Tuesday.
6
Chapter 1
NEL
. if I know the student's name. I drew another mapping diagram for the age and practice day relation.
Two arrows go from 15 to two different days.Each element of the domain corresponds with only one element in the range. Tuesday). I wrote the domain and range by listing what was in each oval. I can predict his or her practice day. In this case. Tuesday). so I can't predict a practice day just by knowing the age. Thursday. Saturday} 15 in the domain corresponds with two different days in the range. The elements in the right oval are the values of the dependent variable and make up the range.
(15. so the relation is a function. (16. I matched the ages to the practice days. Thursday). An element of the domain can't map to two elements in the range.
Each student has only one practice day.
Olivier's Solution: Using a Mapping Diagram
a) Student
Craig Magda Stefani Amit
Practice day
Sat Thurs Tues
I drew a diagram of the relation between students and soccer practice days by listing the student names in an oval and the days in another oval. No name is repeated.
b) {(15. Magda. The relation is a function because each student name has only one arrow leaving it. 16. This cannot be a function.
The ruler crossed the graph in two places everywhere except at the leftmost and rightmost ends of the circle.
Any vertical line drawn on the graph intersects the graph at only one point. This is not the graph of a function. vertical-line test if any vertical line intersects the graph of a relation more than once.
NEL
Introduction to Functions
7
. This is the graph of a function. This showed that there are x-values in the domain of this relation that correspond to two y-values in the range.
5
At least one vertical line drawn on the graph intersects the graph at two points. then the relation is not a function
x 5 0 5
An easy way to do this was to use a ruler to represent a vertical line and move it across the graph.
b) 4 2 4 2 0 2 4 2 4 6 x y
I used the vertical-line test again. This showed that each x-value in the domain corresponds with only one y-value in the range. the vertical line intersected the graph in only one place.1.1
EXAMPLE
2
Selecting a strategy to recognize functions in graphs
Determine which of the following graphs are functions. Wherever I placed my ruler. a) y b) y 5 4
x 5 0 5 4 2 2 0 2 4 2 4 6 x
5
Ken's Solution
a) 5 y
I used the vertical-line test to see how many points on the graph there were for each value of x.
b) This equation defines the graph of a circle centred at
(0. 0) with a radius of 3. I graphed the function and checked it with the vertical-line test. This is the graph of a function.
with a positive slope. a) y 5 2x 2 5 b) x 2 1 y 2 5 9 c) y 5 2x 2 2 3x 1 1 Keith's Solution: Using the Graph Defined by its Equation
a) This equation defines the graph of a linear function
I used my graphing calculator and entered y 5 2x 2 5.
y 5 2x2 2 3x 1 1 is a function. This is not the graph of a function. Then I applied the vertical-line test to check. This graph fails the vertical-line test.
I used my graphing calculator and entered the upper half of the circle in Y1 and the lower half in Y2. This is the graph of a function.
y 5 2x 2 5 is a function. showing that for each x-value in the domain there is only one y-value in the range.EXAMPLE
3
Using reasoning to recognize a function from an equation
Determine which equations represent functions.
x 2 1 y 2 5 9 is not a function. showing that there are x-values in the domain of this relation that correspond to two y-values in the range.
I used my graphing calculator to enter y 5 2x 2 2 3x 1 1 and applied the vertical-line test to check.
8
Chapter 1
NEL
.
This graph passes the vertical-line test. This graph passes the vertical-line test. Its graph is a straight line that increases from left to right. showing that for each x-value in the domain there is only one y-value in the range. c) This equation defines the graph of a parabola that opens upward.
c) Every value of x gives only one value of y in the
I substituted 0 for x in the equation and solved for y. which have the general forms y 5 mx 1 b or Ax 1 By 5 C and whose graphs are straight lines. For example. Linear relations. A graph represents a function if every vertical line intersects the graph in at most one point. are all functions. x • You can recognize whether a relation is a function from its equation. • The range of a relation or function is the set of all values of the dependent A relation that is a function variable.1
Mayda's Solution: Substituting Values
a) For any value of x. I get only one number for y that satisfies the equation. No matter what number I substituted for x. This equation represents a function. a mapping diagram. This is usually represented by the y-values on a coordinate grid.
No matter what number I choose for x. a table of values. so the equation defines a relation.
y
x
Need To Know
• The domain of a relation or function is the set of all values of the independent variable. a graph. (0) 2 1 y 2 5 9 y 5 3 or 23 There are two values for y when x 5 0. Vertical lines are not functions but horizontal lines are. I got two values for y with x 0.
equation y 5 2x 2 2 3x 1 1. y 5 2(1) 2 5 5 23 This equation defines a function. • Functions can be represented in various ways: in words. but not a function.
In Summary
Key Ideas
• A function is a relation in which each value of the independent variable corresponds with only one value of the dependent variable. Quadratic relations. This shows that there is only one element in the range for each element of the domain. which have the general forms y 5 ax2 1 bx 1 c or y 5 a(x 2 h) 2 1 k and whose A relation that is not a function graphs are parabolas. This is usually represented by the x-values on a coordinate grid.
NEL
Introduction to Functions
9
. or an equation. If you can find even one value of x that gives more than one value of y when you substitute x into the equation. y • You can use the vertical-line test to check whether a graph represents a function. are also functions.1. I got only one answer for y when I doubled the number for x and then subtracted 5. I used 0 because it's an easy value to calculate with. a set of ordered pairs. the relation is not a function. the equation y 5 2x 2 5
I substituted numbers for x in the equation.
b) Substitute 0 for x in the equation
x 2 1 y 2 5 9.
produces only one value of y.
50/kg for any order of at least 100 kg.
12
Chapter 1
NEL
. The cost of renting a car depends on the daily rental charge and the number
of kilometres driven.
13. a) Why must this relation be a function? b) What is the domain of this function? What is its range? c) Graph the function. a) Sketch a graph of a function that has the set of integers as its domain and
T
all integers less than 5 as its range. A freight delivery company charges $4/kg for any order less than 100 kg and
$3. c) Is the relation a function? Explain.
Definition: Function Characteristics:
14. d) What suggestions can you offer to the company for a better pricing structure? Support your answer.
y Cost versus Distance Driven
120 100 Cost ($) 80 60 40 20 0
x 100 200 300 Distance (km) 400 500
a) What are the domain and range of this relation? b) Explain why the domain and range have a lower limit. Use a chart like the following to summarize what you have learned about
C
Examples:
Non-examples:
Extending
15.12. b) Sketch a graph of a relation that is not a function and that has the set of real numbers less than or equal to 10 as its domain and all real numbers greater than 25 as its range. functions. A graph of cost versus the distance driven over a one-day period is shown.
If the original position of the rabbit represents the origin and the rabbit's path is along the positive y-axis.
3. is the fox's path the graph of a function? Explain. The resulting curve is called a curve of pursuit.1
Curious Math
Curves of Pursuit
A fox sees a rabbit sitting in the middle of a field and begins to run toward the rabbit. The fox then changes direction to run along line EA. regardless of where the
rabbit is relative to its burrow? Explain.
Burrow
Rabbit's Path
F E B A Rabbit
Fox
If the fox and rabbit are running at the same speed. the rabbit is at B.
a) Draw a curve of pursuit in which i) the rabbit runs faster than the fox ii) the fox runs faster than the rabbit b) Are these relations also functions? How do they differ from the one in
question 1?
NEL
Introduction to Functions
13
. When the fox reaches point F. The rabbit sees the fox and runs in a straight line to its burrow.1. The fox continuously adjusts its direction so that it is always running directly toward the rabbit.
example
ii) the burrow is closer to the rabbit than it is in the first example b) Where does the fox finish in each case? How does the location of the
burrow relative to the rabbit affect the fox's path?
c) Will the path of the fox always be a function.
a) Investigate what happens by drawing a curve of pursuit if i) the burrow is farther away from the rabbit than it is in the first
2. so the fox begins to run along FB.
1. the fox reaches point E when the rabbit reaches point A. and so on.
T 5 11 1 0. T(d ) 5 11 1 0. such as v(t) for velocity as a function of time. so y is equal to f(x). reaches 3585 m into Earth's crust. and write it in function notation. such as f(x).015 ° C>m. Temperature is a function of depth.015d. I used the fact that T starts at 11 °C and increases at a steady rate of 0. b) Use your function to determine the temperature at the bottom of East Rand and Western Deep mines. I wrote the equation again.
I wrote a linear equation for the problem. In function notation. Suppose the temperature at the top of the mine shaft is 11 °C and that it increases at a rate of 0. Since this equation represents a linear relationship between temperature and depth.015d. but other letters are also used. used to represent the value of the dependent variable—the output—for a given value of the independent variable. g(x). The notation f(x) is read "f at x" or "f of x. it is a function.
The equation represents a function. Explain why
function notation notation. and h(x) are often used to name the outputs of functions.015 °C>m as you descend.1.
?
What is the temperature at the bottom of each mine?
EXAMPLE
1
Representing a situation with a function and using it to solve a problem
a) Represent the temperature in a mine shaft with a function.2
YOU WILL NEED
Function Notation
• graphing calculator
GOAL
Use function notation to represent linear and quadratic functions. T(d) makes it clearer that T is a function of d.
14
Chapter 1
NEL
.
LEARN ABOUT the Math
The deepest mine in the world.
Lucy's Solution: Using an Equation
a) An equation for temperature is
Communication
Tip
The notations y and f(x) are interchangeable in the equation or graph of a function. where T represents the temperature in degrees Celsius at a depth of d metres. Another South African mine." The symbols f(x). Western Deep. is being deepened to 4100 m. x—the input
your representation is a function. East Rand mine in South Africa.
775 5 64.
b)
d (m) 0 1000 2000 3000 4000
T(d)(8C) T(0) 5 11 1 0.015(3585)
I found the temperature at the bottom of East Rand mine by calculating the temperature at a depth of 3585 m. I substituted 3585 for d in the equation.
NEL
Introduction to Functions
15
.015(4100) 5 11 1 61.5 5 72.775 T(4100) 5 11 1 0.1.2
b) T(3585) 5 11 1 0.015d
I wrote an equation to show how the temperature changes as you go down the mine.
Stuart's Solution: Using a Graph
a) T(d ) 5 11 1 0. I used d for depth and called the function T(d) for temperature. I substituted the d-values into the function equation to get the T(d)-values. I knew that the relationship was linear because the temperature increases at a steady rate.015(0) 5 11 26 41 56 71
I made a table of values for the function. For the new mine. respectively.
5 11 1 53. so I calculated T(4100).5 The temperatures at the bottom of East Rand mine and Western Deep mine are about 65 °C and 73 °C. I wanted the temperature when d 5 4100.
This is a function because it is a linear relationship.
East Rand mine is 3585 m deep. Then I joined them with a straight line.
b)
I used function notation to write the equation.
I graphed the function by entering Y1 5 11 1 0. B-2 and B-3. (2000. T(d ) 5 11 1 0.
Tech
Support
For help using a graphing calculator to graph and evaluate functions. The temperature at the bottom of Western Deep mine is about 73 °C. see Technical Appendix.Temperature of a Mine y 90 80 Temperature. 72.015X into the equation editor. (m)
4100 x
The other mine is 4100 m deep. Xscl 200. 26). so it is a function of depth.
50 30
T(d
11 +
0.
Eli's Solution: Using a Graphing Calculator
a) Let T(d ) represent the
temperature in degrees Celsius at a depth of d metres. Yscl 10. and 0 # Y # 100.5
I plotted the points (0. 64. The temperature at the bottom is T(3585). I used WINDOW settings of 0 # X # 5000. 11). (3000. 56). It was approximately 65. and (4000. (1000.0
3585
0 1000 3000 5000 Depth.5 Approx.015d Temperature increases at a steady rate. d. T(d). (°C) 70 60 40 20 10
)= 15d
Approx. 41).
The temperature at the bottom of East Rand mine is about 65 °C. 71). I interpolated to read T(3585) from the graph.
16
Chapter 1
NEL
. I found that T(4100) was about 73. By extrapolating.
The temperature at the bottom of Western Deep mine is about 73 °C. This told me that T(3585) 5 64.1. and Eli know that the relationship between temperature and depth is a function? How did Lucy use the function equation to determine the two temperatures? What does T(3585) mean? How did Stuart use the graph to determine the value of T(3585)?
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Introduction to Functions
17
.2
I used the value operation to find the temperature at the bottom of East Rand mine. B.
The temperature at the bottom of the East Rand mine is about 65 °C.775. Stuart. I called up the function on my calculator home screen.
How did Lucy.
Then I used the value operation again to find the temperature at the bottom of the other mine. C. using VARS and function notation to display both answers.5.
As a check. I found that T(4100) 5 72.
Reflecting
A.
they each multiplied the resulting difference by the number they first thought of. The person with the highest final number won the pizza slice. double it. Rhea 22 b) The original numbers chosen by the family members are shown. Who won the pizza slice? Sara 7 c) What would be the best number to choose? Why? Andy 10
Barbara's Solution
a) x input Double the number 2x Subtract from 12 (12 2x)
I used a flow chart to show what happens to the original number in the game.APPLY the Math
EXAMPLE
2
Representing a situation with a function model
A family played a game to decide who got to eat the last piece of pizza.
Multiply by the original number x(12 2x) output
f (x) 5 x(12 2 2x) 5 12x 2 2x 2
The expression for the final answer is quadratic.
18
Chapter 1
NEL
. and subtract the result from 12. I chose x to be the original number. so the final result must be a function of the original number. or output. or input. a) Use function notation to express the final answer in terms of Tim 5 the original number. Finally. Each person had to think of a number. I chose f(x) as the name for the final answer.
y 5 0.
When x 5 3. g(3) 5 2. in function notation. The range is all real numbers greater than or equal to 0. The graph exists only for x $ 21 and y $ 0. I saw that y 5 0 when x 5 21. g(3) 5 2
b) When x 5 21. 0) or below that point.
The y-value was 2. The x-value is 0 at this point. The domain is all real numbers greater than or equal to 21. determine each value.EXAMPLE
3
Connecting function notation to a graph
y 2 1 1 0 1 2 1 y = g(x) x 2 3 4
For the function shown in the graph. so 21 is the x-intercept and g(21) 5 0.
g(21) 5 0
c) g(x) 5 1 when x 5 0 d) The graph begins at the point
(21. y 5 2. a) g(3) b) g(21) c) x if g(x) 5 1 d) the domain and range of g(x)
Ernesto's Solution
a) 2 1 1 0 1 2 1 y = g(x) x 2 3 4 y
I looked at the graph to find the y-coordinate when x 5 3. So the only possible x-values are x $ 21. 0) and continues upward.
20
Chapter 1
NEL
. I drew a line up to the graph from the x-axis at x 5 3 and then a line across from that point of intersection to the y-axis. I saw that there was no graph to the left of the point (21. and the only possible y-values are y $ 0. I saw that the graph crosses the y-axis at y 5 1. so.
with examples. b) Use the function to determine the new marks that correspond to original marks of 95. a) Use function notation to express the final answer in terms of the original number. what function notation is and how it relates to the
C
graph of a function. 175. is supported by
T
15. A function f (x) has these properties:
• The domain of f is the set of natural numbers.15/km. • f (1) 5 1 • f (x 1 1) 5 f (x) 1 3x(x 1 1) 1 1 a) Determine f (2). Ontario. c) Determine how far you can drive in a day for $80. a) Determine a linear function that will convert 285 to 200 and 75 to 60. b) Describe the function. 25.
b) Determine the rental cost if you drive 472 km in one day. f (5). Let f (x) 5 3x 1 1 and g(x) 5 2 2 x. Let f (x) 5 x 2 1 2x 2 15. Determine the values of x for which 17. Each arch is set in concrete foundations that are on opposite sides of the St.
All the marks must be reduced so that the highest and lowest marks become 200 and 60. Clair River. The arches are 281 m apart. a teacher asked her students to think of a
A
number. a)
16. The highest and lowest marks awarded on an examination were 285 and 75. 215. c) Determine the maximum result possible. Include a discussion of the advantages of using function notation. and subtract the resulting number from 24. The top of each arch rises 71 m above the river. Graph the function f (x) 5 3(x 2 1) 2 2 4. As a mental arithmetic exercise. A company rents cars for $50 per day plus $0.
a) Express the daily rental cost as a function of the number of kilometres
travelled.
24
Chapter 1
NEL
. Write a function to model the arch.
20. and 10. Determine values for a such that a) f (a) 5 g(a) b) f (a 2 ) 5 g(2a) 18. The second span of the Bluewater Bridge in Sarnia. Finally.
13. triple it. Explain. they were asked to multiply the resulting difference by the number they first thought of.12. f (4). c) Use the equation to determine i) f (2) 2 f (1) ii) 2f (3) 2 7 iii) f (1 2 x)
a) f (x) 5 0 b) f (x) 5 212 c) f (x) 5 216
14. f (3). b) What does f (21) represent on the graph? Indicate on the graph how you would find f (21). and f (6). and 255. two parabolic arches.
Extending
19. b) Determine the result of choosing numbers 3.
for example. Is it a function? Explain.
You can change to these settings by pressing
ZOOM
26
4
. How is it different from the graphs of linear and quadratic functions? In your table. D.
What are the characteristics of these parent functions that distinguish them from each other?
Make a table like the one shown. sketch the graph and describe its shape. Graph the square root function.
?
A. Explain how you know that these equations are both functions. record the domain and range of each of f (x) 5 x and f (x) 5 x 2.
C. the 1 reciprocal function f (x) 5 x . f (x) 5 !x.
Sketch of Graph Special Features/ Symmetry • straight line that goes through the origin • slope is 1 • divides the plane exactly in half diagonally • graph only in quadrants 1 and 3 • parabola that opens up • vertex at the origin • y has a minimum value • y-axis is axis of symmetry • graph only in quadrants 1 and 2 Domain Range
y
x
f(x) 5 x2
quadratic function
y
x
f(x) 5 "x f(x) 5 1 x f(x) 5 |x|
square root function reciprocal function absolute value function B. 0.
Clear all equations from the equation editor. |3| 5 3 and |23| 5 2 (23) 5 3 Equation of Function f(x) 5 x Name of Function linear function
Three more parent functions are the square root function f (x) 5 !x . equals x when x $ 0 or 2x when x . describes the distance of x from 0.
Tech
Support
Use the following WINDOW settings to graph the functions:
Use your graphing calculator to check the sketches shown for f (x) 5 x and f (x) 5 x 2 and add anything you think is missing from the descriptions.absolute value written as |x|. and the absolute value function f (x) 5 |x|. In your table.
NEL
Chapter 1
.
Using the table of values and the graph.
Explain how each of the following helped you determine the domain and range. Parent functions include.U. and f (x) 5 |x| could you use to x sketch their graphs?
II III
I IV
In Summary
Key Idea
• Certain basic functions. form the building blocks for families of more complicated functions. When you have finished. I.
Tech
Support
G. x
(continued)
NEL
Introduction to Functions
27
. vertices. on some portion of its domain
y
Reflecting
J. M. press
MATH X. but never meets. a) the table of values b) the graph c) the function's equation Which graphs lie in the listed quadrants? a) the first and second quadrants b) the first and third quadrants Which graph has asymptotes? Why?
asymptotes
x
K. see Technical Appendix. called parent functions. and f(x) 5 |x|.
You have used the slope and y-intercept to sketch lines.
Quadrants
L.
Go to the table of values and scroll up and down the table. What characteristics of the new 1 parent functions f (x) 5 !x. Use the table x of values to see what happens to y when x is close to 0 and when x is far from 0.
a line that the graph of a relation or function gets closer and closer to.3
E. and directions of opening to sketch parabolas. f(x) 5 x. but are not 1 limited to. 1 Repeat parts D through F for the reciprocal function f (x) 5 .1.n
asymptote
1 )
. Where are the asymptotes of this graph? Repeat parts D through F for the absolute value function f (x) 5 |x|. F. Explain why the graph is in two parts with a break in the middle.
H. f (x) 5 . f(x) 5 "x.
For help with the TABLE function of the graphing calculator.T. f(x) 5 . Which of the other functions is the resulting graph most like? Explain. determine and record the domain and range of the function. To graph f(x) 5 |x|. B-6. Does ERR: appear in the Y column? Explain why this happens. make sure that your table contains enough information for you to recognize each of the five parent functions. f(x) 5 x2.
This gave me the domain for the function. So d can take only values that lie in between 0 and 346. The speed of the pebble as it falls to the ground is a function of how far it has fallen. Xscl 20 and 0 # Y # 100. The equation for this function is v(d) 5 !2gd .1.
Sally's Solution: Using a Graph
The pebble falls a total distance of 346 m. graphs.4
GOAL
Determining the Domain and Range of a Function
YOU WILL NEED
Use tables.6 x into my graphing calculator.8 metres per second squared (m>s2 ) A gull landing on the guardrail causes a pebble to fall off the edge. in metres.6.
? How can you determine the domain and range of the function v(d)?
EXAMPLE
1
Selecting a strategy to determine the domain and range
Determine the domain and range of v(d ). Yscl 10 for WINDOW settings. and equations to find domains and ranges of functions.
• graph paper • graphing calculator
LEARN ABOUT the Math
The CN Tower in Toronto has a lookout level that is 346 m above the ground. in metres per second (m/s) • g is the acceleration due to gravity—about 9. the pebble's speed.8 5 19. I entered Y= ! 19.
The distance d is 0 m when the pebble first falls off the edge and 346 m when the pebble lands on the ground. so the function is
v(d ) 5 "19. where • d is the distance. the pebble has fallen • v(d ) is the speed of the pebble. So the domain is 0 # d # 346.6d for 0 # d # 346
NEL
Introduction to Functions
29
.
2 3 9. I used 0 # X # 346.
The domain is 5d [ R | 0 # d # 3466 and the range is 5v(d ) [ R | 0 # v(d ) # 82. This happens when d 5 346. and d 5 346 when it lands.
real numbers
numbers that are either rational or irrational. such that x is greater than or equal to 0 and less than 50.
I found the domain by thinking about all the values that d could have. these include positive and negative integers. When the pebble lands. to one decimal place
v(0) 5 "19. 506 is read "the set of all values x that belong to the set of real numbers. and irrational numbers such as ! 2 and p Set notation can be used to describe domains and ranges. I used set notation to write the domain and range. d 5 346.6(0) 5 0
v(346) 5 "19.46."
Domain 5 5d [ R | 0 # h # 3466
Range 5 5v(d ) [ R | 0 # v(d ) # 82. The pebble starts with no velocity. I used the equation as a check. zero.
v(346) 8 82. its speed increases. 5 82. so the speed was zero at the start. I used the function equation to find how fast the pebble was falling when it landed." The symbol "|" stands for "such that.46
David's Solution: Using the Function Equation
d 5 0 when the pebble begins to fall.
As the pebble falls. So d must take values between 0 and 346. I used set notation to write the domain and range. The graph showed me that as the pebble's distance increases. I defined them as sets of real numbers. I knew that the pebble would gain speed until it hit the ground. 5x [ R | 0 # x . I evaluated this using the value operation.Range:
I saw that the graph started at the origin. d is 0 m when the pebble first falls off the edge and 346 m when the pebble lands on the ground. fractions.4
So the range is 0 # v(d ) # 82. so does its velocity.6(346)
30
Chapter 1
NEL
. The pebble must be travelling the fastest when it hits the ground. For example.4. The maximum value of the range is v(346) .
Communication
Tip
The pebble starts with speed 0 m/s. The pebble fell off the edge. So the domain is 0 # d # 346.4. So 0 is the minimum value of the range.
The function has a maximum value at the vertex (21.
b) g(x) 5 23(x 1 1) 2 1 6
Domain 5 5x [ R6
This is a quadratic equation in vertex form. x cannot be 25. x can be any value.
Range 5 52. Therefore. 3). A closed circle means that the endpoint is included. so x can be any number. x # 116
c) Domain 5 5x [ R6
Range 5 5 y [ R | y # 3)6 This is a function. There are only two y-values. This is the equation of a parabola that opens down. 0) and radius of 5. The graph is a parabola with a maximum value at the vertex. The graph is a circle with centre (0. so y can never be more than its value at the vertex. The graph passes the vertical-line test. There are many vertical lines that cross the graph in two places
d) Domain 5 5x [ R | 25 # x # 56
Range 5 5 y [ R |25 # y # 56 This is a function. Range 5 5 y [ R | y # 66
32
Chapter 1
NEL
. So. which is the point (1. x and f(x) can be any numbers. but y cannot be greater than 3.
EXAMPLE
3
Determining domain and range from the function equation
a) f (x) 5 2x 2 3
Determine the domain and range of each function. 6). so x and y can be any value. 66
An open circle on the graph shows that the endpoint of the line is not included in the graph.b) Domain 5 5x [ R | 25 . The graph passes the vertical-line test. I used y instead of f(x) to describe the range. b) g(x) 5 23(x 1 1) 2 1 6
Jeff's Solution
a) f (x) 5 2x 2 3
c) h(x) 5 !2 2 x
This is a linear function.
Domain 5 5x [ R6 Range 5 5 y [ R6
This is the equation of a straight line that goes on forever in both directions. Any value of x will work in the equation. This is not a function. x can be any real number. The graph fails the vertical-line test. but it can be 11.
2 is okay.
Jenny's Solution
a) Let the width of the garden be x m.
EXAMPLE
4
Determining domain and range of an area function
Vitaly and Sherry have 24 m of fencing to enclose a rectangular garden at the back of their house.c) h(x) 5 !2 2 x
1. I realized I had to use values less than or equal to 2.
Domain 5 5x [ R | 0 . To find the length. at 0 and 12. so y is never negative. I thought about different possible values of x.
Domain 5 5x [ R | x # 26 22x$0 Range 5 5 y [ R | y $ 06 2 2 x $ 0 as long as x # 2
!2 2 x means the positive square root.
A(x) 5 22x(x 2 12)
NEL
Introduction to Functions
33
. but 4 is not. Any number # 0 or $ 12 will result in a zero or negative area. A(x) 5 x(24 2 2x)
b) The smallest the width can
Area 5 width 3 length I factored out 22 from 24 2 2x to write the function in factored form. It has two zeros.
Then the length is (24 2 2x) m. The vertex lies halfway in between the zeros. so 2 2 x must be positive or zero. above the x-axis.4
You cannot take the square root of a negative number. The largest the width can approach is 12 m. This is a quadratic function that opens down. a) Express the area of the garden as a function of its width. since 2 2 2 5 0. I subtracted the two widths from 24.
They need fencing on only three sides of the garden because the house forms the last side. so the numbers in the domain have to be between 0 and 12. x . which doesn't make sense. 126 approach is 0 m. since 2 2 4 is negative.
x 24 2x
x
Let the area be A(x). b) Determine the domain and range of the area function.
The range of a function depends on the equation of the function. Linear functions of the form f(x) 5 mx 1 b. or from the function equation. have range 5 y [ R6. They are usually easier to determine from a graph or a table of values. For example. so the domain or range must be restricted to nonnegative values. where m 2 0. • All quadratic functions have domain 5x [ R6 . • The function f(x) 5 ! x has domain 5 5x [ R | x $ 06 and range 5 5y [ R | y $ 06. from a table of values. A(x) # 726 The area ranges from 0 to 72 m2. • The domains of square root functions are restricted because the square root of a negative number is not a real number. The x-coordinate of the vertex is 6. The graph depends on the domain and range.x 5 (0 1 12) 4 2 x56 A(6) 5 22(6) (6 2 12) 5 72
The vertex is halfway between x 5 0 and x 5 12. Constant functions f(x) 5 b have range {b}.
In Summary
Key Ideas
• The domain of a function is the set of values of the independent variable for which the function is defined. For example. The ranges are restricted because the square root sign refers to the positive square root.
Range 5 5A(x) [ R | 0 .
Need to Know
• All linear functions include all the real numbers in their domains. • The function g(x) 5 ! x 2 3 has domain 5 5x [ R | x $ 36 and range 5 5y [ R | y $ 06. I substituted x 5 6 into the area function to find the y-coordinate of the vertex. • When working with functions that model real-world situations.
34
Chapter 1
NEL
. The range of a quadratic function depends on the maximum or minimum value and the direction of opening. • The domain and range of a function can be determined from its graph. Since area must be a positive quantity. consider whether there are any restrictions on the variables. all the output values of the function must lie between 0 and 72. negative values often have no meaning in a real context.
6. The graph shows how 2007 prices for mailing letters in Canada vary with mass.
8.
K
a) Graph the relation. jog. The route for a marathon is 15 km long. Participants may walk. Why is it important for this
to be so?
b) State the domain and range of the function. A relation is defined by x 2 1 y 2 5 36. Write a function to describe coffee dripping into a 10-cup carafe at a rate of
1 mL/s.
Speed (km/h) Time (h) 1 15.PRACTISING
5.
36
Chapter 1
NEL
.00
0.50
Canadian Domestic Letter Prices
2. Copy and complete the table to show times for completing the marathon at different speeds.
7.
y 2. b) State the domain and range of the relation. or
cycle.00
Price ($)
1.0 2 7. c) Is the relation a function? Explain.00
x 100 300 200 Mass (g) 400 500
a) Explain why this relation is a function. run.50
1.5 3 4 5 6 8 10 15 20
Graph the relation in the table and explain how you know that it is a function. State the domain and range of the function.50
0. State the domain and range of the function (1 cup 5 250 mL).
Extending
16. 66. The large square in the diagram has side length 10 units. a) Write an equation to express the total area enclosed as a function of the width.
NEL
Sketch the graph of a function whose domain is 5x [ R6 and range is 5 y [ R | y # 2)6. c) Determine an equation for the function.5. Determine the domain and range of each function. 0. The ball reaches a
A
height of 45 m above the ground after 2 s and hits the ground 5 s after being thrown. b) Determine the domain and range of this area function.5(x 1 3) 2 1 4
2 (x 2 2) 2 2 5 3 5 e) q(x) 5 11 2 x 2 f ) r(x) 5 !5 2 x
d) p(x) 5
11. a) Determine the area of the inscribed square as a function of x. You can draw a square inside another square by placing each vertex of the
inner square on one side of the outer square. Explain the terms "domain" and "range" as they apply to relations and
C
functions. Determine the range of each function if the domain is 523. A ball is thrown upward from the roof of a 25 m building.
x
Introduction to Functions
37
.
a) f (x) 5 4 2 3x b) f (x) 5 2x 2 2 3x 1 1
1 x2 d) p(x) 5 !x 2 2 5
c) h(x) 5
15. c) Determine the perimeter of the inscribed square as a function of x. A farmer has 450 m of fencing to enclose a rectangular area and divide it into
T
14. and an equation. d) Determine the domain and range of this perimeter function. 2. b) State the domain and range of the function. a) Sketch a graph that shows the height of the ball as a function of time. with examples.
10.
two sections as shown. c) Determine the dimensions that give the maximum area. b) Determine the domain and range of this area function.
12. how the domain and range are determined from a table of values. a graph. 21. a)
17. Describe. b) Sketch the graph of a relation that is not a function and whose domain is 5x [ R | x $ 246 and range is 5 y [ R6.1.4
9. a) f (x) 5 !3 2 x 1 2
b) g(x) 5 x 2 2 3x a) f (x) 5 4x 1 1 b) f (x) 5 !x 2 2 c) f (x) 5 3(x 1 1) 2 2 4 d) f (x) 5 22x 2 2 5
c) h(x) 5 !x 2 1
a) f (x) 5 23x 1 8 b) g(x) 5 20. and range. Use a graphing calculator to graph each function and determine the domain
13. Write the domain and range of each function in set notation.
(3. the point (21. the relation 5(22. The equation f (21) 5 3 means "When x 5 21.
When a relation is a function.
3. For example. 4). you can use the vertical-line test. 3). (0.
A
A
y 4 2 4 2 0 2 4 2 4
A3:
x
2 1 0 1 2
B
0 3 4
5 4 3 5
4 3 0 3 4
B
y 4 2 4 2 0 2 4 2 4
A4:
If you have the graph of the relation." in other words.2. 23). y 5 3. 3). you can see if any first elements appear more than once. the relation is not a function. If you have the equation of the relation. f is a name for the function and f (a) is the value of y or output when the input is x 5 a. For example. (21. For example. 0).
38
Chapter 1
NEL
. but the relation 5 (25.
If the relation is shown in a mapping diagram. If you can draw a vertical line that crosses the graph in more than one place.
A2:
If the relation is described by a list of ordered pairs. there must be only one value of the dependent variable for each value of the independent variable. Examples 1.1
Study
Mid-Chapter Review
FREQUENTLY ASKED Questions
Q:
A1:
Aid
How can you determine whether a relation is a function?
• See Lesson 1. you can look at the arrows. If more than one arrow goes from an element of the domain (on the left) to an element of the range (on the right).
• Try Mid-Chapter Review
Questions 1 and 2. (2. you can use function notation to write the equation. graph A shows a function but graph B does not. you can substitute numbers for x to see how many y-values correspond to each x-value. 4). Examples 1. For a relation to be a function. then the relation is not a function.
What does function notation mean and why is it useful?
x
Study
Aid
Q:
A:
• See Lesson 1. If a single x-value produces more than one corresponding y-value. 0). because 24 and 3 each appear more than once as first elements. (3. 24). the equation does not represent a function. diagram A shows a function but diagram B does not. (24.1. 3) belongs to the function. If they do. (1. and 3. you can write the equation y 5 4 2 x 2 in function notation as f (x) 5 4 2 x 2. 0) 6 is not. 0) 6 is a function.
2.
• Try Mid-Chapter Review
Questions 3 and 4. (24. then an element in the domain corresponds to two elements in the range. and 4. For example. For example. The equation x 2 1 y 2 5 25 does not represent a function because there are two values for y when x is any number between 25 and 5. so the relation is not a function. 3). (5. the equation y 5 4 2 x 2 is the equation of a function because you would get only one answer for y by putting a number in for x.
Domain 5 5x [ R | x $ 216. such as v(t) to describe velocity as a function of time. so g(x) is always positive or zero. substitute 21 for x in the function equation: f (21) 5 4 2 (21) 2 5421 53 Or you can read the value from a graph. 3) 2 4 x
How can you determine the domain and range of a function?
Study
and 3. Range 5 5y [ R | y # 46
If g(x) 5 !x 1 1. So x can be any real number greater than or equal to 21 and y can be any real number greater than or equal to 0. Also. You can also determine the domain and range from the equation of a function. then any value of x will work in this equation. f (x) is always less than or equal to 4. 0) and continues forever in the positive x direction and positive y direction. because x 2 is always positive or zero.
Aid
The domain of a function is the set of input values for which the function is defined. the square root sign refers to the positive square root. or the number inside the square root sign would be negative. Range 5 5y [ R | y # 46
A
• See Lesson 1. if f (x) 5 4 2 x 2. You can express these facts in set notation: Domain 5 5x [ R6. For example. and 8. you can work with more than one function at a time by giving each function a different name. 7. then x cannot be less than 21.4. or C(n) to describe the cost of producing n items. you can see the domain and range.
y 4 2 4 2 0 2 4 y = f(x) x 2 4
Graph B starts at the point (21.Mid-Chapter Review
To evaluate f (21). Set notation can be used to describe the domain and range of a function. If you have the graph of a function. t.
Domain 5 5x [ R | x $ 216 . Because this function has a maximum value at the vertex. y cannot have a value greater than this maximum value. 3) 4 2 4 2 0 2 4 y = f(x) (0. The range is the set of output values that correspond to the input values. so x [ R. Examples 2 • Try Mid-Chapter Review
Questions 6. Range 5 5y [ R | y $ 06
B
y 4 2 2 0 2 4 2 4 6 8 y = g(x) x
NEL
Introduction to Functions
39
. Range 5 5 y [ R | y $ 06 Domain 5 5x [ R6. Also. You can choose meaningful names. Function notation is useful because writing f (x) 5 3 gives more information about the function—you know that the independent variable is x—than writing y 5 3. Also. Q:
A:
y ( 1. as in the following examples: Because graph A goes on forever in both the positive and negative x direction. x can be any real number.
40
Chapter 1
NEL
.4
d) f (x) 5 |x |
c) f (x) 5 !x
6.
7. explain why. c) A circle has a centre at (0. b) Determine the outputs for the input numbers 1.2
e) y 5 2 (x 2 3) 1 5 f ) y 5 !x 2 4 5
2
enclosed as a function of the width. a) A parabola has a vertex at (22.
function. Determine the domain and range of each relation in
question 1. d) A circle has a centre at (2. and 7. Then she asked them to multiply the resulting difference by the number they first thought of.
Lesson 1. Use numeric and graphical representations to show
that x 2 1 y 5 4 is a function but x 2 1 y 2 5 4 is not a function. (2. and y 5 4 is its minimum value. A farmer has 600 m of fencing to enclose a
d)
5
y
rectangular area and divide it into three sections as shown.PRACTICE Questions
Lesson 1. a) Graph the function f (x) 5 22(x 1 1) 2 1 3. ii) 3f (2) 2 5. (2. and y 5 5 is its maximum value.
5
0
x 5 a) Write an equation to express the total area
2.
multiply it by 5. Graph each function and state its domain and range. a) Use function notation to express the final answer in terms of the original number. A teacher asked her students to think of a number. c) Determine the maximum result possible. 4).
a) f (x) 5 x 2
1 b) f (x) 5 x
x 2 4
Lesson 1. 0) and a radius of 7. For those
which are. 4). 2). c) Determine the dimensions that give the maximum area. b) A parabola has a vertex at (3. c) What does f (23) represent on the graph of f ? d) Use the equation to determine i) f (1) 2 f (0).3
c) 4 2 4 2 0 2 4
y
5.1
1. and iii) f (2 2 x). 5) and a radius of 4. 5).
Lesson 1. 21.
b) Determine the domain and range of this area
3. Determine the domain and range for each. 8.
b) Evaluate f (23). 5)6
b) 2 0 3 7 1 1 3 4
4. a) 5(1. and subtract the product from 20. Determine which relations are functions. 3). (4.
Is this relation a function? Explain. using the company's prices.1. C. Tom needs to do the reverse of what the company's function does.5
GOAL
The Inverse Function and Its Properties
YOU WILL NEED
Determine inverses of linear functions and investigate their properties.
E
Cost ($) 450 850 1250 1650 2050
Area (sq ft) 40
F. B. The company calculates the cost to the customer as a function of the area to be paved. What is the independent variable in table A? the dependent variable? Is the relation in table A a function? Explain.
inverse of a function the reverse of the original function. Tom wants to express area in terms of cost to see how much of his yard he can pave for different budget amounts.
A
x Area (sq ft) 40 80 120
y Cost ($) 450
What relation can Tom use. Write the equation for f (x) that describes the cost as a function of area. E.
• graph paper • MiraTM (transparent
mirror) (optional)
INVESTIGATE the Math
The Backyard Paving Company charges $10/sq ft for installing interlocking paving stones. undoes what the original function has done
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Introduction to Functions
41
. plus a $50 delivery fee. Use the same scale of 2100 to 2100 on each axis. Graph f (x). D. Copy and complete table E for Tom. Graph this inverse relation on the same axes as those in part D. and how is it related to the function used by the company?
160 200
Copy and complete table A. What is the independent variable? the dependent variable? How does this table compare with table A? The relationship in part E is the inverse of the cost function.
?
A.
O. M. Use inverse operations on the cost function.
Draw the line y 5 x on your graph.
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. Make a list of all the connections you have observed between the Backyard Paving Company's cost function and the one Tom will use. K. or fold your graph paper along that line.
J. to solve for x. Place a Mira along the line y 5 x. the operations on x are as follows: Multiply by 25 and then add 2.
Reflecting
L. I. subtract 2 and then divide the result by 25. Is the inverse a function? Explain.
I wrote down the operations on x in the order they were applied. i) How are the slopes related? ii) How are the y-intercepts related? iii) Use the slope and y-intercept to write an equation for the inverse. What do you notice about the two graphs? Where do they intersect? Compare the coordinates of points that lie on the graph of the cost function with those which lie on the graph of its inverse. Compare this equation with the equation of the inverse you found in part I. f.
Then I worked backward and wrote the inverse operations.
How are the domain and range of the inverse related to the domain and range of a linear function? How could you use inverse operations to determine the equation of the inverse of a linear function from the equation of the function?
APPLY the Math
EXAMPLE
1
Representing the equation of the inverse of a linear function
Find the inverse of the function defined by f (x) 5 2 2 5x.G. What do you notice? Write the slopes and y-intercepts of the two lines. To reverse these operations.
How would a table of values for a linear function help you determine the inverse of that function?
i) How can you determine the coordinates of a point on the graph of the
inverse function if you know a point on the graph of the original function? ii) How could you use this relationship to graph the inverse?
N.
Jamie's Solution: Reversing the Operations
In the equation f (x) 5 2 2 5x.
H.
so I switched x and y in the equation. so it must be a function. then (y.
x 2 2 5 25y 1 2 2 2 x 2 2 5 25y x22 5y 25 f 21 (x) 5 x22 22x or f 21 (x) 5 25 5
I wrote the equation in function notation.5
f 21 (x) 5 x22 or 25 1 2 f 21 (x) 5 2 x 1 5 5
I used these inverse operations to write the equation of the inverse.
I knew the inverse was a line.
The inverse is a function. a) b) y y
y = f(x) 4 2 4 2 0 2 4 2 4 x y = g(x) 8 8 4 4 0 4 8 4 8 x
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Introduction to Functions
43
.1. This use of 21 is different from raising values to the power 21. The graph of y 5 f 21 (x) is a straight 1 line with slope 5 2 . since all linear relations except vertical lines are functions.
The function f is the inverse of the function f. x) is on the inverse graph.
EXAMPLE
2
Relating the graphs of functions and their inverses
Use the graph of each function to obtain the graph of the inverse.
Communication
21
Tip
The inverse is linear. y) is on the graph of f(x). The graph 5 passes the vertical-line test.
Lynette's Solution: Interchanging the Variables
f (x) 5 2 2 5x y 5 25x 1 2 x 5 25y 1 2
I wrote the function in y 5 mx 1 b form by putting y in place of f(x). Is the inverse a function? Explain. I knew that if (x. I solved for y by subtracting 2 from both sides and dividing both sides by 25.
1). I plotted the points in red. 0) of g21 (x). 21).
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Chapter 1
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.
The inverse is not a function: The graph fails the vertical-line test at x 5 1. Then I switched the coordinates to find the two points (0. 3). (1. 23). 0). Plot the points for the inverse and draw the line y 5 x to check for symmetry. 1).
I checked that the points on one side of the line y 5 x were mirror images of the points on the other side.and y-intercepts of g(x): (5. (4. 3).Carlos's Solution
a) f (x) is a function represented by the
So f 21 (x) is 5(1. 22)6 .
y 4 f(x) 4 2 2 0 2 4 2 f –1 y=x x 4
set of points 5 (23. I wrote the coordinates of the x. so a vertical line drawn here would go through three points. I drew the line y 5 x and checked that the graphs of g(x) and g21 (x) crossed on that line.
The inverse is a function. That gave me the inverse. I plotted the two points of g21 (x) and joined them with a straight line. (3. (22.and y-coordinates of each point. (1. (1. 0) and (0. 22).
b) g(x) 4 2 4 2 0 2 4 2 y=x x 4 g –1(x) y
There are three red points for x 5 1. (0. 1).
The inverse is a function because it passes the vertical-line test. 4)6. (22. (0. 1). 5) and (3. 0). I noticed that they were the intercepts. (21.
I wrote the coordinates of the points in the graph and then switched the x.
I wrote the temperature function with y and x instead of T(d) and d.015x x 5 11 1 0. so I chose to end the domain at 5000. 22 °C.5
EXAMPLE
3
Using the inverse of a linear function to solve a problem
Recall from Lesson 1. for example.015y
I realized that d is 0 m on the surface. d(22) 5 22 2 11 0.015d. c) State the domain and range of T 21 (d ). The deeper mine has a depth of 4100 m. Because I had switched the variables.
The inverse function is used to determine how far down a mine you would have to go to reach a temperature of. the depth is about 733 m.015y x 2 11 5 0.015d y 5 11 1 0. I wrote the inverse in function notation.
a) Domain 5 5d [ R |0 # d # 50006
Erynn's Solution
b)
Range 5 5T(d ) [ R | 11 # T(d ) # 866 T(d ) 5 11 1 0.015
I switched x and y and solved for y to get the inverse equation.015 Range 5 5d(T ) [ R | 0 # d(T ) # 50006 function of the temperature. This is the beginning of the domain.2 that the temperature below Earth's surface is a function of depth and can be defined by T(d ) 5 11 1 0.
c) Domain 5T [ R | 11 # T # 866 d) The inverse shows the depth as a
x 2 11 5y 0.015 T 2 11 d(T ) 5 is the inverse function. I calculated the beginning and end of the range by substituting d 5 0 and d 5 5000 into the equation for T(d). 0. and the range of the inverse is the same as the domain of the original function. d) Explain what the inverse represents. b) Determine the inverse of this function. I substituted 22 for T in the equation to get the answer.
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Introduction to Functions
45
.
8 733 When the temperature is 22 °C. The domain of the inverse is the same as the range of the original function.1. Someone planning a geothermal heating system would need this kind of information. I knew that y was now distance and x was temperature. a) State the domain and range of T(d ).
a) is a point on the graph of y 5 f 21 (x). For each graph. (3. (2. rewrite this x28 . Which inverse relations are functions? a) b) y y 5 5
3 5 3 1 10 1 3 5 x 3 5 5 3 3 1 10 1 3 5 x 3 5
46
Chapter 1
NEL
. • If (a. 4). 1). To reverse this function. the inverse is also a linear function. (4. b) is a point on the graph of y 5 f(x). 1)6
identify the points that are common to the function and its inverse. Copy the graph of each function and graph its inverse. Determine the inverse relation for each set of ordered pairs. to apply the function defined by f(x) 5 5x 1 8. This implies that the domain of f is the range of f21 and the range of f is the domain of f21.In Summary
Key Ideas
• The inverse of a linear function is the reverse of the original function.
relation and its inverse. • The graph of the inverse is the reflection of the graph of y 5 f(x) in the line y 5 x. equation as x 5 5y 1 8 and solve for y to get y 5 5 • If the original function is linear (with the exception of a horizontal line). then (b. 5). if y 5 5x 1 8.
Need to Know
• A way to determine the inverse function is to switch the two variables and solve for the previously independent variable. (2. For example. For example. (5. (0. 5). Graph each
2. It undoes what the original has done and can be found using the inverse operations of the original function in reverse order.
CHECK Your Understanding
1. subtract x28 8 from x and then divide the result by 5: f 21 (x) 5 . 21). 6)6 b) 5 (2. Which of the relations and inverse relations are functions? a) 5 (22. 5 • The inverse of a function is not necessarily a function itself. multiply x by 5 and then add 8. 3). • f 21 is the notation for the inverse function of f.
b) Write f 21 as a rule. uses a 5 simpler rule to convert from Celsius to Fahrenheit: Double the Celsius temperature. Repeat parts (a) to (e) for g(x) 5 2 1x 1 3. Explain how you can tell that f 21 is also a linear function. uses this rule to convert
centimetres to inches: Multiply by 4 and then divide by 10. Graph f and f 21 on the same axes. e) Another day. (2. 4). Who might use this rule? c) Determine f 21 (x). The formula for converting a temperature in degrees Celsius into degrees
A
Fahrenheit is F 5 9C 1 32. according to Ben's rule. (3. Ben. and state the domain and range of both the function and its inverse. the temperature was 70 °F. Use function notation to express this temperature in degrees Fahrenheit. sketch the graphs of the function
K
and its inverse. another American visitor to Canada. 2 p(x) 5 6 2 x. d) One day. (1. how do the domain and range of the function compare with the domain and range of the inverse? a) 5 (21. determine each value. a) Determine f
b) c) d) e) f)
for the linear function f (x) 5 5x 2 2. Compare the slopes of the two lines. an American visitor to Canada. Call the function f and let x represent the temperature in degrees Celsius. a) Use function notation to write an equation for this rule. State the coordinates of any points that are common to both f and f 21. the temperature was 14 °C. a) Write g21 as a rule. then add 30.
a) g(13) b) g(7)
c)
g(13) 2 g(7) 13 2 7
e) g21 (7) f) g21 (13) 2 g21 (7) 13 2 7
d) g21 (13)
11.
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Chapter 1
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. h(x) 5 2x 2 1. Use function notation to express this temperature in degrees Celsius. and q(x) 5 2. In each case. Let the function g be the method for converting centimetres to inches. determine the inverse. For g(t) 5 3t 2 2. Shirelle. 2). 8)6 c) f (x) 5 1 2 3x b) d) y 4 4 3 2 2 6 1 x 8 5 0 4 2 2 4 2
4
9. Explain what parts (c) and (f ) represent in question 10. 12. 6). b) Describe a situation in which the rule for g21 might be useful.
21
10.8. For each function.
13.
Then evaluate. Use function notation to represent his height in
centimetres. Explain why the ordered pair
(2. d) One day. Self-inverse functions are their own inverses.
15. Then she would earn $8.05(40) 1 0.5
c) Determine g(x) and g21 (x). tall. The ordered pair (1. Find three linear functions that
are self-inverse. 15 cm of snow fell. Given f (x) 5 k(2 1 x).63x 2 1. For example.
14. plus 5% of her sales over $1000. a) Graph the relation between Tiffany's total pay for a 40 h work week and her sales for that week. b) Write the relation in function notation. Write the correct equation for the relation in the form y 5 mx 1 b.29.05/h. for a 40 h work
week. Tiffany is paid $8. 5) belongs to a function f. Determine the inverse of the inverse of f (x) 5 3x 1 4.
Definition: Inverse of a Linear Function Methods:
Examples:
Properties:
Extending
19. Use function notation to represent this
amount in inches. find the value of k if f
T
21
(22) 5 23.
17. c) Graph the inverse relation. Use a chart like the one shown to summarize what you have learned about
C
the inverse of a linear function.
20. Ali did his homework at school with a graphing calculator.05($800) 5 $362. Once he got home. d) Write the inverse relation in function notation.1. e) Write an expression in function notation that represents her sales if she earned $420 one work week.
18.
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Introduction to Functions
49
.
e) Ben is 5 ft 10 in. he realized he had mixed up the independent and dependent variables. He determined that
the equation of the line of best fit for some data was y 5 2. 1) cannot belong to f 21.
16. suppose Tiffany sold $1800 worth of merchandise.
1 3 2 Repeat part C for y 5 . and for y 5 3|x|.
NEL
1 Graph the parent functions f (x) 5 x 2. and y 5 2 .
When f(x) 5 ! x. and 2 y 5 22x 2.U.
?
A.
Communication
Tip
EXPLORE the Math
Anastasia and Shelby made patterns with parabolas by applying transformations to the graph of the parent quadratic function y 5 x 2. g(x) 5 x2d
The function defined by g(x) 5 af(x 2 d) 1 c describes a transformation of the graph of f.
Anastasia thinks they could make more interesting patterns by applying transformations to other parent functions as well. Sketch and label each graph. Without using a calculator. g(x) 5 a(x 2 d) 2 1 c. Describe the transformations in words.n . Compare the effect of these transformations with the effect of the same transformations on quadratic functions. 1 When f(x) 5 . use what you know about transformations of quadratic functions to sketch the graphs of y 5 3x 2. h(x) 5 .
Use brackets when entering transformed versions of y 5
D.
When f(x) 5 x 2. Predict what the graphs of y 5 3f (x) 1 2 and y 5 3f (x) 2 1 for each of the other parent functions will look like. g(x) 5 !x.6
YOU WILL NEED
Exploring Transformations of Parent Functions
GOAL
• graphing calculator
or graphing software
Investigate transformations of parent functions. y 5 1 !x. press
MATH
1
X.
When f(x) 5 |x|. y 5 1|x|. Shelby wonders whether the transformations will have the same effect on the other functions as they do on quadratic functions. Sketch y 5 3x 2 1 2 and y 5 3x 2 2 1 without a calculator. along with a sketch of the parent function. g(x) 5 a ! x 2 d 1 c.
Do transformations of other parent functions behave in the same way as transformations of quadratic functions?
B. x a 1 c. and x j(x) 5 |x|.
E. 2 x x 2x and y 5 22|x|.1. y 5 1 x 2.T.
Predict what the graphs of y 5 3!x. To enter f(x) 5 |x|. Verify your predictions with a graphing calculator. y 5 . Sketch and label each curve on the same axes.
50
Chapter 1
. and y 5 22!x will look 2 like. Make labelled sketches and compare them with transformations on quadratic functions as before. Use a graphing calculator to verify your predictions. g(x) 5 a|x 2 d| 1 c. Describe the transformations in words.
Tech
Support
1 : x
C.
Experiment with each of the parent functions to create patterns on
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Introduction to Functions
51
. I.6
F. and y 5 3f (x 2 1) 1 2. y 5 1 f (x 1 1). c. How do the
4. K.
Reflecting
H. The graph of y 5 2x 2 is narrower than the graph of y 5 x 2. What do you know about the graphs of the following equations? 1 a) y 5 !x 2 1 1 2 b) y 5 |x 2 1| 1 2 c) y 5 12 x21 would you compare the graphs of the following pairs of equations? a) y 5 !x and y 5 2 !x b) y 5 |x| and y 5 2|x| 1 1 c) y 5 and y 5 2 x x following graphs compare? a) y 5 2!x and y 5 !x b) y 5 2|x| and y 5 |x| 2 1 c) y 5 and y 5 x x a graphing calculator screen. 1 compare with the graph for which 0 . 2).
Repeat part E for y 5 f (x 2 2).
FURTHER Your Understanding
1. The graph of y 5 x 2 opens up and the graph of y 5 2x 2 opens down. G. The shape of the graph of g(x) depends on the graph of the parent function g(x) and on the value of a. the constants a. 1?
In Summary
Key Idea
• In functions of the form g(x) 5 af(x 2 d) 1 c. a .1. J.
3.
How did the effect of transformations on parent functions compare with that on quadratic functions? When you graphed y 5 af (x 2 d ) 1 c. and d each change the location or shape of the graph of f(x). Did the transformations have the same effect on the new parent functions as they had on quadratic functions? Explain. what were the effects of c and d? How did the graphs with a $ 0 compare with the graphs with a # 0? How did the graphs for which a . The graph of the equation y 5 (x 2 1) 2 1 2 is the graph of a parabola that
2. How
opens up and has its vertex at (1. 2 Examine your sketches for each type of transformation.
y 5 "x x y 0 1 4 9 10 y 5 " 2x x 0 0. as a function of its length. 0) for this graph and transformation
y 4 y = x2 4 4 2 2 0 2 y= 4 x2 2 4 4 x
Graph both functions on the same set of axes. (22.5 8 y
What transformation must be applied to the graph of y 5 f(x) to get the graph of y 5 f(kx)?
invariant point a point on a graph (or figure) that is unchanged by a transformation—for example. She wonders 1 what transformation is caused by multiplying x by 10. In this formula. Compressions. C.5 2 4. • p(L) represents the time in seconds • L represents the pendulum's length in metres
Shannon wants to sketch the graph of this function.7
YOU WILL NEED
Investigating Horizontal Stretches.1. 0) and (2. and reflections to parent functions
INVESTIGATE the Math
1 The function p(L) 5 2p"10 L describes the time it takes a pendulum to complete one swing.
52
Chapter 1
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.
How could you transform the graph of y 5 "x to obtain the graph of y 5 "2x?
Compare the position and shape of the two graphs.
?
A. from one side to the other and back. and Reflections
GOAL
• graph paper (optional) • graphing calculator
Investigate and apply horizontal stretches. Are there any invariant points on the graphs? Explain.
B. She knows that the parent function is f (x) 5 "x and that the 2p causes a vertical stretch. State the domain and range of each function. compressions.
Copy and complete tables of values for y 5 "x and y 5 "2x .
1. 0?
APPLY the Math
EXAMPLE
1
Applying horizontal stretches.
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53
. 1 a) f (x) 5 x 2 b) f (x) 5 c) f (x) 5 |x| x Write a summary of the results of your investigations. but for horizontal stretches/ compressions f(kx). the scale factor is a. The
quadratic function f (x) 5 x 2 is the parent function. Compare the points in the tables of values. L. In both cases. describe the transformations required to graph them from the parent function. What happens to the 2 point (x. y 5 a xb
For each pair of functions. y) under this transformation? Describe the transformation in words. a compression occurs. identify the parent function. K. with x multiplied by a number.
H. investigate the effect of varying k in y = f (kx) on the graphs of the given parent functions. and iii) less than 0. In each case.
1 pendulum function p(L) 5 2p "10 L ?
Communication
Tip
In describing vertical stretches/compressions af (x).
Reflecting
I. Using a graphing calculator. 1 5
2
b) y 5 |0. 2 Repeat parts A through D for y 5 "x and y 5 "2 x. try values of k that are i) between 0 and 1.
Repeat parts A through D for y 5 "x and y 5 "1 x. Explain why this is a good description. and sketch all three graphs on the same set of axes. 1? ii) 0 . G. Explain how you would use the graph of y 5 f (x) to sketch the graph of y 5 f (kx). and for a scale factor between 0 and 1. ii) greater than 1. for k a scale factor greater than 1.
E. y 5 |2x 2 3|
Ana's Solution
a) These functions are of the form y 5 f (kx)2.
1 What transformation is caused by multiplying L by 10 in the
J.
How is the graph of y 5 f (2x) different from the graph of y 5 2f (x)? How is the graph of y 5 f (2x) different from the graph of y 5 2f (x)? What effect does k in y 5 f (kx) have on the graph of y 5 f (x) when i) |k| . a stretch occurs. 1? iii) k .25x|. y) under this transformation? This transformation is called a horizontal compression of factor 1. |k| . the scale 1 factor is .7
D. How could you use the first table to obtain the second? What happens to the point (x.
I saw that these functions were y 5 x2. and reflections
a) y 5 (4x) 2. F. compressions.
1). Q 1 . 2). 1) corresponded to the new point (4. 4 R . but is 6 1 4 for y 5 (4x)2. (2. Instead of using an x-value of 61 to get a y-value of 1.25x x
I joined these points to the invariant point (0. (10. stretch the graph of 5 y 5 x 2 horizontally by the factor 5. 9) on y 5 x2 by 1 to find three points. stretch the graph of y 5 |x| horizontally by the factor 4. That makes sense. Then I used symmetry to complete the other half of the graph. 1 R . 5 I knew from the absolute value signs that the parent function was the absolute value function.
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.25x|. I need an x-value of 65. and (12. 4). 0) to graph one-half of the parabola. 9 R . 1). on y 5 (4x)2. since the x-value required to make y 5 1 is 61 for y 5 x2.
To graph y 5 ( 1 x 2 ) . 1). The point that originally was (1. 0) and symmetry to complete the graph of y 5 ( 1 x) 2. 2). and (3. I used the same x-coordinates as before and multiplied by 5. 9) to plot.To graph y 5 (4x)2. 4
y 8 6 4 2 16 12 8 4 0 4 8 12 16 x y = x2 y = (4x)2
I multiplied the x-coordinates of the points (1.
I plotted these points and joined them to the invariant point (0. (8.
3
When x is multiplied by a number greater than 1. 3) on the new graph.
y 8 6 4 2 16 12 8 4 0 4 8 y = x2 y = (4x)2
1 y = ( – x) 2 5
This time. 4). 0) to graph one-half of the stretched absolute value function.
1 I knew that the stretch factor was 0. I used symmetry to complete the graph.
x 12 16
b) The parent function is f (x) 5 |x|. Q 1 . the graph is compressed horizontally. 1).25 5 4. 1). and 4 4 2
Q 4. and (3. So I multiplied the x-coordinates of the points (1. compress the graph of y 5 (x)2 horizontally by the factor 1.
To graph y 5 | 0. I used the invariant point (0. (2. x is multiplied by a number between 0 and 1. so the graph is stretched horizontally. 3) on y 5 |x| by 4 to find the points (4.
y 3 2 1 16 12 8 4 0 4 8 12 16
y= x y = 0. which gave me points (5. and (15.
(4. The equation is y 5 1 .25. 2). and point (3. So I could complete the equation. 1). Since the stretch scale factor is 6. and 0 . Each x-coordinate has been divided by 24. 3). The red graph is a compressed version of the green graph that had been flipped over the y-axis. I divided the corresponding x-coordinates to find k: 1 4 20. 22) corresponds to (23. 4) corresponds to (24. 22). k . 2) corresponds to (3. so k 5 24
The green graph has been compressed horizontally and reflected in the y-axis to produce the red graph. 21). 6
The graph has been stretched horizontally. 1. ( 6 x)
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. The equation is y 5 "24x. 2). 4
b) The green graph is a graph of the square root
The red graph is a stretched-out version of the green graph. 2 Each x-coordinate has been multiplied 1 by 6. 3) corresponds to point (12. The red graph is further away from the asymptotes than the green graph.Point (2. Since the stretch scale factor is 4. 4). (1. (1. (16.
c) The parent function is f (x) 5 . 1) corresponds to (20. it follows that k 5 1. 2). 21) corresponds to (26. function f (x) 5 "x. so k must be between 0 and 1. Therefore. The x-coordinates of points on the red graph are 4 times the ones on the green graph. it follows that k 5 1. 2 (21. 0) and has the shape of a half parabola on its side. 1) corresponds to (6. The equation is y 5 | 1 x|.
1 x
I recognized the reciprocal function because the graph was in two parts and had asymptotes. 2) corresponds to (21. (2 1. 4 The green graph is the square root function because it begins at (0. 2) corresponds to point (8. (1.25 5 24 4 4 21 5 24 16 4 24 5 24. 1). so it must have been stretched. and 0 . The red graph is the green graph stretched horizontally by the factor 4. k . k is negative and less than 21. 1.
the graph is stretched horizontally by a scale 1 factor of 1 5 10. 1) moves to (10. (40. the graph is stretched vertically by a scale factor of 2p. Since 0 .6) .3 (to one decimal place): (10. I multiplied the x-coordinates by 10 to find points on the horizontally stretched graph: (1. k .3). 1.
10
y 15 10 5 0 y 15 y=2 10 5 0 5 y=
1x —– 10
Because a 5 2p and 2p . 2) moves to (40.
Shannon's Solution
The original equation was in the form y 5 af(kx). 2) moves to (40. 1).
Use transformations to sketch the graph of the pendulum function 1 p(L) 5 2p" 10 L.
y= x
1x —– 10
x
10 15 20 25
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Introduction to Functions
57
.
y= x y= 5
1x —– 10
x
10 15 20 25
Then I applied the vertical stretch to the red graph. in metres. I multiplied the y-coordinates by 2p. in seconds. 1) moves to (10. that it takes for a pendulum to complete one swing and L is the length of the pendulum. (4. 2). where p(L) is the time.7
EXAMPLE
3
Using transformations to sketch the graph for a real situation
1 The graph of y 5 2p" 10 x is the graph of the parent function y 5 !x stretched horizontally by the factor 10 and vertically by the factor 2p. I applied the horizontal stretch.1. which is approximately 6. 1. 12. 6.
0. Write the equation of the blue graph.
Need to Know
y 5 y = x2 x 5 0 5
CHECK Your Understanding
1. the graph is compressed horizontally by the factor . the graph is stretched horizontally by the factor . • If g(x) 5 f(kx). 1 2 1 a) y 5 |0. 1.
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Chapter 1
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. then the value of k has the following effect on the graph of f(x): 1 • When |k| . a)
y 4 2 4 2 0 y= x x 2 4
2. y) on the graph of f(x) is the point a . |k| 1 • When 0 . b) The blue graph has been stretched horizontally by the factor 2 relative to the graph of y 5 !x and then reflected in the y-axis.
In Summary
Key Idea
x • The image of the point (x. the graph is also reflected in the y-axis. yb on the k graph of f(kx). identify the parent function and describe how the graph of
the function can be obtained from the graph of the parent function.Period versus Length for a Pendulum y 15 Time (s) p(L) 10 5 0 x 5 10 15 20 25 Length (m)
I drew a correctly labelled graph of the situation. |k| • When k . • Functions of the form g(x) 5 f(kx) have graphs that are not congruent to the graph of f(x). I copied the sketch onto a graph with length L on the x-axis and time p(L) on the y-axis. Then sketch both graphs on the same set of axes. The differences in shape are a result of stretching or compressing in a horizontal direction.5x| b) y 5 a xb c) y 5 !22x d) y 5 4 (5x)
The red graph has been compressed horizontally by the factor 1 relative 3 to the graph of y 5 x 2. |k| . Write the equation of the red graph. For each function. 1.
If you get two different results. Determine the
1 x` 2
value of k for each transformation. a) a horizontal stretch by the factor 4 b) a horizontal compression by the factor 1 2 c) a reflection in the y-axis d) a horizontal compression by the factor 1 and a reflection in the y-axis 5
12. h(x) 5 !25x Å5 1 1 1 c) f (x) 5 . Include examples that show how the transformations vary with the value of k. A quadratic function has equation f (x) 5 x2 2 x 2 6.
b) Apply a horizontal stretch with factor 2. g(x) 5 a x 2 b. g(x) 5 |22x|. ground is a function of the height from which it was dropped. transform the graph of f (x) to sketch g(x) and
h(x). h(x) 5 x 4x (2 1x) 3
d) f (x) 5 |x|. What two transformations
are required? Does the order in which you apply these transformations make a difference? Choose one of the parent functions and investigate. h(x) 5 (24x 2 ) 4 1 b) f (x) 5 !x.9. h(x) 5 `
for this function is t(h) 5 #4. Suppose you are asked to graph y 5 f (2x 1 4). Determine the
T
x-intercepts for each function. When an object is dropped from a height. the time it takes to reach the
A
10. b) Compare the graph of y 5 f (kx) with the graph of y 5 kf (x) for different values of k and different functions f (x). and state the domain and range of each function. What do you notice? d) Write the equations of the functions that result from the transformations
in parts (b) and (c). a) Describe the domain and range of the function.9. 1 a) f (x) 5 x 2. g(x) 5 x. Explain why these equations are the same. How are the transformations alike? How are they different?
Extending
1 14.
a) y 5 f (2x)
13. c) Apply a vertical stretch with factor 2.
60
Chapter 1
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. a) Describe how the graph of y 5 f (kx) can be obtained from the graph of
C
b) y 5 f a xb
1 3
c) y 5 f (23x)
y 5 f (x). a) Graph the function y 5 x . g(x) 5 . use a graphing calculator to verify which graph is correct. where h is in metres and t is in seconds. For each set of functions. The function y 5 f (x) has been transformed to y 5 f (kx).
15. b) Sketch the graph by applying a transformation to the graph of t(h) 5 !h. An equation
h
11.
64
Chapter 1
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. 21). 1) 2 4 x
I sketched the graph of f(x) and labelled the points (1. 1) and (21. c) State the domain and range of both functions. d 5 5 and c 5 4. in order. because there is a vertical stretch by the factor 2.APPLY the Math
EXAMPLE
2
Applying transformations to the equation and the graph 1 : x
Some transformations are applied. A horizontal stretch by the factor 3 and a reflection in the y-axis means that k 5 2 1. to the reciprocal function f (x) 5
• • • •
horizontal stretch by the factor 3 vertical stretch by the factor 2 reflection in the y-axis translation 5 units right and 4 units up
a) Write the equation for the final transformed function g(x). 1) 2 4 (1. a) g(x) 5 af 3k(x 2 d )4 1 c
Lynn's Solution
5
1 5 2f c2 (x 2 5) d 1 4 3 1 a2 (x 2 5) b 3 2 14
I built up the equation from the transformations. The vertical asymptote is x 5 0 and the horizontal asymptote is y 5 0. b) Sketch the graphs of f (x) and g(x). because the translation is 5 units right and 4 units up. 3 a 5 2.
b) Graph of f (x): y 4 2 4 2 0 ( 1.
1) 2 4 (3. 2) 4 2 (1.
4 2 0 ( 1. for g(x). Also. The asymptotes did not change. 2) and (21.
c) For f (x). I drew in the translated asymptotes first. 6) 3 1 y=– x 2 4 (8. The graphs do not meet their asymptotes.1. x cannot be 5 and y cannot be 4. the new horizontal asymptote is y 5 4. x cannot be 0 and y cannot be 0. 1) became (23. the new vertical asymptote is x 5 5. 21) became (3.8
y ( 3.
Domain 5 5x [ R | x 2 06 Range 5 5 y [ R | y 2 06 For g(x). Since all the points moved 5 right. I made a sketch of the stretched and reflected graph before applying the translation.
x= 5
Domain 5 5x [ R | x 2 56 Range 5 5 y [ R | y 2 46
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Introduction to Functions
65
. Then I drew the stretched and reflected graph in the new position after the translation. 5 right and 4 up. (1. since x and y still couldn't be 0. I labelled the graphs and wrote the equations for the asymptotes. 2) x
I applied the stretches and reflection to the labelled points by multiplying the x-coordinates by 23 and the y-coordinates by 2. Since all the points moved up 4. I used the equations of the asymptotes to help determine the domain and range. so for f(x). 1) 2 4
Graph of g(x):
8 6 y= 4 4 2 4 2 0 2 4 y 2 y = ——————— + 4 1 (x 5) – (2. 22). 2) x 6 8 10
To apply the translations.
and shifted 2 units left and 1 unit up. when you apply the translations by 3 subtracting 1 from the x-coordinate and 2 from the y-coordinate. so the answer can't be equation 7 and has to be equation 5. So.
This is the graph of a square root function that has been flipped over the x-axis.
Graph C is a parabola. I checked: The vertex is (4. a will be negative.
This is the graph of an absolute value function. this point
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Chapter 1
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. 1B. in the equation. It starts at (22. c 5 22 and d 5 21. the parabola opens upward. 1 when you multiply the x-coordinate by 0.5 5 2. Equation 5 is the equation of a parabola with vertex (4. stretched vertically. The equations for the asymptotes are x 5 21 and y 5 22.
Also. 21.3. 2). so it has been stretched horizontally or compressed vertically. so it has to match equation 5 or equation 7.3. The parent square root graph has been compressed horizontally or stretched vertically. Then. Graph C is wider than the parent function.
Graph E matches equation 1. and that has been stretched horizontally by the 1 factor 0. This is a transformation of the graph of y 5 becomes A 7. 2). k 5 0.
Graph C matches equation 5. so the x answer has to be equation 1 or equation 6. so it has been translated 2 units left and 1 unit down. Since a . 0. The equation must have a . so d 5 4 and c 5 2. in the equation. 10 The point (1.Graph B matches equation 3. and c 5 1.3 means that the parent 1 graph has been stretched horizontally by the factor 0. so. 21) instead of (0. 21B. 1) on the parent graph becomes A 3 . so d 5 21 and c 5 22. The parent graph has been reflected in the x-axis. 0). that opens up. d 5 22. 1 .
Graph D matches equation 2. This matches equation 1.
Also. compressions. 0. one at a time. • The value of c determines the vertical translation: • For c . the graph is compressed horizontally by the factor . then c and d together. the graph is also reflected in the y-axis. • When a . |k| . the graph is translated d units right. • The value of k determines the horizontal stretch or compression and whether there is a reflection in the y-axis: 1 • When |k| . 0.1. the graph of y 5 f(x) is stretched vertically by the factor |a|. |k| 1 • When 0 . 0. the graph is translated d units left. c 5 21 and d 5 4. and translated 4 units right and 3 units up. 0. so it has to match equation 7 because it is vertically stretched (narrow) and has vertex at (21.
In Summary
Key Ideas
• You can graph functions of the form g(x) 5 af 3k(x 2 d)4 1 c by applying the appropriate transformations to the key points of the parent function. to get the desired graph in fewer steps.
Graph G matches equation 7. 1. Equation 7 has a 5 23. the graph is compressed vertically by the factor |a|. so 21 . 4). 0. 1. 0. |k| • When k .
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Introduction to Functions
69
. It has been stretched horizontally. so k .
This is another square root function.8
Graph F matches equation 4. Graph G is a parabola that opens down. • For d . and reflections) are done before additions and subtractions (translations). 0. • For c . you can apply a and k together. since multiplications (stretches. which means that the vertex is (21. The parent function has been flipped over the y-axis. making sure to apply a and k before c and d. This order is like the order of operations for numerical expressions.
Need to Know
• The value of a determines the vertical stretch or compression and whether there is a reflection in the x-axis: • When |a| . the graph is translated c units down. as in graph G. • The value of d determines the horizontal translation: • For d . which means that the parabola opens down and is vertically stretched by the factor 3. k . the graph is stretched horizontally by the factor . |a| . 1. 1. so d 5 4 and c 5 3. the graph is also reflected in the x-axis. • For 0 . 0. 4). the graph is translated c units up. • When using transformations to graph.
12 8 y=? 16 12 8 4 4 0 4 8
y y = x2 x
4
8
23. a 2 translation 3 units left. and a translation 4 units down are applied to 1 f (x) 5 . and c. a translation 1 unit right. How are they alike? How are they different? Develop a procedure to obtain the graph of g(x) from the graph of f (x).
1. and d. Compare the graphs and the domains and ranges of f (x) 5 x 2 and
g(x) 5 !x. and state the domain and range for each transformation. d.
a) y 5 f (x) b) y 5 24f (x)
C
21. second parabola from the first. x c) A horizontal compression by the factor 1 .
b) Determine a possible equation for the second parabola. 1 3 d) y 5 f (2 (x 1 2) )
c) y 5 f a2 xb
20. a) A vertical stretch by the factor 2. and a translation 4 units right are applied to y 5 !x. k. determine the x-intercepts for each function.19. c. k.8
Determine a. a vertical stretch by the factor 3 3. The function y 5 f (x) has been transformed to y 5 af 3k(x 2 d )4 1 c. a reflection in the y-axis. The graphs of y 5 x 2 and another parabola are shown. and the order in which they would be applied. List the steps you would take to sketch the graph of a function of the form
y 5 af (k(x 2 d ) ) 1 c when f (x) is one of the parent functions you have studied in this chapter. and a translation 6 units down are applied to y 5 |x|. sketch the graph. a reflection in the x-axis.
Extending
22. If f (x) 5 (x 2 2) (x 1 5). Discuss the roles of a. b) A vertical compression by the factor 1 .
a) Determine a combination of transformations that would produce the
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Introduction to Functions
73
.
so the inverse function is f 3
A3:
21
(x) 5
x14 . Whenever k is negative.
• Try Chapter Review
Questions 10 and 11. so the x14 inverse function is f 21 (x) 5 . x 5 3y 2 4 x 1 4 5 3y 2 4 1 4
y 4 ( 4. or reflection. 3
4 2 0 2 4 2 y = f(x) y=x 4 (0. the graph is stretched horizontally by the 1 factor . y) is on the graph of f (x). When k is a number greater than 1 or less than 1 21. compression. the operations on x are as follows: Multiply by 3 and then subtract 4. if f (x) 5 3x 2 4. For example. x) is on the inverse graph. 0) 2 y=f
1(x)
x 1 4 5 3y x14 5 y. the graph is compressed horizontally by the factor . • Try Chapter Review
Questions 12 and 13.7. or reflection to the graph of a function?
• See Lesson 1. It undoes what the original has done. unless the original function represents a horizontal line. When k is a k number between 21 and 1. compression. if f (x) 5 3x 2 4. 3
Input 3 4 Output
Inverse
Input
3
4
A2:
If (x. 4)
x
If you have the graph of a linear function. This means that you can find the equation of the inverse by reversing the operations on x.5. For example. so you can switch x and y in the equation to find the inverse equation. Examples
1 and 2. you add 4 and divide by 3. you can write this as y 5 3x 2 4.
The inverse of a linear function is the reverse of the original function. then ( y. Then switch x and y and solve for y. To reverse these operations.
The graph of y 5 f (kx) is the graph of y 5 f (x) after a horizontal stretch. the graph is also reflected in the y-axis. you can graph the inverse function by reflecting in the line y 5 x.1
Study
Chapter Review
FREQUENTLY ASKED Questions
Q:
A1:
Aid
How can you determine the inverse of a linear function?
• See Lesson 1. Example 1. The inverse of a linear function is another linear function. k
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74
Chapter 1
.
Q:
Study
Aid
A:
How do you apply a horizontal stretch.
For a fundraising event. reflected in the x-axis. and a translation 4 units down. and state the domain and range of each function. and state its domain and range. 4) is on the graph of y 5 f (x). sketch its graph. For each set of functions. In each graph. x-intercepts of each function. a parent function has undergone a
transformation of the form f (kx). a) y 5 2f (x) c) y 5 3f (x 1 1) 1 4 b) y 5 f (2x) d) y 5 22f (2x 1 5) 1 1
b) y 5 22f (x)
1 2 d) y 5 f (2 (x 1 1))
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Introduction to Functions
77
. Determine the equations of the transformed functions graphed in red. a) The graph of f (x) 5 !x is compressed horizontally by the factor 1. The point (1. In each case. 2 and translated 3 units right and 2 units down. a) Explain what you would need to do to the graph
Determine the coordinates of the image of this point on the graph of y 5 3f 324(x 1 1)4 2 2.Chapter Review 11. A function f (x) has domain 5x [ R |x $ 246 and
range 5 y [ R | y . h(x) 5 ` x ` 4
19. transform the graph of
17. Explain your reasoning. Determine the domain and range of each function. a) Use function notation to express the total income from the event as a function of the number of people who attend. 1 2 a) f (x) 5 x 2. a translation 3 units right. y a)
8 4 8 b) 4 0 4 10 6 10 6 2 20 2 6 10
13. c) The organizers want to know how many tickets they need to sell to reach their fundraising goal. a) Is the order of the transformations important? b) Is there any other sequence of these transformations that could produce the same result?
15.8
expects to receive $15 000 from corporate sponsorship. and translated 4 units left and 1 unit up. If f (x) 5 (x 2 4) (x 1 3). Create a function to express the number of people as a function of expected income. h(x) 5 2(2x) 2 2 1 b) f (x) 5 |x|. g (x) 5 |24x|. Three transformations are applied to y 5 x 2: a
vertical stretch by the factor 2.7
14. 3 b) Graph the function in part (a) for f (x) 5 x 2.
Lesson 1. g(x) 5 a xb . function. State the domain of this new function.
x 4 y 8
18. 216. write the equation for the transformed
of y 5 f (x) to graph the function y 5 22f 3( 1 x 1 4)4 2 1. 1 b) The graph of y 5 is stretched vertically by the x factor 3. plus $30 from each person who attends the event. Explain your reasoning. a local charity organization
Lesson 1.
a) y 5 f (x) c) y 5 f a2 xb
12.
16. reflected in the y-axis. b) Suggest a reasonable domain and range for the function in part (a). determine the
x 6 10
f (x) to sketch g(x) and h(x).
A fluorescent bulb costs $3. c) Graph the inverse relation.
1 x22 b) f (x) 5 !3 2 x 2 4
a) f (x) 5
c) f (x) 5 2| x 1 1| 1 3
8. c) After how long is the fluorescent bulb cheaper than the regular bulb? d) Determine the difference in costs after one year. Determine the
value of k for each transformation. translation 2 units 1 left. d. a) Use function notation to write a cost equation for each type of bulb. (0. a) 5(22.
1. Explain your reasoning. (4. An incandescent light bulb costs $0. 3).65 to buy and $0. write the equation.
3. The function y 5 f (x) has been transformed to y 5 f (kx). For each function.50 to buy and $0. 5). and c.1
5 2 0 3 1 1 7
Chapter Self-Test
is a function. How are
5. b) State the domain and range of each function. a) a horizontal stretch by the factor 5 b) a horizontal compression by the factor 1 and a reflection in the y-axis 3
Determine a. and translation 2 units down.004/h for electricity to
run. applied to y 5 |x|
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78
Chapter 1
. and state the domain and range of both the function and its inverse. vertical stretch by the factor 3 . reflection in the y-axis.
7. determine the domain and range and whether the relation
2. and translation 3 units down. 6). Show your steps. 8)6 b) f (x) 5 3 2 4x the domain and range of a linear function related to the domain and range of its inverse? commission on her sales over $2500. sketch the graphs of the function and
6. and 2 translation 2 units right. reflection in the x-axis. d) Use function notation to write an equation of the inverse. applied to y 5 !x b) vertical stretch by the factor 4. Assume the light is on for an average of 6 h/day. a) Graph the relation between monthly earnings and sales. b) Use function notation to write an equation of the relation. a) The function shown at the left.001/h to run.
b) y 5 !x 1 2
4. determine the inverse. The function y 5 f (x) has been transformed to y 5 af 3k(x 2 d ) 4 1 c. Rebecca is paid a monthly salary of $1500. e) Use the equation in part (d) to express Rebecca's sales if she earned $1740 one month. translation 3 units right. Explain what the term inverse means in relation to a linear function. and state the domain and range of each transformed function. sketch the graph. 4 2
reflection in the x-axis. plus 4%
its inverse. Determine the domain and range of each function. At Phoenix Fashions. (2. applied to y 5 x
c) horizontal compression by the factor 1 . For each relation. Then evaluate. a) vertical compression by the factor 1 . k.
x
x
1
x ?
2
Suppose each dimension of the box is doubled. and x 1 2. length l. x 1 1. B. D. Write expressions for the volume and surface area of the new box after the dimensions are doubled.
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Equivalent Algebraic Expressions
83
. and height h? Use the given dimensions to write expressions for the volume and surface area of the box. By what factor has the volume increased? By what factor has the surface area increased? Explain how you know. By what factors will the surface area and volume of the box increase?
A.Getting Started
APPLYING What You Know
Doubling Dimensions
A rectangular box has dimensions x.
What are the formulas for the volume and surface area of a rectangular box of width w. C.
1
YOU WILL NEED
Adding and Subtracting Polynomials
GOAL
• graphing calculator
Determine whether polynomial expressions are equivalent.
EXAMPLE
1
Selecting a strategy to determine equivalence
Determine if f (t) and g(t) are equivalent functions. is given by f (t) 5 (25t 2 1 100t 1 1000) 2 (25t 2 1 75t 1 1200) Fred simplified f (t) to g (t) 5 175t 1 2200 .
Polynomials behave like numbers because.
? Are the functions f(t) and g(t) equivalent?
The numbers 1 and 2 in a1 (t) and a2 (t) are called subscripts. the functions are not equivalent. In this case. subtraction is equivalent to adding the opposite. Anita's Solution: Simplifying the Polynomial in f(t) f (t) 5 (25t 2 1 100t 1 1000) 2 (25t 2 1 75t 1 1200) 5 25t 2 1 100t 1 1000 1 5t 2 2 75t 2 1200 5 25t 2 200 Since g(t) 2 25t 2 200. f (t).2. Then I collected like terms.
LEARN ABOUT the Math
Fred enjoys working with model rockets. in metres. He wants to determine the difference in altitude of two different rockets when their fuel burns out and they begin to coast. they are used to distinguish one function from the other. The altitudes. in seconds. the result is a number. The difference in altitude. I know that with numbers. are given by these equations: a1 (t) 5 25t 2 1 100t 1 1000 and
Communication
Tip
a2 (t) 5 25t 2 1 75t 1 1200 where t is the elapsed time. so I subtracted the polynomials by adding the opposite of the second expression. This distinction is necessary because both functions are named with the letter a.
84
Chapter 2
NEL
. for any value of the variable.
their function values are different. in Y1 and his final function. C.1
If two expressions are not equivalent. g(t) . the functions must be different. the functions are not equivalent. B. Maria's Solution: Evaluating the Functions for the Same Value of the Variable f (t) 5 (25t 2 1 100t 1 1000) 2 (25t 2 1 75t 1 1200) f (0) 5 (25(0) 2 1 100(0) 1 1000) 2 (25(0) 2 1 75(0) 1 1200) 5 1000 2 1200 5 2200 g(0) 5 175(0) 1 2200 5 2200 Since f (0) 2 g(0).
Reflecting
A. then their graphs should be identical.
The functions are not equivalent. The exception is when the functions intersect.
Y2
Y1
Since the graphs are different. in Y2.
How is subtracting two polynomials like subtracting integers? How is it different? If Fred had not made an error when he simplified. Sam's Solution: Graphing Both Functions
I used t 5 0 because it makes the calculations to find f(0) and g(0) easier. If two functions are equivalent. whose method would have shown that his original and final expressions are equivalent? What are the advantages and disadvantages of the three methods used to determine whether two polynomials are equivalent?
Equivalent Algebraic Expressions
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85
. then.
I graphed Fred's original function. I zoomed out until I could see both functions. f(t) .2. for most values of t.
I substituted some values for x. y 5 0. I evaluated the second expression. using the same values for x. I tried x 5 0. 1. Evaluating both functions at a single value is sufficient to demonstrate non-equivalence. but it isn't enough to demonstrate equivalence.2. y. The expressions result in different values.
In Summary
Key Ideas
• Two polynomial functions or expressions are equivalent if • they simplify algebraically to give the same function or expression • they produce the same graph • Two polynomial functions or expressions are not equivalent if • they result in different values when they are evaluated with the same numbers substituted for the variable(s)
Need to Know
• If you notice that two functions are equivalent at one value of a variable. and 21 are often good choices.
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Equivalent Algebraic Expressions
87
. y. and z 5 1 and evaluated the first expression. since they usually make expressions easy to evaluate.1
EXAMPLE
3
Reasoning about the equivalence of expressions
Are the expressions xy 1 xz 1 yz and x 2 1 y 2 1 z 2 equivalent? Dwayne's Solution To check for non-equivalence. and z.
The values 0. • The difference of two polynomial functions or expressions can be determined by adding the opposite of one polynomial and collecting like terms. and z.
xy 1 xz 1 yz 5 0(0) 1 0(1) 1 0(1) 5 0
x 2 1 y 2 1 z 2 5 02 1 02 1 12 5 1
The expressions are not equivalent. it does not necessarily mean they are equivalent at all values of the variable. • The sum of two or more polynomial functions or expressions can be determined by writing an expression for the sum of the polynomials and collecting like terms.
b) Suppose m is a natural number that is greater than 2 and n 5 5m. Express 105 as the sum of five consecutive natural numbers. for the other three orientations of the L. f (3) 5 g(3). c) Express 91 as a sum of seven consecutive natural numbers. Show that the functions must be equivalent. Show that the functions must be equivalent. 70 has 5 as a divisor. Express n as the sum of five consecutive natural numbers. For example.Extending
15. Write expressions for the sum of the five numbers.
90
Chapter 2
NEL
. Sanya noticed an interesting property of numbers that form a five-square
capital-L pattern on a calendar page:
5 12 19
6 7 13 14 20 21
1 8 15
2 9 16
3 10 17
corner number
In each case that she checked. the sum of the five numbers was 18 less than five times the value of the number in the corner of the L. Suppose that f (2) 5 g(2). b) The sum of certain numbers in this pattern is 112.
16. and f (4) 5 g(4). b) Consider the two quadratic functions f (x) 5 ax 2 1 bx 1 c and g(x) 5 px 2 1 q x 1 r.
a)
17. Since 70 5 5 3 14. when x is the corner number. Determine the value c)
of the corner number. Suppose that f (2) 5 g(2) and f (5) 5 g(5). The number 70 can also be
expressed as the sum of five consecutive natural numbers: 70 5 12 1 13 1 14 1 15 1 16 105 5 5 3 21. a)
Consider the linear functions f (x) 5 ax 1 b and g (x) 5 cx 1 d. 5 1 12 1 19 1 20 1 21 5 5(19) 2 18 1 1 8 1 15 1 16 1 17 5 5(15) 2 18
a) Show that this pattern holds for any numbers on the calendar page. in the calendars shown.
2.5) 5 50t 3 2 50t 2 2 25t 2 2 25t 1 25t 1 12. Then I grouped and collected like terms. I multiplied each of the three terms in the trinomial by each of the terms in the binomial. h(t).
LEARN ABOUT the Math
In a physics textbook. ultimately landing in the water below the cliff.5
3 2 2
I used the commutative property for multiplication to create an equivalent expression.5).
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Equivalent Algebraic Expressions
91
.5 5 50t 3 2 75t 2 1 12.
? How can she simplify the expression for v(t) 3 h(t)?
EXAMPLE
1
Selecting a strategy to simplify a product: The distributive property
Simplify the expression v(t) 3 h(t) 5 (210t 1 5) (25t 2 1 5t 1 2. Kristina reads about an experiment in which a ball is thrown upward from the top of a cliff. v(t).5) 1 (5) (25t 2 1 5t 1 2. I used the distributive property to expand the product of the polynomials.
Sam's Solution
v(t)h(t) 5 (210t 1 5) (25t 2 1 5t 1 2.2
GOAL
Multiplying Polynomials
Simplify polynomials by multiplying. and when the ball moves toward. The height of the ball above the cliff.5) 5 (50t 2 50t 2 25t) 1 (225t 1 25t 1 12. and its velocity. at time t are respectively given by h(t) 5 25t 2 1 5t 1 2.5) 5 (210t) (25t 2 1 5t 1 2. Kristina learns that the product of the two functions allows her to determine when the ball moves away from. the top of the cliff.5 and v(t) 5 210t 1 5.
How does the simplified expression differ from the original? 2.
3. Next. Is this always necessary?
Explain.5) (210t 1 5) by multiplying each term in the second factor by each term in the first factor? Explain.
x
3 x x 1 2
Fred's Solution: Starting with the First Two Binomials
V 5 lwh V 5 (x 1 1) (x 1 2) (x 1 3) 5 (x 2 1 3x 1 2) (x 1 3)
I know that multiplication is associative.
WORK WITH the Math
EXAMPLE
2
Selecting a strategy to multiply three binomials
Determine a simplified function that represents the volume of the given box. I simplified by collecting like terms and arranged the terms in descending order. Would Sam have gotten a different answer if he multiplied
(25t 2 1 5t 1 2.Reflecting
1.
5 (x 2 1 3x 1 2) (x) 1 (x 2 1 3x 1 2) (3)
5 x 3 1 3x 2 1 2x 1 3x 2 1 9x 1 6 5 x 3 1 6x 2 1 11x 1 6
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Chapter 2
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. I multiplied the first two binomials together and got a trinomial. Finally. I took the 3 term from (x 1 3) and multiplied it by the trinomial. Sam grouped together like terms in order to simplify. so I can multiply in any order. Then I took the x-term from (x 1 3) and multiplied it by the trinomial.
w
A1 2w
A2 5 (2w 1 1) (w 2 1) 5 2w 2 2 w 2 1
change in area 5 A2 2 A1 5 (2w 2 2 w 2 1) 2 (2w 2 ) 5 2w 2 1 My prediction was wrong. Predict how the area will change if the length of the rectangle is increased by 1 and the width is decreased by 1. to represent the change in area. A1. A2. I increased the length by 1 and decreased the width by 1 by adding and subtracting 1 to my previous expressions. I let w represent the width and 2w the length. A1. A1 5 (2w)w 5 2w 2
w 1 A2 2w 1
To check my prediction. which is always positive. I took the difference of the new area.EXAMPLE
4
Representing changes in area as a polynomial
A rectangle is twice as long as it is wide. w represents width. Write an expression for the change in area and interpret the result. Their product gives the original area.
Kim's Solution
Since we are increasing the length and decreasing the width by the same amount. and the original area. The product gives the area of the new rectangle. I predict that there will be no change in the area. Substituting any positive value for w into 2w 2 1 results in a negative number. A2. This means that the new rectangle must have a smaller area than the original one.
94
Chapter 2
NEL
.
If f (x) has two terms and g(x) has three terms, how many terms will the product of f (x) and g(x) have before like terms are collected? Explain and illustrate with an example. b) In general, if two or more polynomials are to be multiplied, how can you determine how many terms the product will have before like terms are collected? Explain and illustrate with an example.
1 3 1 3 1 cubes. a) How many of the 27 smaller cubes are coloured red on i) three faces? iii) one face? ii) two faces? iv) no faces? b) Answer (i) to (iv) from part (a) for a 10 3 10 3 10 cube divided into one thousand 1 3 1 3 1 cubes. c) Answer (i) to (iv) from part (a) for an n 3 n 3 n cube. d) Check your results for parts (a) and (b) and by substituting 3 and then 10 into the expressions obtained in your answers to part (c).
16. Many tricks in mental arithmetic involve algebra. For instance, Cynthie
claims to have an easy method for squaring any two-digit number whose last digit is 5; for example, 75. Here are her steps: i) Remove the last digit from the number you wish to square. ii) Add the resulting number from part (i) to its square. iii) Write the digits 25 at the end of the number you obtained in part (ii). The number that results will be the answer you want. Choose a two-digit number whose last digit is 5, and determine whether Cynthie's method for squaring works for that number. b) Show algebraically that Cynthie's method always works.
a)
Cynthie's steps for 752 75 7 7 1 72 5 56 5625 5 75 2
Curious Math
Pythagorean Triples
A Pythagorean triple is three natural numbers that satisfy the equation of the Pythagorean theorem, that is, a 2 1 b 2 5 c 2. An example of a Pythagorean triple is 3, 4, 5, since 32 1 42 5 52. The numbers 6, 8, 10 also work, since each number is just twice the corresponding number in the example 3, 4, 5. 1. Show that 5, 12, 13 and 8, 15, 17 are Pythagorean triples. 1 1 2 1 1 2 2. Show that, for any value of p, p 2 1 a p 2 2 b 5 a p 2 1 b . 2 2 2 2 3. Use the relationship in question 2 to produce three new Pythagorean triples. 4. Show that, for any values of p and q, (2pq)2 1 ( p 2 2 q 2 )2 5 ( p 2 1 q 2 ) 2. This relationship was known to the Babylonians about 4000 years ago! 5. Use the relationship in question 4 to identify two more new Pythagorean triples.
Sometimes an expression that doesn't appear to be factorable directly can be factored by grouping terms of the expression and dividing out common factors.
98
Chapter 2
NEL
2.3
EXAMPLE
2
Selecting a factoring strategy: Grouping
Factor f (n) 5 n3 1 3n2 1 2n 1 6 by grouping.
Noah's Solution
f (n) 5 n3 1 3n2 1 2n 1 6 5 (n 3 1 3n 2 ) 1 (2n 1 6) 5 n 2 (n 1 3) 1 2(n 1 3) 5 (n 1 3) (n 2 1 2) Both factors produce numbers greater than 1, so f (n) can never be expressed as the product of 1 and itself. So Mai's claim is true.
I separated f(n) into two groups: the first two terms and the last two terms. I factored each group by dividing by its common factor. Then I factored by dividing each term by the common factor n 1 3.
Reflecting
A. B.
Why is Noah's method called factoring by grouping? What are the advantages and disadvantages of Sally's and Noah's methods of factoring?
Winnie's Solution
a) x 2 2 x 2 30
This is a trinomial of the form ax 2 1 bx 1 c, where a 5 1. I can factor it by finding two numbers whose sum is 21 and whose product is 230. These numbers are 5 and 26.
First I divided each term by the common factor, 2. This left a difference of squares. I took the square root of 9x 2 and 25 to get 3x and 5, respectively. This is a trinomial of the form ax2 1 bx 1 c, where a 2 1, and it has no common factor. I used decomposition by finding two numbers whose sum is 21 and whose product is (10) (23) 5 230. These numbers are 5 and 26. I used them to "decompose" the middle term. I factored the group consisting of the first two terms and the group consisting of the last two terms by dividing each group by its common factor. I divided out the common factor of 2x 1 1 from each term. I noticed that the first and last terms are perfect squares. The square roots are 3x and 5, respectively. The middle term is double the product of the two square roots, 2(3x) (5) 5 30x. So this trinomial is a perfect square, namely, the square of a binomial. Trinomials of this form may be factored by decomposition. I tried to come up with two integers whose sum is 1 and whose product is 6. There were no such integers, so the trinomial cannot be factored.
Fran's Solution
g(x) 5 x 2 2 6x 1 9 2 4y 2 5 (x 2 3) 2 2 (2y) 2
I recognized that the group consisting of the first three terms was the square of the binomial x 2 3 and the last term was the square of 2y. I factored the resulting expression by using a difference of squares.
5 (x 2 3 2 2y) (x 2 3 1 2y)
In Summary
Key Ideas
• Factoring a polynomial means writing it as a product. So factoring is the opposite of expanding.
factoring
x 2 1 3x 2 4 5 (x 1 4) (x 2 1)
expanding
(continued)
NEL
Equivalent Algebraic Expressions
101
• If a polynomial has more than three terms, you may be able to factor it by grouping. This is only possible if the grouping of terms allows you to divide out the same common factor from each group.
Need to Know
• To factor a polynomial fully means that only 1 and 21 remain as common factors in the factored expression. • To factor polynomials fully, you can use factoring strategies that include • dividing by the greatest common factor (GCF) • recognizing a factorable trinomial of the form ax 2 1 bx 1 c, where a 5 1 • recognizing a factorable trinomial of the form ax 2 1 bx 1 c, where a 2 1 • recognizing a polynomial that can be factored as a difference of squares: a 2 2 b 2 5 (a 1 b) (a 2 b) • recognizing a polynomial that can be factored as a perfect square: a 2 1 2ab 1 b 2 5 (a 1 b) 2 and a 2 2 2ab 1 b 2 5 (a 2 b) 2 • factoring by grouping
CHECK Your Understanding
1. Factor.
a) x 2 2 6x 2 27 b) 25x 2 2 49
c) 4x 2 1 20x 1 25 d) 6x 2 2 x 2 2
2. Each expression given can be factored by grouping. Describe how you would
number that is divisible by an odd number greater than 1, for any natural number, n. of length a and b and hypotenuse of length c. a) Write an expression in factored form for a 2. b) The hypotenuse is 3 m longer than b, and the sum of the lengths of the hypotenuse and b is 11 m. What are the lengths of the sides of the pond?
11. Sedna has designed a fishpond in the shape of a right triangle with two sides
12. Saturn is the ringed planet most people think of, but Uranus and Neptune
A
also have rings. In addition, there are ringed planets outside our solar system. Consider the cross-section of the ringed planet shown.
outer ring inner ring planet r1 r2 r3
Write factored expressions for i) the area of the region between the planet and the inner ring ii) the area of the region between the planet and the outer ring iii) the difference of the areas from parts (i) and (ii) b) What does the quantity in part (iii) represent?
a)
NEL
Equivalent Algebraic Expressions
103
13. Create a flow chart that will describe which strategies you would use to try
C
to factor a polynomial. For each path through the flow chart, give an example of a polynomial that would follow that path and show its factored form. Explain how your flow chart could describe how to factor or show the non-factorability of any polynomial in this chapter.
Extending
14. The polynomial x4 2 5x 2 1 4 is not factorable, but it can be factored by a
So f (x) and g(x) are equivalent, but h(x) is not equivalent to either of them.
A2:
If the domains of two functions differ in value for any number in both domains, then the functions are not equivalent. For example, for the functions above, f (0) 5 4 while h(0) 5 24. So f (x) and h(x) are not equivalent.
A3:
You can graph both functions. If the graphs are exactly the same, then the functions are equivalent; otherwise, they are not.
How do you add, subtract, and multiply polynomials? Study
LEARN ABOUT the Math
Adonis has designed a game called "2 and 1" to raise money at a charity casino. To start the game, Adonis announces he will draw n numbers from a set that includes all the natural numbers from 1 to 2n. The players then pick three numbers. Adonis draws n numbers and announces them. The players check for matches. Any player who has at least two matches wins.
rational function any function that is the ratio of two polynomials. A rational function can be expressed as R(x) , where R and S are f(x) 5 S(x) polynomials and S 2 0; for example, x 2 2 2x 1 3 1 ,x2 4x 2 1 4 A rational expression is a quotient of polynomials; for example, f(x) 5 2x 2 1 , x20 3x
The probability of a player winning is given by the rational function P(n) 5 3n3 2 3n2 8n3 2 12n2 1 4n
The game is played at a rapid pace, and Adonis needs a fast way to determine the range he should use, based on the number of players and their chances of winning.
? What is the simplified expression for the probability of a player winning at "2 and 1"?
NEL
108
Chapter 2
2.4
EXAMPLE
1
Simplifying rational functions
Write the simplified expression for the function defined by 3n3 2 3n2 P(n) 5 3 . 8n 2 12n2 1 4n Faez's Solution P(n) 5 3n2 (n 2 1) 4n(2n2 2 3n 1 1)
I knew that I could simplify rational numbers by first factoring numerators and denominators and dividing each by the common factor 24 3(8) 8 ae.g., 5 5 b. 27 3(9) 9 So I tried the same idea here. I factored the numerator and denominator of P(n).
3n2 (n 2 1) 5 4n(2n 2 1) (n 2 1)
1
1
Then I divided by the common factor, (n 21).
5
3n 4n(2n 2 1)
Since I cannot divide by zero, I determined the restrictions by calculating the values of n that make the factored denominator zero. I solved 4n(2n 2 1) (n 2 1) 5 0 by setting each factor equal to 0. restrictions the values of the variable(s) in a rational function or rational expression that cause the function to be undefined. These are the zeros of the denominator or, equivalently, the numbers that are not in the domain of the function.
How is working with rational expressions like working with rational numbers? How is it different? How do the restrictions on the rational expression in P(n) relate to the domain of this rational function? How does factoring help to simplify and determine the restrictions on the variable?
NEL
Equivalent Algebraic Expressions
109
5
25y 2 . This gives the restrictions x.
5
5x 2 4x 1 2 . y 2 0. y 2 0 x3
EXAMPLE
3
Selecting a strategy for simplifying the quotient of a polynomial and a monomial
Simplify and state any restrictions on the variables.APPLY the Math
EXAMPLE
2
Selecting a strategy for simplifying the quotient of a monomial and a monomial
Simplify and state any restrictions on the variables. x. Then I divided both the numerator and denominator by 2x. I determined the restrictions by finding the zeros of the original denominator by solving 26x7y 5 0. The only restriction is x 2 0. Then I divided the numerator and denominator by the GCF.x20 x
3
110
Chapter 2
NEL
. I determined the restrictions by solving 2x 2 5 0 to get the zeros of the original denominator. 30x 4y 3 26x 7y Tanya's Solution 26x4y (25y 2 ) 30x 4y 3 7 5 26x y 26x4y (x 3 )
1 1
I factored the numerator and denominator by dividing out the GCF 26x 4y. 10x 4 2 8x 2 1 4x 2x 2 Lee's Solution 10x 4 2 8x 2 1 4x 2x 2 2x (5x 3 2 4x 1 2) 5 2x (x)
1 1
I factored the numerator and denominator by dividing out the GCF 2x.
• The restrictions are found by determining all the zeros of the denominator.x21 x21 • Both rational functions and rational expressions are undefined for numbers that make the denominator zero. the domain is the set of all real numbers. Simplify. 6x 1 2 . Simplify. State any restrictions on the variables. If the denominator contains two or more terms. These numbers must be excluded or restricted from being possible values for the variables. except those numbers that make the denominator equal zero.
CHECK Your Understanding
1. f(x) 5 6x 1 2 .
a)
b)
c)
112
Chapter 2
NEL
.x21 x21
A rational expression is the ratio of two polynomials.
a)
b)
c)
3. For example.
a)
6 2 4t 2 5(x 1 3) (x 1 3) (x 2 3) (x 2 1) (x 2 3) (x 1 2) (x 2 1)
b)
9x 2 6x 3 6x 2 9 2x 2 3 5x 2 1 x 2 4 25x 2 2 40x 1 16
c)
7a 2b3 21a4b 4a 2b 2 2ab2 (2a 2 b) 2 x 2 2 7xy 1 10y 2 x 2 1 xy 2 6y 2
2.
Need to Know
• Rational functions and rational expressions can be simplified by factoring the numerator and denominator and then dividing both by their greatest common factor. Simplify. State any restrictions on the variables. As a result. For example. State any restrictions on the variables.In Summary
Key Ideas
• A rational function can be expressed as the ratio of two polynomial functions. for all rational functions. the zeros can be determined from its factored form before the function or expression is simplified.
1 . and zoom in and trace near x 5 2.
Explore some features of rational functions.
Determine another rational function with domain 5x [ R | x 2 26 that can't be simplified. Graph your function. for example.5
GOAL
Exploring Graphs of Rational Functions
YOU WILL NEED
1 The graph of the rational function f (x) 5 is shown at the right. Some rational functions cannot be simplified.
B.
Equivalent Algebraic Expressions
E. Determine the equation of a simplified rational function that has two vertical asymptotes: x 5 21 and x 5 2. or horizontal asymptotes.
NEL
115
. x Apply a transformation to h(x) that will result in a rational function that has the horizontal asymptote y 5 2. For example." Graph your function. Determine the equation of this function and graph it. g(x) 5 x23 Graph g(x) and zoom in near x 5 3." vertical asymptotes. Graph your function. where x 2 2. F.
• graphing calculator
EXPLORE the Math
6 4 2 6 4 2 0 2 4 6
y f(x) 2 4 1 x x 6
?
A. Describe what happens to the graph near x 5 3.
D.
C. Determine the equation of a rational function without any "holes. at or near numbers that are not in their domain?
Some rational functions simplify to polynomials. 1 The rational function h(x) 5 has a horizontal asymptote y 5 0.
What are some features of the graphs of rational functions. Its domain is x 5x [ R | x 2 06.
Determine another rational function that simplifies to a polynomial with domain 5x [ R | x 2 16.2. Describe what happens to the graph at x 5 1. Determine the equation of a rational function that has both a vertical asymptote and a "hole.
H. Describe what happens to the graph at x 5 2. Graph x22 f (x) prior to simplifying it. x2 2 4 f (x) 5 can be simplified by factoring from x22 (x 1 2) (x 2 2) f (x) 5 to f (x) 5 x 1 2. I. Graph your function and describe what happens to the graph at x 5 2. Review what you have discovered and summarize your findings. and it has a vertical asymptote at x 5 0 and a horizontal asymptote at y 5 0. G.
h(x) 5 has a vertical asymptote at x 5 8. For example. and some have both. Graph the function.
3.
L. others have holes.
2. which graphical features a rational function will have.
In Summary
Key Idea
• The restricted values of rational functions correspond to two different kinds of graphical features: holes and vertical asymptotes. K. g(x) 5 x 2 1 7x 1 12 x13
has a hole at x 5 23. Graph the function.
Need to Know
• Holes occur at restricted values that result from a factor of the denominator that is also a factor of the numerator. without graphing. Explain how you can identify.Reflecting
J.
What determines where a rational function has a hole? A vertical asymptote? When does a rational function have the horizontal asymptote y 5 0? When does a rational function have another horizontal line as a horizontal asymptote? Some rational functions have asymptotes. since g(x) can be simplified to the polynomial g(x) 5 (x 1 3) (x 1 4) 5x14 (x 1 3)
• Vertical asymptotes occur at restricted values that are still zeros of the denominator after simplification. Identify a rational function whose graph has the horizontal asymptote y 5 2
and two vertical asymptotes. For example.
116
Chapter 2
NEL
. Identify a rational function whose graph is a horizontal line except for two
holes. 5 x28
FURTHER Your Understanding
1. Identify a rational function whose graph lies entirely above the x-axis and has
a single vertical asymptote. Graph the function.
Daisy: Tell me what the units of the three quantities are. mol L . All of the units are like rational expressions. and Z are related. Daisy: Hold on. Tulia: I guess I'll have to look for help elsewhere. Tulia: They were # .
Tulia: I didn't catch how X.6
GOAL
Multiplying and Dividing Rational Expressions
Develop strategies for multiplying and dividing rational expressions. L s mol Daisy: I have no idea what any of those mean. so maybe there is some operation that relates them.
LEARN ABOUT the Math
Tulia is telling Daisy about something that her chemistry teacher was demonstrating. and s 21. Tulia: Like what?
? How are the three quantities related?
NEL
Equivalent Algebraic Expressions
117
. and Z.2. Y. respectively. It is about the variables X. Y.
and s21 are related. For x 5 1. and Z were rational expressions? Explain why Daisy's method for multiplying the rational expressions was correct. I wrote the quantities as a product and then simplified the expression.
s 21 5 So.
Was Daisy correct in saying that the units of X. Y. .
APPLY the Math
EXAMPLE
2
Selecting a strategy for multiplying simple rational expressions
Simplify and state the restrictions. L # s mol
Use multiplication to show how the expressions
Luke's Solution
mol L 1 # s 3 mol 5 s L
1 1 1 1
Daisy suggested that the quantities are related by multiplication.EXAMPLE
1
Selecting a strategy for multiplying rational expressions mol L . so I multiplied the rational expressions the same way as when I multiply fractions. B. This showed that the quantities are related by multiplication. y 5 1 the expression becomes 6(1) 2 15(1) (1) 3 6 15 9 3 5 3 5 5(1) (1) 8(1) (1) 4 5 8 4
118
Chapter 2
NEL
.
1 s
mol L 21 # s 3 mol 5 s L
Reflecting
A. I wrote the variable with a negative exponent as a rational expression with a positive exponent. the result was a fraction. 6x 2 15xy 3 3 5xy 8xy 4
Buzz's Solution
6x 2 15xy 3 3 5xy 8xy 4
When I substituted values for the variables.
x 1 y 5 0. then y 5 1. since x 5 y. Pluto is 102. Is this true? Explain. G is a constant (the universal gravitational constant).
m2 2 n2 m 2 2 mn 24 2 6m 1 11mn 1 3n 2m 2 mn 2 6n 2 4m 2 2 7mn 2 2n 2 3m 2 1 7mn 1 2n 2
2
14.
x1y50 Sarit says that's impossible because if x 5 1.6
10. and r is the separation distance between the centres of objects. where r 5 . An object has mass m 5
p2 2 1 p11 and density r 5 2 . x2 5 y2 So x 2 2 y 2 5 0. gives 1 1 1 5 2.
12. m1 and m 2 are the masses of the objects. I factored and simplified.
11.
Extending
13. c) Sam says that dividing two rational functions and multiplying the first function by the reciprocal of the second will produce the same function. a)
C
Why do you usually factor all numerators and denominators before multiplying rational functions? b) Are there any exceptions to the rule in part (a)? Explain. Newton's law of gravitation states that any two objects exert a gravitational force
m1m2 . How many times greater is the gravitational force between the Sun and Mercury than the gravitational force between the Sun and Pluto? on each other due to their masses.2.1 times as far from the Sun as Mercury. The mass of Mercury is 2. Explain the error in Liz's reasoning. Simplify. Substituting into Liz's final equation. Fg 5 G
NEL
Equivalent Algebraic Expressions
123
. where F is the gravitational r2 force. I divided both sides by x 2 y.2 times greater than the mass of Pluto. 3p 1 1 9p 1 6p 1 1 m Determine its volume v. I rearranged terms in the equation. State any restrictions on the variables. State the restrictions on any v variables. she can show that x 1 y 5 0 by following these steps:
T
Since x 5 y. x2 2 y2 0 5 x2y x2y (x 2 y) (x 1 y) 0 5 x2y x2y
1 1
I squared both sides of the equation. not 0. Liz claims that if x 5 y.
v 540 is the time elapsed from Toronto to Montreal. Determine which trip takes less time.
distance . respectively. since its value is not given.7
Adding and Subtracting Rational Expressions
GOAL
Develop strategies for adding and subtracting rational expressions. I determined speed expressions for the elapsed time for each way of the trip at each airspeed. it made the round trip when there was a constant wind blowing from Toronto to Montreal at 80 km/h.
LEARN ABOUT the Math
A jet flies along a straight path from Toronto to Montreal and back again. So the speed with the wind from Toronto and the speed against the wind from Montreal are v 1 80 and v 2 80. Using the relation time 5
124
Chapter 2
NEL
. v 1 80 540 is the time elapsed from Montreal to Toronto. to the jet's airspeed in still air. 540 is the time elapsed when there is no wind.
Basil's Solution
v is the jet's airspeed in still air. v. On Friday. The straight-line distance between these cities is 540 km. v 2 80
I assigned a variable. v 1 80 is the jet's airspeed from Toronto to Montreal. v 2 80 is the jet's airspeed from Montreal to Toronto. the jet made the round trip when there was no wind.2.
?
EXAMPLE
Which round trip takes less time?
1
Selecting a strategy for adding and subtracting rational expressions
Write expressions for the length of time required to fly from Toronto to Montreal in each situation. the jet travels at constant speed. While travelling in still air. On Monday.
I need to have the same denominator. I found the round-trip times by adding the times for each way. To compare T1 with T2. C.
Reflecting
A. Since I am dealing with division. in this case T1.
Why were the expressions for time rational expressions? How can you determine a common denominator of two rational functions? How do the methods for adding and subtracting rational expressions compare with those for adding and subtracting rational numbers?
NEL
Equivalent Algebraic Expressions
125
.
T2 has a smaller denominator because 6400 is subtracted from v 2.
5
1080v v2
The trip without wind took less time.
1080 5 v Wind T2 5 5 5 T1 5 540 540 1 v 1 80 v 2 80 540(v 2 80) 1 540(v 1 80) (v 1 80) (v 2 80) 1080v v 2 2 6400 v 1080 3 v v
I noticed that T1 has the denominator v while T2's denominator contains v 2. with no wind. I let T2 represent the time on Friday. the lesser of the two expressions is the one with the greater denominator. B. so I rewrote T1 by multiplying its numerator and denominator by v.7
No wind T1 5 540 540 1 v v
I let T1 represent the time on Monday. Now the numerators are both the same. with wind.2.
Extending
15. you encountered an equation of the form
1 1 1 1 5 . x 1 100 km in 2 h. he decides to move farther from the stage. a) Show that this method always produces a triple. 37 is a triple. k Sound intensity is given by the formula I 5 2 . a) For two rational numbers in simplified form.
13. So 12. b) Determine the values of x for which the speed was greater for the second trip. To avoid
14. Matthew is attending a very loud concert by The Discarded. Suppose you want to determine natural-number x z y 1 1 1 1 1 1 solutions of this equation. where k is a constant and d d is the distance in metres from the listener to the source of the sound. 122 1 352 5 1369. Fred drove his car a distance of 2x km in 3 h. For example. 5 2 and 5 2 .
130
Chapter 2
NEL
. One way to produce a Pythagorean triple is to add the reciprocals of any two consecutive even or odd numbers. time a) Write a simplified expression for the difference between the first speed and the second speed. the lowest common
C
denominator is always one of the following: i) one of the denominators ii) the product of the denominators iii) none of the above Give an example of each of these.12. 35. 1 1 1 5 2 n n(n 1 1) n11 1 1 1 b) State two more solutions of the equation 5 2 . x z y can be written as
16. Illustrate with examples. which it is: 1369 5 372. A Pythagorean triple is a triple of natural numbers satisfying the equation
a 2 1 b 2 5 c 2. In question 11. b) Explain how you would determine the LCD of two simplified rational functions with different quadratic denominators. 5 7 35
Now. Determine an expression for the decrease in sound intensity if Matthew moves x metres farther from the stage. This is a triple if 1369 is a square. Use the equation speed 5 permanent ear damage. which x y z
1 1 1 5 2 . he drove a distance of
A
distance . Later. b) Determine a triple using the method. 6 2 3 20 4 5 a) Show that the difference between reciprocals of consecutive positive integers is the reciprocal of their product. for example. 1 1 12 1 5 .
?
A. • When a polynomial and a different-coloured rational expression are side by side. then one or both of the polynomials must be perfect squares. their product must simplify. will make filling the table and counting your score as easy as possible? Why should you avoid reusing a factor unless it is necessary? Play the game by completing the table. including numerators and denominators. Check your answers. Rules: 1. After you have completed the table. their quotient must simplify. The expressions you write must satisfy each of these conditions:
1R 1P 6P 6R
2R 3R 3P 8P
2P 5R 5P 8R
4P 4R 7R 7P
• Polynomials and numerators and denominators of each rational function must be quadratics without a constant common factor. D. E.
How can you maximize your score?
Task
Checklist
What form for the polynomials. You get one point for every different linear factor that remains in your table. simplify the products and quotients
wherever possible. • When two rational expressions are side by side. Count the linear factors and write your score next to your table.
2. Write a polynomial in each square marked P and a rational function in each square marked R. their quotient must simplify.
Does each square contain a
polynomial and rational function of the right type?
Are all of the rules
satisfied?
Did you check to see if you
could make changes to improve your score?
NEL
Equivalent Algebraic Expressions
135
. • When a polynomial is on top of a different-coloured rational expression (or vice versa). Tally your score. C. • When a rational expression is on top of a rational expression. What could you do to increase your score? List some strategies you can use to maximize your score.
4. B.
3. their quotient must simplify. F. their product must simplify. • When two polynomials are side by side. • When a polynomial is on top of another polynomial. • Restrictions on the variable of each rational function must be stated in its square.24
Chapter Task Chapter Self-Test
The Algebraic Dominos Challenge
The game shown at the right consists of eight pairs of coloured squares called dominos.
136
NEL
.
Chapter
3
Quadratic Functions
GOALS
You will be able to
• • • •
Graph and analyze the properties of quadratic functions Determine the zeros of quadratic functions Calculate the maximum or minimum values of quadratic functions Solve problems involving quadratic functions
? What role does the parabola play
in the construction of the bridge in the photograph?
NEL
137
.
Draw a curve of good fit. Is the relation between width and perimeter linear or nonlinear? How do you know? Is the relation between width and area linear or nonlinear? How do you know? Determine the second differences for the Area column.
NEL
Quadratic Functions
139
. F.
YOU WILL NEED
• centimetre grid paper • graph paper
If this pattern continues.
?
A. what function can be used to model the relationship between width and area?
Use centimetre grid paper to draw the next four rectangles in the pattern. What is the relationship between length and width for each rectangle? Write the function that models the relationship between width and area. D. Martina draws the first rectangle with dimensions 1 cm 3 2 cm.Getting Started
APPLYING What You Know
Building Rectangles
A set of rectangles can be formed out of 1 cm squares on centimetre grid paper.
Calculate the first differences for the Perimeter and Area columns.
C.
Perimeter (cm) 6 12
Shape
Width (cm) 1 2
Length (cm) 2 4
Area (cm)2 2 8
3
6
18
B. Use your diagrams to extend and complete the table. Does the shape of your graph support your answer to part D? Explain. G. and the third 3 cm 3 6 cm. the second 2 cm 3 4 cm. E. What do they tell you about the relationship between width and area? Create a scatter plot of area versus width.
This confirms that the graph is a parabola that opens downward.
NEL
Quadratic Functions
141
. To make sure. the vertex is (5. The parabola also passes through (2.1
Develop an algebraic expression for the function that models Francisco's profit from selling snowboards. I used the vertex form of the quadratic function and substituted the coordinates of the vertex from the graph. The second differences are all equal and negative.
From the graph. so I substituted (2. 18). Since the first differences are not constant. the function is nonlinear. So the function is quadratic. Once I did that. I solved for a. f (x) 5 a(x 2 h) 2 1 k 5 a(x 2 5) 2 1 18
I could determine the quadratic function model if I knew the vertex and at least one other point on the graph. 0). f (2) 5 0. since its graph appears to be a parabola. I checked the first and second differences.EXAMPLE
1
Selecting a strategy to describe the algebraic model
3. Kelly's Solution: Using the Vertex Form of the Quadratic Function
Snowboards Sold (1000s) 0 1 2 3 4 5 6 7 8 9 Profit First ($10 000s) Differences 232 214 0 10 16 18 16 10 0 214 18 14 10 6 2 22 26 210 214 Second Differences 24 24 24 24 24 24 24 24 This function looks quadratic.
0 5 a(2 2 5) 2 1 18 0 5 9a 1 18 218 5 9a 22 5 a The function f (x) 5 22(x 2 5) 2 1 18 models Francisco's profit. 0) into the function.
10) from the table of values and substituted its coordinates into f(x) to help me find the value of a. 0).
f (x) 5 a(x 2 r) (x 2 s) f (x) 5 a(x 2 2) (x 2 8) 10 5 a(3 2 2) (3 2 8) 10 5 a(1) (25) 10 5 25a 225a
The function f (x) 5 22(x 2 2) (x 2 8) models Francisco's profit.Jack's Solution: Using the Factored Form of the Quadratic Function The graph is a parabola. I read these from the graph and used the table of values to check. B. I took the factored form of a quadratic function and substituted the values of the x-intercepts for r and s. Explain. C. The graph is symmetric about the line x 5 5.
Reflecting
A. I then chose the point (3. so it has to be a quadratic function.
What information do you need to write the vertex form of the quadratic function? What information do you need to write the factored form of the quadratic function? Use the graph to state the domain and range of the function that models Francisco's profit. 0) and (8. or zeros. opening down with axis of symmetry x 5 5. D.
The x-intercepts are the points (2. Will both of the models for Francisco's profit lead to the same function when expressed in standard form?
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Chapter 3
NEL
. I could find the factored form of the quadratic function if I knew the x-intercepts.
The graph looks like a parabola.
The axis of symmetry passes through the midpoint of the two zeros of the function. but the negative answer is not possible.
Range 5 5h [ R 0 0 # h # 456 5 220 1 40 1 25 5 45
NEL
h(2) 5 25(2) 2 1 20(2) 1 25
Quadratic Functions
143
. The least possible value of h(t) is zero. where h(t) is height in metres and t is the time in seconds after the toss.
5 25 The window is 25 m above the ground. The height of the tool above the ground is modelled by the quadratic function h(t) 5 25t 2 1 20t 1 25.3. I found the value of h at t 5 2.
b) 0 5 25t 2 1 20t 1 25
0 5 25(t 2 2 4t 2 5) 0 5 25(t 1 1) (t 2 5) t 5 21 or t55
If the partner missed the tool. and the tool will stop when it hits the ground. so its equation is t 5 2. The height at the ground is zero. where would the axis of symmetry be? d) Determine the domain and range of the function in this situation.
The tool will hit the ground after 5 s.1
APPLY the Math
EXAMPLE
2
Determining the properties of a quadratic function
A construction worker repairing a window tosses a tool to his partner across the street.
c) t 5
21 1 5 2 4 2
d) Domain 5 5t [ R 0 0 # t # 56
t5
t52
The axis of symmetry is t 5 2. André's Solution
a) h(0) 5 25(0) 2 1 20(0) 1 25
The height of the window must be the value of the function at t 5 0 s. Then I factored the quadratic. a) How high above the ground is the window? b) If his partner misses the tool. since the greatest value is the y-coordinate of the vertex. I found two values for t. which is always on the axis of symmetry. In this situation. t must be 0 or greater. I added the zeros together and divided by two to find that t-value. when will it hit the ground? c) If the path of the tool's height were graphed. it would hit the ground. since time must be positive. I set h(t) equal to zero. The axis of symmetry is a vertical line.
f (2) 5 23 f (3) 5 3 f (4) 5 13 The values had to be the same for 0. axis of symmetry. I found the values of f(x) when x 5 2. Because the vertex has a y-value of 25 and the parabola opens up. and 4. state the vertex. direction of opening. and 22 because the graph is symmetric about the line x 5 1. the parabola opens up. 25) Axis of symmetry: x 5 1 Direction of opening: up
The x-coordinate of the vertex is 1 and the y-coordinate is 25. 3. Sacha's Solution Vertex: (1. the y-values have to be greater than or equal to 25. domain. I substituted x 5 0 to calculate the y-intercept and solved the equation. 25). 21.
f (0) 5 2(0 2 1) 2 2 5 Domain: 5x [ R6 5225 5 23 5 2(21) 2 2 5
y-intercept: 23
Range: 5 y [ R 0 y $ 256
f (x) 14 12 10 8 6 4 2 2 0 2 4 6 2 4 6 8 1)2 x x=1
There are no restrictions on the values for x. I plotted the vertex and the axis of symmetry. To graph the function. I plotted the points and joined them with a smooth curve. The axis of symmetry is a vertical line through the vertex at (1.EXAMPLE
3
Graphing a quadratic function from the vertex form
Given f (x) 5 2(x 2 1) 2 2 5. Since a is positive. y-intercept. and range. Graph the function.
f(x) = 2(x
5
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Chapter 3
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.
The height of a rocket above the ground is modelled by the quadratic
A
function h(t) 5 24t 2 1 32t. A 20 m tree is between his ball and the hole. write an algebraic expression to model the height of the ball if it just clears the top of the tree. and two ways in which they are different. Determine the y-intercept of this parabola. in thousands of dollars. Calculate the maximum profit the business can earn. A quadratic function has these characteristics:
T
13. If f (22) 5 19. where x is in thousands of computers sold. The first differences and second differences of a quadratic function with
domain ranging from x 5 22 to x 5 3 are given. Describe two ways in which the functions f (x) 5 2x 2 2 4x and
C
Extending
14. • y 5 32 is the maximum value. With the base of the tree as the origin. Jim has a difficult golf shot to make. He
wants the ball to land 5 m in front of the hole. The company's profit. a) Graph the quadratic function. b) How long will the rocket be in the air? How do you know? c) How high will the rocket be after 3 s? d) What is the maximum height that the rocket will reach? • x 5 21 is the equation of the axis of symmetry. His ball is 100 m from the hole. so it can roll to the hole. where x is in thousands of stereo systems sold. A company's profit. 40 m from the hole and 60 m from Jim's ball.
16.
x f(x) First Differences Second Differences 22 19 210 2 6 4 4 22 4 2 4 6 21 0 1 2 3
15.
12. on sales of stereo systems is modelled by the function P(x) 5 2 (x 2 2) (x 2 7).3. copy the table and complete the second row by determining the missing values of the function. where h(t) is the height in metres t seconds after the rocket was launched. in thousands of dollars. • x 5 3 is the x-intercept.1
11. g(x) 5 2 (x 2 1) 2 1 2 are alike.
NEL
Quadratic Functions
147
. on sales of computers is modelled
by the function P(x) 5 22(x 2 3) 2 1 50.
a is negative.2
Determining Maximum and Minimum Values of a Quadratic Function
GOAL
Use a variety of strategies to determine the maximum or minimum value of a quadratic function.
148
Chapter 3
NEL
. The function that models the height of the ball is h(t) 5 25t 2 1 40t 1 100. I changed the value of the expression. The graph of h(t) is a parabola that opens down. determine whether the ball will hit the power line.
Jonah's Solution: Completing the Square
h(t) 5 25t 2 1 40t 1 100
I needed to find the maximum height of the ball to compare it to the height of the power lines. Its maximum value occurs at the vertex. I subtracted 16. then squared it to create a perfect-square trinomial. There are power lines 185 m above the ground.
5 25(t 2 2 8t 1 16 2 16) 1 100
I divided the coefficient of t in half.
?
EXAMPLE
Will the golf ball hit the power lines?
1
Selecting a strategy to find the vertex
Using the function for the golf ball's height. I factored 25 from the t 2 and t terms. By adding 16. where h(t) is the height in metres at time t seconds after contact.3. To make up for this.
h(t) 5 25(t 2 2 8t) 1 100
I put the function into vertex form by completing the square.
LEARN ABOUT the Math
A golfer attempts to hit a golf ball over a gorge from a platform above the ground.
I factored the perfect square and collected like terms. I found them by setting each factor equal to 0 and solving the resulting equations. To find the zeros. This was also the x-coordinate. The zeros are the values that make h(t) 5 0.
h(t) 5 25(t 2 2 8t 2 20) h(t) 5 25(t 2 10) (t 1 2) 0 5 25(t 2 10) (t 1 2) t 5 10 or t 5 22
For the axis of symmetry. I substituted t 5 4 into the factored form of the equation.
Sophia's Solution: Factoring to Determine the Zeros h(t) 5 25t 2 1 40t 1 100
The maximum height of the golf ball is at the vertex of the parabola. The maximum height will be 180 m after 4 s. Since the power lines are 185 m above the ground.
Since the vertex is at the maximum height. which is always in the middle of the two zeros of the function. The vertex is located on the axis of symmetry. 180). I added the zeros and divided the result by 2 to locate the axis of symmetry. the ball goes up only 180 m.3. h(4) 5 25(4 2 10) (4 1 2) 5 25(26) (6) 5 180 The vertex is (4.2
5 25(t 2 2 8t 1 16) 1 80 1 100
I grouped the first 3 terms that formed the perfect square and moved the subtracted value of 16 outside the brackets by multiplying by 25. t5 t5 10 1 (22) 2 8 2
t54 The t-coordinate of the vertex is 4. or height h.
5 25(t 2 4) 2 1 180 The vertex is (4. or in this case. the ball will not hit them. the ball will not hit them. after 4 s. Alternatively. Since the power lines are 185 m above the ground. I divided 25 out as a common factor. Inside the brackets was a simple trinomial I could factor. The maximum height will be 180 m.
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Quadratic Functions
149
. the t-coordinate of the vertex. I could have substituted into the function in standard form.
To find the y-value. 180). I factored the quadratic.
I used a graphing calculator because the numbers in the question were decimals and it would not have been easy to complete the square or factor.
150
Chapter 3
NEL
.5. I picked a maximum x of 10 and an x scale of 1. C.Reflecting
A. How are they the same? How are they different? Not all quadratic functions have zeros. I chose both the x.
How can you tell from the algebraic form of a quadratic function whether the function has a maximum or a minimum value? Compare the two methods for determining the vertex of a quadratic function. Which method allows you to find the vertex without finding the zeros? Explain.
APPLY the Math
EXAMPLE
2
Using the graphing calculator as a strategy to determine the minimum value
The cost. I chose a WINDOW that would display the graph. where x is the speed in kilometres per hour. B-9. B. the minimum value will be at the vertex.
Tech
Support
For help using the graphing calculator to determine the minimum value of a function.8x 2 2 14. c(x). see Technical Appendix. since neither cost nor speed could be negative. I estimated the corresponding maximum y from the function. Therefore. in dollars per hour of running a certain steamboat is modelled by the quadratic function c(x) 5 1.and y-values to have a minimum of 0.4x 1 156. At what speed should the boat travel to achieve the minimum cost?
Rita's Solution
This parabola opens up.
EXAMPLE
3
Solving a problem to determine when the maximum value occurs
Communication
Tip
The demand function for a new magazine is p(x) 5 26x 1 40. Levi's Solution 5 3 p(x)4 (x)
Revenue 5 Demand 3 Number sold P(x) 5 3 p(x)4 (x) 2 C(x)
I found the revenue function by multiplying the demand function by the number of magazines sold. I started by factoring the common factor 26x from 26x 2 and 36x. Profit is the difference between revenue and cost.70/h. The cost function C(x) is the total cost of making x items. The coefficient of x 2 is negative. so the parabola opens down with its maximum value at the vertex. Revenue is the money brought in by selling x items. Revenue is the product of the demand function and the number sold.2
I used the minimum operation to locate the vertex. x. of the magazine and x is the number sold. I determined two points symmetrically opposite each other.
The demand function p(x) is the relation between the price of an item and the number of items sold. in thousands.3.70). To find the profit function. I subtracted the cost function from the revenue function and simplified. Instead of completing the square.
Profit 5 Revenue 2 Cost
5 (26x 1 40) (x) 2 (4x 1 48) 5 26x 2 1 40x 2 4x 2 48 5 26x 2 1 36x 2 48
P(x) 5 26x(x 2 6) 2 48
NEL
Quadratic Functions
151
.
The minimum cost to operate the steamboat is $127. Calculate the maximum profit and the number of magazines sold that will produce the maximum profit.127. in thousands of dollars. The cost function is C(x) 5 4x 1 48.
The vertex is (4. when the boat is travelling at about 4 km/h. where p(x) represents the selling price.
if possible. factored form. or vertex form. I found the axis of symmetry. and averaging the x-coordinates of these points to determine the x-coordinate of the vertex • by using a graphing calculator
152
Chapter 3
NEL
. P(3) 5 26(3) 2 1 36(3) 2 48 5 254 1 108 2 48 56 The maximum profit is $6000. then the parabola opens down.
016
I knew that the x-intercepts of the graph of y 5 26x(x 2 6) would help me find the two points I needed on the graph of the profit function. So the x-coordinate of the vertex is 3. since both functions have the same axis of symmetry. factored form. or vertex form. • If a . I remembered that x is in thousands of magazines sold. and P(x) is in thousands of dollars.
Need to Know
• If a . The quadratic has a minimum value. then the parabola opens up. 0 in standard form.
The axis of symmetry is x 5 2 or x 5 3.
In Summary
Key Idea
• The maximum or minimum value of a quadratic function is the y-coordinate of the vertex. which gave me the x-coordinate of the vertex.26x(x 2 6) 5 0 x 5 0 or x 5 6 Points on the graph of the profit function are (0. when 3000 magazines are sold. • The vertex can be found from the standard form f(x) 5 ax 2 1 bx 1 c algebraically in several ways: • by completing the square to put the quadratic in vertex form f(x) 5 a(x 2 h) 2 1 k • by expressing the quadratic in factored form f(x) 5 a(x 2 r) (x 2 s). 248) and (6. The quadratic has a maximum value. and averaging the zeros at r and s to locate the axis of symmetry. 0 in standard form. 248). This will give the x-coordinate of the vertex 2 • by factoring out the common factor from ax 1 bx to determine two points on the parabola that are symmetrically opposite each other. I substituted x 5 3 into the function to determine the profit.
in metres. A rock is thrown straight up in the air from an initial height h0. The profit P(x) of a cosmetics company.7. Can both coaches be satisfied? Explain your answer. C(x) 5 12x 1 28 5 22x 2 1 32x.
15. 30 fewer students attend the dance. C(x) 5 8x 1 18 5 22x 2 1 25x. A ticket to a school dance is $8. Usually. function f (x) 5 3x 2 2 7x 1 2. For each pair of revenue and cost functions.9t 2 1 v0t 1 h0. The football coach wants the rectangular infield area to be as large as possible. where h(t) is the ball's height above the ground. determine
i) the profit function ii) the value of x that maximizes profit a) b) c) d)
R(x) R(x) R(x) R(x)
5 2x 2 1 24x. A high school is planning to build a new playing field surrounded by a
T
13. a) Determine the maximum height of the ball. Determine the minimum production cost. Which method would you choose for this particular function? Give a reason for your answer. where x is the amount spent on advertising. c) What amount must be spent on advertising to obtain a profit of at least $4 000 000? running track. C(x) 5 3x 1 17
8. The track coach wants two laps around the track to be 1000 m. The cost function in a computer manufacturing plant is
C(x) 5 0. 11. The height in metres above the ground after t seconds is given by h(t) 5 24. Find an expression for the time it takes the rock to reach its maximum height. a) Determine the maximum profit the company can make. What ticket price will maximize the revenue?
154
Chapter 3
NEL
.
10.28x 2 2 0.
12. The dance
committee knows that for every $1 increase in the price of a ticket. at time t seconds after the throw. is given by
A
P(x) 5 25x 2 1 400x 2 2550. Show that the value of 3x 2 2 6x 1 5 cannot be less than 1. 300 students attend. in thousands of dollars. Compare the methods for finding the minimum value of the quadratic
C
Extending
14. in thousands of dollars. b) Determine the amount spent on advertising that will result in the maximum profit. where C(x) is the cost per hour in millions of dollars and x is the number of items produced per hour in thousands. b) How long does it take for the ball to reach its maximum height? c) How high is the rooftop?
9. C(x) 5 14x 1 45 5 23x 2 1 26x. in metres.
with initial velocity v0. in metres per second.7x 1 1. The height of a ball thrown vertically upward from a rooftop is modelled by
h(t) 5 25t 2 1 20t 1 50.
She has 864 cm2 of cardboard to use. State the domain and range of the inverse. F.
Cube Side Length (cm) Area of Each Face (cm2) Surface Area (cm2) 1 1 6 2 4 24 3 4 5 6 7 8 9 10
B.
How are the surface area function and its inverse related a) in the table of values? b) in their graphs? c) in their domains and ranges?
Quadratic Functions
NEL
155
. Determine the equation that represents the cube's surface area as a function of its side length. C. H. Write the equation that represents the cube's side length for a given surface area. Use function notation and state the domain and range.3. given different representations. E. Compare the graph of the inverse with the original graph. Is the inverse a function? Explain.3
GOAL
The Inverse of a Quadratic Function
YOU WILL NEED
Determine the inverse of a quadratic function. Use your equation from part H to determine the largest cube Suzanne will be able to construct.
Draw a graph of surface area versus side length. What type of function is this? Explain how you know. D. She wants to use all of the material provided. I. Draw a graph of the inverse.
?
A.
• graph paper • ruler • graphing calculator
INVESTIGATE the Math
Suzanne needs to make a box in the shape of a cube.
Reflecting
J.
How long will each side of Suzanne's box be?
Copy and complete this table. G. How would you calculate the inverse of this function to describe the side length of the cube if you know its surface area? Make a table of values for the inverse of the surface area function.
How is any quadratic function related to its inverse a) in their domains and ranges? b) in their equations?
APPLY the Math
EXAMPLE
1
Determining the domain and range of the inverse of a quadratic function
y 8 6 4 2 6 4 2 0 2 2 4 6 x f(x) = x2
Given the graph of f (x) 5 x 2. Give a reason for your answer. The range c)
The inverse of f (x) 5 x 2 is not a function.
156
Chapter 3
NEL
. (2.
Paul's Solution
a) 8 6 4 2 2 0 2 4 6 2 4 6 8 x y f(x) = x2 y=x
To graph the inverse of f(x) 5 x 2. Therefore.
of f (x) 5 x 2 is 5 y [ R | y $ 06. and the range is 5 y [ R6. and vice versa. a) graph the inverse relation b) state the domain and range of f (x) 5 x 2 and the inverse relation c) determine whether the inverse of f (x) 5 x 2 is also a function. the domain of f 21 is 5x [ R | x $ 06. I took the coordinates of each point on the original graph and switched the x. The graph of the inverse is a reflection of the original function about the line y 5 x.K. 2).and y-coordinates. I had to do this because the input value becomes the output value in the inverse. 4) became (4. The range of the inverse is the domain of the original function.
f -1: + x -
The domain of the inverse is the range of the original function.
b) The domain of f (x) 5 x 2 is 5x [ R6. I knew this because the inverse fails the vertical-line test: For each number in the domain except 0. there are two values in the range. For example.
I could tell from the graph of the inverse that there were two parts to the inverse function. (23. respectively. I interchanged x and y in the original function.3
EXAMPLE
2
Determining the equation of the inverse of a quadratic function
Given the quadratic function f (x) 5 2(x 1 3) 2 2 4.
NEL
Quadratic Functions
157
. the lower from taking the negative square root. an upper branch and a lower branch.and y-coordinates of the points on the original function. which are also the same values of f(24) and f(25). because y represents the output value in the function. a) graph f (x) and its inverse b) determine the equation of the inverse
Prashant's Solution
a) 8 6 4 2 6 4 2 0 2 4 6 8 f -1: 3+ x + 4 2
First I wrote the equation with y replacing f(x).3. since the inverse is not a function. To graph the inverse. I interchanged the x. The upper branch came from taking the positive square root of both sides. I then rearranged the equation and solved for y by using the inverse of the operations given in the original function. and I could write it in function notation.
y f(x) = 2(x + 3)2 4
I graphed f(x) by plotting the vertex. f 21 (x) 5 23 1
x14 5y Å 2
x14 5y13 2 x14 Å 2
I couldn't write f 21 (x) for y. But if I restricted the original domain to x $ 23. The parabola opens up because the value of a is positive. I found f(22) 5 22 and f(21) 5 4.
x 2 4 6 8
b) f (x) 5 2(x 1 3) 2 2 4
y 5 2(x 1 3) 2 2 4 x 5 2( y 1 3) 2 2 4 Å x 1 4 5 2( y 1 3) 2 x14 5 ( y 1 3) 2 2
6 23 6
For f (x) restricted to x $ 23. 24). then I would use only one branch of the inverse. To find the equation of the inverse.
This function is a parabola that opens down.18). then the value of t is 6.005(t 2 6) 2 S 2 0.18. the range values are between 0 and 12.18 Å 0.18 5 (t 2 6) 2 20.3. The vertex is (6. so the value under the square root would be negative.18 5t26 Å 20. so 0 is the minimum value. Surface area cannot be less than zero.18 5 20. the numerator would be positive. b) Determine the model that describes time in terms of the surface area. so S must be at least 0.18. Therefore. I solved the given equation for t by using the inverse operations.18. For values greater than 0. so the greatest value S can have is 0. If S 5 0.
Thomas' Solution
a) Domain 5 5t [ R | 0 # t # 126
Range 5 5S [ R | 0 # S # 0.3
EXAMPLE
4
Solving a problem by using the inverse of a quadratic function
The rate of change in the surface area of a cell culture can be modelled by the function S(t) 5 20.18 S 2 0. c) Determine the domain and range of the new model. so the maximum value is 0. To find the inverse of the original function. and 0 # t # 12. If S 5 0.005 S 2 0.186 Range 5 5t [ R | 0 # t # 126
2S 1 0.005
The domain is given in the problem as part of the model.
NEL
Quadratic Functions
159
.005
The value under the square root sign has to be positive. a) State the domain and range of S(t).18 Å 20. where S(t) is the rate of change in the surface area in square millimetres per hour at time t in hours. 0.186 S 5 20.18.
6
I did not interchange S and t in this case because S always means surface area and t always means time.005(t 2 6) 2 1 0.
The domain and range of the new model: Domain 5 5S [ R | 0 # S # 0.18. The surface area also cannot be negative.005 t566 t566 S 2 0. then t 5 6 6 6. so t 5 0 or t 5 12.005(t 2 6) 2 1 0.
(1. 4). (0. (2. • In the equation of the inverse of a quadratic function.In Summary
Key Ideas
• The inverse of the original function undoes what the original function has done. • The inverse of a quadratic function can be a function if the domain of the original function is restricted. (1. 3). while the negative root represents the lower branch. If the original quadratic opens down (a .
CHECK Your Understanding
1. 4). (2. • The inverse of a quadratic function defined over all the real numbers is not a function. 0). 24)6
a) y b) 8 6 4 2 6 4 2 0 2 2 4 6 x 8 6 4 4 2 2 0 2 4 6 8
y x 2
3. It is a parabolic relation that opens either to the left or to the right. the inverse opens to the left. graph the inverse relation. the positive square root function represents the upper branch of the parabola. Given the graph of f (x). If the original quadratic opens up (a . the inverse opens to the right.
160
Chapter 3
NEL
.
Need to Know
• The equation of the inverse of a quadratic can be found by interchanging x and y in vertex form and solving for y. 0). 0). (3. a) 5(0. 27)6 b) 5(23. Given f (x) 5 2x 2 2 1. 5). It can be used to determine which values of the original dependent variable produce given values for the original independent variable. Each set of ordered pairs defines a parabola. (22. 24). (21.
its inverse. Graph the relation and
2. determine the equation of the inverse. 12). 1). (3. 1).
in metres. b) Sketch the graph of its inverse on the same axes. Given f (x) 5 7 2 2(x 2 1) 2. Time (s) Height (m) 0 0 0. determine 5. The height of a ball thrown from a balcony can be modelled by the function
A
h(t) 5 25t 2 1 10t 1 35.0 2. determine the equation for
K
f
21
(x).5 1. For 22 . Graph the data and a curve of good fit for the relationship. a) Write h(t) in vertex form.3
PRACTISING
4.
7.5 12. x . Is the inverse a function? Explain. at time t seconds after it is thrown. d) Determine the equation for g 21 (x). determine
10. Determine the vertex of the parabola. c) State the domain and range of g 21 (x). x . 3
9. Determine the domain and range of f 21 (x) for y $ 0. Consider f (x) 5 22x 2 1 3x 2 1.375
a) b) c) d)
T
Use first and second differences to extend the table. b) Determine the domain and range of h(t). 3 and f (x) 5 3x 2 2 6x.75. d) What are the domain and range of the new model?
11.5 39.
a) b) c) d) e)
NEL
Quadratic Functions
161
. Given f (x) 5 2 (x 1 1) 2 2 3 for x $ 21. a)
Graph g(x) 5 2 !x for x $ 0.
a)
f (3)
b) f
21
(x)
c) f
21
(5)
2
d) f
21
(2a 1 7)
Sketch the graph of f (x) 5 3(x 2 2) 2 2.3. Graph f (x). x $ 1. find the equation for f 2
(x) for the part of the function where x # 5.5 30. Graph the function and its inverse on the same axes.
a) the domain and range of f (x) b) the equation of f 21 (x) if f (x) is further restricted to 1 . Graph f 21 (x) for y $ 0. Why were the values of x restricted in parts (c) and (d)?
12. Given f (x) 5 (x 2 5) 2 1 3.375 2 36.
21
1 8. a) 6. where h(t) is the height above the ground. c) Determine the model that describes time in terms of the height. b) Graph its inverse on the same axes. Use a graphing calculator to graph f 21 (x). Graph the inverse relation and its curve of good fit.375 1 22.75. The height of a golf ball after Lori Kane hits it is shown in the table.
NEL
162
Chapter 3
. a) Determine an algebraic expression for P(x). Interpret its meaning.13. You are given the relation x 5 4 2 4y 1 y 2.
14. A meat department manager discovers that she can sell m(x) kilograms of
ground beef in a week. will its inverse be a function? Explain. Graph f 21. b) If the domain of the quadratic function is 5 x [ R 6. She pays her supplier $3. b) Find the equation for the inverse relation. Determine the domain and range of the relation. c) Write an expression in function notation to represent the price that will earn $1900 in profit. a)
C
Extending
If you are given a quadratic function in standard form. Each graph shows a function f that is a parabola or a branch of a parabola. Determine the equation of the inverse. Is the inverse a function? Explain.
a)
4 2 0 –2 –4 –6
y x 2 4 6 8
c)
6 4 2 –6 –4 –2 0 –2 –4
y
x 2
b)
6 4 2 –4 –2 0 –2 –4
y
d)
8 6 4 2 –6 –4 –2 0 –2
y
x 2 4 6
x 2 4
i) ii) iii) iv)
Determine f (x). e) The supply cost drops to $3. where P(x) represents the total profit in dollars for 1 week. Determine the equation(s) for f 21. explain how you could determine the equation of its inverse. d) Determine the price that will maximize profit. real numbers also to be a function?
16. where m(x) 5 14 700 2 3040x. Evaluate and explain.10/kg.21/kg for the beef. What price should the manager set? How much profit will be earned at this price?
17. State restrictions on the domain or range of f to make its inverse a function.
a) b) c) d)
Graph the relation. What must happen for the inverse of a quadratic function defined over all the
15. if she sells it at x dollars per kilogram.
H. Compare the results in each pair of products. From the preceding results.
Reflecting
E.
?
A.4
GOAL
Operations with Radicals
Simplify and perform operations on mixed and entire radicals. is given by s(t) 5 10!4 1 t. the distance between the particle and the given point at 20 s?
Copy and complete these products of radicals: !100 3 !9 5
B. explain why using the factors !4 3 !6 is a better choice than using !3 3 !8.3. such as ! 4 5 2 or "27 5 3. Determine s(20). s(t). describe how !ab is related to !a 3 !b. express this product as a) an entire radical b) a mixed radical Use your calculator to verify that your products are equivalent. in millimetres of a particle from a certain point at any time. If a . Why would the decimal answer for !24 not be considered exact?
!16 3 !9 5 !4 3 !36 5
!25 3 !4 5 5 3 2 5 10
!100 3 9 5 ! 5
! 4 3 36 5 ! 5
!16 3 9 5 ! 5
!25 3 4 5 !100 5 10
a square. His answer must be in simplest form and he is not permitted to use a decimal. for example.
a radical with coefficient other than 1. 2 ! 3 mixed radical
a radical with coefficient 1. Don needs to find the exact distance between the point and the particle after 20 s. G.
INVESTIGATE the Math
The distance. ! 12 entire radical
F. why is b!a a simpler form of "ab 2 ? Determine s(20) as a decimal. C. If a and b are positive whole numbers. 0. for example. Use what you observed in parts A to C to simplify the expression so that your answer uses the smallest possible radical.
What is the exact value of s(20). t. cube. ! is called the radical symbol radical
3
D. or higher root.
Consider !4 3 !6. To express !24 as a mixed radical.
NEL
Quadratic Functions
163
.
but to put the mixed radical in lowest terms. I could have chosen 4 or 9.
164
Chapter 3
NEL
.
EXAMPLE
2
Changing mixed radicals to entire radicals
a) 4!5 5 4 3 !5 a) 4!5
Express each of the following as entire radicals. So I wrote 26 as the product of 21 and 6. I evaluated ! 36. b) 26!3 5 !16 3 !5 5 !80 5 2 !108
Sami's Solution
b) 26!3 5 (26) 3 !3
5 (21) 3 6 3 !3
5 (21) 3 !36 3 !3
To create an entire radical. I evaluated the square root of 9 and multiplied it by the coefficient 5. I had to choose the greatest perfect square. I had to change 4 into a square root. Then I was able to multiply the numbers under the radical signs.APPLY the Math
Express each of the following as a mixed radical in lowest terms. I expressed 6 as ! 36 so that I could multiply the radical parts together to make an entire radical. which was 36. I knew that the negative sign would not go under the radical. since squares of real numbers are always positive. I found the largest perfect square that would divide evenly into 27. it was 9. a) !72 b) 5!27
EXAMPLE
1
Simplifying radicals by using a strategy involving perfect-square factors
a) !72 5 !36 3 !2
Jasmine's Solution
5 6!2
b) 5!27 5 5 3 !9 3 !3
5 5 3 3 3 !3 5 15!3
I needed to find a perfect square number that divides evenly into 72. I expressed 4 as the square root of 16. Once I expressed ! 72 as ! 36 3 ! 2.
so my answer was in lowest terms.236
I used my calculator to evaluate two radicals that were not like each other: ! 3 and ! 5.732 1 2.968 3!2 5 !2 1 !2 1 !2 5 !2 3 4 5 4!2 5 4!2 !8 8 2. When I added ! 2 to this sum. he may also have to add expressions that contain radicals. I used 3 ! 2 and ! 2. 5 3. 55 was not divisible by a perfect square. 3!2 1 !2 5 !2(3 1 1)
Then I tried two like radicals. It makes sense that I can add like radicals by adding the integers in front of the radicals together.Simplify. I was able to simplify.732 !5 8 2. When I calculated ! 8. I expressed 3 ! 2 as the sum of three ! 2 s. or 4 ! 2.
EXAMPLE
4
Adding radicals
radicals that have the same number under the radical symbol. so I grouped together the integer products and the radical products.
So
3!2 1 !2 5 !2 1 !2 1 !2 1 !2 Also. I had 4 of them altogether. It looks like I cannot add radicals together if the numbers under the radical signs are different. I found that it was not equal to ! 3 1 ! 5.236 !3 1 !5 2 !8.4
Multiplying radicals
3
a) !5 3 !11 5 !55
b)
24!6 3 2!6
Caleb's Solution
b) 24!6 3 2!6 5 (24) 3 2 3 !6 3 !6
5 (28) 3 6 5 248
5 (28) !36
I multiplied the numbers under the radical signs together. Since 36 is a perfect square. such as 3 ! 6 and 22 ! 6 like radicals
In Don's research.828 !3 8 1. Can he add radicals that are like radicals? What about other radicals? Marta's Solution !3 1 !5 5 1. a) !5 3 !11
EXAMPLE
3.
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Quadratic Functions
165
. A mixed radical is the product of the integer and the radical. I rounded each value to 3 decimal places and then performed the addition.
0) and the minimum point if the parabola opens up (a . graph the function on a graphing calculator and determine the maximum or minimum value. 10. Examples
2 and 3. and 12.
• See Lesson 3.1. • If the equation is of degree 2.
How can you determine the maximum or minimum value of a quadratic function?
• See Lesson 3.4. If two points are known that are the same distance from the vertex and opposite each other on the graph. The maximum or minimum value occurs at the x-value that is the average of the two x-intercepts. Remember that when you take the square root of an expression. If the equation is in factored form f (x) 5 a(x 2 r) (x 2 s). find the x-intercepts of the function.
• Try Mid-Chapter Review
Questions 9. then the maximum or minimum value occurs at the x-value that is the average of the x-coordinates of the two points.
A2:
A3:
A4:
Q:
A:
Study
Aid
Interchange the values of x and y in the equation that defines the function. the function is quadratic.
How can you determine the equation of the inverse of a quadratic function?
• See Lesson 3. • Try Mid-Chapter Review
Question 13. Example 1.2.
How do you know if a radical can be simplified?
• See Lesson 3.
Q:
A:
Study
Aid
A radical can be simplified if the number under the radical sign is divisible by a perfect-square number other than 1. • If the graph is a parabola opening either up or down. which is the maximum point if the parabola opens down (a . one positive and one negative. then the function is quadratic. and then solve the new equation for y. If the equation is in standard form and the values of the coefficients are decimals. then the function is quadratic. complete the square to find the vertex.3
Q:
A:
Mid-Chapter Review
Study
FREQUENTLY ASKED Questions
How can you tell if a function is quadratic from its table of values? its graph? its equation?
Aid
• If the second differences are constant. Example 2. • Try Mid-Chapter Review Questions 5 and 8.
Q:
A1:
Study
Aid
If the equation is in standard form f (x) 5 ax 2 1 bx 1 c. Examples 1 to 3. • Try Mid-Chapter Review
Questions 1 and 2.3.
NEL
Quadratic Functions
169
. 0) . there are two possible solutions.
where x is the number of items sold.1
8.
!7 3 !14 3!5 3 2!15 !12 1 2!48 2 5!27 3!28 2 2!50 1 !63 2 3!18 (4 2 !3) (5 1 2!3) (3!5 1 2!10) (22!5 1 5!10)
d) 23!75 e) 5!98 f ) 28!12
170
Chapter 3
NEL
.2x 1 2. b) List the domain and range of f (x) and its inverse.
equation P(x) 5 24x 1 16x 2 7. Express the number sold in terms of the revenue. where x is the number of items sold.
Lesson 3. where C(x) is the cost per hour in thousands of dollars.
possible product between these numbers?
Lesson 3.3
Determine whether the function is quadratic. The sum of two numbers is 16. and P(x) is dollars in thousands.
14. in thousands. a) f (x) 5 x 2 2 6x 1 2 b) f (x) 5 2(x 2 4) (x 1 6) c) f (x) 5 22x 2 1 10x d) f (x) 5 3. For each function in question 2. For each table. calculate the second differences. in thousands. The profit function for a business is given by the
13. Express each radical in simplest form.2
12. a)
Determine the equation of the inverse of the quadratic function f (x) 5 x 2 2 4x 1 3. Almost all linear functions have an inverse that is a
5. The revenue for a business is modelled by the
2. c) Sketch the graphs of f (x) and its inverse. Simplify. the
equation of the axis of symmetry. Explain why?
a) the domain and range of f (x) b) the equation of f 21 a) !48 b) !68 c) !180 a) b) c) d) e) f)
10. function R(x) 5 22.2x 2 1 15x 2 7
6. The cost per hour of running an assembly line in a
2
manufacturing plant is a function of the number of items produced per hour.4
11. Determine the most economical production level.
4. Determine the maximum or minimum value of each
quadratic function. state the vertex.3x 2 2 1. Graph each function. in thousands.8(x 2 10) 2 1 15.
7. and x is the number of items produced per hour.PRACTICE Questions
Lesson 3. and R(x) is the revenue in thousands of dollars. a) b)
x y x y 22 21 0 1 2 28 22 0 22 28 22 21 0 1 2 0 1 4 9 16
9. Calculate the maximum profit and how many items must be sold to achieve it. but quadratic functions do not. and the domain and range. The cost function is C(x) 5 0. Express each function in question 2 in standard form.
a) f (x) 5 23(x 2 2) 2 1 5 b) f (x) 5 2(x 1 4) (x 2 6)
3. What is the largest
1. function. Graph f (x) 5 2 !x 1 3 and determine
Lesson 3.
What point is the focus of this parabola?
NEL
Quadratic Functions
171
.
5. (2. 4). 1). and the parabola represents
the dish.
1
YOU WILL NEED
• ruler • protractor
1 f(x) 5 4 x2 y 6
4 2 6 4 2 0 2 4 6 tangent lines 2 4 6 x
1. 3. With a protractor. Mark these points on a copy of the graph: (24. 2.
4. The reflected ray will make the same angle with the tangent line. Draw a line parallel to the y-axis through each of the points. Each vertical line represents a satellite signal ray. measure the angle made between the ray and the tangent line. called the
focus of the parabola. All reflected rays should pass through the same point on the y-axis. which is positioned at a fixed point called the focus.4
Curious Math
Investigating a Special Property of Parabolas
In a parabolic satellite reception dish. Use the
protractor to draw the line representing the reflected ray. You can locate this 1 point by using math. 1). Each signal ray will be reflected.) You can use the tangent lines to determine the focus. The graph of f (x) 5 x 2 shows four tangent lines. (22. (4. 4. 6. (Each tangent line just touches 4 the curve at one point. all of the signal rays are directed to a feed horn.3. Let the function f (x) 5 x 2 model the shape of a satellite 4 dish.). Extend the lines representing the reflected ray until they cross the y-axis.
LEARN ABOUT the Math
Anthony owns a business that sells parts for electronic game systems.
? How many parts must Anthony sell in order for his business to break even?
EXAMPLE
1
Selecting a strategy to solve a quadratic equation
Determine the number of parts Anthony must sell to break even.3. Because x is measured in thousands.
Tech
Support
Derek's Solution: Using Graphing Technology
For help using the graphing calculator to graph functions and find their x-intercepts.
I used the zero operation and found the x-intercepts from my graph.5
YOU WILL NEED
Quadratic Function Models: Solving Quadratic Equations
GOAL
• graphing calculator
Solve problems involving quadratic functions in different ways. where x is the quantity sold. and P(x) is the profit in thousands of dollars. in thousands. see Technical Appendix.5x 2 1 8x 2 24. B-2 and B-8. I knew that the break-even values were 4000 and 12 000.
Anthony must sell 4000 parts or 12 000 parts. The profit function for his business can be modelled by the equation P(x) 5 20. so I looked for the x-intercepts of my function.
172
Chapter 3
NEL
.
I graphed P(x). Breaking even means that the profit is zero. There were two possible values. at x 5 4 and x 5 12.
25(t 2 3) 2 1 55 5 0
Range 5 5h(t) [ R | 0 # h(t) # 556 5 25(t 2 2 6t 1 9 2 9) 1 10 5 25(t 2 3) 2 1 55 5 25(t 2 2 6t 1 9) 1 45 1 10 t 5 3 1 "11 or t 5 3 2 "11 25(t 2 3) 2 5 255 (t 2 3) 2 5 11 t 5 3 1 3.
How are the three methods for calculating the break-even points for Anthony's business the same? How are they different? Will there always be two break-even points for a profit function? Why or why not? If break-even points exist. The vertex is (3. I calculated the values of t that would make the height 0.32 or t 5 3 2 3.
174
Chapter 3
NEL
.326 t 5 6. a) What are the domain and range of this function? b) When will the balloon reach a height of 30 m? Brian's Solution
a) h(t) 5 25(t 2 2 6t) 1 10
The graph must be a parabola opening down because the value of a is negative. To get the range. in metres.
Domain 5 5t [ R | 0 # t # 6. h(t).32 or
One value of t is negative.Reflecting
A. of the balloon after t seconds is h(t) 5 25t 2 1 30t 1 10.32 t 5 20. B. C. which method may not work to determine where they are?
APPLY the Math
EXAMPLE
2
Solving a problem involving a quadratic equation
A water balloon is catapulted into the air from the top of a building. The time t must be greater than 0. so the domain must start at 0 and go to the positive value of t that I found. 55). I found the vertex by completing the square. so the maximum height is 55 m and the minimum height is 0.32 t 2 3 5 6 "11
The domain is the interval of time the balloon was in flight. It stops when it hits the ground. The height.
764 s or 5.b) 30 5 25t 2 1 30t 1 10
0 5 25t 2 1 30t 2 20 t5 5 8 t8
230 6 "500 210 230 6 22.
EXAMPLE
3
Representing and solving a problem by using a quadratic equation
80 m
Communication
Tip
A factory is to be built on a lot that measures 80 m by 60 m. How wide is the strip of lawn.236 s.36 210
or
t8
230 2 22. The ball reaches 30 m going up and again coming down. equal to the area of the factory.
t 8 0. must surround it. I replaced h(t) with 30 and solved for t.5
To know when the ball reached 30 m.
60 – 2x
60 m
80 – 2x
NEL
Quadratic Functions
175
.36 210
Both answers are possible in this question.
80 m x x
x x
I chose a variable for the width of the lawn. A lawn of uniform width.236 The ball will reach a height of 30 m after 0.36 210
230 6 " (30) 2 2 4(25) (220) 2(25)
3.
60 m
Rachael's Solution Let the width of the lawn be x metres. and what are the dimensions of the factory?
Any solution of an equation that does not work in the context of a problem is said to be an inadmissible solution.
230 1 22.764 or t 8 5.
I used the quadratic formula to solve for t.
This equation was quadratic. I found two possible values of x.
The area of the lawn is equal to the area of the factory. I wrote down and simplified an expression for the area of the factory.
The area of the lawn is the difference between the area of the lot and the area of the factory. so I rearranged it so that it was equal to zero and solved it by factoring. Area of factory 5 length 3 width 5 (60 2 2x) (80 2 2x) 5 4800 2 120x 2 160x 1 4x 2 5 4800 2 280x 1 4x 2 Area of lawn 5 Area of lot 2 Area of factory 5 4800 2 (4800 2 280x 1 4x 2 ) 5 24x 2 1 280x 24x 2 1 280x 5 4800 2 280x 1 4x 2 28x 2 1 560x 2 4800 5 0 28(x 2 2 70x 1 600) 5 0 28(x 2 60) (x 2 10) 5 0 x 5 60 or x 5 10
Since the lawn was the same width all around. Since a width of 60 for the strip made the dimensions of the factory negative. I had to subtract 2x from 60 and 2x from 80 to get the dimensions of the factory.
But x 5 60 is inadmissible in this problem.
176
Chapter 3
NEL
. the width of the lawn had to be 10 m. so x 5 10. I then substituted x 5 10 into the expressions for the length and width of the factory to find its dimensions.
60 2 2(10) 5 40 80 2 2(10) 5 60 The lawn is 10 m wide.The dimensions of the factory are (60 2 2x) m and (80 2 2x) m. so I set the two expressions equal to each other. and the dimensions of the factory are 60 m by 40 m.
6 m tall can be modelled by the
function h(t) 5 24. The length of the hypotenuse is
6 cm more than twice the length of one of the other sides. the company estimates that it will lose 40 passengers per day. Find the zeros of the function f (x) 5 3x 2 1 1
1 . A right triangle has a height 8 cm more than twice the length of the base. If
the area of the triangle is 96 cm2. For each
$0. The height of the flare above
14. How long will it take for the ball to hit the ground? P(t) 5 0.
11. what fare should be charged?
15. what are the dimensions of the triangle?
12. where x is the
number sold.
What are the dimensions of the rectangle?
10. Jackie mows a strip of uniform width around her 25 m by 15 m rectangular
T
lawn and leaves a patch of lawn that is 60% of the original area.9t 2 1 92t 1 9. A small flare is launched off the deck of a ship. a) When will the flare's height be 150 m? b) How long will the flare's height be above 150 m?
13.4t 2 1 10t 1 50.
8.
17. The population of a region can be modelled by the function
9. One side is 7 m longer than the other.6.6. A rectangle has an area of 330 m2.
Extending
16. in thousands. A bus company has 4000 passengers daily. a) What was the population in 1995? b) What will be the population in 2010? c) In what year will the population be at least 450 000? Explain your answer. What is the width of the strip? the water is given by the function h(t) 5 24. If the company needs to take in $10 450 per day to stay in business. Determine the break-even quantities for each profit function.
a) P(x) 5 2x 2 1 12x 1 28 b) P(x) 5 22x 2 1 18x 2 40
A
c) P(x) 5 22x 2 1 22x 2 17 d) P(x) 5 20. The perimeter of a right triangle is 60 cm. where h(t) is measured in metres and t is time in seconds. Find the lengths of all three sides. What could the
integers be? List all possibilities. The flight of a ball hit from a tee that is 0.15 increase. Describe three possible ways that you could determine the zeros of the
C
quadratic function f (x) 5 22x 2 1 14x 2 24. x11
178
Chapter 3
NEL
. The sum of the squares of two consecutive integers is 685.5x 2 1 6x 2 5
7. each paying a fare of $2. where P(t) is the population in thousands and t is the time in years since the year 1995. where h(t) is the height in metres at time t seconds.9t 2 1 6t 1 0.
so I factored it. or 2 zeros. then factored the trinomial inside the brackets. Tara's Solution: Using Properties of the Quadratic Function f (x) 5 22x 2 1 12x 2 18 f (x) 5 22(x 2 2 6x 1 9) 5 22(x 2 3) 2 Vertex is (3.
LEARN ABOUT the Math
Samantha has been asked to predict the number of zeros for each of three quadratic functions without using a graphing calculator. g(x) 5 2x 2 1 6x 2 8 g(x) 5 2(x2 1 3x 2 4) 5 2(x 1 4) (x 2 1) This function has two zeros.
I decided to find the vertex of the first function.6
GOAL
The Zeros of a Quadratic Function
Use a variety of strategies to determine the number of zeros of a quadratic function. This function has one zero. I used the factors to find the zeros. Because the vertex is on the x-axis. This put the function in vertex form. so this function has two. there is only one zero.
NEL
Quadratic Functions
179
. Samantha knows that quadratics have 0. I factored 22 out as a common factor. at x 5 24 and x 5 1. The three functions are: f (x) 5 22x 2 1 12x 2 18 g(x) 5 2x 2 1 6x 2 8 h(x) 5 x 2 2 4x 1 7
? How can Samantha predict the number of zeros each quadratic has without graphing?
EXAMPLE
1
Connecting functions to their graphs
Determine the properties of each function that will help you determine the number of x-intercepts each has. 1.3. 0) and the parabola opens down. I factored 2 out as a common factor in the second function. The trinomial that was left was a perfect square.
How can finding the vertex help determine the number of zeros? Why is the factored form useful in determining the number of zeros of a quadratic function? Explain how the quadratic formula can be used to predict the number of zeros of a quadratic function. C.6
26 1 10 4
x 5 24 or
This function has two zeros. and c 5 7.
Describe the possibilities for the number of zeros of a quadratic function.
4 6 "212 2
Reflecting
A. The discriminant was negative. B. D.
NEL
Quadratic Functions
181
. b 5 24. The function has no zeros. so there were no real-number solutions.
This function has no zeros.5 x5
26 6 "100 4 26 2 10 4 or x51 x5
3. h(x) 5 x 2 2 4x 1 7 0 5 x 2 2 4x 1 7 x5 5 5 5 4 6 " (24) 2 2 4(1) (7) 2 4 6 "16 2 28 2
2b 6 "b2 2 4ac 2a
I used the quadratic formula with a 5 1.
I expressed the values of k using a mixed radical in simplest form.
NEL
Quadratic Functions
183
. where x is based on the number of items produced. Ruth's Solution b 2 2 4ac 5 0 (2k) 2 2 4(1) (3) 5 0 k 2 2 12 5 0 k 2 5 12 k 5 6!12 k 5 62!3
Since I had to take the square root of both sides to solve for k.3x 2 1 3x 2 15. there were two possible values. Therefore.3) (215) 5 9 2 18 5 29 Since b2 2 4ac .
EXAMPLE
4
Solving a problem by using the discriminant
A market researcher predicted that the profit function for the first year of a new business would be P(x) 5 20. b.3x 2 1 3x 2 15 5 0 b 2 2 4ac 5 (3) 2 2 4(20.3. there are no zeros for this function. and c into the equation b2 2 4ac 5 0 and solved for k. it is not possible for the business to break even in its first year. then the discriminant is zero. 0.
At a break-even point. I used the value of the discriminant to decide. If there is only one zero. Will it be possible for the business to break even in its first year? Raj's Solution P(x) 5 0 20. I put the values for a.6
EXAMPLE
3
Solving a problem involving a quadratic function with one zero
Determine the value of k so that the quadratic function f (x) 5 x 2 2 kx 1 3 has only one zero. so I only needed to know if the profit function had any zeros. the profit is zero. I just wanted to know if there was a break-even point and not what it was.
0. there are two zeros. and the vertex is above the x-axis. You can determine the number of zeros either by graphing or by analyzing the function.In Summary
Key Idea
• A quadratic function can have 0. and the vertex is below the x-axis. there are no zeros. and the vertex is below the x-axis.
Need to Know
• The number of zeros of a quadratic function can be determined by looking at the graph of the function and finding the number of x-intercepts. 0. 0. and the vertex is above the x-axis. • If a .
7 5 3 5 3
y
a. see the table below: Value of the Discriminant b2 2 4ac . and the direction of opening: • If a . • For a quadratic equation ax 2 1 bx 1 c 5 0 and its corresponding function f(x) 5 ax2 1 bx 1 c. 0
2 2
Number of Zeros/Solutions 2 1 0
• The number of zeros can be determined by the location of the vertex relative to the x-axis. 1. there are no zeros. • If a .0
a. 0. • If the vertex is on the x-axis. there is one zero.0
2 0 1 1 3 5 7
x 3 5 5 3
2 0 1 1 3 5 7
2 0 1 1 3 5 7
x 3 5
no zeros
one zero
two zeros
a.0
a.0
7 5 3
y
7 5 3 x 3 5 5 3
y
a.0
a. 0 b 2 4ac 5 0 b 2 4ac . • If a .0
184
Chapter 3
NEL
. there are two zeros. or 2 zeros.
Machine A: C(x) 5 4. For what values of k does the function have no zeros? one zero? two zeros?
186
Chapter 3
NEL
. The cost function for each machine is shown.8x 1 55. Which machine would you recommend to the company?
13. Explain your thinking.36 Machine C: C(x) 5 8. determine
T
the value of k so that there is exactly one point of intersection between the two parabolas. without solving the related quadratic equation or graphing. Describe how each transformation or sequence of transformations
of the function f (x) 5 3x 2 will affect the number of zeros the function has. where x is
the quantity sold in thousands and p is the price in dollars.5.9x 1 19. The company that manufactures the product is planning to buy a new machine for the plant.1x 1 92.16 Machine B: C(x) 5 17.12. 18. The demand function for a new product is p(x) 5 24x 1 42. If f (x) 5 x 2 2 6x 1 14 and g(x) 5 2x 2 2 20x 2 k.4 Investigate the break-even quantities for each machine. a) a vertical stretch of factor 2 b) a horizontal translation 3 units to the left c) a horizontal compression of factor 2 and then a reflection in the x-axis d) a vertical translation 3 units down e) a horizontal translation 4 units to the right and then a vertical translation 3 units up f ) a reflection in the x-axis. Investigate the number of zeros of the function f (x) 5 (k 1 1)x 2 1
2kx 1 k 2 1 for different values of k. then a horizontal translation 1 unit to the left. There are three different types of machine. Determine the number of zeros of the function f (x) 5 4 2 (x 2 3) (3x 1 1)
16.
15. and then a vertical translation 5 units up
14. Show that (x 2 2 1)k 5 (x 2 1) 2 has one solution for only one value of k. Describe how you can determine the number of zeros of a quadratic function
C
if the equation of the function is in a) vertex form b) factored form
c)
standard form
Extending
17.
F. yet their graphs can have similar characteristics. How are the graphs the same? How are they different? Write each of the functions in Group 2 in vertex form. C.7
GOAL
Families of Quadratic Functions
Determine the properties of families of quadratic functions. What do you notice? Clear all functions.
G.
What characteristics do the graphs in each of these groups have in common?
A.
B. What do you notice? Clear all functions.5x 2 2 1x 1 3. and then graph each of the functions in Group 3 on a graphing calculator.
NEL
Quadratic Functions
187
.
C. F. E. D. What do these functions have in common? Summarize your findings for each group. Use the window settings shown.5x 2 1 1.
INVESTIGATE the Math
Equations that define quadratic functions can look quite different.
Group 1 f(x) 5 x 2 2 3x 2 10 g(x) 5 22x 2 1 6x 1 20 h(x) 5 4x 2 2 12x 2 40 k(x) 5 20.5 p(x) 5 26x 2 1 12x 2 3 q(x) 5 10x 2 2 20x 1 13 Group 3 r(x) 5 23x 2 1 5x 2 2 s(x) 5 2x 2 1 x 2 2 t(x) 5 7x 2 2 2x 2 2 u(x) 5 24x 2 2 4x 2 2
?
A. How are the graphs the same? How are they different? Write each of the functions in Group 1 in factored form. Use the window settings shown.
Graph each of the functions in Group 1 on a graphing calculator.5x 1 5 Group 2 m(x) 5 22x 2 1 4x 1 1 n(x) 5 0.3. Use the window settings shown. and then graph each of the functions for Group 2 on a graphing calculator.
but not horizontally or vertically translated. 20) if its zeros are 2 and 21. then simplified by expanding. In each of Groups 1 and 2.
To get another quadratic function with the same vertex. I. Describe the common characteristics of each of the groups. I needed to change the value of a because parabolas with the same vertex are vertically stretched or compressed. 21).
I wrote the general function of all parabolas that have zeros at 2 and 21. Ian's Solution f (x) 5 23(x 1 2) 2 2 1 Vertex is (22.
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. what single value was varied to create the family? What transformation is this parameter associated with? What common characteristic appears in all quadratic functions in the same family if the equation is in factored form? vertex form? standard form?
APPLY the Math
EXAMPLE
1
Looking for quadratics that share a vertex
Given the function f (x) 5 23(x 1 2) 2 2 1. J. Family of parabolas is of the form f (x) 5 a(x 1 2) 2 2 1 So another quadratic in the family is g (x) 5 2(x 1 2) 2 2 1. determine another quadratic function with the same vertex.
EXAMPLE
2
Determining a specific equation of a member of the family
f (x) 5 a(x 2 2) 3x 2 (21) 4 Preet's Solution 5 a(x 2 2) (x 1 1) 5 a(x 2 2 x 2 2) 5 a 3x 2 2 2x 1 x 2 24
Determine the equation of the quadratic function that passes through (23.
Each of the three groups of functions forms a family of parabolas. I identified the vertex.Reflecting
family of parabolas a group of parabolas that all share a common characteristic H.
EXAMPLE
3
Solving a problem by applying knowledge of the vertex form of a quadratic
A highway overpass has a shape that can be modelled by the equation of a parabola. f (x) 5 2(x 2 2 x 2 2). so I substituted those values into the equation and solved for a. Once I had the value of a. If the edge of the highway is the origin and the highway is 10 m wide. then one of the zeros is 0. 20). I wrote the equation in factored form. If the edge of the highway is at the origin. If the highway is 10 m wide. I substituted the point into the equation and solved for a. I wrote the equation in factored form.3.7
f (x) 5 a(x 2 2 x 2 2) 20 5 a3 (23) 2 2 (23) 2 2)4 20 5 10a a52 Therefore. Since I had the zeros. (2. 13) is a point on the curve. what is the equation of the parabola if the height of the overpass 2 m from the edge of the highway is 13 m? Elizabeth's Solution
y
I drew a sketch. the equation that models the overpass is h52 13 x(x 2 10) 16
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189
.
x 0 2 4 6 8 10
h 5 ax(x 2 10)
13 5 a(2) (2 2 10) 13 5 216a a52 13 16
Therefore. then the other zero is at (10.
To determine the equation passing through (23. I had to find the correct value of a. so I needed to find the value of a. 0). The equation that would model the overpass would be in the same family.
to 19-year-old males who smoke has been tracked by Health Canada.
Bryce's Solution
a) Percent of 15.6 22.3 1995 1996 28. Since the values were 22.6 in both 1989 and 1991. Determine a quadratic function that will model the data.7 25.
Year Smokers (%) 1981 1983 1985 1986 1989 1991 1994 43. I used (1990. 22) as my estimated vertex.EXAMPLE
4
Selecting a strategy to determine the quadratic function from data
The percent of 15.6 26.
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Chapter 3
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.1
a) b) c) d)
Draw a scatter plot of the data. The data from 1981 to 1996 are given in the table.5 29. Estimate the location of the vertex.4 39. Draw a curve of good fit. They could be represented by a quadratic function.
I drew a curve of good fit by hand so that I could estimate the vertex.to 19-year old Males that Smoke 50 40 % 30 20 10 0 1980 1985 1990 1995 2000 Year b) Percent of 15.6 27.to 19-year old Males that Smoke 50 40 % 30 20 10 0 1980 1985 1990 1995 2000 Year
I plotted the points on a graph.2 22.
3. write in vertex form with a unknown.
In Summary
Key Ideas
• If the value of a is varied in a quadratic function expressed in vertex form.
Need to Know
• The algebraic model of a quadratic function can be determined algebraically. a family of parabolas with the same x-intercepts and axis of symmetry is created. I substituted the point into the equation and solved for a. I chose the point (1995.5 5 a(1995 2 1990) 2 1 22 28. substitute another known point.7
c) The graph models a parabola
I used vertex form with the vertex I knew. a family of parabolas with the same vertex and axis of symmetry is created.5 5 25a 1 22 6.5 5 25a 65 5a 25 a 8 0. Estimated vertex: (1990. and solve for a. • If the vertex is known. • If the values of a and b are varied in a quadratic function expressed in standard form.26
Therefore. and solve for a. 22)
d) f (x) 5 a(x 2 1990) 2 1 22
28.
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Quadratic Functions
191
. substitute a known point.
with the vertex above the x-axis. • If the value of a is varied in a quadratic function in factored form. • If the zeros are known. f(x) 5 a(x 2 r) (x 2 s). I needed to find the value of a to approximate the data. write in factored form with a unknown.26(x 2 1990) 2 1 22.5) as the point on the curve. f(x) 5 ax 2 1 bx 1 c. f(x) 5 a(x 2 h) 2 1 k. a model for the data is f (x) 5 0. a family of parabolas with the same y-intercept is created.5 5 a(25) 2 1 22 28. 28.
Determine the equation of the parabola with vertex
a) (22. What characteristics will two parabolas in the family
f (x) 5 a(x 2 3) (x 1 4) share?
2.
7. Determine the equation of the quadratic function that passes through 10. 5) and that passes through (4. 25) and that passes through (21.
192
Chapter 3
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. 7) 0 and 8. and that passes through (25. 3) 1 2 ! 2 and 1 1 ! 2. and that passes through (2. 27) c) (4. If the height of the arch 4 m
T
from the left edge is 6 m. 23) d) (4. A tunnel with a parabolic arch is 12 m wide.
b) Use your graph to sketch the graph of g(x) 5 22(x 2 2) (x 1 6). b) Draw a curve of good fit. c) Sketch the graph of h(x) 5 3(x 2 2) (x 1 6).CHECK Your Understanding
1.
8. 5) if its zeros are 2 1 ! 3 and 2 2 ! 3. How are the parabolas f (x) 5 23(x 2 2) 2 2 4 and
g(x) 5 6(x 2 2) 2 2 4 the same? How are they different? g(x) 5 5x 2 1 3x 2 7 have in common?
3. 28) b) (1. a)
Sketch the graph of f (x) 5 (x 2 2) (x 1 6). 0) and that passes through (11. 6) and that passes through (0. 4)
6.5 m wide pass through the tunnel? Justify your decision. A projectile is launched off the top of a platform. 6). The table gives the height
A
a) Draw a scatter plot of the data. Determine the equation of the quadratic function
f (x) 5 ax 2 2 6x 2 7 if f (2) 5 3. What point do the parabolas f (x) 5 22x 2 1 3x 2 7 and
PRACTISING
a) b) c) d)
4.
11.
Time (s) Height (m) 0 11 1 36 2 51 3 56 4 51 5 36 6 11
(24.
9. Determine the equation of the parabola with x-intercepts 64 and passing
K
through (3. c) Determine the equation that will model this set of data. of the projectile at different times during its flight. 8)
24 and 3. Determine the equation of the parabola with x-intercepts
5. and that passes through (2. 26) !7 and 2 !7. can a truck that is 5 m tall and 3. and that passes through (23.
60
0. if the height 5 m in from the outside edge is 8 m. c) Determine an algebraic model for the data.45
0. A family of cubic equations with zeros 23.00 5 1. Express the
function in standard form.35
1.90
1. 29) is on
the graph?
15.
Definition: Families of Parabolas Characteristics:
Examples:
Non-examples:
Extending
16. c) Determine an algebraic expression that models the data.1875 0. and factored form.50
35
104
198
287
348
401
427
442
418
a) Draw a scatter plot and a curve of good fit. x. in hundreds of kilograms per hectare (100 kg/ha). vertex. Which equation describes the cubic in the family that passes through the point (3.25 4.3. x (ha • m) Yield.05
1. and 5 can be represented by
the function f (x) 5 a(x 1 3) (x 2 1) (x 2 5).75 0.75 1. 1.50 3. A parabolic bridge is 40 m wide.7
12. 6)?
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Quadratic Functions
193
.6875 1.
Amount of Water.75 4. Determine the height of the bridge 12 m in
from the outside edge.75
0. b) Draw a curve of good fit. Students at an agricultural school collected data showing the effect of
different annual amounts of water (rainfall plus irrigation). on the yield of broccoli.
13.
14.75 2. What is the equation of the parabola at the right if the point (24.20
1. y.6875 1. Complete the chart shown.1875 2. Include what you know about families of
C
4 2 8 6 4 2 0 2 4 6 8
y x 2
parabolas in standard.
17.50 3.30
0. b) Estimate the location of the vertex. y (100 kg/ha)
0. Jason tossed a ball over a motion detector and it recorded these data.25 2.00 0
a) Draw a scatter plot of the data. in hectaremetres (ha ? m). Time (s) Height above Ground (m) 0 0 0.
t seconds after jumping out.
For help using the graphing calculator to find points of intersection. how long after jumping out of the airplane should Adam release his parachute? At what height will this occur?
EXAMPLE
1
Selecting a strategy to solve a linear–quadratic system of equations
Determine when the two functions have the same height values. • His height after jumping from the airplane before he opens his parachute can be modelled by the quadratic function h1 (t) 5 24.3.
? According to these models. B-12.
LEARN ABOUT the Math
Adam has decided to celebrate his birthday by going skydiving. He loves to freefall. so he will wait for some time before opening his parachute. he begins falling at a constant rate. where t is time in seconds and h1 (t) is the height above the ground. in metres. • After he releases his parachute.
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. see Technical Appendix.
Tech
Support
Kobi's Solution: Using a Graphical Approach
I graphed both functions on my graphing calculator.9t 2 1 5500. I looked for the point of intersection of the two graphs.8
Linear–Quadratic Systems
GOAL
Solve problems involving the intersection of a linear and a quadratic function. His height above the ground can be modelled by the linear function h2 (t) 5 25t 1 4500. I was looking for the time at which both function values were the same.
and c 5 1000 to solve for t.8. I put the equation into standard form and then used the quadratic formula with a 5 24.
I got two possible values for t.9t 1 5t 1 1000 5 0 t5 5 5 5
25 6 "52 2 4(24.8 or t 8 213.8 s.8 25 6 "19 625 29. so the only answer was t 5 14.
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Quadratic Functions
195
.78 d inadmissible Adam should open his parachute after 14. He will be 4426 m above the ground at that time.9) (1000) 2(24.9)
2
25 6 "25 1 19 600 29.8
I adjusted the window settings until both graphs and the point of intersection on the right was visible.9. the heights represented by each equation will be the same. 4426). This resulted in a quadratic equation that I needed to solve.9t 2 1 5500 5 25t 1 4500
t 8 14. I was only interested in this point since time must be positive. b 5 5. so I set the right-hand sides of the equations equal to each other.8 s after jumping out of the airplane.8 s. I used the intersection operation to locate the point of intersection: (14.9t 2 1 5500 Just after parachute opens: h(t) 5 25t 1 4500
At the moment the parachute is opened.
Christina's Solution: Using Algebra Just before parachute opens: h(t) 5 24. That means that Adam should release his parachute 14.3.8
2b 6 "b 2 2 4ac 2a
24. but the time after jumping cannot be negative.
24.
I put the resulting quadratic function into standard form.
Explain how you would determine the point of intersection of a linear function and a quadratic function graphically and algebraically. I chose the second one because it was an easier calculation. What are an advantage and a disadvantage of each method? Do you think that a linear function always intersects a quadratic function in two places? Why or why not?
APPLY the Math
EXAMPLE
2
Selecting a strategy to predict the number of points of intersection
Determine the number of points of intersection of the quadratic and linear functions f (x) 5 3x 2 1 12x 1 14 and g(x) 5 2x 2 8.8) 1 4500 5 4426 Adam will be about 4426 m above the ground when he opens his parachute. the values of f(x) and g(x) will be equal at that point. Julianne's Solution 3x 2 1 12x 1 14 5 2x 2 8 3x 2 1 10x 1 22 5 0 b2 2 4ac 5 (10) 2 2 4(3) (22) 5 100 2 264 5 2164
If there is a point of intersection. so I set the expressions for f(x) and g(x) equal to each other. B.
Reflecting
A.h2 5 25(14.
To find the height above the ground. I substituted my value for t into one of the equations. Since 2164 . 0. C.
196
Chapter 3
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. there are no real roots.
The line and the parabola do not intersect. I then calculated the value of the discriminant.
Then I put the resulting quadratic equation into standard form.5) 5 31.5 6 " (0.8
EXAMPLE
3
Solving a problem involving the point of intersection
Justin is skeet shooting. The height of the skeet is modelled by the function h(t) 5 25t 2 1 32t 1 2.4
20.5(0. but time cannot be negative.5 2 4.5) 2 2 4(25) (1) 210
20.5 t5 210 t 5 0. g(0.75 m off the ground when it is hit.5t 1 1 25t 2 1 0.75 The skeet will be 16.4.5 t5 210 t 5 20.5t 1 1.5) 1 1 5 16.
The bullet will hit the skeet after 0.25 210 or or 2b 6 "b2 2 4ac 2a
I needed to find the point of intersection of the quadratic and the linear functions. I set them equal to each other. with the same units. so I couldn't use the solution t 5 20. I used the quadratic formula to solve for t.5 s.5 1 4.5
I got two possible values for t.
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Quadratic Functions
197
.
I substituted the value of t into g(t) to solve for the height.5t 1 1 25t 2 1 32t 1 2 5 31. How long will it take for the bullet to hit the skeet? How high off the ground will the skeet be when it is hit? Stephanie's Solution h(t) 5 25t 2 1 32t 1 2 g(t) 5 31. where h(t) is the height in metres t seconds after the skeet is released.5t 1 1 5 0 t5 5 5 20.5 6 "20.3. The path of Justin's bullet is modelled by the function g(t) 5 31.
20.
and h2 (t) 5 24t 1 142 after he released his parachute. (ii) two points. a)
2
Copy the graph of f (x) 5 (x 2 2) 2 2 3.
f(x) = (x
8. Given a quadratic function f (x) and a linear function g(x). A linear function
T
is defined by g(x) 5 mx 2 5. using the same t. in metres. describe two
C
Extending
14. Determine the equation of the line that passes through the points of
intersection of the graphs of the quadratic functions f (x) 5 x 2 2 4 and g(x) 5 23x 2 1 2x 1 8.43t 1 4.3.
7. A punter kicks a football. Can the blocker knock down the punt? If so. What value(s) of the slope of the line would make it a tangent to the parabola? kick is given by the equation h(t) 5 24. Determine the value(s) of k such that the linear function g(x) 5 4x 1 k does 10. His height.
15. In how many ways could the graphs of two parabolas intersect? Draw a
sketch to illustrate each possibility. at what point will it happen? ways you could determine the number of points of intersection of the two functions without solving for them. How long after jumping did the daredevil release his parachute?
11. t seconds after the
13.
16. Determine the value of k such that g(x) 5 3x 1 k intersects the quadratic 9. A daredevil jumps off the CN Tower and falls freely for several seconds before
A
releasing his parachute. t seconds after jumping can be modelled by h1 (t) 5 24. Determine the coordinates of any points of intersection of the functions
x 2 2 2x 1 3y 1 6 5 0 and 2x 1 3y 1 6 5 0.
12. The height of an approaching blocker's hands is modelled by the equation g(t) 5 21. b) Write the equations of the lines from part (a). not intersect the parabola f (x) 5 23x 2 2 x 1 4. The cost function for the production is C(t) 5 600 2 50t. h(t) . h(t) .
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Quadratic Functions
199
. and (iii) no points.24t 1 0. Its height.8
6.9t 2 1 t 1 360 before he released his parachute. in metres.8. Then draw lines with slope 24 that intersect the parabola at (i) one point.26. c) How are all of the lines with slope 24 that do not intersect the parabola related? function f (x) 5 2x 2 2 5x 1 3 at exactly one point. Determine the ticket price that will allow the production to break even.9t 2 1 18. A quadratic function is defined by f (x) 5 3x 2 1 4x 2 2. where t is the ticket price in dollars. The revenue function for a production by a theatre group is
y 8 6 4 2 2 0 2 4 2 4 6 2)2 3 x
R(t) 5 250t 1 300t.
A:
The members of the first family of parabolas will all have the same x-intercepts.
Arrange the equation into standard form ax 2 1 bx 1 c 5 0. y 5 27. The members of the third family will have the same y-intercept.7. (2. Example 2.3
Study
Chapter Review
FREQUENTLY ASKED Questions
Q:
A:
Aid
How can you determine the solutions to a quadratic equation?
• See Lesson 3.6. 25). Examples 2. • Try Chapter Review Questions 12.
What characteristics do the members of the family of parabolas f(x) 5 a(x 2 2) (x 1 6) have in common? the family of parabolas g(x) 5 a(x 2 2) 2 2 5? the family of parabolas h(x) 5 ax 2 1 bx 2 7?
Study
Aid
Q:
• See Lesson 3. determines the number of roots of a quadratic equation.5. 2.
y
x
200
Chapter 3
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. there are no roots. If b 2 2 4ac . there are two distinct roots. 13.
In how many ways can a line intersect a parabola?
Q:
A:
A line can intersect a parabola in at most two places. 2 and 26. • Try Chapter Review Questions 15 and 16.
How can you use the discriminant to determine the number of solutions of a quadratic equation?
The value of the discriminant. 3. If b 2 2 4ac . Examples 1. If b 2 2 4ac 5 0. and 4. the numbers that make each factor zero are the solutions to the original equation • graph the corresponding function f (x) 5 ax 2 1 bx 1 c 5 0 and determine its x-intercepts. The members of the second family will have the same vertex. 0. there is one root. then: • try to factor. b 2 2 4ac. • Try Chapter Review Questions 18 and 19. 0. and 3. and the same axis of symmetry. these points are the solutions to the original equation • use the quadratic formula x 5
2b 6 "b 2 2 4ac 2a
Study
Aid
Q:
A:
• See Lesson 3. x 5 22. and 14.
Examples 1 and 3.Chapter Review
It may intersect a parabola at only one point.
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Quadratic Functions
201
. You could equate the two functions.
y
x
Q:
A1:
How can you determine the points of intersection between a linear and a quadratic function?
Study
Aid
You can graph both functions on the same set of axes and determine the point(s) of intersection from the graphs.
y
x
There may be no points of intersection.8. • Try Chapter Review Questions 21 and 23. This results in a quadratic equation whose solutions are the x-coordinates of the points of intersection.
A2:
• See Lesson 3.
and 10.
10.4
2
9. and h(x) 5 2!x.3
6 cm and 3 cm? Leave your answer in simplest radical form. c) State the domain and range. Determine the equation of the axis of symmetry of the
parabola with points (25.
8. a) State the direction of opening and the zeros of the function.
2. The height. f (x) 5 x 2. What is the perimeter of a right triangle with legs
a1b1c
A 5 "s(s 2 a) (s 2 b) (s 2 c). Consider the quadratic function
Give a reason for your answer. g(x) 5 !x. Determine the maximum height of the football and the time when that height is reached.1
7. Leave your answer in simplest radical form. Describe the relationship between
202
NEL
. and the axis of symmetry. a) State the direction of opening. Is the inverse of a quadratic function also a function?
1. Express each number as a mixed radical in simplest
football is given by h(t) 5 2 1 28t 2 4.
where a. 7.9t .
c) 4!12 2 3!48 d) (3 2 2! 7 ) 2
Lesson 3.
Chapter 3 11. a)
f (x) 5 23(x 2 2) 2 1 5. d) Graph the function.PRACTICE Questions
Lesson 3. c) Graph the function. of the trajectory of a
Lesson 3. For each quadratic function. b) State the domain and range of the inverse relation.
Lesson 3. b) State the domain and range. c) Is the inverse relation a function? Why or why not?
6 4 2 2 0 2 4 6 8 2 4 6 8 y y 5 f(x) x
f (x) 5 4(x 2 2) (x 1 6).
3. h(t). 3) and (3. The area of a triangle can be calculated from Heron's
form. b. Calculate the area of a triangle with s5 2 side lengths 5. a) !98 b) 25!32 formula. sketch the graph of the inverse relation.
6. state the maximum or
minimum value and where it will occur. Consider the quadratic function
Given the graph of f (x). a) f (x) 5 23(x 2 4) 2 1 7
b) f (x) 5 4x(x 1 6)
5.2
4. where t is the time in flight. in seconds. 3) equally distant from the vertex on either side of it. in metres. the vertex. b) Determine the coordinates of the vertex. and c are the side lengths and .
can be
modelled by the equation h(t) 5 14t 2 5t 2. in metres. a) Determine a quadratic function that satisfies these conditions. Will the paintball hit the baseball? If so. h(t). Can the projectile ever reach a height of 9 m? Explain.7
18.
13. 6)?
20.
Lesson 3. Which member passes through the point (22. A rectangular field with an area of 8000 m2 is enclosed
Lesson 3. in thousands. Determine the x-intercepts of the quadratic function
f (x) 5 2x 1 x 2 15. b) Change the slope of the line so that it will intersect the parabola in two locations. The population of a Canadian city is modelled by
family of parabolas f (x) 5 a(x 1 3) 2 2 4 have in common. function P(x) 5 22x 2 1 7x 1 8. The height. at time
15. b) What is the width of the arch at its base?
14. Describe the characteristics that the members of the
2
12.
17. h(t). The trajectory of the paintball is given by the function g(t) 5 3t 1 3. A boy shoots at the baseball with a paintball gun.6
21. where t is the time in years. and 6 m wide at a height of 8 m. in metres. when? At what height will the baseball be?
23. and passing through the point (2. When t 5 0. The arch
P(t) 5 12t 2 1 800t 1 40 000. The height. Calculate the point(s) of intersection of
f (x) 5 2x 2 1 4x 2 11 and g(x) 5 23x 1 4. 2 1 !3 and 2 2 !3. of a baseball. Check these values in the original equation. 5).8
by 400 m of fencing. what will the population be in 2020? b) In what year is the population predicted to be 300 000?
must be 15 m high. where x is the number of dirt bikes produced. Determine the dimensions of the field to the nearest tenth of a metre.Chapter Review
Lesson 3.
22. a) According to the model. the year is 2007.
Lesson 3.
t seconds after it is tossed out of a window is modelled by the function h(t) 5 25t 2 1 20t 1 15. Determine the values of k for which the function
f (x) 5 4x 2 2 3x 1 2kx 1 1 has two zeros. of a projectile. a)
16. Determine the equation of the parabola with roots
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Quadratic Functions
203
. An engineer is designing a parabolic arch. Determine the break-even points of the profit
Will the parabola defined by f (x) 5 x 2 2 6x 1 9 intersect the line g(x) 5 23x 2 5? Justify your answer.5
19. where t is the time in seconds after the projectile is released.
Does the linear function g(x) 5 6x 2 5 intersect the quadratic function
f (x) 5 2x 2 2 3x 1 2? How can you tell? If it does intersect.
function in standard form. Determine the equation in standard form of the parabola shown below. c) Explain why the answer to part (a) has fewer terms than the answer to
4. Identify the zeros. Graph the function.
2. If you needed to know the information listed. Determine the maximum area of a rectangular field that can be enclosed by
2400 m of fencing.
4 2 0 –2 –4 –6
y x 2 4 6 8
204
Chapter 3
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. which form would you choose and why? a) the vertex b) the y-intercept c) the zeros d) the axis of symmetry e) the domain and range
b) Simplify (3 1 "5) (5 2 "10). State the domain and range. determine the point(s) of intersection. factored form. You can choose whether you are provided the equation of a quadratic
6.
9.
part (b). 8.
a) b) c) d)
Express the function in factored form and determine the vertex. For each function. a) f (x) 5 22x 2 2 8x 1 3 b) f (x) 5 3(x 2 1) (x 1 5)
3. state whether it will have a maximum or a minimum
value. Describe the method you would choose to calculate the maximum or minimum value. the axis of symmetry. You are given f (x) 5 25x 2 1 10x 2 5.
5.
7. and the direction of opening. Calculate the value of k such that kx 2 2 4x 1 k 5 0 has one root. or vertex form. Determine the equation of the inverse of f (x) 5 2(x 2 1) 2 2 3. a) Simplify (2 2 !8) (3 1 "2).3
Chapter Self-Test
1.
Why did you approach the problem this way? Use the model you created to graph the flight of Vernon's ball. and h(x) is the height of the ball at that distance.5 m above the ground when it was hit. but you want to find one that is close to reality. and show all of your steps. b) The ball reached a maximum height of 17 m when it was approximately 70 m away from Vernon.5x(x 2 142) b) h(x) 5 20.0015x 2 1 0.5x 2 1 71x 1 1 c) h(x) 5 20.
What is the function that will model the height of Vernon's ball accurately over time?
Assume that the ball was between 0.6 m and 1. Explain the method you are using to get the equation.
Task
Checklist
Did you state your reasons
that the given models were not reasonable?
Did you draw a welllabelled graph.2 m off the ground when it was hit.
D. including some values?
B.3
Chapter Task
Baseball Bonanza
Vernon Wells hits a baseball that travels for 142 m before it lands. and support your claim with reasons and a well-labelled sketch.
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Quadratic Functions
205
. given this additional information: a) The ball was 1. The flight of the ball can be modelled by a quadratic function in which x is the horizontal distance the ball has travelled away from Vernon. a) h(x) 5 20.2 Determine an equation that models the path of the ball.
?
A.
Did you show your work in
your choice of method for part C?
Did you support your choice
of method in part C?
C. There are many quadratic equations you could use to model the distance and height.213x 1 1. a) What would h(142) be? b) What happens when x 5 0? c) What are the possible values for h(x) when x 5 0? d) What would be a good range of values for the height of the ball? Are some values for the height unreasonable? Explain why each function is not a good model of the situation.
Even with this delay. This year he will increase the price and knows that for each $50 price increase. Studying Functions
Analyze two of the following functions in depth. Last year he had 25 students go and each paid $550. including all applied transformations iii) a sketch of the function
34.4 km/h faster than Jill. but sees a friend on the route and stops to talk for 20 min. who has trained all year for this event. Charity Walk
Sacha and Jill set off at the same time on a 30 km walk for charity. Sacha finishes the walk 2 h ahead of Jill. 2 fewer students will go on the trip. The bus costs a flat fee of $5500. Determine a) the number of students who must go for Josh to break even b) the cost of the trip that will maximize his profit
NEL
Cumulative Review
209
. and how long did it take for each person to finish the walk?
35. Ski Trip
Josh is running a ski trip over March Break.Cumulative Review
Investigations
33. a) f (x) 5 3x 2 2 24x 1 50 b) g(x) 5 5 2 2"3x 1 6 1 c) h(x) 5 22 1 (x 2 6) 3 Include: i) the domain and range ii) the relationship to the parent function. How fast was each person walking. and hotel and lift tickets cost $240 per person. Sacha. walks 1.
210
NEL
.
Do you think a linear relation would model their growth? Explain.Chapter
4
Exponential Functions
GOALS
You will be able to
• • • •
Describe the characteristics of exponential functions and their graphs Compare exponential functions with linear and quadratic functions Evaluate powers with integer and rational exponents and simplify expressions involving them Use exponential functions to solve problems involving exponential growth and decay
? Yeast cells grow by dividing at
regular intervals.
NEL
211
.
E. Binary numbers are numbers that are written in base 2. 011. 11 000. G. 010. 1 00.
?
A. Write the two numbers and their quotient as powers of two. 111 Number of Possible Codes 2 4 8 Possible Codes as a Power of Two 21 22
B. Repeat this several times with different pairs of numbers. What is the relationship between the exponents of the powers that you divided and the exponent of the resulting quotient? What rule for dividing powers with the same base does this suggest? How can you predict the possible number of codes if you know the number of digits?
Exponential Functions
C. 235 means 2 3 102 1 3 3 101 1 5 3 100.
Select any two numbers (other than the last two) from the third column. number. 101.
Is there a relationship between the number of digits and the possible number of codes?
Copy and complete the table shown.Getting Started
APPLYING What You Know
Binary Code
The numbers we work with every day are written in base 10. H. Each number consists of only the digits 0 and 1.
F. For example. 01. 10. D. 110. For example. 100. Each letter. Computers use binary code to do their calculations. or symbol needs a separate binary code. What is the relationship between the exponents of the powers that you multiplied and the exponent of the resulting product? What rule for multiplying powers with the same base does this suggest? Select any two numbers (other than the first two) from the third column. Write the two numbers and their product as powers of two. 001.
NEL
213
. and calculate their product. Repeat this several times with different pairs of numbers. 1011 means 1 3 23 1 0 3 22 1 1 3 21 1 1 3 20. and divide the greater by the lesser.
Number of Digits 1 2 3 4 5 6 7 Possible Codes 0.
?
A.
Is the height of each bounce related to the height of the previous bounce?
Set the CBR to "Ball Bounce" mode. Use the trace key to determine the height of each bounce. golf ball) • graphing calculator with a motion detector (CBR)
Collect data and study the characteristics of rapidly decaying functions. One of you holds the ball. When the CBR is triggered.4. while the other holds the CBR 0. Work with a partner.
EXPLORE the Math
When you drop a ball it will bounce several times. basketball. drop the ball. soccer ball.1
YOU WILL NEED
Exploring Growth and Decay
GOAL
• two different types of balls
that bounce (e. racquetball. Let the ball bounce at least 5 times while you collect the data.5 m over the ball.
214
Chapter 4
NEL
..g.
B.
record the original height of the ball. plot the bounce height versus bounce number on the same graph. What happens to the bounce height as the bounce number increases? If you continue the pattern indefinitely. Repeat parts A to D for two additional starting heights.
Reflecting
J. Does each set of data represent a function? How do you know? State the domain and range of each graph.
Bounce Number 0 1 2 Bounce Height First Differences
D. and explain how the height of each bounce is related to its previous bounce.
G.
Need to Know
• The domain and range of a function should be considered in terms of the situation it is modelling.
L. F. After recording data for three different starting heights. M. Did the initial height of the drop influence the graph? Explain. In the first row. Describe how the bounce height changed from one bounce to the next. K. Draw a dashed curve through each set of points. H. will the bounce height ever reach zero? Explain. Describe the shape of each graph.
Calculate and complete the first-differences column. I. Was this pattern the same for each type of ball? Explain.
In Summary
Key Ideas
• Some real-world situations can be modelled by functions whose first differences follow a multiplicative pattern. Did the type of ball you used influence the graph? Explain.
NEL
Exponential Functions
215
.4.1
C. • The scatter plots for these situations show increasing or decreasing nonlinear patterns.
Why was a dashed curve used in part F instead of a solid one? Look at the first-differences column. Use a different colour for each set of data. E. then record the bounce number and bounce height for next bounces.
Copy the table. Repeat the exploration with a different type of ball.
the king
offers the man anything he desires. A folktale tells of a man who helps a king solve a problem. in minutes. How are they the same and how are they different?
216
Chapter 4
NEL
. In return. The graph shows the relationship between the temperature of the cocoa. a) Complete the table of values for the first 10 squares. and time.
Number of Squares on the Chessboard 1 2 3 Number of Grains on that Square 1 2 4
First Differences
b) Create a scatter plot of the data in the first two columns. c) Compare this graph and the first differences of the data with your graphs
and data for the ball-bounce experiment. in degrees Celsius.FURTHER Your Understanding
1.
Cooling Curve for Cocoa 90 Temperature (degrees Celsius) 80 70 60 50 40 30 20 10 0 10 20 30 40 50 60 70 80 90 Time (min)
What characteristics of this graph are the same as the graph(s) you drew in the ball-bounce experiment? b) What was the temperature of the cocoa at the start of the experiment? c) What is the temperature of the classroom?
a)
2. The man asks for one grain of rice on a square of a chessboard and then double the number of grains of rice for each subsequent square. A cup of hot cocoa left on a desk in a classroom had its temperature
measured once every minute.
Consider the chart listing the prefix names and their factors for the unit of measure for length.000 000 001 0.001 0.1 0.000 000 000 001
?
How can powers be used to represent metric units for lengths less than 1 metre?
NEL
Exponential Functions
217
. it is easy to convert from one unit to another.000 1 0.
LEARN ABOUT the Math
The metric system of measurement is used in most of the world.000 01 0.2
GOAL
Working with Integer Exponents
Investigate powers that have integer or zero exponents.4.
Multiple as a Power of 10 1012 109 106 103 102 101
Name terametre gigametre megametre kilometre hectometre decametre metre decimetre centimetre millimetre micrometre nanometre picometre femtometre attometre
Symbol Tm Gm Mm km hm dam m dm cm mm mm nm pm fm am
Multiple of the Metre 1 000 000 000 000 1 000 000 000 1 000 000 1 000 100 10 1 0. A key feature of the system is its ease of use. the metre. Since all units differ by multiples of 10.01 0.000 001 0.
EXAMPLE
2
Connecting the concept of an exponent of 0 to the exponent quotient rule
Use the quotient rule to show that 100 51.1. I rewrote each decimal as a fraction and each denominator as a power of 10.
Jemila's Solution
As I moved down the table. When you divide powers with the same base. I used a power of 10. To come up with the next row in the table. I'll get 100 5 1. 1021 5 0. while the multiples were divided by 10. David's Solution 106 51 106 106 5 10626 5 100 106
I can divide any number except 0 by itself to get 1.
10
I don't think it mattered that the base was 10. 1022 5 0. The relationship would be true for any base. the powers of 10 decreased by 1.
Therefore. you subtract the exponents. I noticed that 100 5 1 and 1 102n 5 n. 100 5 1.
218
Chapter 4
NEL
.01. If I continue this pattern. I applied the rule to show that a power with zero as the exponent must be equal to 1. I divided the multiples and the powers by 10.EXAMPLE
1
Using reasoning to define zero and negative integer exponents
Use the table to determine how multiples of the unit metre that are less than or equal to 1 can be expressed as powers of 10. etc.
the square of the number is positive. B. I evaluated the power.2
Reflecting
A. a) 523
Stergios's Solution
a) 523 5
Rational numbers can be written in a variety of forms. In this case. Since the negative sign is in the parentheses.4."
1 53 1 125 1 (24) 2 1 16 1 34 1 81
523 is what you get if you divide 1 by 53. so the entire power is negative. or as a fraction. it can be evaluated in a similar manner.
APPLY the Math
EXAMPLE
3
Representing powers with integer bases in rational form
b) (24) 22 c) 2324
Communication
Tip
Evaluate. 1? How is 102 related to 1022? Why do you think this relationship holds for other opposite exponents? Do you think the rules for multiplying and dividing powers change if the powers have negative exponents? Explain. the negative sign is not inside the parentheses. The term rational form means "Write the number as an integer.
What type of number results when x2n is evaluated if x is a positive integer and n . I knew that 324 5 34.
1
5
c) 2324 5 2
52
If the base of a power involving a negative exponent is a fraction. C.
NEL
Exponential Functions
219
.
5
b) (24) 22 5
(24) 22 is what you get if you divide 1 by (24) 2.
3
Q R
2 3
23
is what you get if you divide
Dividing by a fraction is the same as multiplying by its reciprocal. multiplied exponents for the denominator. B-15.
EXAMPLE
5
Selecting a strategy for expressions involving negative exponents .
Tech
Support
Derek's Solution: Using a Calculator
I entered the expression into my calculator.
Evaluate
35 3 322 (323 ) 2
Kayleigh's Solution: Using Exponent Rules
35 3 322 351 (22) 5 2332 (323 ) 2 3 33 5 26 3 5 332 (26) 5 39 5 19 683
I simplified the numerator and denominator separately.EXAMPLE
4
Representing powers with rational bases as rational numbers
Evaluate ( 2 ) 23. Then I divided the numerator by the denominator.
220
Chapter 4
NEL
.
For help with evaluating powers on a graphing calculator. see Technical Appendix. 3
Sadira's Solution
2 23 a b 5 3
1 2 3 a b 3 1 5 8 a b 27 27 513 8 27 5 8
3 1 by Q 2R . I added exponents for the numerator. so I used this to evaluate the power. I made sure I used parentheses around the entire numerator and denominator so that the calculator would compute those values before dividing. and subtracted exponents for the final calculation.
4.2
In Summary
Key Ideas
• An integer base raised to a negative exponent is equivalent to the reciprocal of the same base raised to the opposite exponent. b2n 5 1 , bn where b 2 0
• A fractional base raised to a negative exponent is equivalent to the reciprocal of the same base raised to the opposite exponent. a 2 0, b 2 0
a 2n a b 5 b
• A number (or expression), other than 0, raised to the power of zero is equal to 1. b0 5 1, where b 2 0
1 b n 5 a b , where a n a a b b
Need to Know
• When multiplying powers with the same base, add exponents. bm 3 bn 5 bm1n • When dividing powers with the same base, subtract exponents. bm 4 bn 5 bm2n if b 2 0 • To raise a power to a power, multiply exponents. (bm ) n 5 bmn • In simplifying numerical expressions involving powers, it is customary to present the answer as an integer, a fraction, or a decimal. • In simplifying algebraic expressions involving powers, it is customary to present the answer with positive exponents.
CHECK Your Understanding
a) b)
1. Rewrite each expression as an equivalent expression with a positive exponent.
The volume of this cube is V(x) 5 x 3 and the area of its base is A(x) 5 x 2. In this cube, x is the side length and can be called • the square root of A, since if squared, the result is A(x) • the cube root of V, since if cubed, the result is V(x)
LEARN ABOUT the Math
? What exponents can be used to represent the side length x as the square root of area and the cube root of volume?
EXAMPLE
1
Representing a side length by rearranging the area formula
Express the side length x as a power of A and V. Ira's Solution A 5 x2 x 5 An A 5 (x) (x) A 5 An 3 An A5A
n1n
I used the area formula for the base. Since I didn't know what power to use, I used the variable n to write x as a power of A. I rewrote the area formula, substituting An for x. Since I was multiplying powers with the same base, I added the exponents. I set the two exponents equal to each other. I solved this equation. The exponent that represents a square root is 1 . 2
I used the volume formula for a cube. I represented the edge length x as a power of the volume V. I used the variable n. I rewrote the volume formula, substituting V n for x. I added the exponents.
I set the two exponents equal to each other. I solved this equation. The exponent that represents a cube root is 1. 3
Reflecting
A. B. C.
Why could x be expressed as both a square root and a cube root? Make a conjecture about the meaning of x n. Explain your reasoning. Do the rules for multiplying powers with the same base still apply if the exponents are rational numbers? Create examples to illustrate your answer.
1
I know that the exponent 1 3 indicates a cube root. So I used the power-of-a-power rule to separate the exponents:
5 (!27) 2 5 27 3 32
1 1
5 (27 3 )2
3
5 (3) 2 59
3 5 !272
2 3
5231 3
and
2 3
5132 3
5 27 23 3
5 (27 2 ) 3
59
3 5 !729
1
1
To see if the order in which I applied the exponents mattered, I calculated the solution in two ways. In the first way, I evaluated the cube root before squaring the result. In the other way, I squared the base and then took the cube root of the result. Both ways resulted in 9.
226
Chapter 4
NEL
4.3
EXAMPLE
5
Evaluating a power with a rational exponent
b) (16) 20.75
Evaluate. 4 a) (227) 3
Casey's Solutions
4
a) (227) 3 5 ( (227) 3 ) 4
3 4
5 (!227) 5 (23) 4 5 81
3
1
3 I represented (227) 3 as ! 227. I calculated the cube root of 227. I evaluated the power.
I rewrote the exponent as 4 3 1. 3
1
b) 1620.75 5 1624
1 5 3 2 5 1 8
1 5 4 (!64) 3 5 1 16 4
3
I rewrote the power, changing the exponent from 20.75 to its equivalent fraction. I expressed 1624 as a rational number, using 1 as the numerator and 16 4 as the denominator. I determined the fourth root of 64 and cubed the result.
3 3
The rules of exponents also apply to powers involving rational exponents.
EXAMPLE
6
Representing an expression involving the same base as a single power
5
Simplify, and then evaluate Lucia's Solution
5
86 ! 8 83
5
86 !8 83
5
.
5
8682 83
5 5 1 6 12 5
5 1
To simplify, I converted the radical into exponent form. Since the bases were the same, I wrote the numerator as a single power by adding exponents, then I subtracted exponents to simplify the whole expression.
5
8
83
NEL
Exponential Functions
227
5
83 83
5 5 4
4
5 83 2 3 58 5 5
23
1
1 83 1 2
1
Once I had simplified to a single power of 8, the number was easier to evaluate.
I checked my work on my calculator.
In Summary
Key Ideas
exponent 1 indicates the nth root of the base. If n . 1 and n [ N, then n 1 n bn 5 ! b, where b 2 0. • If the numerator of a rational exponent is not 1, and if m and n are positive m n n integers, then bn 5 (! b) m 5 ! bm, where b 2 0. • A number raised to a rational exponent is equivalent to a radical. The rational
Need to Know
• The exponent laws that apply to powers with integer exponents also apply to powers with rational exponents. Included are the product-of-powers rule a an3 bn 5 (ab) n and the quotient of powers rule an 4 bn 5 Q b R n. • The power button on a scientific calculator can be used to evaluate rational exponents. • Some roots of negative numbers do not have real solutions. For example, 216 does not have a real-number square root, since whether you square a positive or negative number, the result is positive. • Odd roots can have negative bases, but even ones cannot.
13. The power 43 means that 4 is multiplied by itself three times. Explain the
meaning of 42.5.
1 1 1
14. State whether each expression is true or false.
6 1 1 1 1 21 f ) c a x 3 b a y 3 b d 5 x 2y 2 a 1 b 5a1b a b m 15. a) What are some values of m and n that would make (22) n undefined? m T b) What are some values of m and n that would make (6) n undefined?
a)
92 1 42 5 (9 1 4) 2
1 2 1 2 1 2
b) 9 1 4 5 (9 3 4) c)
2 2 e) a x 3 1 y 3 b 5 x 1 y
d) a 3 b
f ) (27776) 1.6
e) " (0.0016) 3
4
1 a
1 b
21
5 ab
1
1 6
Extending
16. Given that x y 5 y x, what could x and y be? Is there a way to find the answer
each dollar she invested for a year at the interest rate i per year. Her friend Bindu suggests that she begin by taking the 12th root of each side of the equation. Will this work? Try it and solve for the variable i. Explain why it does or does not work.
18. Solve.
a) b)
1 4 4 2 " x4 1 15 5 " 16 Å8
1
1 4 8 a b 2 3 5 "x 2 16 Å 27
3
230
Chapter 4
NEL
4.4
GOAL
Simplifying Algebraic Expressions Involving Exponents
Simplify algebraic expressions involving powers and radicals.
LEARN ABOUT the Math
The ratio of the surface area to the volume of microorganisms affects their ability to survive. An organism with a higher surface area-to-volume ratio is more buoyant and uses less of its own energy to remain near the surface of a liquid, where food is more plentiful. Mike is calculating the surface area-to-volume ratio for different-sized cells. He assumes that the cells are spherical. For a sphere, SA(r) 5 4pr 2
Radius (mm) 1 Surface Area/ Volume 4p 4 a pb 3
1.5
9p 4.5p
and
V(r) 5 4pr 3. 3
2
He substitutes the value of the radius into each formula and then divides the two expressions to calculate the ratio.
16p 32 a pb 3
2.5
3
36 p 36 p
25p 125 a pb 6 49p 343 a b 6p
3.5
?
How can Mike simplify the calculation he uses?
NEL
Exponential Functions
231
EXAMPLE
1
Representing the surface area-to-volume ratio
Simplify
SA(r) , given that SA(r) 5 4pr 2 and V(r) 5 4 pr 3. 3 V(r)
Bram's Solution
SA(r) V(r)
I used the formulas for SA and V and wrote the ratio. The numerator and denominator have a factor of p, so I divided both by p. I started to simplify the expression by dividing the coefficients. a4 4
4pr 2 5 4 3 pr 31 5 3r 21 5 3 r
1
4 3 5 4 3 5 3b 3 4
The bases of the powers were the same, so I subtracted exponents to simplify the part of the expression involving r.
I used a calculator and substituted r 5 2 in the unsimplified ratio first and my simplified expression next. Each version gave me the same answer, so I think that they are equivalent, but the second one took far fewer keystrokes!
Reflecting
A. B. C. D.
How can you use the simplified ratio to explain why the values in Mike's table kept decreasing? Is it necessary to simplify an algebraic expression before you substitute numbers and perform calculations? Explain. What are the advantages and disadvantages to simplifying an algebraic expression prior to performing calculations? Do the exponent rules used on algebraic expressions work the same way as they do on numerical expressions? Explain by referring to Bram's work.
232
Chapter 4
NEL
4.4
APPLY the Math
EXAMPLE
2
Connecting the exponent rules to the simplification of algebraic expressions
Simplify
(2x23y 2 ) 3 . (x 3 y24 )2
Adnan's Solution
(2 x23y 2 ) 3 (2) 3 (x23 ) 3 ( y 2 ) 3 5 (x 3y24 ) 2 (x 3 ) 2 ( y24 ) 2
I used the product-of-powers rule to raise each factor in the numerator to the third power and to square each factor in the denominator. Then I multiplied exponents. I simplified the whole expression by subtracting exponents of terms with the same base. One of the powers had a negative exponent. To write it with positive exponents, I used its reciprocal.
I substituted 23 for x and 2 for n and evaluated. I did the same in the denominator.
5 x(5n) 2 (2n25)
5 x 3n15 5 (23) 3(2) 15 5 (23) 11 5 2177 147
EXAMPLE
4
1 1
Simplifying an expression involving powers with rational exponents .
Simplify
(27a 23b 12 ) 3 (16a 28b 12 ) 2
Jane's Solution
1 1 23 12 3 3 3 1 28 12
(27a23b12 ) 3 (16a b )
1 28 12 2
5
27 a b
162a 2 b 2 3a21b4 5 24 6 4a b
In the numerator. using the power-of-a-power rule.Alana's Solution: Simplifying. I applied the 1 exponent 3 to each number or variable inside the parentheses. I added the exponents in the numerator to express it as a single power.
3 5 a2114b426 4 3 5 a 3b22 4 5 3a 3 4b 2
I expressed the answer with positive exponents. so I simplified by using exponent rules before I substituted. Then I subtracted the exponents in the denominator to divide the powers. Once I had a single power. 1 applying the exponent 2 to the numbers and variables. I simplified by subtracting the exponents. then Substituting
(x 2n11 ) (x 3n21 ) x 2n25 5 5 x (2n11) 1 (3n21) x 2n25 x 5n x 2n25
Each power had the same base.
234
Chapter 4
NEL
.
I couldn't write it as a single radical.4. I changed the radical expressions to exponential form and used exponent rules to simplify. Simplify. I converted it to radical form. • Algebraic expressions involving radicals can often be simplified by changing the expression into exponential form and applying the rules for exponents. !x 3
5
Representing an expression involving radicals as a single power
Albino's Solution
8
5 !x8 3 x5 3 a 3 b 5 a 3b !x x2
Since this is a fifth root divided by a square root.4
Sometimes it is necessary to express an expression involving radicals using exponents in order to simplify it. simplify prior to substituting. but the number of calculations will be reduced. Simplify a
EXAMPLE
5 ! x8 3 b.
5 (x5
8
2
3 2 3
)
5 (x10 5 x10
10
3
5 (x )
5 !x 3
16
2
15 10 3
)
1 10 3
When I got a single power.
Need to Know
• When evaluating an algebraic expression by substitution.
CHECK Your Understanding
1. The answer will be the same if substitution is done prior to simplifying.
In Summary
Key Idea
• Algebraic expressions involving powers containing integer and rational exponents can be simplified with the use of the exponent rules in the same way numerical expressions can be simplified.
a) b)
NEL
x 4 (x 3 ) ( p23 ) ( p) 5
m5 m23 a24 d) 22 a
c)
e) ( y 3 ) 2 f)
(k 6 ) 22
Exponential Functions
235
. Express each answer with positive exponents.
C. where 23 # x # 5 In each of your tables. y 5 5x. g(x) 5 x.
y 10 y = 2x
x y = 1 10 2
()
y
x –10 0 10 –10 0 10
x
?
What are the characteristics of the graph of the exponential function f(x) 5 b x. Use the WINDOW settings shown. Describe the difference patterns for each type of function. State the domain and range of each function. x. h(x) 5 x 2.
Create a tables of values for each of the following functions. Graph all three functions on the same graph.4. increase and decrease. change as the values of the independent variable. describe how values of the dependent variable. calculate the first and second differences. F. Label each curve with the appropriate equation. Graph each function on graph paper and draw a smooth curve through each set of points.
EXPLORE the Math
exponential function a function of the form y 5 a(bx )
Functions such as f (x) 5 2x and g(x) 5 ( 1 ) x are examples of exponential 2 functions.
240
Chapter 4
NEL
. and k(x) 5 2 x. For each function. y. and y 5 10 x.5
YOU WILL NEED
Exploring the Properties of Exponential Functions
GOAL
• graphing calculator • graph paper
Determine the characteristics of the graphs and equations of exponential functions. including population growth and the cooling of a liquid. E. Use a graphing calculator to graph the functions y 5 2x. and how does it compare with the graphs of quadratic and linear functions?
A. B. These types of functions can model many different phenomena. D.
Compare the features of the graphs of f (x) 5 bx for each group. and replace 1 them with y 5 ( 1 ) x and y 5 ( 10 ) x (or y 5 0.
H.4. compared with values of b . 1 ii) different values of b when 0 . state • the domain and range • the intercepts • the equations of any asymptotes
1 Describe how each of the graphs of y 5 ( 1 ) x and y 5 ( 10 ) x differs from 2 y 5 2x as the x-values increase and as they decrease. intercepts.
How do the differences for exponential functions differ from those for linear and quadratic functions? How can you tell that a function is exponential from its differences? The base of an exponential function of the form f (x) 5 b x cannot be 1.
L. i) different values of b when b . 0
Reflecting
N. Which curve increases faster as you trace to the right? Which one decreases faster as you trace to the left? Delete the second and third functions ( y 5 5x and y 5 10x ). B-2. b . Think about the domain. J. Explain how you can distinguish an exponential function from a quadratic function and a linear function by using • the graphs of each function • a table of values for each function • the equation of each function
Exponential Functions
O. Explain why this restriction is necessary. if present Examine the y-values as x increases and decreases.
K. P. 1 iii) values of b when 0 . 1.1x ). range. 0 compared with values of b .
For each function.5x and y 5 0. see Technical Appendix. Discuss your findings.5
G.
Investigate what happens when the base of an exponential function is negative.
For help tracing functions on the graphing calculator. state • the domain and range • the intercepts • the equations of any asymptotes.
Tech
Support
I. M. 1 iii) values of b . and asymptotes. 2 For each new function.
NEL
241
. b . Try y 5 (22) x.
Use the window settings shown. d. k.
You can adjust to these settings by pressing ZOOM and
Use your graphing calculator to graph the function f (x) 5 2x. • It is an increasing function.
ZDecimal . • It has a y-intercept of 1.
y 10 y = 2x
x –10 0 10
?
If f(x) 5 2x. • Its asymptote is the line y 5 0.
INVESTIGATE the Math
Recall the graph of the function f (x) 5 2x.
Predict what will happen to the function f (x) 5 2x if it is changed to • g(x) 5 2x 1 1 or h(x) 5 2x 2 1 • p(x) 5 2x11 or q(x) 5 2x21
244
Chapter 4
NEL
.
B.4.
4
. how do the parameters a. and c in the function g(x) 5 af(k(x 2 d)) 1 c affect the size and shape of the graph of f(x)?
Tech
Support
A.6
YOU WILL NEED
Transformations of Exponential Functions
GOAL
• graphing calculator
Investigate the effects of transformations on the graphs and equations of exponential functions.
Exponential Functions
E. Keep the graph of f (x) 5 2x as Y1 for comparison.
NEL
245
. Predict what will happen to the function f (x) 5 2x if it is changed to • g(x) 5 3(2x ) • h(x) 5 0. complete the table of values. and describe how its points and features have changed. Graph each function one at a time. one at a time.6
C. sketch the graph on the same grid.5(2x ) • j(x) 5 2 (2x ) Create a table like the one in part C using the given functions in part E. Keep the graph of f (x) 5 2x as Y1 for comparison. Comment on how the features and points of the original graph were changed by the transformations. For each function.
Description of Changes of New Graph y 5 2x 1 1
Function g(x) 5 2x 1 1
Sketch
Table of Values x 21 0 1 y 5 2x
h(x) 5 2x 2 1
x 21 0 1
y 5 2x
y 5 2x 2 1
p(x) 5 2x11
x 21 0 1
y 5 2x
y 5 2x11
q(x) 5 2x21
x 21 0 1
y 5 2x
y 5 2x21
D. sketch the graph on the same grid and describe how its points and features have changed. as Y2.
F. as Y2.
Copy and complete the table by graphing the given functions. In your table.4.
Describe the types of transformations you observed in part C.
1 f (x) 5 x 2. and describe how its points and features have changed.G. Are the changes in the function affected by the value of the base? Summarize your findings by describing the roles that the parameters a. as Y2. Experiment with different kinds of transformations.
J. Keep the graph of f (x) 5 2x as Y1 for comparison.
L. Graph each function one at a time. d. and c play in the function defined by f (x) 5 ab k(x2d ) 1 c.
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.
H. Describe the types of transformations you observed in part I. Choose several different bases for the original function. N. sketch the graph.
Do the transformations affect f (x) 5 b x in the same way they affect f (x) 5 x. f (x) 5 x . and f (x) 5 |x|? Explain. Which transformations change the location of the asymptote? Explain how the equation is changed by these transformations.
I.
Describe the types of transformations you observed in part F.5x • j(x) 5 22x Create a table like the one in part C using the given functions in part H. O. k. Comment on how the features and points of the original graph were changed by such transformations. Predict what will happen to the function f (x) 5 2x if it is changed to • g(x) 5 22x • h(x) 5 20.
K. f (x) 5 !x.
Reflecting
M. Comment on how the features and points of the original graph were changed by the transformations. In your table.
Which transformations change the shape of the curve? Explain how the equation is changed by these transformations. complete the table of values.
4.6
APPLY the Math
EXAMPLE
1
Using reasoning to predict the shape of the graph of an exponential function
Use transformations to sketch the function y 5 22(3x24 ). State the domain and range.
J.P.'s Solution
10 8 6 4 2 4 2 0 2 4 2 4 y = 3x
The function I really want to graph is y 5 22(3x24 ). The base function, y 5 3x, was changed by multiplying all y-values by 22, resulting in a vertical stretch of factor 2 and a reflection in the x-axis. Subtracting 4 from x results in a translation of 4 units to the right. I could perform these two transformations in either order, since one affected only the x-coordinate and the other affected only the y-coordinate. I did the stretch first.
y
I began by sketching the graph of y 5 3x. Three of its key points are (0, 1), (1, 3), and (21, 1 ). The asymptote is 3 the x-axis, y 5 0.
The domain of the original function, 5x [ R6 , was not changed by the transformations. 1 2
y = 22( 3x)
The range, determined by the equation of the asymptote, was y . 0 for the original function. There was no vertical translation, so the asymptote remained the same, but, due to the reflection in the x-axis, the range changed to 5 y [ R | y , 06.
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Exponential Functions
247
EXAMPLE
2
Connecting the graphs of different exponential functions
Use transformations to sketch the graph of y 5 422x24 1 3.
Ilia's Solution
I began by sketching the graph of the base curve, y 5 4x. It has the line y 5 0 as its asymptote, and three of its key points are (0, 1), (1, 4), and (21, 1 ). 4 I factored the exponent to see the different transformations clearly: y 5 422(x12) 1 3 The x-values were multiplied by 22, resulting in a horizontal compression of factor 1, as well as a reflection in the y-axis. 2
10 8 6 4 2 4 2 0
y y = 4x
x 2 4
There were two translations: 2 units to the left and 3 units up. I applied the transformations in the proper order. The table shows how the key points and the equation of the asymptote change:
10 8 6 4 2 4 2 0
y y = 4x
Point or Asymptote
4
Horizontal Stretch and Reflection (0, 1) (2 1, 2 4)
Horizontal Translation (22,1) (221, 2 4) (211, 1 ) 2 4 y50
Vertical Translation (22, 4) (221, 7) 2 (211, 31 ) 2 4 y53
(0, 1)
2 y = 422x 1 y = 422x x
2 4
3
(1, 4) (21, 1 ) 4 y50
( 1, 1 ) 2 4 y50
1 There was one stretch and one reflection, each of which applied only to
the x-coordinate: a horizontal compression of factor 1 and a reflection 2 in the y-axis (shown in red).
2 There were two translations: 2 units to the left and 3 units up (shown
in black).
Pinder's Solution: Using Exponent Rules
f (x) 5 9x 5 (32 ) x 5 32x 5 g(x) Both functions are the same.
9 is a power of 3, so, to make it easier to compare 9x with 32x, I substituted 32 for 9 in the first equation. By the power-of-a-power rule, f(x) has the same equation as g(x).
Kareem's Solution
f(x) 5 9x is an exponential function with a y-intercept of 1 and the line y 5 0 as its asymptote. Also, f(x) 5 9x passes through the points (1, 9) and (21, 1 ). 9 g(x) 5 32x is the base function y 5 3x after a horizontal compression of factor 1. This 2 means that the key points change by multiplying their x-values by 1. 2 The point (1, 3) becomes (0.5, 3) and (2, 9) becomes (1, 9). When I plotted these points, I got points on the curve of f(x).
f(x) = 9x y y = 3x 10 8 6 4 2 4 2 0 2 4 x
Both functions are the same.
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Exponential Functions
249
EXAMPLE
4
Connecting the verbal and algebraic descriptions of transformations of an exponential curve
An exponential function with a base of 2 has been stretched vertically by a factor of 1.5 and reflected in the y-axis. Its asymptote is the line y 5 2. Its y-intercept is (0, 3.5). Write an equation of the function and discuss its domain and range.
Louise's Solution
y 5 a2k(x2d) 1 c
I began by writing the general form of the exponential equation with a base of 2. Since the function had been stretched vertically by a factor of 1.5, a 5 1.5. The function has also been reflected in the y-axis, so k 5 21. There was no horizontal translation, so d 5 0. Since the horizontal asymptote is y 5 2 the function has been translated vertically by 2 units, so c 5 2. I substituted x 5 0 into the equation and calculated the y-intercept. It matched the stated y-intercept, so my equation seemed to represent this function.
The range changed, since there was a vertical translation. The asymptote moved up 2 units along with the function, so the range is 5 y [ R | y . 26.
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4.6
In Summary
Key Ideas
• In functions of the form g(x) 5 af(k(x 2 d) ) 1 c, the constants a, k, d, and c change the location or shape of the graph of f(x). The shape is dependent on the value of the base function f(x) 5 bx, as well as on the values of a and k. • Functions of the form g(x) 5 af(k(x 2 d) ) 1 c can be graphed by applying the appropriate transformations to the key points of the base function f(x) 5 bx, one at a time, following the order of operations. The horizontal asymptote changes when vertical translations are applied.
Need to Know
• In exponential functions of the form g(x) 5 a bk(x2d ) 1 c: • If |a| . 1, a vertical stretch by a factor of |a| occurs. If 0 , |a| , 1, a vertical compression by a factor of |a| occurs. If a is also negative, then the function is reflected in the x-axis. 1 • If |k| . 1, a horizontal compression by a factor of Z k Z occurs. If 0 , |k| , 1, a 1 horizontal stretch by a factor of Z k Z occurs. If k is also negative, then the function is reflected in the y-axis. • If d . 0, a horizontal translation of d units to the right occurs. If d , 0, a horizontal translation to the left occurs. • If c . 0, a vertical translation of c units up occurs. If c , 0, a vertical translation of c units down occurs. • You might have to factor the exponent to see what the transformations are. For example, if the exponent is 2x 1 2, it is easier to see that there was a horizontal stretch of 2 and a horizontal translation of 1 to the left if you factor to 2(x 1 1). • When transforming functions, consider the order. You might perform stretches and reflections followed by translations, but if the stretch involves a different coordinate than the translation, the order doesn't matter. • The domain is always 5x [ R 6 . Transformations do not change the domain. • The range depends on the location of the horizontal asymptote and whether the function is above or below the asymptote. If it is above the asymptote, its range is y . c. If it is below, its range is y , c.
CHECK Your Understanding
1. Each of the following are transformations of f (x) 5 3x. Describe each
transformation.
a)
g(x) 5 3x 1 3
c) g(x) 5
1 x (3 ) 3
x
b) g(x) 5 3x13
2. For each transformation, state the base function and then describe the
state the transformations that must be applied to f (x) state the y-intercept and the equation of the asymptote sketch the new function state the domain and range
c) g(x) 5 22f (2x 2 6) d) h(x) 5 f (20.5x 1 1)
a) g(x) 5 0.5f (2x) 1 2 b) h(x) 5 2f (0.25x 1 1) 2 1
6. Compare the functions f (x) 5 6x and g(x) 5 32x.
C
7. A cup of hot liquid was left to cool in a room whose temperature was 20 °C.
T(t) 5 80 Q R 30 1 20. Use your knowledge of transformations to sketch this 2 function. Explain the meaning of the y-intercept and the asymptote in the context of this problem. The temperature changes with time according to the function
1
t
8. The doubling time for a certain type of yeast cell is 3 h. The number of cells
after t hours is described by N(t) 5 N023, where N0 is the initial population. a) How would the graph and the equation change if the doubling time were 9 h? b) What are the domain and range of this function in the context of this problem?
9. Match the equation of the functions from the list to the appropriate graph at
The regional municipality of Wood Buffalo, Alberta, has experienced a large population increase in recent years due to the discovery of one of the world's largest oil deposits. Its population, 35 000 in 1996, has grown at an annual rate of approximately 8%.
Regional Municipality of Wood Buffalo
?
How long will it take for the population to double at this growth rate?
LEARN ABOUT the Math
EXAMPLE
1
Selecting a strategy to determine the doubling rate
Carter's Solution: Using a Table of Values and a Graph 0.08(35) 1 35 5 35(0.08 1 1) 5 35(1.08)
When you add 8% of a number to the number, the new value is 108% of the old one. This is the same as multiplying by 1.08, so I created the table of values by repeatedly multiplying by 1.08. I did this 10 times, once for each year, and saw that the population doubled to 70 000 after 9 years of growth. Time (year from 1996) Population (thousands)
0 35.0
1 37.8
2 40.8
3 44.1
4 47.6
5 51.4
6 55.5
7 60.0
8 64.8
9 70.0
10 75.6
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4.7
Population of Wood Buffalo 80 70 Population (thousands) 60 50 40 30 20 10 0 1 2 3 4 5 6 7 8 9 10 Years
I plotted the points and drew a smooth curve through the data. I drew a horizontal line across the graph at 70 000 and saw that it touched the curve at 9 years.
Sonja's Solution: Creating an Algebraic Model P(1) 5 35(1.08) 5 37.8 P(2) 5 37.8(1.08) 5 40.8 Substituting P(1) into P(2): P(2) 5 35(1.08) (1.08) 5 35(1.08) 2 So, P(3) 5 35(1.08) (1.08) 5 35(1.08) 3 Therefore, P(n) 5 35(1.08) n P(6) 5 35(1.08) 6 5 35(1.586 874 323) 5 55.540 601 3 P(9) 5 35(1.08) 9 5 35(1.999 004 627) 5 69.965 161 95 8 70 The population would double in approximately 9 years at an 8% rate of growth.
Since population is a function of time, I expressed the relationship in function notation. I used P (n), where the exponent, n, would represent the number of years after 1996 and P (n) would represent the population in thousands. I guessed that it would take 6 years for the population to double. I substituted n 5 6 into the expression for the function, but it was too low. I tried values for n until I got an answer that was close to the target of 70; n 5 9 was pretty close.
2
To calculate the population after 1 year, I needed to multiply 35 by 1.08. For each additional year, I repeatedly multiply by 1.08. Repeated multiplication can be represented with exponents. The value of the exponent will correspond to the number for the year. This led to an algebraic model.
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Exponential Functions
255
Reflecting
A. B. C.
Which features of the function indicate that it is exponential? Describe what each part of the equation P(n) 5 35(1.08) n represents in the context of the problem and the features of the graph. Compare Carter's and Sonja's solutions. Which approach do you think is better? Why?
APPLY the Math
EXAMPLE
2
Solving an exponential decay problem, given the equation
A 200 g sample of radioactive polonium-210 has a half-life of 138 days. This means that every 138 days, the amount of polonium left in a sample is half of the original amount. The mass of polonium, in grams, that remains after 1 t t days can be modelled by M(t) 5 200 Q R 138. 2 a) Determine the mass that remains after 5 years. b) How long does it take for this 200 g sample to decay to 110 g? Zubin's Solution: Using the Algebraic Model
a) 5 years 5 5(365) days
1 1825 M(1825) 5 200a b 138 2 5 1825 days 8 0.021
Since the half-life is measured in days, I converted the number of years to days before substituting into the function. I used my calculator to determine the answer.
8 200(0.000 104 5)
1 138 b) M(t) 5 200a b 2
t
There is approximately 0.02 g of polonium-210 left after 5 years. 1 138 110 5 200a b 2
t t
I began by writing the equation and substituting the amount of the sample remaining. I needed to isolate t in the equation, so I divided each side by 200. I didn't know how to isolate t, so I used guess and check to find the answer.
110 1 138 5a b 200 2
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1 M(100) 5 200a b 138 2
t 100
1 138 0.55 5 a b 2 8 121 g
4.7
I knew that if the exponent was 1 (t 5 138 days), the original amount would be halved, but the amount I needed to find was 110 g, so the exponent needed to be less than 1. I guessed 100 days, which I substituted into the original equation. I calculated the answer. It was too high, which meant that my guess was too low. I guessed and checked a few more times until I found the answer of approximately 119 days.
1 138 M(119) 5 200a b 2
119
8 110 g
Barry's Solution: Using a Graphical Model
a)
I graphed M(t), then used the value operation. I had to change 5 years into days.
Tech
Support
For help determining the point(s) of intersection between two functions, see Technical Appendix, B-12.
There is about 0.02 g remaining after 5 years.
b)
I graphed M(t), and I graphed a horizontal line to represent 110 g. I knew that the point where the line met the curve would represent the answer.
I used the "Intersect" operation on the graphing calculator to find the point. The x-value represents the number of days.
It takes approximately 119 days for the sample to decay to 110 g.
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Exponential Functions
257
EXAMPLE
3
Solving a problem by determining an equation for a curve of good fit
A biologist tracks the population of a new species of frog over several years. From the table of values, determine an equation that models the frog's population growth, and determine the number of years before the population triples.
Year Population 0 400 1 480 2 576 3 691 4 829 5 995
Tech
Support
Fred's Solution
I used my graphing calculator to create a scatter plot. The equation is of the form P(t) 5 abt, where • P(t) represents the population in year t • a is the initial population • b is the base of the exponential function Since the function is increasing, b . 1. The initial population occurs when x 5 0. That means that a 5 400. If b 5 2, then the population would have doubled, but it went up by only 80 in the first year, so the value of b must be less than 2.
For help creating scatter plots on a graphing calculator, see Technical Appendix, B-11.
I tried b 5 1.5 and entered the equation P(t) 5 400(1.5) t into the equation editor. The graph rose too quickly, so 1.5 is too great for b.
I changed the equation to P(t) 5 400(1.1) t. I graphed the equation on the calculator. I checked to see if the curve looked right. It rose too slowly, so b must be between 1.1 and 1.5.
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4.7
I continued this process until I found a good fit when b 5 1.2.
The equation that models this population is P(t) 5 400(1.2) t
To determine the year that the population tripled, I graphed the line P 5 1200 and found the intersection point of the curve and the line.
From this graph, I determined that the frog population tripled in approximately 6 years.
EXAMPLE
4
Representing a real-world problem with an algebraic model
A new car costs $24 000. It loses 18% of its value each year after it is purchased. This is called depreciation. Determine the value of the car after 30 months. Gregg's Solution y 5 abx
The car's value decreases each year. Another way to think about the car losing 18% of its value each year is to say that it keeps 82% of its value. To determine its value, I multiplied its value in the previous year by 0.82. The repeated multiplication suggested that this relationship is exponential. That makes sense, since this has to be a decreasing function where 0 , b , 1.
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Exponential Functions
259
608 884 097) 814. as a decimal. For example.
n 5 30 months 5 30 4 12 years 5 2. For example." then the number of decay periods in the equation is measured in metres.5 years V(2.08. then the base of the power in the equation can be obtained by adding the rate. then the base of the power in the equation is obtained by subtracting the rate. Note that • f(x) is the final amount or number • a is the initial amount or number • for exponential growth.82) 2. if light intensity decreases "per metre. b 5 1 1 growth rate. a growth rate of 8% involves multiplying repeatedly by 1. b 5 1 2 decay rate • x is the number of growth or decay periods
Need to Know
• For situations that can be modeled by an exponential function: • If the growth rate (as a percent) is given.5 5 24(0.g. as a decimal. light intensity.5) 5 24(0.82) n
I used V and n to remind me of what they represented. which is the value of a and the exponent n is measured in years.. • One way to tell the difference between growth and decay is to consider whether the quantity in question (e.
I converted 30 months to years to get my answer. The initial value is $24 000.92.
In Summary
Key Ideas
• The exponential function f(x) 5 ab x and its graph can be used as a model to solve problems involving exponential growth and decay. The base of the exponential function that models the value of the car is 0. to 1. dollar value) has increased or decreased. For example.6 The car is worth about $14 600 after 30 months. population.V(n) 5 24(0.
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Chapter 4
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. from 1. • The units for the growth/decay rate and for the number of growth/decay periods must be the same. • If the decay rate (as a percent) is given. for exponential decay. a decay rate of 8% involves multiplying repeatedly by 0. too.82.
Solve each exponential equation.8) n A(x) 5 0.25) 1.02) t 5 w Q(w) 5 600a b 8 P(n) 5 (0.
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Exponential Functions
261
. Express answers to the nearest hundredth of
b) P 5 9000a b
a unit.
a) b) c) d)
What is the growth rate? What is the initial amount? How many growth periods are there? Write an equation that models the growth of the investment. a) What is the initial value of the computer? Explain how you know. Its value as a
function of time.95) m. b) What is the growth rate? Explain how you know. b) What is the rate of depreciation? Explain how you know.4.7
CHECK Your Understanding
1. Complete the table. in months. a) What is the initial population? Explain how you know. A computer loses its value each month after it is purchased. and use it to determine the value of the investment after 15 years. c) Determine the value of the computer after 2 years. is modelled by V(m) 5 1500(0.25 d) 625 5 P(0.05) 10 1 8 2 c) 500 5 N0 (1.71) 9
2. a) A 5 250(1. In 1990.5(3) x
3. c) Determine the population in the year 2007. a sum of $1000 is invested at a rate of 6% per year for 15 years. The growth in population of a small town since 1996 is given by the function
P(n) 5 1250(1.03) n. d) In which year does the population reach 2000 people?
4. d) In which month after it is purchased does the computer's worth fall below $900?
PRACTISING
5. Exponential Growth or Decay? Growth or Decay Rate
Function
Initial Value
a) b) c) d)
V(t) 5 20(1.
c) How much time did it take for the sandwich to reach an internal temperature of 30 °C?
t
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Chapter 4
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. The function that models the town's growth is P(n) 5 12(1. c) Use this equation (or another method) to determine the number of years ago that the population was 8000. P. a) What was the temperature of the sandwich when she began to record its temperature? b) Determine the temperature. A species of bacteria has a population of 500 at noon.5% for the last 10 years. Determine the population at noon the next day. of the sandwich after 20 min. d) What are the domain and range of the function?
9.
P(t) 5 500 a 2 10 b .
t
7. t.025n ) where P(n) represents the population (in thousands) and n is the number of years from now.
a) g(x) 5 24(3) x b) h(x) 5 0. at any hour.6. Suppose this growth rate will be maintained in the future.9) 2 3
8. Determine the time at which the population first exceeds 2000. to the nearest degree.2) x c) j(x) 5 3(0.
K
The function that models the growth of the population. It doubles every 10 h. A student records the internal temperature of a hot sandwich that has been
A
left to cool on a kitchen counter. A town with a population of 12 000 has been growing at an average rate of
2. is
t
a) b) c) d) e) f)
Why is the exponent 10? Why is the base 2? Why is the multiplier 500? Determine the population at midnight. b) Determine the number of years until the population doubles. An equation that models this situation is T(t) 5 63(0. a) Determine the population of the town in 10 years.8(1.8) 2x d) k(x) 5
x 1 (0. Answer to the nearest year. The room temperature is 19 °C.5) 10 1 19 where T is the temperature in degrees Celsius and t is the time in minutes. Which of these functions describe exponential decay? Explain.
b) Explain how each part of your equation is related to the given information. a) Write an equation that represents the percent of germs left with n applications. write an equation that models the situation described. Explain
what each part of each equation represents. of this colony of yeast cells? b) Write an equation that can be used to determine the population of cells at t hours.7
10.
13.
a)
14. Assume that the intensity is 100% at the surface.. b) Suppose a kitchen countertop has 10 billion (1010 ) germs.5 m. b) Determine the increase in value of the card in the 4th year after it was purchased (from year 3 to year 4).5% each year after that for t years c) the population of a colony if a single bacterium takes 1 day to divide into two. c) Determine the increase in value of the card in the 20th year after it was purchased.4. A collector's hockey card is purchased in 1990 for $5. Fifteen years later. d) Use your equation to determine the population after 90 min. How many applications are required to eliminate all of the germs?
15. At 9 a.m. b) Determine the intensity of light at a depth of 7. In each case. The value increases by
6% every year. A town has a population of 8400 in 1990. a) the percent of colour left if blue jeans lose 1% of their colour every time they are washed b) the population if a town had 2500 residents in 1990 and grew at a rate of 0. given the number of years since 1990. Assume an initial
population of 80 cells. A population of yeast cells can double in as little as 1 h. its population
T
grew to 12 500. a) Write an equation that models the number of cells.m. there are
C
NEL
263
.
Exponential Functions
16. a) What is the growth rate.
Write an equation that models the intensity of light per metre of depth. A group of yeast cells grows by 75% every 3 h. Determine the average annual growth rate of this town's population. A disinfectant is advertised as being able to kill 99% of all germs with
each application. c) Use your equation to determine the population after 6 h. a) Write an equation that models the value of the card. in percent per hour. the population is P after t days
11. 200 yeast cells. Light intensity in a lake falls by 9% per metre of depth relative to the surface. e) Approximately how many hours would it take for the population to reach 1 million cells? f ) What are the domain and range for this situation?
12. given the number of hours after 9 a.
Curious Math
Zeno's Paradox
Zeno of Elea (c.08T 0. there was a new. after a period of time. and create a possible equation to model the growth in the popularity of this name. a Greek philosopher and mathematician.
A C D B
He illustrated his point of view with a story. Achilles travelled half the distance between himself and Tortoise (point C ). In the year 2002.21
where R(T ) is the percent of words remembered after T hours. a) Investigate whether or not this is an example of exponential growth. is famous for his paradoxes that deal with motion.
• Two years later. Each time he arrived at the halfway point. but may actually be true. there is an infinite number of halfway points and Achilles would never catch up to Tortoise. So if you look at it this way. Wed. Describe its features and decide whether or not it is an example of exponential decay. and smaller. What is the function that models this problem. there were 70 girls with the name (National Post. Ebbinghaus performed experiments in which he had people
memorize lists of words and then tested their memory of the list. After the race started. He found that the percent of words they remembered can be modelled by R(T ) 5 100 1 1 1. Who will win the race between Achilles and Tortoise? Explain.
18. Psychologist H.
2.. he travelled half the remaining distance between himself and Tortoise (point D).Extending
17. Tortoise was given a head start (point B). (A paradox is a statement that runs counter to common sense.
2006. b) Determine what the growth rate might be.
1. there were 18 girls (including the first one) with that name. a single baby girl born in Alberta was given the name Nevaeh. This equation is now known as the "forgetting curve. c) Discuss any limitations of your model. 490–425 BCE).
264
Chapter 4
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. b) Predict the percent of words remembered after 24 h. May 24. p." even though it actually models the percent of words remembered! a) Graph this function with technology.) Zeno suggested that it is impossible to get to point B from point A. • By 2005. And again. A2). Achilles (point A) and Tortoise agreed to have a race. halfway point. if Tortoise was given a head
start of 1000 m? Does this function support Zeno's paradox? Explain.
1.
First Differences 2 6 18 54 162 486 Second Differences
x 0 1 2 3 4 5 6
y 5 3x 1 3 9 27 81 243 729
4 12 36 108 324
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Exponential Functions
265
.
y 10 y = 2x y = 1 10 2 x –10 0 10 –10 0 10
Study
Aid
• See Lesson 4. A differences table for an exponential function shows that the differences are never constant. They are related by a multiplication pattern.
()
x
y
x
If b . the function has the x-axis (the line y 5 0) as its horizontal asymptote. • Try Chapter Review
Questions 9 and 10. The shape of its graph depends upon the parameter b.
In each case. then the curve decreases as x increases. 1. then the curve increases as x increases. where the variable is an exponent. as they are for linear and quadratic functions.4
Q: • its equation? • its graph? • a table of values?
A:
Chapter Review
FREQUENTLY ASKED Questions
How can you identify an exponential function from
The exponential function has the form f (x) 5 b x.
If 0 .5. b .
This will help you organize the information and create the equation you require to solve the problem. The shape of the graph of g(x) depends on the value of the base of the function. then reflections. If a . then the function has also been reflected in the x-axis. and 3. list these four elements of the equation and fill in the data as you read the problem. 1.1
When solving problems. following an appropriate order—often. then amount decay 0.
Study
Aid
Q:
A:
• See Lesson 4. and finally translations. Examples 1.): P(t) 5 P0 (2)D Population growth: P(n) 5 P0 (1 1 r) n Growth in money: A(n) 5 P(1 1 i) n
t
1 t Radioactivity or half-life: N(t) 5 100a b H 2 Decay Depreciation of assets: V(n) 5 V0 (1 2 r) n Light intensity in water: V(n) 5 100(1 2 r) n
266
Chapter 4
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.b. Each formula is modelled after the exponential function y 5 ab x
The amount in the future The present The growth/decay The number factor: if growth. d. • Try Chapter Review Questions 13 to 17.7. • a represents the vertical stretch or compression factor. 3 and 4.
2. If k . 0.Study
Aid
Q:
A:
• See Lesson 4. 0. and c change the location or shape of the graph of f (x).
How can transformations help in drawing the graphs of exponential functions?
Functions of the form g (x) 5 af (k(x 2 d )) 1 c can be graphed by applying the appropriate transformations to the key points and asymptotes of the parent function f (x) 5 b x. Here are some examples:
Growth Cell division (doubling bacteria.6. etc.
How can exponential functions model growth and decay? How can you use them to solve problems?
Exponential functions can be used to model phenomena exhibiting repeated multiplication of the same factor. f (x) 5 b x. of periods of value or then b . • d represents the number of units of horizontal translation right or left. • k represents the horizontal stretch or compression factor. In functions of the form g (x) 5 ab k(x2d ) 1 c. k. yeast cells. Examples 1. • Try Chapter Review Questions 11 and 12.
2. then the function has also been reflected in the y-axis. the constants a. stretches and compressions. if initial growth or decay. • c represents the number of units of vertical translation up or down.
Complete the table. a) Write an equation that models the growth of the city.
case.5 3 109 years c) the intensity of light if each gel used to change the colour of a spotlight reduces the intensity of the light by 4%
17. Explain how the equation would change if the coffee cooled faster.95) n
n
14.875) n where V(n) is the car's value in the nth year since it was purchased.08)t P(n) 5 32(0. b) Use your equation to determine the population of the city in 2007. A hot cup of coffee cools according to the equation
t
where T is the temperature in degrees Celsius and t is the time in minutes.
1 30 T(t) 5 69a b 1 21 2
a) b) c) d) e) f)
What is the purchase price of the car? What is the annual rate of depreciation? What is the car's value at the end of 3 years? What is its value at the end of 30 months? How much value does the car lose in its first year? How much value does it lose in its fifth year?
16. a) the percent of a pond covered by water lilies if they cover one-third of a pond now and each week they increase their coverage by 10% b) the amount remaining of the radioactive isotope U238 if it has a half-life of 4. the population was 45 000. In each
a) b) c) d) e) f)
Which part of the equation indicates that this is an example of exponential decay? What was the initial temperature of the coffee? Use your knowledge of transformations to sketch the graph of this function.Chapter Review
Lesson 4.
Function
a) c)
d) Q(n) 5 600a b b)
A(x) 5 5(3) x 5 8
V(t) 5 100(1. Explain what each part of the equation represents.or Decay Decay? intercept) Rate
according to the formula V(n) 5 28 000(0. d) Suppose the population took only 10 years to double. after 48 min. describe each part of your equation. In 1990. What growth rate would be required for this to have happened?
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Exponential Functions
269
. Exponential Initial Growth Growth or Value (y. c) Determine the year during which the population will have doubled. The value of a car after it is purchased depreciates
13. to the nearest degree. Write the equation that models each situation.7
15. The population of a city is growing at an average rate
of 3% per year. Explain how the graph would change if the coffee cooled faster. Determine the temperature of the coffee.
b) Use your equation to determine the percent of light left if three gels are used. Express answers as rational numbers. What are the restrictions on the value of n in an if a . c) Describe the transformations necessary (in the proper order) that map g(x) onto f (x). a) Write an equation that models the intensity of light. Evaluate.
5.
y 4 2 0 2 4 6 8 10 2 4 x
6. Each gel reduces the original intensity of the light by 3. b) Use your equation to determine when the population will double (assuming that the growth rate remains stable). I. A spotlight uses coloured gels to create the different colours of light required
for a theatrical production. Use only positive exponents in your final answers.6%. a) Explain how you can tell what type of function f (x) represents just by looking at the equation.
a) f (x) 5 2(32x ) 1 5 b) g(x) 5 (322x24 ) 2 5
7. 1024(x21 ) 10 Å (2x23 ) 5
2
2.
d) p(x) 5 22 a 3 2
1
1
c) h(x) 5 20.
a) (23x 2y) 3 (23x23y) 2
c)
5
(5a21b 2 ) 22 b) 125a 5b23
(8x6y23 ) 3 d) (2xy) 3
1
4. 0? Explain. Simplify. a) Write an equation that models the population. P. since 1990.4
Chapter Self-Test
1. b) Create a table of values for f (x). of this country as a function of the number of years.8(3x23 )
x 21
b22
270
Chapter 4
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. as a function of the number of gels used. The function f (x) 5 2 2 (32x14 ) 1 5 is the transformation of the function
1
g(x) 5 3x. A small country that had 2 million inhabitants in 1990 has experienced an
average growth in population of 4% per year since then. n.
a)
(25) 23
b) 273
3. Describe how to tell the type of function it is from its table of values. Sketch f (x) and state the equation of its asymptote. Which of these equations correspond to the graph? Explain how you know. c) Explain why this is an example of exponential decay.
Does the model you found predict these same values? Write a report that summarizes your findings.
Task
Checklist
Did you include all the
required elements in your report?
Did you properly label the
graphs including some values?
Did you show your work in
your choice of the equation for part E?
Did you support your
decision in part G?
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Exponential Functions
271
. How well does your graph fit the data? Make any necessary adjustments to your equation. What type of function is this? Explain.5 4.7 6. until you are satisfied with the fit. Use the data in the table to estimate the average growth rate for a 5-year period and the y-intercept of the function. and national population size and growth.
What is the equation of the function you could use to model the world's population?
Use the data from the table to create a scatter plot on graph paper. C.
Chapter Task
Modelling Population
Every two years.2 billion by 2075 and decline slightly to 8. Graph your equation. D. F. Draw a curve of good fit.97 billion by 2300. The estimated population of the world since 1950 is given in the table. E. B. supported by your reasons. Using a graphing calculator. create a scatter plot. the UN predicted that the world's population will peak at 9.1
H.3 5. G.8 3 3. the United Nations Population Division prepares estimates and projections of world.7 4 4.55 2. Discuss • the estimated average 5-year growth rate and the y-intercept of the function • the original equation you determined • the changes you made to your original equation and the reasons for those changes • the calculations you used to check whether your model matches the future population predictions made by the UN • why the UN predictions may differ from those of your model
Year 1950 1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 Years since 1950 0 5 10 15 20 25 30 35 40 45 50 World Population (billions) 2. In your report. In 2004. regional. include your graph-paper scatter plot and the prediction of the type of function you thought this might be. Do so as often as needed.4
?
A. Use the values you found in part D to write an equation for this function.3 3.85 5.
272
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.
Chapter
5
Trigonometric Ratios
GOALS
You will be able to
• • •
Relate the six trigonometric ratios to the unit circle Solve real-life problems by using trigonometric ratios. properties of triangles. and the sine and cosine laws Prove simple trigonometric identities
? How would changes in the boat's speed and the wind's speed affect the angles in the vector diagram and the speed and direction of the boat? N speed of wind
2
speed of boat
1
NEL
273
.
h 42 32 40 m
6. determine the sine. he estimates the angle of depression to the bottom of the tower as 32°.
Aid
• For help.
Use a graphic organizer to show how to use the sine law to calculate an unknown angle.8 m C b a A Tech
8. see Technical Appendix. He is in the basket
of a repair truck 40 m from the tower. cosine. a) /A b) /D
3. Use a calculator to evaluate to the nearest thousandth.2 m long.3 m parking meter that is nearby casts a shadow 1.8 m long. B-13. determine each given angle to the
nearest degree.
a b c 5 5 sin A sin B sin C
B
Support
c
For help using the inverse trigonometric keys on a graphing calculator. a) /B
a) sin 31° a) cos u 5 0.
NEL
274
Chapter 5
. a tower casts a shadow 35. a
1. he estimates the angle of elevation to the top of the tower as 42°. see Essential Skills
Appendix.1145
4. At the same time. When he looks up. Using the triangles in question 1.2 m 1. 5.3 m 35. When he looks down.3312
b) /F b) cos 70° b) sin u 5 0. The sine law states that in any triangle. How high is the tower to the nearest tenth of a metre?
1. Use the Pythagorean theorem to determine each unknown side length. On a sunny day. Question 1 2–7 8 Appendix A-4 A-16 A-17
a) c A 12 m
B 5m C
b)
E f D
8m
F
11 m
2. Mario is repairing the wires on a radio broadcast tower. the side lengths are proportional to
the sines of the opposite angles. How high is the tower to the nearest metre?
7. Using the triangles in question 1. Use a calculator to determine u to the nearest degree. and tangent
ratios for each given angle.7113 c) tan u 5 1.5
Study
Getting Started
SKILLS AND CONCEPTS You Need
1.
D. Is ^ ABC a right triangle? Justify your decision. B. F. Which triangular stone would you use for the corner of the patio? Justify your decision. The side opposite each angle is labelled with the lower case letter corresponding to that angle. They need one tile that has a right angle for the corner of the patio.
C 66 cm 38 cm B 77 cm A J 30 cm K F 86 cm 91 cm 35 cm 40 cm Communication
Tip
It is common practice to label the vertices of a triangle with upper case letters. The measurements are shown. E. Repeat parts A to D for the remaining triangles. They don't have a protractor. Check to see if ^ ABC is a right triangle by evaluating each side of the relationship you wrote in part B. C.
A b
c
B
a
C
D ?
A.
20 cm
E
L
Which of these triangles can be used for the corner of the patio?
In ^ ABC. which angle is most likely a right angle? Justify your decision. Assuming that ^ ABC is a right triangle.Getting Started
APPLYING What You Know
Finding a Right-Angled Triangle
Raymond and Alyssa are covering a patio with triangular pieces of stone tile. so they use a tape measure to measure the side lengths of each triangle.
NEL
Trigonometric Ratios
275
. Compare both sides. write down the mathematical relationship that relates the three sides.
1
Trigonometric Ratios of Acute Angles
GOAL
Evaluate reciprocal trigonometric ratios.7°
I used my calculator to evaluate. determine the length of MN. I multiplied both sides of the equation by MN.
MN(tan 16.0 m.0 MN 5 3.5.
LEARN ABOUT the Math
From a position some distance away from the base of a tree.0 m
EXAMPLE
1
Selecting a strategy to determine a side length in a right triangle
In ^MNP. to the nearest tenth of a metre.
A clinometer is a device used to measure the angle of elevation (above the horizontal) or the angle of depression (below the horizontal).7° 5 3.0 m Monique is about 10. is Monique from the base of the tree?
P M 16.
276
Chapter 5
NEL
.
? N How far.7° 3. So I used tangent. Monique uses a clinometer to determine the angle of elevation to a treetop.
The symbol 8 means "approximately equal to" and indicates that a result has been rounded.
Communication
Tip
MN 8 10. then divided by tan 16.
Communication
Tip
Clive's Solution: Using Primary Trigonometric Ratios
tan 16.7º.0 MN
I knew the opposite side but I needed to calculate the adjacent side MN. Monique estimates that the height of the tree is about 3.0 tan 16.0 m away from the base of the tree.7°) 5 3.
1
Cot u is the short form for the cotangent of angle u.7° 5 MN 3.7°.
Most calculators do not have buttons for evaluating the reciprocal ratios. use cos 20° 1 • cot 20°.5. use sin 20° 1 • sec 20°. sec u is the short form for the secant of angle u.
Reflecting
A. use tan 20°
APPLY the Math
EXAMPLE
2
Evaluating the six trigonometric ratios of an angle
C 3 cm A B
^ABC is a right triangle with side lengths of 3 cm.7° angle.0 m away from the base of the tree. 4 cm. and MN is adjacent.0
NP is opposite the 16.
Tech
Support
10.
What was the advantage of using a reciprocal trigonometric ratio in Tony's solution? Suppose Monique wants to calculate the length of MP in ^MNP. This gave me an equation with the unknown in the numerator. Which one would be better? Explain. I used the reciprocal trigonometric ratio cot 16. and csc u is the short form for the cosecant of angle u.0. I multiplied both sides by 3. To solve for MN. B.7 ° cot 16.0) cot 16. For example.0 m 8 MN Monique is about 10. and 5 cm.7°. If CB 5 3 cm and /C 5 90°.7° 5 MN
I evaluated to get tan 16.1
reciprocal trigonometric ratios the reciprocal ratios are defined as 1 divided by each of the primary trigonometric ratios csc u 5 sec u 5 cot u 5 1 hypotenuse 5 sin u opposite 1 hypotenuse 5 cos u adjacent 1 adjacent 5 tan u opposite
Tony's Solution: Using Reciprocal Trigonometric Ratios cot 16. making the equation easier to solve. which trigonometric ratio of /A is the greatest?
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Trigonometric Ratios
277
. State the two trigonometric ratios that she could use based on the given information. to evaluate 1 • csc 20°.
(3.
D 8.67
5 1.33
I expressed these ratios as decimals to compare them more easily.
EXAMPLE
3
Solving a right triangle by calculating the unknown side and the unknown angles
a) Determine EF in ^DEF to the nearest tenth of a centimetre.Sam's Solution
C 3 cm
I labelled the sides of the triangle relative to /A.75 cot A 5 5
Then I evaluated the reciprocal trigonometric ratios for /A. If the side opposite /A is 3 cm.
The greatest trigonometric ratio of /A is csc A.
5 0. The hypotenuse is the longest side. I wrote the reciprocal of each primary ratio to get the appropriate reciprocal ratio. first in words and then with the side lengths.25
8 1. and tangent of /A. then the side adjacent to /A is 4 cm.0 cm 20.
8 1. Then calculate the unknown angles to the nearest degree.0 cm E
F
278
Chapter 5
NEL
. b) Express one unknown angle in terms of a primary trigonometric ratio and
the other angle in terms of a reciprocal ratio. cosine.80 sec A 5 5
5 0.
t en jac ad 4 cm
A
e sit po op
B
5 cm hypotenuse
sin A 5 5
opposite hypotenuse 3 5 hypotenuse opposite 5 3
cos A 5 5
adjacent hypotenuse 4 5 hypotenuse adjacent 5 4
tan A 5 5
opposite adjacent 3 4 adjacent opposite 4 3
First. I used the definitions of the primary trigonometric ratios to determine the sine. so its length must be 5 cm.60 csc A 5 5
5 0.
secant. The wires are attached to the antenna 3. r.7512 d) sin A 5 0.11 m from the door.
19. c.2703 b) cos A 5 0.1515 of 55° with the ground.
20. Describe the appearance of a triangle that has a secant ratio that is greater
than any other trigonometric ratio. The maximum grade (slope) allowed for highways in Ontario is 12%. describe what
this triangle would look like. In right ^PQR.35 m tall at an angle of elevation of 25°. From a position some distance away from the base of a flagpole. sine cotangent secant cosine hypotenuse angle of depression tangent cosecant angle opposite adjacent angle of elevation
Extending
18. Given a right triangle with an acute angle u. needed to fly the kite using a) a primary trigonometric ratio b) a reciprocal trigonometric ratio run of 7. to the nearest tenth of a metre. associated with this slope. to the nearest hundredth of a metre. to the nearest degree. The two guy wires supporting an 8.
16. If Julie is 1. A kite is flying 8.
A
Calculate the length of string.
12.
17.
List all the angles between 0° and 90° (if any) for which cosecant. including explanations where
C
appropriate. The hypotenuse.0 cm long. Which of these triangles has the greatest area? Justify your decision. Calculate the length of the ramp to the nearest hundredth of a metre.10.51. c) Determine the six trigonometric ratios for angle u. Organize these terms in a word web.
282
Chapter 5
NEL
. if tan u 5 cot u. Calculate
side lengths p and q to the nearest centimetre and all three interior angles to the nearest degree. of right ^ABC is 7. A trigonometric ratio for
T
angle A is given for four different triangles.6 m above the ground at an angle of elevation of 41°.
a) Predict the angle u.71 m above ground.7105 c) csc A 5 2. the hypotenuse. is 117 cm and tan P 5 0. Using a reciprocal trigonometric ratio. a) sec A 5 1.5 m TV antenna each form an angle
15. and cotangent are undefined.55 m tall. Julie estimates
that the pole is 5. b) Calculate the value of u to the nearest degree. A wheelchair ramp near the door of a building has an incline of 15° and a
13. The tangent ratio is undefined for angles whose adjacent side is equal to zero.
11. What assumption did you make?
14. calculate the length of each wire to the nearest tenth of a metre. use a reciprocal trigonometric ratio to calculate how far she is from the base of the flagpole.
I used the Pythagorean theorem to calculate the length of the hypotenuse. 458. The height of an equilateral triangle of side length 2 units creates two congruent right scalene triangles. The triangle is isosceles with equal sides of length 1. and 608 angles?
EXAMPLE
1
Evaluating exact values of the trigonometric ratios for a 458 angle
Use ^ ABC to calculate exact values of sine.5. cosine.
Carol's Solution
B 45 1 2 adjacent
A
opposite 45 1
C
BC 2 5 AB 2 1 AC 2 BC 2 5 12 1 12 BC 2 5 2 BC 5 !2
I labelled the sides of the triangle relative to /B.
• ruler • protractor
LEARN ABOUT the Math
The diagonal of a square of side length 1 unit creates two congruent right isosceles triangles. cosine.
B 45 2 30 E
A ?
45 1
C
D
60 F
How can isosceles ^ ABC and scalene ^ DEF be used to determine the exact values of the primary trigonometric ratios for 308. and tangent for specific angles.2
GOAL
Evaluating Trigonometric Ratios for Special Angles
YOU WILL NEED
Evaluate exact values of sine.
hy po n te
e us
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Trigonometric Ratios
283
. and tangent for 45°.
2.2
I noticed that sin E and cos E are equal to cos D and sin D.
Tina's Solution
2 !3 1 4 4 2 1 !3 4
(sin 45°) (cos 45°) 1 (sin 30°) (sin 60°)
I substituted the exact values of each trigonometric ratio. sin 30° 5 cos 60° cos 30° 5 sin 60° tan 30° 5 cot 60°
sin E 5 cos D
cos E 5 sin D
tan E 5 cot D
1 "3
!3 3
1 3 !3 !3 3 !3
"3
1 !3
If I multiplied both the numerator and denominator by !3.57. and tangent for 30° are 2.sin 30° 5
1 2
cos 30° 5
!3 2
tan 30° 5 5 5
The exact values of sine. so a third of it is about 0. I evaluated the expression by multiplying. cosine. would you get the same results if you used /C for the 45° angle instead of /B? Explain.
Reflecting
A.
5
The exact value is
NEL
Trigonometric Ratios
285
. How can remembering that a 30°260°290° triangle is half of an equilateral triangle and that a 45°245°290° triangle is isosceles help you recall the exact values of the primary trigonometric ratios for the angles in those triangles?
APPLY the Math
EXAMPLE
3
Determining the exact value of a trigonometric expression
Determine the exact value of (sin 45°) (cos 45°) 1 (sin 30°) (sin 60°).
In Example 1. In Example 2. since ! 3 is about 1.7. respectively. D. and !3 .
5. C. This is an easier number to estimate. then adding the numerators. B. respectively. respectively "3 1 and for 60° are 2 . Explain how sin 30° and cos 60° are related. explain why the reciprocal ratios of tan 30° and cot 60° are equal. and 3 .
5a 5
!2 !2 1 !3 ba b 1 a ba b 2 2 2 2
2 1 !3 4 . 2 . I would get an equivalent number with a whole-number denominator. I also noticed that tan E is equal to the reciprocal of tan D.
Label the sides using the lengths 1. and 60° angles can be found by using the appropriate ratios of sides in isosceles right triangles and half-equilateral triangles with right angles. 1. and 1.5 2 1 5 0. • If a right triangle has one side that is half the length of the hypotenuse.
2.In Summary
Key Idea
• The exact values of the primary trigonometric ratios for 30°. Label the sides using the lengths ! 3.5 2
tan u
! 3 8 1."
B 45 2 1 2 30
E
3
A
45 1
u 30° 45° 60° sin u
60 C
!3 !3 8 0. a)
Draw a right triangle that has one angle measuring 30°. the angle opposite that one side is always 30°.
Draw a right triangle that has one angle measuring 45°.
CHECK Your Understanding
1. and angles between 45° and 90° have tangent ratios greater than 1.7321
Need to Know
• Since tan 45° 5 1. then the angles opposite those sides are always 45°. c) Identify the adjacent and opposite sides relative to the 60° angle. • If a right triangle has two equal sides.7071 8 0. Explain your reasoning. angles between 0° and 45° have tangent ratios that are less than 1. b) Identify the adjacent and opposite sides relative to one of the 45° angles.8660 2 1 5 0. and ! 2.7071 1 2 2 cos u !3 8 0. 2.5774 2 3
D
1
F
!2 !2 8 0. a)
3.8660 8 0. State the exact values. Explain your reasoning. 45°. b) Identify the adjacent and opposite sides relative to the 30° angle. These are often referred to as "special triangles.
a)
sin 60°
b) cos 30°
c) tan 45°
d) cos 45°
NEL
286
Chapter 5
.
14.
a) For each puzzle piece.
a)
u 5 30°
b) u 5 45°
c) u 5 60°
Curious Math
The Eternity Puzzle
Eternity. show that 1 1 cot 2 u 5 csc 2 u for each angle.
Copy these pieces and see if you can find a solution. Monckton's solution remains unknown.
2.
b) Calculate the area of each puzzle piece. respectively. Suppose one such triangle has side lengths of 1. The seven puzzle pieces shown can be fit together to form a convex shape. consists of 209 different pieces. A second solution was found by Guenter Stertenbrink shortly afterwards.
1. Round your answer to the
piece 1
piece 2
piece 3
nearest tenth of a square unit. It turned out that the puzzle didn't take an eternity to solve after all! Alex Selby and Oliver Riordan presented their solution on May 15. determine the perimeter. The puzzle was introduced in Britain in June 1999. If csc b 5 2. calculate
tan b 2 sin2 b exactly. 2000.Extending
13. Using exact values.
288
Chapter 5
NEL
. and collected the prize. calculate (sin a) (cot a) 2 cos2 a exactly. Interestingly. Consider the first three pieces of the Eternity puzzle. Each piece is made up of twelve 30°260°290° triangles. and a £1 000 000 award (about $2 260 000 Canadian dollars) was offered for the first solution. and 2. Write your answer in
radical form. and the goal was to arrange the pieces into the shape of a dodecagon (12-sided polygon). Monckton provided six clues to solve his puzzle. all three mathematicians ignored Monckton's clues and found their own answers. Each contains twelve
30°260°290° triangles. If cot a 5 ! 3. sec b
15. a puzzle created by Christopher Monckton. !3.
or 4 principal angle the counterclockwise angle between the initial arm and the terminal arm of an angle in standard position. 3.
y
m l ar ina m ter
Next he performed a series of reflections in the y. What is its angle measure? What is the size of the principal angle u and in which quadrant does the terminal arm lie?
P y
related acute angle the acute angle between the terminal arm of an angle in standard position and the x-axis when the terminal arm lies in quadrants 2.
0 Which angles in the Cartesian plane. have primary trigonometric ratios related to those of a 308 angle?
initial arm
x
vertex
Use Raj's sketch of a 30° angle in standard position in the Cartesian plane to record the lengths of all sides and the primary trigonometric ratios for 30° to four decimal places.5.
?
A.3
GOAL
Exploring Trigonometric Ratios for Angles Greater than 90°
YOU WILL NEED
Explore relationships among angles that share related trigonometric ratios. /PrOrQr is now called the related acute angle b. Angle u is measured from the initial arm to the terminal arm (the arm that rotates). He drew one of the special triangles on a Cartesian grid as shown.
• graph paper • dynamic geometry software
(optional)
EXPLORE the Math
Raj is investigating trigonometric ratios of angles greater than 90°.
y 2 30 0 3 Q 60 P 1 x
standard position an angle in the Cartesian plane whose vertex lies at the origin and whose initial arm (the arm that is fixed) lies on the positive x-axis. y quadrant 1 quadrant 2 terminal arm related acute angle vertex 0 quadrant 3
B.and x-axes. if any. Reflect the triangle from part A in the y-axis. Its value is between 0° and 360°.
x Q O
principal angle initial arm quadrant 4
x
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Trigonometric Ratios
289
.
Round your answers to four decimal places and record them in a table similar to the one shown. Use negative angles for some of your trials.
y O Q x
negative angle an angle measured clockwise from the positive x-axis F. but this time. but this time start with a 45° and then a 60° angle in quadrant 1. Based on your observations. which principal angles and related acute angles in the Cartesian plane have the same primary trigonometric ratio?
x 0 210
290
Chapter 5
NEL
. How are the primary trigonometric ratios for the related acute angle related to the corresponding ratios for the principal angle?
y Q O x
P
E.
Repeat part D.
Reflect the triangle from part B in the x-axis. G. What is the size of the related acute angle b? What is the size of the principal angle u.
Angles principal angle related acute angle Quadrant Sine Ratio Cosine Ratio Tangent Ratio
How are the primary trigonometric ratios for the related acute angle related to the corresponding ratios for the principal angle?
D.
Use a calculator to determine the values of the primary trigonometric ratios for the principal angle and the related acute angle.C. and in which quadrant does the terminal arm lie? Use a calculator to complete your table for each of these angles.
P
y
Repeat parts A to E. reflect the triangle from part D in the y-axis.
• An angle in standard position is determined by a counterclockwise rotation and is always positive. An angle determined by a clockwise rotation is always negative. and tangent of an acute principal angle and the resulting reflected principal angles? How could you have predicted the relationships you described in part I?
J. why is the resulting principal angle 180° 1 u? iii) When you reflect an acute principal angle u in the x-axis. the ratios for the related acute angle.
When you reflect an acute principal angle u in the y-axis. the values of the primary trigonometric ratios are either the same as. or the negatives of. These relationships are based on angles in standard position in the Cartesian plane and depend on the quadrant in which the terminal arm of the angle lies. • If u is an acute angle in standard position. • If the terminal arm of an angle in standard position lies in quadrants 2.5. 3. why is the resulting principal angle 180° 2 u? ii) When you reflect an acute principal angle u in the y-axis and then in the x-axis. or 4.
What does your table tell you about the relationships among the sine.
Need to Know
• An angle in the Cartesian plane is in standard position if its vertex lies at the origin and its initial arm lies on the positive x-axis. why is the resulting principal angle 360° 2 u (or 2u)?
i)
I. cosine.3
Reflecting
H. then • the terminal arm of the principal angle (180° 2 u) lies in quadrant 2
y
sin (180° 2 u) 5 sin u cos (180° 2 u) 5 2cos u tan (180° 2 u) 5 2tan u
180 x
(continued)
NEL
Trigonometric Ratios
291
.
In Summary
Key Idea
• For any principal angle greater than 908. there exists a related acute angle and a principal angle.
4) on the circumference of
the circle. Since r is the radius of the circle. 4) on the circumference. determine the primary trigonometric ratios for the principal angle.
• graph paper • protractor • dynamic geometry software
(optional)
LEARN ABOUT the Math
Miriam knows that the equation of a circle of radius 5 centred at (0. I also knew that r 2 5 x 2 1 y 2. In ^OPQ. b) Determine the principal angle to the nearest degree. so I labelled it r. She also knows that a point P(x. Angle u is the principal angle and is in standard position. Then I formed a right triangle with the x-axis. 4) y 5 x 3 Q 5 4 x
I drew a circle centred about the origin in the Cartesian plane and labelled the point P(3. In this case.5. y) r 0 x y x
EXAMPLE
1
Relating trigonometric ratios to a point in quadrant 1 of the Cartesian plane
a) If Miriam chooses the point P(3.
? For any point on the circumference of the circle. Flavia's Solution
a) y r 0 5 P(3. y) on its circumference can rotate from 0° to 360°. From the Pythagorean theorem. it will always be positive. I noticed that the side opposite u has length y 5 4 units and the adjacent side has length x 5 3 units. 0) is x 2 1 y 2 5 25. r 5 5 units.
x2
y2
25 5
5
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293
. The hypotenuse is equal to the radius of the circle. how can Miriam determine the size of the corresponding principal angle?
x2
y2
y 25 P(x.4
GOAL
Evaluating Trigonometric Ratios for Any Angle Between 08 and 3608
YOU WILL NEED
Use the Cartesian plane to evaluate the primary trigonometric ratios for angles between 08 and 3608.
I used the Pythagorean theorem to confirm this. Angle b is the related acute angle. 0
In ^OPQ. and r in the Cartesian plane. b) Determine the principal angle to the nearest degree.
EXAMPLE
2
Relating trigonometric ratios to a point in quadrant 2 of the Cartesian plane
a) If Miriam chooses the point P(23. since r . cosine. and tangent to write each ratio in terms of x. I knew that the lengths of the two perpendicular sides were 0x 0 5 023 0 5 3 and y 5 4. Gabriel's Solution
a) P( 3.
y 5 r 4 u 5 sin 21 a b 5 4 5 5 4 5
x 5 r 5 3 5
y 5 x 5 4 3
b) sin u 5
I used the inverse sine function on my calculator to determine angle u. y.
y x2 1 y2 5 25
Q x 0
r2 5 x2 1 y2 r 2 5 32 1 42 r 2 5 9 1 16 r 2 5 25 r 5 5. determine the primary trigonometric ratios for the principal angle to the nearest hundredth. Angle u is the principal angle and is in standard position.
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Chapter 5
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. so r 5 5. 4) on the circumference
of the circle. 4) y r x
I drew a circle centred about the origin in the Cartesian plane and labelled the point P(23.sin u 5
opposite hypotenuse
cos u 5
adjacent hypotenuse
tan u 5
opposite adjacent
I used the definitions of sine. 4) on the circumference. Then I formed a right triangle with the x-axis. The radius of the circle is still 5.
u 8 53°
The principal angle is about 53°.
4
sin u 5 sin b 5
cos u 5 2cos b 52 3 5
tan u 5 2tan b 52 4 3
b) sin b 5
4 b 5 sin21 a b 5 4 5 8 53°
u 1 b 5 180° u 5 180° 2 b 5 180° 2 53° 5 127° The principal angle is about 127° because the related acute angle is about 53°. y. the sine ratio is positive while the cosine and tangent ratios are negative.sin b 5 5 5
opposite hypotenuse y r 4 5 4 5
cos b 5 5 5
adjacent hypotenuse
3 5
0x0 r
tan b 5
opposite adjacent
5 5
4 3
y 0x0
In ^OPQ. If Miriam chose the points (23.
I knew that u and b add up to 180°. cos u 2 cos b. B.
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. I used the definitions of sine. Then I took into account the relationship among the trigonometric ratios of the related acute angle b and those of the principal angle u. explain how the coordinates of that point vary from quadrants 1 to 4.
Reflecting
A. How does this variation affect the size of the principal angle (and related acute angle. and tan u 2 tan b. what would each related acute angle be? How would the primary trigonometric ratios for the corresponding principal angles in these cases compare with those in Examples 1 and 2? Given a point on the terminal arm of an angle in standard position. 24). the side opposite b has length y and the adjacent side has length 0x 0.
In Example 2. I used a calculator
5. and r in the Cartesian plane. to evaluate sin21 A 4 B directly. Since the terminal arm of angle u lies in quadrant 2. explain why sin u 5 sin b. if it exists) and the values of the primary trigonometric ratios for that angle?
C. So I subtracted b from 180° to get u. cosine. 24) and (3. and tangent to write each ratio in terms of x. 5 To determine angle b.
Since x 5 0 and it is in the denominator. cosine. I used the definitions of sine. and r to write each ratio. and r 5 1.
x2 1 y 2 5 1
This meant that I couldn't use the trigonometric definitions in terms of opposite. I couldn't draw a right triangle by drawing a line perpendicular to the x-axis to P. Charmaine's Solution
y P(0. 1) x 0
I drew a circle centred about the origin in the Cartesian plane and labelled the point P(0. Angle u is the principal angle and is 90°. tan 90° is undefined. 1) defines a principal angle of 90°.
sin u 5 5
y r 1 1
cos u 5 5
x r 0 1
tan u 5 5
y x 1 0
Since P(0. I knew that x 5 0.
sin 90° 5 1 cos 90° 5 0 tan 90° is undefined The point P(0. and hypotenuse. adjacent. 1) to determine the values of sine. 1).APPLY the Math
EXAMPLE
3
Determining the primary trigonometric ratios for a 908 angle
Use the point P(0. respectively. y. and tangent for 90°. y 5 1. 1) on the circumference. cosine.
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Chapter 5
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. and tangent in terms of x. The sine and cosine of 90° are 1 and 0. In this case. The tangent of 90° is undefined.
There were two cases where a point on the terminal arm has a negative y-coordinate: one in quadrant 3 and the other in quadrant 4. 3
EXAMPLE
4
Determining all possible values of an angle with a specific trigonometric ratio
Jordan's Solution 2!3 3 3 2!3
Since 0° # u # 360°.
The angle 260° corresponds to a related acute angle of 60° of clockwise rotation and has its terminal arm in quadrant 4. I added 360° to 260° to get the equivalent angle using a counterclockwise rotation. In quadrant 3. y must be 23 in this case. Then I took the inverse sine of
One angle is 260°. I used my calculator to evaluate
csc u 5 2 sin u 5 2
the result to determine the angle. u can be either 240° or 300°.5. I had to use the Cartesian plane to determine u. I found the reciprocal ratio by switching r and y. Given csc u 5 2 2 ! 3 and 3 0° # u # 360°.
23 2 !3 . The angle in quadrant 3 must have a related acute angle of 60° as well. which is equivalent to 360° 1 (260°) 5 300° in quadrant 4. Cosecant is the reciprocal of sine. the angle is 180° 1 60° 5 240°.
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297
.4
Determine the values of u if csc u 5 2 2 ! 3 and 0° # u # 360°. Since r is always positive. So I added 180° to 60° to determine the principal angle.
28)
a) (5.5. Calculate the value of u to the nearest degree. only Cosine (C) is positive. Since r is always positive. only Tangent (T) is positive because both x and y are negative. 3) c) (25. the sign of each primary ratio depends on the signs of the coordinates of the point. only Sine (S) is positive. a) sin 315° b) tan 110° c) cos 285° d) tan 225°
2. • In quadrant 4. 28) d) (6.4
• The CAST rule is an easy way to remember which primary trigonometric ratios are positive in which quadrant. use a sketch to determine in which quadrant the
terminal arm of the principal angle lies.
i) ii) iii) iv)
Draw a sketch of each angle u. Each point lies on the terminal arm of angle u in standard position. • In quadrant 3. Use the related acute angle to state an equivalent expression.
2
y
1
S T
3
0
A C
4
x
CHECK Your Understanding
1. All (A) ratios are positive because both x and y are positive.
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299
. • In quadrant 2. a) 180° b) 270°
a) sin 160° b) cos 300°
c) 360° c) tan 110° d) sin 350°
4. since x is negative and y is positive. the value of the related acute angle b. Use the method in Example 3 to determine the primary trigonometric ratios
for each given angle. Determine the value of r to the nearest tenth.
b) (28. • In quadrant 1. For each trigonometric ratio. Determine the primary trigonometric ratios for angle u. 11)
3. since x is positive and y is negative. and the sign of the ratio.
y). where r is the distance from the origin to P.4. The trigonometric ratios can then be expressed in terms of x.
• Try Mid-Chapter Review
Questions 10. 11. y. Use this value of b to determine all possible values of the principal angle u. and r. and 12.
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303
.
y x x y
y
r
P(x.
Examples 1 to 4. r 2 5 x 2 1 y 2 from the Pythagorean theorem and r . Interpret the calculator result in terms of your sketch. 0 sin u 5 csc u 5 y r r y cos u 5 sec u 5 x r r x tan u 5 cot u 5
y x x 0
• See Lesson 5.4. Use the appropriate inverse trigonometric function on your calculator to determine a value for u.
• See Lesson 5. Then sketch the angle(s) in standard position. • Try Mid-Chapter Review Questions 9 to 13.Mid-Chapter Review
Q:
A:
How can you determine the trigonometric ratios for any angle u.3 and
Lesson 5. An angle in standard position is determined by a counterclockwise rotation and is always positive. A negative angle is determined by a clockwise rotation. provided that P lies on the terminal arm of the angle. y)
Q:
A:
How can you determine all possible values of the principal angle u in the Cartesian plane associated with a given trigonometric ratio?
Study
Aid
Use the sign of the ratio to help you decide in which quadrant(s) the terminal arm of angle u could lie. Example 4. and determine the value of any related acute angle b. where 0° # u # 360°?
Study
Aid
Any angle in standard position in the Cartesian plane can be defined using the point P(x.
A trigonometric ratio is 5 . and with her eyes level with the base of the mast. Determine the value of u to the nearest degree if
8. a) cos u 5 2. The mast is 8. ii) If 0° # u # 360°.1
a) determine the exact measure of each unknown
1.3151 f ) sin u 5 2.1011
7
c) cot 10° d) csc 81°
side if sin a 5 1 2 b) determine the exact values of the primary trigonometric ratios for /A and /DBC
Lesson 5. Given ^ABC as shown. determine
the value of u to the nearest degree. a) sin 60° b) tan 45°
7. What ratio could it be. a) cot u 5 0. Use the sketch to determine the exact value of the given trigonometric ratio. a) Sketch the principal angle u.PRACTICE Questions
Lesson 5. a) csc 20° b) sec 75° 0° # u # 90°.2
Is that possible if 0° # u # 360°? Explain. P(29.
15
6.
11. Determine the exact value of each trigonometric
ratio. Without using a
B 14 D 45 A
calculator. If sin u 5 20.
c) csc 30° d) sec 45°
a) state the other five trigonometric ratios as
fractions
b) determine the value of u to the nearest degree
12.3151 d) csc u 5 2. If csc u .3151 b) tan u 5 2.
5.3151
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.3 m long.8190 and 0° # u # 360°. Claire is attaching a rope to the top of the mast of her
sailboat so that she can lower the sail to the ground to do some repairs.6406 d) sec u 5 2. i)
c) csc u 5 1.4312
3. How much rope does Claire need if 0. which ratios must be false? Justify your reasoning.
Sketch each angle in standard position. Given tan u 5 2 8 . b) What is the value of the related acute angle b to the nearest degree? c) What is the value of the principal angle u to the nearest degree?
10.3
2. Evaluate each reciprocal trigonometric ratio to four
decimal places.3151 c) sec u 5 2. Angle u lies in quadrant 2. sec u and u is acute. 4) lies on the terminal arm of an angle in
standard position.5 m of rope is required to tie to the mast? Round your answer to the nearest tenth of a metre. Jeff said he found three angles for which cos u 5 5.8701 b) sec u 5 4.
C
13.4
9.
4
about u?
Lesson 5.
a) sin 120° b) cos 225° c) tan 330° d) cos 300°
and what angle might it be referring to?
4. what do you know
Lesson 5. state all values of u that have the same given trigonometric ratio.3151 e) cot u 5 2. where 90° # u # 180°. the top forms an angle of 31° with the ground.
where 0° # u # 360°.
identity a mathematical statement that is true for all values of the given variables. The object of the game is to lay down pairs of equivalent expressions so that each pair forms an identity. Any restrictions on a variable must be stated.
LEARN ABOUT the Math
Trident Fish is a game involving a deck of cards. the denominators cannot be zero. If the identity involves fractions. and r
sin u
Prove the quotient identity tan u 5 cos u for all angles u.S. y.5
GOAL
Trigonometric Identities
Prove simple trigonometric identities. Jinji's Solution tan u 5 sin u cos u R. Suppose you have the cards shown. 5 sin u cos u
I separated the left and the right sides so that I could show that both expressions are equivalent. each of which has a mathematical expression written on it.S.
sin ——— cos sec2 tan
csc2 1 tan2 sin2 cos2
1 1 cot2
?
What identities can you form with these cards?
EXAMPLE
1
Proving the quotient identity by rewriting in terms of x.5. 5 tan u
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1
Trigonometric Ratios
cot2 sin2 cos2 tan
305
.
L.
tan u. Tan u is undefined when cos u 5 0. 5
x a b r
y a b r
I wrote sin u. Lisa's Solution sin2 u 1 cos2 u 5 1 y 2 x 2 5a b 1a b r r 5 y2 x2 1 2 r2 r y 2 1 x2 5 r2 5 r2 r2
I separated the left and the right sides so that I could show that both expressions are equivalent. and cos u in terms of x. the original equation is an identity. I wrote sin u and cos u in terms of x. 5 1
I knew that r 2 5 x 2 1 y 2 from the Pythagorean theorem. where 0° # u # 360° and u 2 90° or 270°. where 0° # u # 360°. and r.y L. y. the original equation is an identity. 6 sin2 u 1 cos2 u 5 1 for all angles u. and r
Prove the Pythagorean identity sin2 u 1 cos2 u 5 1 for all angles u. 5 sin2 u 1 cos2 u
R. 5 x
R.S. since u can be greater than 90°. This occurs when u 5 90° or 270°. So u cannot equal these two values.S. and r. y.S.
51 5 R.S.S.S.
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. Since the left side works out to the same expression as the right side. where 0° # u # 360°. Then I simplified. since u can be greater than 90°. I simplified the right side by multiplying the numerator by the reciprocal of the denominator.
L. 6 tan u 5 cos u for all angles u. y.
sin u
EXAMPLE
2
Proving the Pythagorean identity by rewriting in terms of x.
5
y r1 3 r1 x
5
y x
5 L. I used this equation to further simplify the left side. Since the left side works out to the same expression as the right side.
sin f 1 sin2 f
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. 5 tan f sin f cos f
I separated the left and the right sides so that I could show that both expressions are equivalent. where cos f 2 0. Jamal's Solution tan f 5 sin f 1 sin2 f (cos f) (1 1 sin f) R. I knew that tan f could be written as cos f. C. so I factored out sin f from the numerator.
What strategy would you use to prove the identity 1 1 tan2 u 5 sec2 u? What restrictions does u have? When is it important to consider restrictions on u? How might you create new identities based on Examples 1 and 2?
APPLY the Math
EXAMPLE
4
Proving an identity by factoring
sin f 1 sin2f
Prove that tan f 5 (cos f) (1 1 sin f) for all angles f between 0° and 360°. the original equation is an identity. 6 tan f 5 (cos f) (1 1 sin f) for all angles f between 0° and 360°. the denominator will not be 0. The right side is more complicated.S. 5 sin f 1 sin2 f (cos f) (1 1 sin f) sin f(1 1 sin f ) (cos f) (1 1 sin f)
1 1
L.Reflecting
A. Since the left side works out to the same expression as the right side.S. where cos f 2 0.
sin f
5
5 5
sin f cos f
5 L. Since cos f 2 0.S. B. I divided the numerator and denominator by the factor 1 1 sin f.
and r • rewriting all expressions involving tangent and the reciprocal trigonometric ratios in terms of sine and cosine • applying the Pythagorean identity where appropriate • using a common denominator or factoring as required
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309
. y. Identities Based on Definitions Reciprocal Identities csc u 5 sec u 5 cot u 5 1 . where sin u 2 0 sin u Pythagorean Identities sin2 u 1 cos2 u 5 1 1 1 tan2 u 5 sec 2 u 1 1 cot 2 u 5 csc 2 u
• To prove that a given trigonometric equation is an identity.5
In Summary
Key Ideas
• A trigonometric identity is an equation involving trigonometric ratios that is true for all values of the variable. where sin u 2 0 sin u 1 . while others are derived from relationships that exist among trigonometric ratios.
Need to Know
• Some trigonometric identities that are important to remember are shown below (0° # u # 360°) . where cos u 2 0 cos u 1 . where cos u 2 0 cos u cos u .5. This can be done by • simplifying the more complicated side until it is identical to the other side or manipulating both sides to get the same expression • rewriting all trigonometric ratios in terms of x. where tan u 2 0 tan u Identities Derived from Relationships Quotient Identities tan u 5 cot u 5 sin u . both sides of the equation need to be shown to be equivalent. • Some trigonometric identities are a result of a definition.
8 5 sin 36° sin B
1
1
(sin B) (5. Round only after the very last calculation.9 to isolate sin B.
5. and the balloon are positioned relative to each other. The team is about to launch a weather balloon into an active part of a cloud. So I first calculated /B using the sine law. I knew one angle and two sides. to the nearest tenth of a metre. Albert's rope is 7.9 m long.
? How far. save intermediate answers by using the memory keys of your calculator.5.
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Chapter 5
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. Belle.9
1
(sin 36 °) (7.9 ) (sin 36°) (7. I multiplied both sides of the equation by the lowest common denominator (sin 36° 3 sin B) to eliminate the fractions.
A
c
B
1 5. To determine AB (side c).9 m a
7. it is not clear how Albert. Adila's Solution: Assuming that Albert and Belle are on Opposite Sides of the Balloon
C b 5.6
YOU WILL NEED
The Sine Law
• dynamic geometry software
(optional)
GOAL
Solve two-dimensional problems by using the sine law.
LEARN ABOUT the Math
Albert and Belle are part of a scientific team studying thunderclouds.9 5. I assumed that Albert and Belle are on opposite sides of the balloon. Then I divided both sides by 5. Belle's rope is 5.8 m
36
From the problem. is Albert from Belle?
EXAMPLE
1
Using the sine law to calculate an unknown length
Determine the distance between Albert and Belle.
Communication
Tip
To perform a calculation to a high degree of accuracy.9 7.8 (sin 36° 3 sin B) a b 5 (sin 36° 3 sin B ) a b sin 36° sin B
b a 5 sin A sin B
In ^ ABC.9 7. I drew a sketch of this situation.8) 5 5. I needed to know /C.9
1 1
To solve for /B.8) b /B 5 sin21 a 5.8 m long and makes an angle of 36° with the ground.
and the balloon are positioned relative to each other.
10. I rounded to the nearest tenth.
To solve for c. I calculated /C by using the fact that all three interior angles add up to 180°.9 m a
b 7. they are about 10.8 m
36
The problem did not state how Albert.6
I used my calculator to evaluate /B.0 m 8 c If Albert and Belle are on opposite sides of the balloon.
Reuben's Solution: Assuming that Albert and Belle are on the Same Side of the Balloon
C 5.0 m apart. Belle.
/B 8 51° /A 1 /B 1 /C 5 180° 36° 1 51° 1 /C 5 180° /C 5 180° 2 (36° 1 51°) 5 93° a c 5 sin A sin C c 5.5. I assumed that Albert and Belle are on the same side of the balloon.
A
c
B
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Trigonometric Ratios
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. I multiplied both sides of the equation by sin 93°.
Then I used the sine law again to determine c.9 3 sin 93° 5 3 sin 93° sin 36° sin 93° 1
I rounded to the nearest degree. I drew a sketch of this situation.9 5 sin 36° sin 93°
1 c 5.
but I wanted the obtuse angle /CBA in the triangle.
/CBX 8 51° /CBA 5 180° 2 51° 5 129° /C 5 180° 2 (/A 1 /CBA) 5 180° 2 (36° 1 129°) 5 15°
51° is the value of the related acute angle /CBX.C
b
7.9 m
A
36 c
B
X
In ^ ABC.
I used my calculator to evaluate /B.8 5 sin 36° sin B
1
(sin 36°) (7. If I knew /C. I knew one angle and two sides.
314
Chapter 5
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.8) (sin B) (5.8 (sin 36° 3 sin B) a b 5 (sin 36° 3 sin B ) a b 1 sin 36° sin B
b a 5 sin A sin B
I used the sine law to calculate /B. I first multiplied both sides of the equation by the lowest common denominator (sin 36° 3 sin B) to eliminate the fractions.
5. I could determine c using the sine law.9 7.9
1
1
To solve for /B.8 m
a
5.9
1
(sin 36°) (7.9 7.8) /B 5 sin21 a b 5.9 5. I calculated /C by using the fact that all three interior angles add up to 180°.
1 5.9 to isolate sin B.9 ) 5 5. First I had to determine /B in order to get /C. I divided both sides by 5. I rounded to the nearest degree.
1. or 2 triangles can be drawn given the information in a problem.
Why is the situation in Example 1 called the ambiguous case of the sine law? What initial information was given in this problem? What is the relationship between sin B in Adila's solution and sin B in Reuben's solution? Explain why both values of sine are related. 0 or 1 triangle is possible (see the In Summary box for this lesson).
APPLY the Math
EXAMPLE
2
Using the sine law in the ambiguous case to calculate the only possible angle
Karl's campsite is 15. If the given angle is acute. What is the bearing of the lookout from Karl's position (/NAC)?
N B C 140
N
1
A
5. I multiplied both sides of the equation by sin 15°.
Reflecting
A.0 m from a scenic lookout as shown. This occurs when you know two side lengths and an angle opposite one of the sides rather than between them (an SSA triangle).9 c 5 sin 15° 3 sin 36° sin 15° 1
I used the sine law again to calculate side c.6
c a 5 sin A sin C c 5. For example. or 2 triangles are possible. If the given angle is obtuse. D. 0. What do you notice? Compare this value with the length of a and b. Calculate the height of ^ ABC in both solutions.
To solve for c. I rounded to the nearest tenth. the bearing of the lighthouse shown is 335°. bearing the direction in which you have to move in order to reach an object.0
E W
25
E 335 S
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Trigonometric Ratios
315
.6 m from a lake and 36.5. B. the angle formed between the campsite and the lookout is 140°. From the lake. 1.6
m
m 36.
the ambiguous case of the sine law a situation in which 0. A bearing is a clockwise angle from magnetic north.6 m apart. C.6 m 8 c If Albert and Belle are on the same side of the balloon.
2.9 5 sin 36° sin 15° sin 15° 3
1 5. Karl starts hiking from his campsite to go to the lookout. they are about 2.
the lookout has a bearing of about 74°.
u 8 16°
N C 16 A E
I rounded to the nearest degree.6 36. so I set up the sine law equation with the angles in the numerators. I needed to know the complementary angle of 16° .
316
Chapter 5
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.Sara's Solution
Based on the given information. So I subtracted 16° from 90° . they are equal. this is an SSA triangle. Since /C and u are alternate angles between parallel lines. But since the given angle is obtuse.0 15. I multiplied both sides of the equation by 15.6 3
1
sin u sin 140° 5 15.6 3 15. I knew that AE is parallel to BC. To solve for sin u.0
E
sin C sin B 5 c b sin 140° sin u 5 15.6) (sin 140°) b 36.
B
C
6m 15 .0
1
u 5 sin21 a
(15.
A
140
m 36. only one situation had to be considered. I needed to calculate an angle.
In order to state the bearing of the lookout.6.6 36.
/NAC 5 90° 2 16° 5 74° From Karl's campsite. In ^ ABC.0
I used the inverse sine function on my calculator to determine /C (angle u).
a. and b are given and /A is obtuse. and b are given and /A is acute. b. no triangle exists.
a b A A
If /A is acute and a . the sine law can be used if you know • two sides and one angle opposite a given side (SSA) or • two angles and any side (AAS or ASA) • The ambiguous case arises in a SSA (side. the sine law calculation may lead to 0. If /A is acute and a . no triangle exists. there are two cases to consider.
Need to Know
• In the ambiguous case.6
In Summary
Key Ideas
• The sine law states that in any ^ ABC. h. a . b c a 5 5 sin A sin B sin C or sin B sin C sin A 5 5 a b c
C b A c B a
• Given any triangle. 1. a. b. one triangle exists. the height of the triangle is h 5 b sin A. one triangle exists. if /A. one right triangle exists. or 2 solutions. b or a 5 b.
a b b A
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a A
Trigonometric Ratios
317
.
h
b
h5a
b A
h
a
b
a
h
a
A
If /A. If /A is acute and h . b.5. two triangles exist. the ratios of each side to the sine of its opposite angle are equal. In this situation. If /A is obtuse and a . angle) triangle. there are four cases to consider. side. If /A is obtuse and a . In each case. depending on the size of the given angle and the lengths of the given sides. If /A is acute and a 5 h.
An observer in
a hot-air balloon some distance away from the aqueduct determines that the angle of depression to each end is 54° and 71° . If the slope of the hill is 13° from the horizontal.0 m apart as shown. a 35. Calculate the length of the aqueduct to the nearest tenth of a metre. To
prevent the tree from falling over. The trunk of a leaning tree makes an angle of 12° with the vertical. that are
A
105. Calculate the height. The closest end of the aqueduct is 270. at those two points?
32 9750 m 45 A B
9. h. A building of height h is observed from two points. France.0 m rope is attached to the top of the tree and is pegged into level ground some distance away.5. to the nearest tenth of a metre. A surveyor in an airplane observes that the angle of depression to two points
on the opposite shores of a lake are 32° and 45°. The Pont du Gard near Nîmes. is a Roman aqueduct.0 m
8.0 m from its base to its top.0 m from the balloon. respectively. A wind tower at the top of a hill casts a shadow 30 m long along the side of
T
the hill. respectively.
h 40
32 Q P 105. P and Q.
10. An observer at the farthest edge of the shadow from the tower estimates the angle of elevation to the top of the tower to be 34°. to the nearest metre. as shown. respectively.
7.6
6. how high is the tower to the nearest metre?
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Trigonometric Ratios
319
. If the tree is 20. What is the width of the lake. The angles of elevation at P and Q are 40° and 32°. calculate the angle the rope makes with the ground to the nearest degree.
Determine the other side length of this lot to the nearest metre and the interior angles to the nearest degree. The upper guy wires form an angle of 59° with the ground and attach to the mast 350 m above ground.
14. b) Two triangles are possible. However. on
which a 273. The captain of the Algomarine calls you from the bridge. is the sailor from both lighthouses? b) What is the shortest distance. Your neighbour claims that his lot is triangular. are you from the stern? b) Are you in danger of being swamped?
17. Two friends standing some distance from Carol see the kite at angles of elevation of 66° and 35° . with one side 430 m long and
the adjacent side 110 m long. to the nearest metre. From lighthouse B. A sailor out in a lake sees two lighthouses 11 km apart along the shore and
gets bearings of 285° from his present position for lighthouse A and 237° for lighthouse B. The Algomarine is a cargo ship that is 222. c) No triangle is possible. small
watercraft have the right of way. Using a sketch.5 m guyed mast is mounted. to the nearest kilometre.5 m long. There is a specific point on the ground where you can be equidistant from both the top and the bottom of the tower. When the tower was first
built. a) How far.11.
Extending
15. Carol is flying a kite on level ground.8°. explain the relationship between
C
/L. One friend is 11 m from Carol. For each question below. The Huqiu Tower in China was built in 961 CE. /L is acute. How far is this point from the base of the tower? Round your answer to the nearest metre. and the string forms an angle of 50°
with the ground. The bow and stern of the carrier appear separated by 12°. a) Only one triangle is possible. a) How high is the kite above the ground? b) How long is the string? c) How far is the other friend from Carol?
12. Since then it has tilted 2. its height was 47 m. and says that you are 8° off his bow. lighthouse A has a bearing of 45°. from the sailor to the shore?
16. located at the stern. On the water. a) How far.
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. bulk carriers cruise at nearly 30 km/h. your boat could get swamped! Suppose you spot the Algomarine on your starboard (right) side headed your way. to the nearest kilometre. so it is best to stay out of their way: If you pass a cargo ship within 40 m. In ^LMN. and the height of the triangle for each situation. so it is called China's Leaning Tower. respectively. The angle opposite one of these sides is 35°. sides l and m. The lower guy wires form an angle of 36° with the ground and attach to the tower 155 m above ground. The Gerbrandy Tower in the Netherlands is an 80 m high concrete tower. How long are the upper and lower guy wires? Round your answers to the nearest metre.
13. state all possible answers to the nearest metre.
I labelled the point where the side of the building touches the ground C.
/AXC 1 /CXP 5 180° 135° 1 /CXP 5 180° /CXP 5 180° 2 135° 5 45° 6 ^XCP is a 45° 245° 290° special triangle.
• dynamic geometry software
(optional)
LEARN ABOUT the Math
A barn whose cross-section resembles half a regular octagon with a side length of 10 m needs some repairs to its roof. The base of the ramp is 15.
10 m top of ramp 22.6 m
?
How far.9 m P
26
B
I labelled the top of the ramp A and the bottom of the ramp B. each interior angle is 135°. to the nearest tenth of a metre. forming an angle of 26° with the ground.9 m ramp against the side of the building. The ramp will be used to transport the materials needed for the repair.5.6 m from the side of the building. is the top of the ramp from the flat roof of the building?
EXAMPLE
1
Using the cosine law to calculate an unknown length
Determine the distance from the top of the ramp to the roof by using the cosine law.
Tina's Solution
10 m A X C d ? 22. /C is 90°. Since the octagon is regular.7
GOAL
The Cosine Law
YOU WILL NEED
Solve two-dimensional problems by using the cosine law. I subtracted 135° from 180°. To determine /CXP. Then I drew a line from A along the sloped part of the building to X and extended the line to the ground at P.
In ^ XCP.
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. The roofers place a 22.9 m 5m
26
15. So / AXC is 135°.
9 m The top of the ramp is about 2.
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Chapter 5
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.9)cos 26° b2 5 200.6 PB 5 15.16 AX 5 14. Then I subtracted AX from 10 m to get the distance from the top of the ramp to the roof.6) 2 1 (22.6 m B
b 2 5 a 2 1 p 2 2 2ap cos B b2 5 (10. I used the cosine law instead.9) 2 2 2(10. I knew that XP is a multiple of ! 2 because ^XCP is a 45° 245° 290° special triangle.16 m XP 5 5!2 AX 1 XP 5 b AX 1 5!2 5 14. So I couldn't use the sine law to determine AP.42 m2 b 8 14. I then subtracted CP from CB to determine the length of PB.6 m
A x p 22. So I subtracted XP from b to determine AX. I needed to calculate AX first. since the triangle is isosceles.
CP 1 PB 5 15. so CP 5 5 m.16 2 5!2 8 7.9 m
26
P 15. I substituted the values of a.6 5 1 PB 5 15.42
To determine the distance from the top of the ramp to the roof. p.9 m from the flat roof of the building.09 m required distance 5 10 2 AX 5 10 2 7. I calculated b by evaluating the right side of the equation and determining its square root.10 m A 135 X C 45 22.6 m
B
From the given information.6 2 5 5 10.
b 5 !200.9 m
26
b
P
a
10. and /B into the formula.6) (22. In ^APB. I knew two side lengths and the contained angle formed by those sides. I knew that XC 5 5 m.09 8 2.
Will Mitchell be able to install this system by himself ?
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. The Pythagorean theorem is a special case of the cosine law.1 m long. determine /A to the nearest degree if a 5 55 cm. b. So I used the cosine law to determine /A. Since I knew all three sides (SSS) but no angles.0 wide house to be heated with a solar hot-water system. the array must be installed on the south side of the roof and the roof needs to be inclined by 60°.
Claudio's Solution
C b 26 cm A
2 2 2
a
55 cm c B 32 cm
I drew a diagram of the triangle and labelled all sides. C. What conditions would have to exist in a triangle in order for the cosine law to simplify to the Pythagorean theorem?
APPLY the Math
EXAMPLE
2
Using the cosine law to determine an angle
In ^ABC. the roof will be too steep for Mitchell to install the system himself. If the north side of the roof is inclined more than 40°. B. b 5 26 cm. and c into the formula. The tubes form an array that is 5. I substituted the values of a. instead of the cosine law. and c 5 32 cm. In order for the system to be effective. /A is about 143°.7
Reflecting
A.
Given ^ABC.
EXAMPLE
3
Solving a problem by using the cosine and the sine laws
Mitchell wants his 8.
a 5 b 1 c 2 2bc cos A 552 2 (262 1 322 ) /A 5 cos 21 a b 22(26)(32) 552 2 (262 1 322 ) 22(26)(32) /A 8 143° 552 5 262 1 322 2 2(26)(32)cos A
cos A 5
I used the inverse cosine function on my calculator to determine /A. I couldn't use the sine law. to solve the problem? Explain your reasoning.5.
Why did Tina draw line AP on her sketch as part of her solution? Could Tina have used the sine law.
0 m b 5 !49. I wanted to use the sine law to determine angle u to solve the problem. c.21
Since I knew two sides and the angle between them.0)(5.0) 2 1 (5.
sin C sin B 5 c b sin 60° sin u 5 5.0 5.21 m2 b 8 7.0 m a b
I drew a sketch of the situation.0
I used the inverse sine function on my calculator to determine angle u.1 7. and /B into the formula. I wrote the sine law with the angles in the numerators. I substituted the values of a.1 3
sin 60° 7. I calculated b by evaluating the right side of the equation and determining its square root. I needed to determine the length of side b. I multiplied both sides of the equation by 5.Serina's Solution
A 5.1) 2 2 2(8. But before I could do that. Since I needed to solve for an angle.1 3
1
sin u sin 60° 5 5.0
1
u 5 5. I determined /C (u) by using the sine law.1)cos 60° b 2 5 49. I couldn't use the sine law to determine b. So I used the cosine law.
u 8 39° Since Mitchell's roof is inclined about 39° on the north side. he will be able to install the solar hot-water system by himself.
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Chapter 5
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.1 7.
B
C
b 2 5 a 2 1 c 2 2 2ac cos B b 2 5 (8.1 3 5.1 to solve for sin u.1 m c 60 8.
For each situation. is the fire from each tower?
9. each at a different
location.3 km apart. to the nearest tenth of a kilometre.5° from the vertical.0 cm. If
11. are 20. The Leaning Tower of Pisa is 55. a) What is the distance separating the balloons. a) Create a question that requires using the sine law to solve it.
14.0 cm. The side lengths and the interior angles of any triangle can be determined by
C
using the cosine law. and 10. Darryl is standing on one of the roads
T
B
45
D 2. Two hot-air balloons are moored to level ground below. what is the distance from the top of the tower to the top edge of its shadow? Assume that the ground around the tower is level. Two roads intersect at an angle of 15°. Round your answer to the nearest metre.0
30
C
270 m from the intersection. An observer at each location determines the angle of elevation to the opposite balloon as shown at the right. How far.0 and D is the midpoint of BC. its shadow is 90. The bearing from tower A is 25° and from tower B is 345°. or a combination of both. Round your answer to the nearest metre.5. 7.
A
B
80
20 25 2.0 km apart. Determine the perimeter of the triangle to the nearest centimetre. to the nearest tenth. determine all unknown side lengths to the nearest tenth of
a centimetre and/or all unknown interior angles to the nearest degree.
10.9 m tall and leans 5. The ranger in each tower observes a fire and radios the bearing of the fire from the tower.5 cm. From tower A. A and B. Two forest fire towers. and 20°. BC 5 2. b) Create a question that requires using the cosine law to solve it. 40°. b) An isosceles triangle has at least one interior angle of 70° and at least one side of length 11.
8.
Extending
12. if /ADB 5 45° and /ACB 5 30° . use the algebraic representation of the law to show how to determine all unknown quantities. If more than one solution is possible. the sine law. Sketch a triangle and state the minimum information required to use a) the cosine law b) both laws Under each sketch. state all possible answers.
13. Given ^ ABC at the right. The longest side is
10 cm longer than the shortest side.7
7. a) A triangle has exactly one angle measuring 45° and sides measuring 5.
A
Determine AB.0 m long. to the nearest tenth of a kilometre? b) Determine the difference in height (above the ground) between the two balloons. The interior angles of a triangle are 120°. Include a complete solution and a sketch.4 cm. The observers are 2. the bearing of
tower B is 70°.0 km
75
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Trigonometric Ratios
327
. Include a complete solution and a sketch.
himself. Joe estimates the angle between the base of the cliff.5. In ^ DBC. and his friend Joe to be 63°. himself. and Manny to be 42°. while Manny estimates the angle between the base of the cliff. I don't have enough information to calculate h.8
YOU WILL NEED
Solving Three-Dimensional Problems by Using Trigonometry
GOAL
• dynamic geometry software
(optional)
Solve three-dimensional problems by using trigonometry. Manny uses a clinometer to determine the angle of elevation to the top of a cliff as 38°. I needed to determine /C.
328
Chapter 5
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.5 m away from Manny. but BC is also in ^ DBC. I knew two angles and a side length. 68. Before I could calculate BC. From point D. In ^ ABC. I used the fact that the sum of all three interior angles is 180° .
Matt's Solution
In ^DBC: /C 5 180° 2 (63° 1 42°) 5 75°
BC is in ^ ABC.
A h C 38 63 42 D ? 68.5 m B
What is the height of the cliff to the nearest tenth of a metre?
EXAMPLE
1
Solving a three-dimensional problem by using the sine law
Calculate the height of the cliff to the nearest tenth of a metre.
LEARN ABOUT the Math
From point B.
So I used tangent.45.5 sin 75°
I used my calculator to evaluate.45
1 h 3 47.1 m 8 h The height of the cliff is about 37.45 m tan 38° 5 tan 38° 5 h BC h 47. C.45 47.45 5
37. B. Why did Matt begin working with ^ DBC instead of ^ ABC? What strategies might Matt use to check whether his answer is reasonable?
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. I multiplied both sides of the equation by 47. I knew that ^ ABC is a right triangle and that h is opposite the 38° angle while BC is adjacent to it.
BC 68.1 m.8
BC BD 5 sin D sin C BC 68.
BC 5 sin 42° 3
68. I multiplied both sides of the equation by sin 42° .5 5 sin 42° sin 75° sin 42° 3
1
Using ^ DBC and the value of /C.5 5 sin 42° 3 sin 42° sin 75°
1
To solve for BC.
tan 38° 3 47.
Was the given diagram necessary to help Matt solve the problem? Explain.
BC 8 47. To evaluate h.45 1
Then I used ^ ABC to calculate h. I used the sine law to calculate BC.5.
Reflecting
A.
Using the same reasoning. I used these angles to determine the values of /EAQ and /EBQ.
/Q 5 230° 2 120° 5 110°
E 38 50 m Q b 110 q a 35
In ^ AQB. From her position. boat A has a bearing of 230° and boat B has a bearing of 120°. So I subtracted 120° from 230°. I determined that /EBQ is equal to 35°. I knew only one angle and no side lengths. How far apart are the boats to the nearest metre?
120
230 A q B
Kelly's Solution
In ^ AQB: In ^ AQB. I knew that the value of /Q is equal to the difference of the bearings of boats A and B. I drew a sketch that included the angles of depression. So /EAQ is equal to 38°. I needed to determine AQ (side b) and BQ (side a) first.
A
B
The angle of depression to A is measured from a line parallel to AQ.
/EAQ 5 38°
E
/EBQ 5 35°
50 m Q 38 A b 110 q 35 a B
I included the values of /EAQ and /EBQ in my sketch.
330
Chapter 5
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.APPLY the Math
EXAMPLE
2
Solving a three-dimensional problem by using the sine law
N E 50 m
Q
Emma is on a 50 m high bridge and sees two boats anchored below. In order to calculate q. Emma estimates the angles of depression to be 38° for boat A and 35° for boat B.
such as facts about parallel lines.0 m
a 8 71. use the sine law and/or the cosine law. Then use trigonometry to solve the original problem. use the primary or reciprocal trigonometric ratios. • In right triangles.
q 2 5 b 2 1 a 2 2 2ba cos 110° q 2 5 (64. Since ^ AEQ and ^ BEQ are right triangles. and so on.
In Summary
Key Ideas
• Three-dimensional problems involving triangles can be solved using some combination of these approaches: • trigonometric ratios • the Pythagorean theorem • the sine law • the cosine law • The approach you use depends on the given information and what you are required to find. So I used the cosine law to calculate q. always start with a sketch of the given information.8
In ^AEQ: tan 38° 5 b5 50 b 50 tan 38° In ^BEQ: tan 35° 5 a5 50 a 50 tan 35°
In ^ AQB. I now knew two side lengths and the angle between those sides. • In all other triangles.
Need to Know
• When solving problems.4) 2 2 2(64.5. Given Information SSA ASA or AAS SAS SSS Required to Find angle side side side Use sine law sine law cosine law cosine law
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Trigonometric Ratios
331
. Determine any unknown angles by using any geometric facts that apply. I expressed AQ in terms of tan 38° and BQ in terms of tan 35°. I substituted the values of b and a into the equation and evaluated q.4)cos 110° q 8 111 m
q 5 !12 320.0) 2 1 (71. Revise your sketch so that it includes any new information that you determined.6
b 8 64.4 m
The boats are about 111 m apart.0) (71. interior angles in a triangle. Then I solved for b and a.
5 m overhang on each side of the shed. a) Determine h to the nearest tenth of a metre.0 m
are equal in length and must meet at an angle of 80°.5 m
1. Morana is trolling for salmon in Lake Ontario. b) How could Josh use the primary trigonometric ratios to calculate x ? Explain. The two sides of the roof
4.CHECK Your Understanding
35 1m B x A 0.
PRACTISING
3.0 m 66 C
4. Describe the steps you would use to calculate the length of line she must let out. Explain your reasoning for each step of your solution. She knows that there are fish at a depth of 45 m. She sets the fishing rod so
45 m
that its tip is 1 m above water and the line forms an angle of 35° with the water's surface. Explain each of your steps. a group of students was asked to determine the altitude. Determine the value of x to the nearest centimetre and u to the nearest
K
degree. As a project. h. Josh is building a garden shed that is 4. There will be a 0. of a
promotional blimp.5 m 80 x C 0. a) Should he use the sine law or the cosine law? Explain.
332
Chapter 5
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. The students' measurements are shown in the sketch at the left. b) Is there another way to solve this problem? Explain.0 m wide. Josh wants to determine the length of each side of the roof. a) c) E D 65 55
x F 15 cm 45 x G A A 27 18 cm 70 D B 15 cm B d) 70 115 B A 15 cm D 95 14 cm
D 10 cm C
35 E b)
in
ABC
x
C
A h D 74 B 50 175.
2.
A coast guard helicopter hovers between an island and a damaged sailboat. • From the helicopter. the angle of elevation to the helicopter is 73°. • They measured the angle between the lines of sight to the two towns as 80°. From this position. Paris. Justify your reasoning with calculations. While Travis and Bob were flying a hot-air balloon from Beamsville to
Vineland in southwestern Ontario. • the bearing of boat A is 180° at an angle of depression of 40° • the bearing of boat B is 250° at an angle of depression of 34° Calculate the distance between the two boats to the nearest metre. Ontario. facing north and sees the balcony at an angle of elevation of 20°. • From an altitude of 226 m. From the tourist's position. to the nearest metre.
• From the island. A tourist in the observation deck notices two boats on the water.
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Trigonometric Ratios
333
. • A police rescue boat heading toward the sailboat is 800 m away from the scene of the accident. the angle between the island and the sailboat is 35°.8
5. Suppose Romeo is serenading Juliet while she is on her balcony. Romeo is
h 20 R 100 m 18 P
8. from the helicopter to the sailboat. Determine the height of Juliet's balcony above the ground. between the two towns. is
A
166 m above the Niagara River. to the nearest metre. the island and the sailboat are 40° apart. Is there enough information to calculate the distance between the two towns? Justify your reasoning with calculations. Juliet's other suitor. to the nearest metre. is observing the situation and is facing west. an observer on the island notices that the sailboat and police rescue boat are 68° apart.
J
7.
6. Paris sees the balcony at an angle of elevation of 18°. • At the same moment. Explain how you would calculate the straight-line distance. they decided to calculate the straight-line distance. Romeo and Paris are 100 m apart as shown. they simultaneously measured the angle of depression to Beamsville as 2° and to Vineland as 3°.5. The observation deck of the Skylon Tower in Niagara Falls.
1 m 2.
10. Bert wants to calculate the height of a tree on the opposite bank of a river. Calculate the distance between the two friends. Justify your reasoning with calculations.5 m 4. which runs north to south. The angle of elevation from A to the top of the tree is 28°. each piece must fit exactly inside the vehicle.0 m 2. Justify your reasoning with calculations.
2.0 m 4.7 m 4. In order to prevent damaging the platform. they see a hot-air balloon toward the east at angles of elevation of 41° and 55°.0 m
4. When the boat is equidistant between both girls. Justify your reasoning with calculations. a stage platform has been dismantled
T
into three triangular pieces as shown. Describe how you would calculate the angle. if any. b) What additional information. To
85 A
30 80 m
B
do this.5 m C 2.
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Chapter 5
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.9 m 2.0 m mast is 51° for both observers. Brit and Tara are standing 13.7 m 42 J
A 130 B 4.
G 3. When both of them stand at their front doors.9.0 m I
4. Explain how you would match each piece of the platform to the best-suited vehicle. a) Chandra says she doesn't have enough information to answer the question. Chandra's homework question reads like this:
C
Bill and Chris live at different intersections on the same street. between Tara and the boat as viewed from Brit's position.5 m 1.
12. Explain if this information helps Bert to calculate the height of the tree to the nearest metre. would you need to solve the problem? Justify your answer. he lays out a baseline 80 m long and measures the angles as shown at the left. respectively.5 m apart on a dock when they observe a
sailboat moving parallel to the dock.6 m 6. In setting up for an outdoor concert. to the nearest degree.6 m 1. Justify your reasoning with calculations. the angle of elevation to the top of its 8.1 m 2.5 m
H
L
K
There are three vehicles available to transport the pieces. Evaluate Chandra's statement.8 m
11.
0 m away at an angle of elevation of 34°. Simone is facing north at the entrance of a tunnel through a mountain. Mario notices that his friend Thomas's window on the west side of the building appears 71. From Mario's position. Two cars leave the intersection on different roads
at speeds of 80 km/h and 100 km/h.
14. After she exits the tunnel. a) Calculate the straight-line distance.8
Extending
13. a) If a rope were pulled taut from one window to the other. Calculate the distance between the two planes to the nearest tenth of a kilometre. to the nearest tenth of a metre. how long. Explain your reasoning for each step of your solution.8 km. After 2 h. The
first plane. Assuming that the tunnel is perfectly level and straight. The other plane.
Trigonometric Ratios
NEL
335
. a traffic helicopter that is above and between the two cars takes readings on them. Assume that both cars are travelling at constant speed. The angle of depression to the slower car is 20°. b) Determine the altitude of the helicopter to the nearest kilometre.5 m away at an angle of elevation of 55°. from the helicopter to the faster car.
16. b) What is the straight-line distance through the building between the two windows? Round your answer to the nearest tenth of a metre. Q. P. the same mountain has a bearing of 258° and its peak appears at an angle of elevation of 31°. She
notices that a 1515 m high mountain in the distance has a bearing of 270° and its peak appears at an angle of elevation of 35°. at a bearing of 70° and with an altitude of 2. around the outside of the building. is 180 km away on a bearing of 125° and with an altitude of 1. how long is it to the nearest metre?
15. Mario is standing at ground level exactly at the corner where two exterior
walls of his apartment building meet. his apartment window on the north side of the building appears 44.5.7 km. would the rope need to be? Explain your reasoning. to the nearest kilometre. and the straight-line distance from the helicopter to that car is 100 km. is 120 km from the airport. A. An airport radar operator locates two planes flying toward the airport. Two roads intersect at 34°.
5 tan2 u 1 1
5a
5 5 5
sin2 u 11 cos2 u
sin u 2 b 11 cos u
R.
336
Chapter 5
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.5. 6 tan2 u 1 1 5 sec2 u for all angles u. Find a common denominator.
sin u
5 R.
EXAMPLE
tan2 u 1 1 5 sec2 u L. there are restrictions on the variable because the denominator cannot equal zero.
sin2 u 1 cos2 u cos2 u 1 cos 2 u
Write tan u as cos u.
A trigonometric identity is an equation involving trigonometric ratios that is true for all values of the variable. • Try Chapter Review Questions 6 and 7.S. y. so cos u 2 0. Since the denominator cannot equal 0.S. so simplify it. the solution below is one way to prove that tan2 u 1 1 5 sec2 u is an identity. The left side of the equation is more complicated.S.5
Study
Chapter Review
FREQUENTLY ASKED Questions
Q:
A:
Aid
What steps would you follow to prove a trigonometric identity?
• See Lesson 5. there is a restriction on u. Then use the Pythagorean identity. or you may rewrite each side of the equation in terms of sine and cosine and then use the Pythagorean identity sin2 u 1 cos2 u 5 1. where appropriate. If a trigonometric ratio is in the denominator of a fraction. Examples 1 to 4. and r and then simplify. You may rewrite the trigonometric ratios in terms of x. 5 sec 2 u
5a 5
1 cos 2 u
1 2 b cos u
First separate both sides of the equation. For example. where cos u 2 0.
Q:
A:
Study
Aid
Always start with a sketch of the given information because the sketch will help you determine whether the Pythagorean theorem. use the sine law.
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Trigonometric Ratios
337
. or 2 triangles are possible given the information in a problem.4 10
A 36
a C
6.0 cm 60 C A
• See Lesson 5. 6. Examples 1 and 2.and threedimensional trigonometric problems?
• See Lesson 5. This situation occurs when you know two side lengths and an angle (SSA).
What approaches are helpful in solving two.1 cm. if you know two sides and the angle between those sides. For all other cases. or all three sides. then ^ ABC is a right triangle and 6. Examples 1.7. Examples 1 and 2.0 cm. • Try Chapter Review Questions 8 and 9.0 cm 60
If a 5 6.
cm . If you have right triangles. a 5 7.4 10
A 36
cm . or the cosine law is the best method to use. there are two possible triangles:
B B 7.1 cm
If a . and c 5 10.
• See Lesson 5. use the sine law. and 3. use the cosine law. a triangle cannot be drawn.1 cm is the shortest possible length for a:
B
cm . If you know an angle opposite a side.4 10
36 C
7. 1. • Try Chapter Review Questions 10 and 11. • Try Chapter Review Questions 12 and 13. 2. use the cosine law.1 cm. where /A 5 36°. For example. Q:
A:
How do you decide when to use the sine law or the cosine law to solve a problem?
Study
Aid
Given any triangle.6. If you know three sides or two sides and the contained angle in an oblique triangle.8.Chapter Review
Q:
A:
How do you know when you are dealing with the ambiguous case of the sine law?
Study
Aid
The ambiguous case of the sine law refers to the situation where 0. use the Pythagorean theorem and/or trigonometric ratios. given ^ ABC. the sine law.4 cm.
5 cm. P and Q.7
takes two measurements.8
given each set of information. the base of the pole and point C are 90. State any restrictions on the
6.6
11. Sketch all possible triangles where appropriate.0 km apart. a) K
11. How high. then label. Justin
ranger at station Q sees a fire 15. From her position. From point A.5 30 m
10.9 m wide and 4.1 cm. to the nearest tenth of a metre. to the nearest degree. c 5 8.5
variables if all angles vary from 0° to 360°. to the top of the left exterior wall?
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Trigonometric Ratios
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. are 20. Two spotlights.2 cm. State any
restrictions on angle b.2 55 M
m Q 4. A stationary observer standing on the ballroom floor notices that the angle of elevation is 45° to the blue spotlight and 70° to the white one. /C 5 34° c) a 5 11. Two forest fire stations. Calculate. how far.
h B 15.3 108 j
J 7.0
B
72 C
front of her school. what is the angle of elevation. a) b 5 3. Suzie is 12. To determine the height of a pole across a road. From her position. to the nearest tenth of a kilometre. is station P from the fire?
Lesson 5. are
8. /B 5 30° b) b 5 12. all side lengths to the nearest tenth of a centimetre and all angles to the nearest degree.0 m from the base of the right exterior wall.2 cm.0° apart.0 km away. one blue and the other white. He stands at point A directly across from the base of the pole and determines that the angle of elevation to the top of the pole is 15. a) tan a cos a 5 sin a 1 b) 5 sin f sec f cot f sin x cos x c) 1 2 cos2 x 5 cot x d) sec u cos u 1 sec u sin u 5 1 1 tan u
Lesson 5. She determines that the left and right exterior walls appear to be 39° apart. /C 5 33°
9.2 cm.0 m apart on a track on the ceiling of a ballroom. c 5 5.7 L
A
C
13. c 5 5. is the ceiling of the ballroom?
Lesson 5.7 m high. Determine each unknown side length to the nearest
tenth. where he sees that the base of the pole and point A are 57.
7. A
12. Determine whether the equation 1 cos b cot b 5 sin b 2 sin b is an identity.Chapter Review
Lesson 5. If the angle between the line PQ and the line from P to the fire is 25°. Calculate the height of the pole to the nearest metre.0
c 8. Determine whether it is possible to draw a triangle
placed 6.5° apart.5
c)
N
6. Suzie estimates that the front face is 8.3 57. He then walks 30 m parallel to the freeway to point C. While standing at the left corner of the schoolyard in
b) A 6.3°. Prove each identity.0 cm.
a) P(23. a) sin u cos u restrictions on the variables.8 cm. Use a different method for parts (a) and (b). and / A 5 41° b) a 5 2. then label. For each triangle. where appropriate.1 cm. The angle formed at the base of the tree between points A and B is 90°. 0) b) S(28. Given cos u 5 2 13 . b 5 2.1 cm.
b) cot u tan u
4. Given angle u. respectively.
5. a)
b) 1 1 cot 2 a 5 csc 2 a
Sketch a triangle of your own choice and label the sides and angles. i) Prove each identity. Given each set of information.5
Chapter Self-Test
1. and A and B are 30 m apart. b) State all forms of the cosine law that apply to your triangle. a) tan2 f 1 1 5 sec2 f
5. a) a 5 1.0 m B 49 D w 53 C G 40 48 H w b) I
6. she determined that the angles of elevation for a certain tree were 41° and 52°. i)
For each point.5 m J
7. where appropriate.5 cm. If the tree is perpendicular to the ground. ii) Determine the value of the principal angle and the related acute angle. To estimate the amount of usable lumber in a tree. to the nearest degree. calculate the value of w to the nearest tenth of a metre. c) State all forms of the sine law that apply to your triangle. determine all possible angles for u. Chitra must first estimate
the height of the tree. From points A and B on the ground. determine how many triangles can be drawn. where the terminal arm of angle u lies in quadrant 2. and / A 5 20°
8. what is its height to the nearest metre?
340
Chapter 5
NEL
.
Calculate. 26) 1 2 !3 b) cos u 5 2
a) sin u 5 2
2. State any
ii) Explain why these identities are called Pythagorean identities.
a) A 60. sketch the angle in standard position to determine all six trigonometric ratios.
c) cot u 5 21 d) sec u 5 22
5 3. all side lengths to the nearest tenth and all interior angles to the nearest degree. c 5 6.
evaluate each trigonometric expression. where 0° # u # 360°.
UK) AC (Montréal. took a digital photo of the Moon during a lunar eclipse at exactly the same time. 2004.
Shortest Distance on Earth's Surface Between Two Locations AB (Montréal. Canada to Montevideo.384°
?
A. and for A and C. B. is the Moon from either Montréal or Selsey? Repeat parts A and B for locations B and C. Label all the given angles and distances.
parallax angle Moon apparent image of the Moon ruler Earth apparent image of the Moon
Astronomers measure the parallax of celestial bodies to determine how far those bodies are from Earth. Calculate the relative error.
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341
. Which of your results is most accurate? What factors contribute most to the error in this experiment?
Task
Checklist
Did you draw the correct
sketches?
Did you show your work? Did you provide appropriate
reasoning?
C. for all three distances you calculated. The data related to these photos is shown.
Did you explain your
thinking clearly?
E. Pete Lawrence. 2004. Uruguay) 5 220 km 9 121 km 10 967 km
Parallax Angle 0. D. from the given data?
Sketch a triangle with the Moon and locations A and B as the vertices. On October 28.
What is the most accurate method to determine the distance between the Moon and Earth.189° 1. On October 28.7153° 1. to the nearest tenth of a percent. the Moon was about 391 811 km from Earth (surface to surface). UK to Montevideo. three astronomers (Peter Cleary. Uruguay) BC (Selsey.5
Parallax
A B
Chapter Task
Parallax is the apparent displacement of an object when it is viewed from two different positions. How far. What kind of triangle do you have? Determine all unknown sides to the nearest kilometre and angles to the nearest thousandth of a degree. to the nearest kilometre. Canada to Selsey. and Gerardo Addiègo) each at a different location on Earth.
342
NEL
.
where x is measured in degrees Determine the equations of sinusoidal functions in real-world situations and use those equations to solve problems
•
? This picture of NASA's mission
control shows the flight path of the space shuttle as it orbits Earth.Chapter
6
Sinusoidal Functions
GOALS
You will be able to
• • •
Identify situations that can be modelled using sinusoidal and other periodic functions Interpret the graphs of sinusoidal and other periodic phenomena Understand the effect of applying transformations to the functions f(x) 5 sin x and g(x) 5 cos x. What type of function would model this path?
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343
.
3. A survey
Aid
• For help. see Essential Skills
Appendix.6
Study
Getting Started
SKILLS YOU NEED
1.50
back. he will sell 20 more shirts per week. How tall is the building?
7.25 s? d) State the domain and range of the function.
describe how a graph of f (x) 5 x 2 would change if the transformation is applied to it.
a) 15 cm 40° 22 cm
b) 9 cm 52° 11 cm
35° 7m 9m x
4.
344
Chapter 6
NEL
. Marcus sells 100 T-shirts per week at a price of $30 per shirt. b) How many times will the price have to be dropped for the total revenue to be 0? c) How many times will the price have to be dropped to reach the maximum revenue? d) What is the maximum revenue? e) What price will the T-shirts sell for to obtain the maximum revenue? f ) How many T-shirts will be sold to obtain the maximum revenue?
2. An air hockey puck is shot to the opposite end of the table and ricochets
t 0 0.25 Time (s) 0. 5 6. List all the different types of transformations that you know. An aerial photograph shows that a building casts a shadow 40 m long when
40 m
the angle of elevation of the Sun was 32°. Determine u to the nearest degree. Determine the value of x in the triangle at the left to the nearest tenth
of a metre. Question 3. Use transformations of the graph f (x) 5 2x to sketch the graphs of
the following: a) y 5 2f (x) b) y 5 3f (x)
c) y 5 f (x) 1 4 d) y 5 22f (x 2 3)
6.
5. If x represents the number of times the price is reduced by $2. a) How far did the puck travel? b) When was the puck farthest away from where it was shot? c) How fast was the puck travelling in the first 0. 4. then the revenue generated from T-shirt sales can be modelled by the function R(x) 5 (30 2 2x) (100 1 20x) . 7 Appendix A-16 A-14
180 Distance (cm) 120 60
d(t)
Puck Motion
indicates that if he reduces the price of each shirt by $2. For each one. The puck's distance in centimetres from where it was shot in terms of time in seconds can be modelled by the graph shown at the left. a) Explain what the factors (30 2 2x) and (100 1 20x) represent in R(x) .
Explain what this value represents in this situation. where the height. F. When will the arrow strike the ground? When will the arrow reach its maximum height? What is the maximum height reached by the arrow? State the domain and range of the function in this situation.
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Sinusoidal Functions
345
. C. G. measured in seconds. Summarize what you determined about the relationship between the height of the arrow and time. B. t. E.
How can you describe the flight of the arrow using this function?
What is the initial height of the arrow? Calculate h(2). D. The height of the arrow above the ground is a function of time and can be modelled by h(t) 5 25t 2 1 20t 1 25.
?
A. h(t) .Getting Started
APPLYING What You Know
Flying Arrows
An arrow is shot into the air from the edge of a cliff. is measured in metres at time.
The number of hours of daylight for day 943 is about 17 h. and 531 2 166 5 365.6.9 349 5.5 714 5. I used the pattern to extend the graph to year 3.
LEARN ABOUT the Math
The number of hours of daylight at any particular location changes with the time of year. Note: Day 15 is January 15 of year 1.9
?
EXAMPLE
How many hours of daylight will there be on August 1 of year 3?
1
Representing data in a graph to make predictions
Jacob's Solution
Hours of Daylight in Hudson Bay
I drew a scatter plot with the day as the independent variable and the hours of daylight as the dependent variable. Day 74 is March 15 of year 1.2 684 7. I drew a smooth curve to connect the points. I can tell because the greatest number of hours of daylight occurs on days 166 and 531.9 411 9. Day 411 is February 15 of year 2.9 653 10.
346
Chapter 6
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.
Hours of daylight
24 18 12 6 0
365 730 Day number
Aug 1 1095 day: 943
The least number of hours of daylight occurs on days 349 and 714.8 258 12. The table shows the average number of hours of daylight for approximately a two-year period at Hudson Bay.
I used the graph to estimate the number of hours of daylight for day 943.
Day Hours of Daylight 15 6.2 166 18. also 365 days apart. Nunavut.7 135 17. The data and graph repeat every 365 days.6 531 18.2 470 14. That would be 1095 days.1
YOU WILL NEED
Periodic Functions and Their Properties
GOAL
• graph paper
Interpret and describe graphs that repeat at regular intervals.8 561 18.7 74 11.1 623 12.
During manufacturing.0 Height (cm) Cutting Blade Motion
0.0 7. B.0 4.
periodic function a function whose graph repeats at regular intervals.0 3.6.0 6. Tanya and her brother Norman accompany their mother to work.0 Time (s)
How can Norman interpret the graph and relate its characteristics to the manufacturing process? Norman's Solution
It's a periodic function because the graph repeats in exactly the same way at regular intervals.0 5. C. D.0 2. a cycle of a periodic function is a portion of the graph that repeats
Why does it make sense to call the graph of the hours of daylight a periodic function? How does the table help you predict the period of the graph? Which points on the graph could you use to determine the range of this function? How does knowing the period of a periodic graph help you predict future events?
APPLY the Math
EXAMPLE
2
Interpreting periodic graphs and connecting them to real-world situations
y 1 cycle x 0 period
Part A: Analyzing a Cutting Blade's Motion Tanya's mother works in a factory that produces tape measures. the y-values in the table of values show a repetitive pattern when the x-values change by the same increment period the change in the independent variable (typically x) corresponding to one cycle.0 9.5
0
1. a metal strip is cut into 6 m lengths and is coiled within the tape measure holder.
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Sinusoidal Functions
347
. Tanya's mother shows a graph that models the motion of the cutting blade on the machine in terms of time.1
Reflecting
A.0 8. One day.
1.
This is a periodic function. A cutting machine chops the strips into their appropriate lengths.
5 cm or less.
Part B: Analyzing the Motion of the Tape as It Is Spooled Farther down the assembly line. like the ones from 0 to 3. show that the blade takes 0.
The y-value is always 0. like from t 5 3.5
The cutting blade cuts a new section of metal strip every 4 s because the graph has a pattern that repeats every 4 s.5 s. must mean that the blade stops for these intervals.
1.0 to 7.0. show that the blade takes 0. the metal strip is raised and spooled onto a rotating cylinder contained within the tape measure. Other parts of the graph. When the height is 0 cm.0 Height (cm) 4 0.5 s to go up.0 3.
The blade takes 1 s to go up and down.5 s to go down. Parts of the graph. Since the graph repeats every 4 s and the blade hits the surface at 3.0 to t 5 3.
348
Chapter 6
NEL
.5 cm.0 Time (s)
The maximum height of the blade is 0.5 s and 7. The minimum height is 0 cm.
The blade will strike the cutting surface again at 11.5 s and every 4 s after that.
The blade stops for 3 s intervals. I can figure out the next time it will hit the surface. the blade is hitting the cutting surface.0 and 4.
0
1.0 2.0.0 4. The machine is probably pulling the next 6 m section of metal strip through before it's cut. Flat sections. Tanya notices that the height of the end of the metal strip that attaches to the spool goes up and down as the rest of the strip is pulled onto the cylinder. like from t 5 3.5.The period of this function is 4 s. so the blade can't be higher than this.5 to t 5 4.
0 0 0.50 0.25 t 1.
Height of the End of the Metal Strip 10. trough the minimum point on a graph
This is a periodic function. The next trough is at t 5 0.0 6. The highest the graph goes is 9 mm.25.75 1.25 0.
The range for this function is 5h [ R | 1 # h # 96.
peak the maximum point on a graph
y peak
trough
x
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Sinusoidal Functions
349
.00 Time (s) 1. The first trough is at t 5 0.1
Tanya's mother shows them a graph that models the height of the end of the strip in terms of time. I could also have measured the distance between the first two peaks to get that value.0 4.0 2. This time the action is smooth. The period represents the time it takes for the rotating cylinder to make one complete revolution.50
end of strip
How can Tanya interpret the graph and relate its characteristics to the manufacturing process? Tanya's Solution
It's a periodic function because the graph repeats in exactly the same way at regular intervals. and the lowest is 1 mm.25 s.6.
The period of this function is 0. The distance between the two troughs gives the period. The heights are always at or between these two values.0 h Height (mm) 8.
equation of the axis.
I calculated the halfway point between the maximum and minimum values of the graph. giving me the equation of the axis. and amplitude. The waves repeat at regular intervals. so the function is periodic.
350
Chapter 6
NEL
. That is the period of the function. I calculated the halfway point between the maximum and minimum values of the height. The amplitude of a function is the vertical distance from its axis (h 5 5) to its maximum value (9 mm). state the period.equation of the axis the equation of the horizontal line halfway between the maximum and the minimum.
amplitude half the difference between the maximum and minimum values.
a) the average number of b) the motion of a piston on an c) a student is moving a metre stick
hours of daylight over a three-year period
Daylight Hours 18
automated assembly line
Piston Motion Height (cm)
back and forth with progressively larger movements
Metre Stick Motion Distance (cm) 18 6 2 0 2 6 Time (s) 2 4 6 8 10
Hours
12 6 0 365 730 Days 1095
1 0 1 3 5
6
12
Tina's Solution
a) periodic
Time (s)
The graph looks like a series of waves that are the same size and shape. If so. 92554 The amplitude of this function is 4 mm. it is determined by y5
maximum value 1 minimum value 2
911 55 2 The equation of the axis for this function is h 5 5.
The amplitude is the vertical distance from its axis (h 5 12) to the maximum value (18 h) or minimum value (6 h). To get the equation of the axis. it is also the vertical distance from the function's axis to the maximum or minimum value
EXAMPLE
3
Identifying a periodic function from its graph
Determine whether the term periodic can be used to describe the graph for each situation.
period 5 1 year 18 1 6 5 12 2 equation of the axis: h 5 12 18 2 12 5 6 amplitude 5 6 h
The graph repeats its pattern every 365 days.
5 1 2 (22. or equation of the axis.6. repeating in the same way over and over.
In Summary
Key Ideas
• A function that produces a graph that has a regular repeating pattern over a constant interval is called a periodic function. The equation of the axis is halfway between the maximum of 1 and the minimum of 26. It describes something that happens in a cycle.5 2 equation of the axis: h 5 22.1
b) periodic
The shape of the graph repeats over the same interval. amplitude.
The distance between the maximum and the axis is 3. so the function is not periodic.5. so the function is periodic.5 cm
c) nonperiodic
The graph repeats every 6 s. and equation of the axis.
The shape of the graph does not repeat over the same interval.5) 5 3.5 amplitude 5 3.
Need to Know
• Extending the graph of a periodic function by using the repeating pattern allows you to make reasonable predictions by extrapolating.
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Sinusoidal Functions
351
. so that's the period of the function.
This means that the function does not have a period. • The graph of a periodic function permits you to figure out the key features of the repeating pattern it represents. such as the period. amplitude.
period 5 6 s 1 1 (26) 5 22.
Height on a Ferris Wheel Height above ground (m) 6 4 2 0 2 4 6 8 10 12 14 16 18 20 22 Time (s)
• A function that produces a graph that does not have a regular repeating pattern over a constant interval is called a nonperiodic function.
e) Determine the equation of the axis. determine the domain of the function. period. Determine the range."
352
Chapter 6
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. and amplitude of the
function shown. The motion of an automated device for attaching bolts to a household
1.0 0. Which of the following graphs are periodic? Explain why or why not. Explain what each part could mean in the context of "attaching the bolt. a) What is the period of one complete cycle? b) What is the maximum distance between the device and the appliance? c) What is the range of this function? d) If the device can run for five complete cycles only before it must be turned off. f ) Determine the amplitude. equation of the axis. g) There are several parts to each complete cycle of the graph.5 0
1
2 Time (s)
x 3
appliance on an assembly line can be modelled by the graph shown at the left.CHECK Your Understanding
1.
a)
20 0 –20
f(x) x 1 2
c)
4
f(x) x
0 –4 d) 1 f(x)
5
10
b)
f(x)
5
0
x 5 0 5 10
x
2.
f(x)
10 8 6 4 2 0 Distance between device and appliance (cm) Bolt Machine Motion y
x 2 4 6 8 10 12 14 16 18 20
3.5 1.
7 1. • independent variable: time • dependent variable: intensity of the signal
6. Which of the tables of values might represent periodic functions? Justify. but she is so far away that the ball
bounces on the ground four times.5 7 7.9 1. Chantelle has a submersible pump in her basement. The graph models the depth of the water below her basement floor in terms of time.
water pumped to surface pump intake motor
cable
the pump turned off and on to drain water collecting under her house's foundation. • independent variable: distance • dependent variable: bounce height f ) The antenna on a radar tower is rotating and emitting a signal to track incoming planes.1 1.5
d)
x 1 2 4 7 11 16 22 29
y 5 6 5 6 5 6 5 6
7.7 0. The depth of the water decreased when the pump was on and increased when the pump was off.3 1.
Depth of Water under Basement Floor
Depth (cm)
water enters here
10 8 6 4 2 0 2 4 6 8 10 12 14 16 18 20 22 Time (min)
a) b) c) d) e) f) g) h)
Is the function periodic? At what depth does the pump turn on? How long does the pump remain on? What is the period of the function? Include the units of measure.9 2.e) You throw a basketball to a friend. What is the range of the function? What will the depth of the water be at 3 min? When will the depth of the water be 10 cm? What will the depth of the water be at 62 min?
NEL
354
Chapter 6
.
a)
x 25 24 23 22 21 0 1 2
y 9 4 1 0 1 4 9 16
b)
x 0.1
y 5 6 7 5 6 7 5 6
c)
x 23 26 29 32 35 38 41 44
y 6 6. During a heavy rain.5 8 8.5 9 9.5 1.
Trevor's height above the ground in terms of
A
time can be represented by the graph shown. she picks up a stone in her tire.6.6 km/h. The spacecraft's distance
Plot the data. Is the graph periodic? What is the period of the function. Sketch the graph of a periodic function with a period of 20. While riding on a Ferris wheel.
10. when will Trevor be at the lowest height again? At what times is Trevor at the top of the wheel? When will his height be 4 m between 24 s and 30 s?
12. what is the domain of the function?
a) b) c) d)
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Sinusoidal Functions
355
.
Trevor's Height above the Ground Height above ground (m) 6 4 2 0 a) b) c) d) e) f) g) 2 4 6 8 10 12 14 16 18 20 22 Time (s)
9.1
8. and what does it represent? What is the approximate distance between the spacecraft and Earth at 8 min? e) At what times is the spacecraft farthest from Earth? f ) If the spacecraft completes only six orbits before descending to Earth. and whose equation of the axis is y 5 7. and what does it represent? What is the equation of the axis? What is the amplitude? What is the range of the function? After 24 s.
Time (min) Distance (km) 0 550 6 869 12 1000 18 869 24 550 30 232 36 100 42 232 48 550 54 869 60 1000 66 869 72 550 78 232
5 y [ R |22 # y # 56. an amplitude
of 6. Draw a graph that shows the stone's height above the ground as she continues to ride at this speed for 2 s more. As she rides at a speed of
T
21. and draw the resulting curve. Maria's bicycle wheel has a diameter of 64 cm. Sketch the graph of a periodic function whose period is 4 and whose range is 11.
What is the period of this function. A spacecraft is in an elliptical orbit around Earth. above Earth's surface in terms of time is recorded in the table.
and other times water is added. the equation of the axis.
16. Include an example. Include a scale.5 s. The data are stored in the calculator.13. He holds the paddle there for 2 s and then. When the CBR is activated.
14. within 0. A Calculator-Based Ranger (CBR) is a motion detector that can attach to a
graphing calculator. Water is stored in a cylindrical container. moves the paddle so that it is 30 cm from the detector. d) How fast is the depth of the water increasing when the container is being filled? e) How fast is the depth of the water decreasing when the container is being drained? f ) Is the container ever empty? Explain. Write a definition of a periodic function.
70 60 50 40 30 20 10 0 Distance (cm)
1
2 3 Time (s)
4
356
Chapter 6
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. and the amplitude of the
function.
Time (min) Depth (cm) 0 10 1 20 2 30 3 40 4 40 5 40 6 7 8 20 9 30 10 40 11 40 12 40 13 25 14 10 15 20 16 30 17 40 18 40 19 40
25 10
a) Plot the data. it records the distance an object is in front of the detector in terms of time.
Extension
15. and time is the independent variable. Denis repeats this process three times.5 s. within 0. and draw the resulting curve. moves the paddle back to the 60 cm location. Describe the motion of the paddle in front of a CBR that would have
produced the graph shown. Sometimes water is removed from
the container. a) Draw a sketch of the resulting graph. Denis holds the paddle of the CBR at 60 cm for 3 s and then. Distance is the dependent variable. and use your
C
definition to explain why it is periodic. A scatter plot based on those recorded distances and times can then be drawn using the graphing calculator. The table records the depth of the water at specific times. b) Is the graph periodic? c) Determine the period. b) What is the period of the function? c) Determine the range and domain of the function.
the wheel rotates counterclockwise to power the generator.
C.
How can you describe the position of the nail using an equation?
Construct a scale model of the water wheel. Attach the cardboard wheel to the centre of the rectangular piece of cardboard with a thumbtack. On a rectangular piece of cardboard about 100 cm long and 30 cm wide. The height of the nail changes as the wheel rotates.
INVESTIGATE the Math
Paul uses a generator powered by a water wheel to produce electricity. draw a horizontal line to represent the water level and a vertical line both through the centre. As the current flows down the river. Draw a dot to represent the nail on the circumference of the circle at one of the lines you drew to divide the wheel.6. The wheel has a radius of 1 m.
30° initial position of nail
NEL
Sinusoidal Functions
357
.
nail height 1m angle of rotation
• • • • • •
cardboard ruler protractor metre stick thumbtack graphing calculator
current
?
A. with the rectangle behind the wheel. On a piece of cardboard. Half the water wheel is submerged below the surface of a river. Use a protractor to divide your cardboard wheel into 30° increments through the centre. Locate the centre of the circle.2
GOAL
Investigating the Properties of Sinusoidal Functions
YOU WILL NEED
Examine the two functions that are associated with all sinusoidal functions. A nail on the circumference of the wheel starts at water level. cut out a circle with a radius of 10 cm to represent the water wheel's 1 m radius. B.
Use your graphing calculator to determine the cosine and sine of each rotation angle.
Angle of Rotation. record the height as a negative value.
Use your data to graph height versus angle of rotation. u (º) Actual Distance from Vertical Line. and measure the distance the nail is from the vertical line. of the nail: the perpendicular distance from the nail to the horizontal line. Continue to rotate the wheel in 30° increments. Copy the table. If the nail goes below the horizontal line. Measure the height. Make sure your calculator is in DEGREE mode and evaluate to the nearest hundredth."
u cos u u sin u
0° 30° 60° 90° 120° 150° 180°
210°
240°
270°
300° 330° 360°
358
Chapter 6
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. Continue to rotate the wheel in 30° increments. and record the actual distance the nail is above the horizontal line at 30° by multiplying the scale height by 10 and converting to metres. replacing L2 with "cos(L1)" and L3 with "sin(L1). Start with the nail initially positioned at water level. u (º) Actual Height of Nail. If the nail goes to the left of the vertical line.
0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
Tech
Support
H. h. Copy the table. d (m) G.
d
Rotate the cardboard wheel 30° counterclockwise. h (m) 0 0 30 60 90 120 1 ••• 690 720
E.
1
h
Rotate the cardboard wheel 30° counterclockwise.
You can generate the tables using the List feature on your graphing calculator. Use your model of the water wheel to examine the horizontal distance. measuring h and recording the actual heights. and record the actual distances. Continue until the nail has rotated 720°. and record the actual distance the nail is from the vertical line at 30°. 0 0 30 60 90 1 120 ••• 690 720
Use your data to graph horizontal distance versus angle of rotation. F. Try putting degrees in L1. the nail is from a vertical line that passes through the centre of the water wheel. again adjusting for the scale factor.
Angle of Rotation. record the distance as a negative value. Continue until the nail has rotated 720°.D. d.
What transformation can you apply to the cosine curve that will result in the sine curve? What ordered pair could you use to represent the point on the wheel that corresponds to the nail's location in terms of u. B-14. see Technical Appendix. and compare these graphs to the graphs from parts E and G.
Support
Use your graphing calculator to graph y 5 sin x and y 5 cos x. If it does. where any portion of the wave can be horizontally translated onto another portion of the curve. domain. period. d(u) 5 sin u h(u) 5 sin u d(u) 5 0. select the appropriate equation that describes the height. h.
Based on the tables you created in parts D. equation of the axis. and range in your comparison. increasing intervals.5 u h(u) 5 0. the nail is from the vertical line in terms of the rotation.
K. decreasing intervals. State the coordinates of five key points that would allow you to draw the sinusoidal function y 5 sin x quickly over the interval 0° to 360°. the angle of rotation?
For help graphing trigonometric functions on your graphing calculator.6. of the nail on the water wheel in terms of the rotation. N. M. Use words such as amplitude.5 u u(d ) 5 sin d u(d ) 5 sin h d(u) 5 cos u h(u) 5 cos u
Tech
Reflecting
J.
sinusoidal function a periodic function whose graph looks like smooth symmetrical waves. d. determine whether it represents a sinusoidal function. graphs of sinusoidal functions can be created by transforming the graph of the function y 5 sin x or y 5 cos x
APPLY the Math
EXAMPLE
4 2 0 2
y
1
Identifying the function
Determine whether the graph represents a periodic function. L. where 0° # x # 360°. Also.2
I. a) c) f(x) f(x) 20 x 0 2 4 6 8 10 2 10 4
6 0 b) 2 0 2 4 f(x) x 2 4 6 8 10 d) 2 0 2 4 5 f(x) x 10 20 30 40 50 x 10 15 20 25
x 2 4 6 8 10
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Sinusoidal Functions
359
. F. and H. identify another equation that describes the distance. State the coordinates of five key points that would allow you to draw the sinusoidal function y 5 cos x quickly over the interval 0° to 360°.
a) Is the function periodic? If it is.
360
Chapter 6
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. where any portion of the wave can be horizontally translated onto another portion of the curve. The pattern repeats but the waves aren't symmetrical.Bridget's Solution
a) periodic and sinusoidal
The function repeats. the amplitude. Since it forms a series of identical.
b) periodic
c) neither periodic nor sinusoidal
d) periodic
EXAMPLE
2
Identifying the properties of a sinusoidal function
Graph the function f (x) 5 4 sin(3x) 1 2 on a graphing calculator using the WINDOW settings shown in DEGREE mode. The pattern repeats but the waves aren't smooth curves. I can't horizontally translate any portion of the wave onto another portion of the curve. so it's periodic.
Beth's Solution
a)
Because it repeats. It looks like smooth symmetrical waves. the equation of the axis. it is sinusoidal. however. and the range. It looks like smooth symmetrical waves. symmetrical smooth waves. determine the period. is it sinusoidal? b) From the graph. c) Calculate f (20°). the graph is periodic.
866) 1 2 5 5. To get the amplitude. 0) (3.2
b) period 5 120°
The graph completes three cycles in 360°. y) 70° x (0.6.
equation of the axis: y 5
22 1 6 2
range: 5 y [ R | 22 # y # 66 62254 amplitude 5 4
c)
y52
f (x) 5 4 sin(3x) 1 2 f (20°) 5 4 sin(3(20°) ) 1 2 5 4 sin(60°) 1 2 8 4(0. The axis is halfway between the minimum of 22 and the maximum of 6. the greatest y-value on the graph (the maximum) is 6. y) resulting from a rotation of 70° centred at the origin and starting from the point (3. which is the period. f(20°) means find y when x 5 20°. 0). 0)
Determine the coordinates of the point P(x. For the range. It's 4. I substituted 20 for x and then calculated y. and the least y-value (the minimum) is 22. I calculated the vertical distance between a maximum and the axis.
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. must be 120°. so one cycle.464
EXAMPLE
3
Determining the coordinates of a point from a rotation angle
y P(x.
and the minimum value is 21. • The zeros are located at 0°.
P(x. r sin u) and got the coordinates of the image point. The coordinates of the nail after a rotation of u were (cos u. I substituted the radius and angle of rotation into the ordered pair (r cos u. 2. y) 5 (r cos u. 180°.Anne's Solution
sin u 5 opp hyp
The water wheel solution was based on a circle of radius 1. y) 5 (3 cos 70°. • The domain is 5u [ R6. P(x. y) on a circle of radius r are P(x.
In Summary
Key Idea
• The function f(u) 5 sin u is a periodic function that represents the height (vertical distance) of a point from the x-axis as it rotates u ° about a circle with radius 1. 3 sin 70°) 8 (1.03.82)
This means that the coordinates of the new point after a rotation of u from the point (r. sin u). 0) about (0. • The amplitude is 1.
Need to Know
• The graph of f(u) 5 sin u has these characteristics: • The period is 360°.. 0) can be determined from (r cos u. 360°. r sin u). and the range is 21 # f(u) # 1. But this circle doesn't have a radius of 1. .
f( ) 1 0 1 90° 180° 270° 360° f( ) = sin
(continued)
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... r sin u). Its radius is 3. y) 5 (3 cos u. 3 sin u)
3 y 1 cos sin x
The coordinates for any point P(x. the maximum value is 1.
y sin u 5 r 3 x 5 1 cos u x 5 3 cos u and y 3 5 1 sin u y 5 3 sin u
I used similar triangles to figure out the coordinates of the larger triangle. • The function f(u) 5 cos u is a periodic function that represents the horizontal distance of a point from the y-axis as it rotates u ° about a circle with radius 1.
Use DEGREE mode. a) g(90°)
K
b) h(90°)
8. 450°. and equation of the axis for each.
PRACTISING
5.5x) 1 2 e) y 5 2 sin(0. From the graph. If so.005x 1 sin x f ) y 5 sin 90°
6. in 2. 0 90° 180° 270° 360° • The zeros are located at 90°.5x) 1 2
Sinusoidal Functions
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. 1 • The amplitude is 1. graph each
function. list the values of x where 0° # x # 360°. From the graph. decreasing intervals. a) y 5 3 sin(2x) 1 1 b) y 5 4 cos(0. does the displacement first reach 20. a) y 5 3 sin x 1 1 c) y 5 cos (2x) 2 sin x e) y 5 0. and the range is 21 # f(u) # 1. to the nearest second. 270°. c) What is the displacement at 35 s? d) At what time. b) Determine the period of the function from the graph. what can you conclude about any
function that possesses sine or cosine in its equation?
7.
6. period.• The graph of f(u) 5 cos u has these characteristics: f( ) • The period is 360°. in metres at t seconds. 1 f( ) = cos • The sine function and cosine function are congruent sinusoidal curves. . Determine the coordinates of the new point after a rotation of 50° about
(0.25x) b) y 5 3 sin x 1 1 d) y 5 sin(2x) 2 1 f ) y 5 3 sin(0. a) Graph the displacement from 0 s to 20 s. y) on a circle centred at (0. where 0° # x # 360°. the maximum value is 1.5x) 2 2
2.5 cos x 2 1 b) y 5 (0. a) y 5 2 sin x 1 3 c) y 5 sin(0. and equation of the axis for each. calculate each and
explain what it means. A buoy rises and falls as it rides the waves.8 m?
4.. increasing intervals. Using a graphing calculator and the WINDOW settings shown. 0). 0) with radius r and rotated through an angle u can be expressed as an ordered pair (r cos u.
Use the WINDOW settings shown. • The domain is 5u [ R6. Based on your observations in question 5. h(t). 0) from the point (2. state whether they are sinusoidal. and the minimum value is 21. calculate h(25°). • Any point P(x.
3. a) If h(x) 5 sin(5x) 2 1. the cosine curve is the sine curve translated 90° to the left. r sin u). Using a graphing calculator in DEGREE mode.. state the amplitude. state the amplitude.
b) If f (x) 5 cos x and f (x) 5 0.2
CHECK Your Understanding
1. If g(x) 5 sin x and h(x) 5 cos x. State whether the resulting functions are periodic. graph each sinusoidal function.004x)sin x d) y 5 0.5 s intervals. Using a graphing calculator in DEGREE mode. The equation h(t) 5 cos(36t)°
models the displacement of the buoy. Use the WINDOW settings shown.. period. graph each sinusoidal
function.
0). c) Determine the coordinates of the new point after a rotation of 120° about (0.
C
Extending
15. Jim is riding a Ferris wheel. 0) from the point (1. d) Determine the coordinates of the new point after a rotation of 230° about (0. calculate x for 0° # x # 360°.9. c) Explain why the function models periodic behaviour.
b) Determine the coordinates of the new point after a rotation of 80° about
(0.5 cos (120t)°. calculate x for 0° # x # 360°. 0). 0) from the point (4. If h(x) 5 cos(3x) 1 1. Explain what each of
A
the following represents. If g(x) 5 sin(2x). where t is time in seconds. Sketch the sinusoidal graphs that satisfy the properties in the table.
where d(t) is the displacement in centimetres from the rest position and t is time in seconds. d) What is the relationship between the amplitude of the function and the displacement of the spring from its rest position?
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Chapter 6
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. where h(t) 5 5 cos(18t)° b) h(10).
16. b) Draw the graph.
12. 0). and how are they different?
14. calculate g(10°). Use 0. where h(t) 5 5 sin(18t)° How are they the same. Equation of the Axis y55 y54 y 5 22
Period
Amplitude 3 6 5
Number of Cycles 2 3 2
a) b) c)
4 20 80
13. 11. If the water level in the original water wheel situation was lowered so that
three-quarters of the wheel was exposed. where 0° # x # 360°. determine the equation of the sinusoidal function that describes the height of the nail in terms of the rotation. If f (x) 5 cos x and f (x) 5 21. calculate h(20°). The model does not consider the effects of gravity. 0).5 s intervals. a) h(10). 0) from the point (5. a)
b) c) d) e)
T
If f (x) 5 cos x. 0) from the point (3. If f (x) 5 sin x and f (x) 5 21. Compare the graphs for y 5 sin x and y 5 cos x. Determine all values where sin x 5 cos x for 2360° # x # 360°. A spring bounces up and down according to the model d(t) 5 0.
10. a) Make a table for 0 # t # 9. calculate f (35°). a) Determine the coordinates of the new point after a rotation of 25° about
(0.
6 s to complete one revolution.
LEARN ABOUT the Math
Two students are riding their bikes. The diameter of the wheel on Bike B is 50 cm. is 0. The diameter of the wheel on Bike A is 60 cm.
Glen's Solution: Comparing Periods The wheel on Bike A takes 0.5 s to complete one revolution. A pebble is stuck in the tire of each bike. The peak for Bike A is at h 5 60. the pebble was initially at its highest height of 60 cm.
For Bike A. The period of Bike B is 0. the graph starts at a peak. I noticed that the peaks on the graph are different. so the period is 0. the pebble was initially at its lowest height of 0 cm. the graph starts at a trough. or length of one cycle. so the period.
The graph for Bike A completes 5 cycles in 3 s.6 s.6. h 5 0.6 s. The two graphs show the heights of the pebbles above the ground in terms of time. which is greater than the peak for Bike B. The wheels have different diameters. For Bike B. The graph for Bike B completes 2 cycles in 1 s.
h Height of a Pebble Bike A Bike B
60 50 40 30 20 10 0
Height (cm)
t 1 Time (s) 2 3
?
What information about the bikes can you gather from the graphs of these functions?
EXAMPLE
1
Connecting the graph of a sinusoidal function to the situation
Joanne's Solution: Comparing Peaks of a Sinusoidal Function For Bike A.5 s. The troughs.5 s. The period of Bike A is 0. are the same. The wheel on Bike B takes 0. which is at h 5 50. For Bike B. however.3
GOAL
Interpreting Sinusoidal Functions
Relate details of sinusoidal phenomena to their graphs.
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.
50 1 0 Bike B: 5 25 2 The equation of the axis for Bike A is h 5 30. Speed is equal to distance divided by time.14 m>s
sB 8 3. I added the maximum and the minimum and then divided by 2. so first I had to figure out how far each bike travels when the wheel completes one revolution. The equation of the axis for Bike B is h 5 25.14 m>s
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Chapter 6
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.885 0. This distance is the circumference. The axle for the wheel on Bike B is 25 cm above the ground.
Karen's Solution: Comparing Speeds Circumference: Bike A CA 5 2prA CA 5 2p(30) CA 5 60p CA 8 188. I calculated the two circumferences.571 m sB 5 sB 5 d t 1.1 cm CB 8 1.885 m sA 5 sA 5 d t 1.
sA 8 3.6 Bike B CB 5 2prB CB 5 2p(25) CB 5 50p CB 8 157. I divided each circumference by the time taken to complete one revolution.5
The bikes are travelling at the same speed. The axle for the wheel on Bike A is 30 cm above the ground.5 cm CA 8 1.571 0.Scott's Solution: Comparing Equations of the Axes in Sinusoidal Functions Bike A: 60 1 0 5 30 2
The axis is halfway between a peak (or maximum) and a trough (or minimum). To calculate the speed.
08 0.12 2 4 6 8 10 Time (s) Height of cutting tooth (inches) Height of cutting tooth (inches)
Table Saw B 2 h(t) t 0 0. In each case.08 0.10 0.6.10 0.3
Reflecting
A.04 0.12 2 4 6 8 10 Time (s)
What information about the table saws can Annette gather from the graphs?
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.02 0. the graphs show the height of one tooth on the circular blade relative to the cutting surface of the saw in terms of time.04 0.
APPLY the Math
EXAMPLE
2
Comparing graphs and situations
Annette's shop teacher was discussing table saws.02 0. how would the graph of the resulting sinusoidal function compare with that for Bike A and Bike B? What type of information can you learn by examining the graph modelling the height of a pebble stuck on a tire in terms of time?
C. B. Table Saw A Table Saw B
Table Saw A 2 h(t) t 0 0. The teacher produced two different graphs for two different types of saw.
How would changing the speed of the bike affect the sinusoidal graph? For a third rider travelling at the same speed but on a bike with a larger wheel than that on Bike A.06 0.06 0.
I got the amplitude by taking the difference between 2 and 23. For graph B.03 s to complete one revolution.03 s. On graph A.02 s.
On graph B.
The blade on Table Saw B takes 0.03 s. The axle for the blade on Table Saw A is 3 in. The radius of the circular cutting blade on Table Saw B is 6 in.02 s to complete one revolution. The equation of the axis for graph A is h 5 23.
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. below the cutting surface. the peak for B is at h 5 1. In both cases. and the next is at 0. The blade on Table Saw A takes 0.
The peaks on the graph are different. and the next is at 0. The peak for A is at h 5 2. The amplitude for graph A is 5.02 s. the distance from the axis to a peak represents the radius of the circular cutting blade.
The axle for the blade on Table Saw B is 5 in. The equation of the axis for graph B is h 5 25. For graph A. The amplitude for graph B is 6. For graph B. the first peak is at 0 s. the first peak is at 0 s. the amplitude is the difference between 1 and 25. The radius of the circular cutting blade on Table Saw A is 5 in. below the cutting surface. The period of graph B is 0. That gave me 25. That gave me 23. One of the easiest ways to find the period is to figure out how long it takes to go from one peak on the graph to the next.Repko's Solution The blade on Table Saw A is set higher than the blade on Table Saw B. For graph A. I found the equation of the axis by adding 2 and 28 and then dividing by 2. This means that the period of graph A is 0. I added 1 and 211 and then divided by 2.
a) Using graphing technology in DEGREE mode and the WINDOW settings shown.34 5 5.9 2 1. Karl's Solution
Job applications/ week (hundreds) a)
j(t) 12 8 4 0 5 10 Time (years) 15 t
I sketched all the cycles the window showed.
per week is 170.3
EXAMPLE
3
Using technology to understand a situation
The function j(t) 5 4. I looked for a trough on the graph and read the j-coordinate. I looked for the place where the t-coordinate was 10.
In Summary
Key Idea
• The sine and cosine functions can be used as models to solve problems that involve many types of repetitive motions and trends.56
The employment cycle is 5. b) How long is the employment cycle? Explain how you know.7t)° 1 5. • One cycle of motion corresponds to one period of the sine function.1 sin(64. graph the function and then sketch the graph.8. and explain what it represents in terms of the situation.
Need to Know
• If a situation can be described by a sinusoidal function. models demand for employment in a particular city. the graph of the data should form a series of symmetrical waves that repeat at regular intervals. the distance between peaks or troughs. • The distance of a circular path is calculated from the circumference of the path. I calculated the x-interval between the first and second peak. the time to complete one rotation.
c) The minimum number of applications
To calculate the cycle.
b) 6. c) What is the minimum number of applications per week in this city? d) Calculate j(10).
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369
.56 years.
d) j(10) 5 1. where t is time in years since May 1992 and j(t) is the number of applications for jobs each week (in hundreds).88
There were 188 applications in May 2002.6. The amplitude of the sine or cosine function depends on the situation being modelled. The speed of an object following a circular path can be calculated by dividing the distance by the period.
Olivia's Motion Distance (m) 14 10 6 2 0 t 4 8 Time (s) 12 16 d(t)
a) What is the equation of the axis.
Height of paddle (m) 7 6 5 4 3 2 1 0 1 y Paddle Wheeler A Paddle Wheeler B
x 24 Time (s) 48
3. Olivia was swinging back and forth in front of a motion detector when the
detector was activated.
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Chapter 6
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. Refer to the radius of each wheel. the height of the axle relative to the water. and explain how the wheels differ. Her distance from the detector in terms of time can be modelled by the graph shown. and the speed of each wheel. Marianna collected some data on two paddle wheels on two different boats
and constructed two graphs.CHECK Your Understanding
1. and what does it represent in this situation? How close did Olivia get to the motion detector? At t 5 7 s. and what does it represent in this b) c) d) e) f)
situation? What is the amplitude of this function? What is the period of this function. the time taken to complete one revolution.) Would the resulting graph be sinusoidal? Why or why not?
2. Analyze the graphs. If the motion detector was activated as soon as Olivia started to swing from at rest. Draw two sinusoidal functions that have the same period and axes but have
different amplitudes. how would the graph change? (You may draw a diagram or a sketch. would it be safe to run between Olivia and the motion detector? Explain your reasoning.
with waves 0.6
6.)
258 12.6.8 135 13. a) What is the period of the function? Include the units of measure. and what does it represent in this situation? d) How fast is a tooth on the circular cutting blade travelling in inches per second? curve on its display. An oscilloscope hooked up to an alternating current (AC) circuit shows a sine
Height (in.2 349 8.
Timmins.5 y 3. and what does it represent in this situation? c) What is the amplitude of the function. Evan's teacher gave him a graph to help him understand the speed at which a
K
tooth on a saw blade travels.4 288 10. located at a latitude of 25°.5
Current (amperes)
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371
. The graph shows the height of one tooth on the circular blade relative to the cutting surface relative to time.9 319 9. b) What is the equation of the axis of the function? Include the units of measure.3
PRACTISING
4.7 46 11.04 0. located
Oscilloscope Display 4.
Height of a Saw Tooth 2 h(t) t 0 0. whose axle is 8 m above the ground.5 m in height that occur at 7 s intervals
7.5 Time (s)
at a latitude of 48°. Ontario. Florida.3 74 12. at latitude 488
Day of Year Hours of Daylight 15 8. at latitude 258
Day of Year Hours of Daylight 15 10. c) What is the amplitude of the function? Include the units of measure.9 349 10. and that rotates once every 15 s c) A bicycle tire with a radius of 40 cm and that rotates once every 2 s d) A girl lying on an air mattress in a wave pool that is 3 m deep. Draw at least three cycles.6 319 10.0 1.8 196 13.5 x 0 0. Sketch a height-versus-time graph of the sinusoidal function that models each
situation.0 4.0 105 12.6 227 13. and Miami. a) How high above the cutting surface is the blade set? b) What is the period of the function. and that rotates once every 40 s b) A water wheel with a radius of 3 m. The tables show the varying length of daylight for Timmins.7 135 15.8 46 10.1 196 15.08 1.3 288 11.9 105 13.5 3.12 2 4 6 8 10 Time (s)
5.6 165 13.04 0.1 258 12. a) A Ferris wheel with a radius of 7 m.3
Miami.08 0. whose centre is at water level. The length of the day is calculated as the interval between sunrise and sunset.7 227 14.2 74 11.2 165 16. Assume that the first point plotted on each graph is at the lowest possible height.
d) How far does the buoy drop from its highest point to its lowest point? Explain how you know.a) Plot the data on separate coordinate systems. graph h(t) and sketch the graph. b) How high above the ground will the nail be after the car has travelled 0. amplitude. c) What is the radius of the wheel? Explain how you know. and equation of the axis?
9. of a basket on a water wheel at time t can be modelled by
h(t) 5 2 sin(12t) 1 1. If the wind speed decreases by 20 km/h. c) How many waves will cause the buoy to rise and fall in 1 min? Explain how you know. a) Draw a graph of the height of the nail above the ground in terms of the distance the car has travelled since the tire picked up the nail. Using graphing technology in DEGREE mode and the WINDOW settings shown. The distance the
10.5 sin(72t)° models the displacement of a buoy in
A
metres at t seconds. where t is in seconds and h(t) is in metres. While the car is being driven. The height. amplitudes.
b) Compare the two curves. a) Using graphing technology in DEGREE mode and the WINDOW settings shown. a) Using graphing technology in DEGREE mode and the WINDOW settings shown.5°. b) How long does it take for the wheel to make a complete revolution? Explain how you know.
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Chapter 6
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. and explain what it represents in terms of the situation.
c) What might you infer about the relationship between hours of daylight
and the latitude at which you live?
8. produce the two graphs. e) Calculate h(10). The diameter of a car's tire is 52 cm. h(t) . d) Where is the centre of the wheel located in terms of the water level? Explain how you know. How does the reduced wind speed affect the period. the top of a signpost vibrates back and forth. The equation h(t) 5 2. and draw a smooth curve
through each set of points. b) How long does it take for the buoy to travel from the peak of a wave to the next peak? Explain how you know.
11. Refer to the periods. graph h(t) and sketch the graph. where d(t) represents the distance in centimetres at time t seconds.1 km? c) How far will the car have travelled when the nail reaches a height of 20 cm above the ground for the fifth time? d) What assumption must you make concerning the driver's habits for the function to give an accurate height? tip of the post vibrates to the left and right of its resting position can be defined by the function d(t) 5 3 sin(1080t)°. and equations
of the axes. the vibration of the tip can be modelled by the function d(t) 5 2 sin(1080t)°. the tire
T
picks up a nail. In high winds.
e) Compare the motions of the two wrecking balls. d.2))° 1 5. The average monthly temperature. a) Using graphing technology in DEGREE mode and the WINDOW settings shown. What pieces of information could they be?
Extending
15. a) In which direction is the larger gear turning? b) If the period of the smaller gear is 2 s. A gear of radius 1 m turns counterclockwise and drives a larger gear of radius
4 m. record convenient intervals for each gear.3
12. can be modelled by the function T(t) 5 14.
4 Distance (m) 2 0 2 4 Time (s) a) What is the period of each function. and what does it represent in
this situation?
d) Determine the range of each function. d) What is the displacement of the point on the large wheel when the drive wheel first has a displacement of 20. Both gears have their axes along the horizontal. what is the period of the larger gear? c) In a table. For t 5 1. the month is January. b) What does the period represent in this situation? c) What is the average temperature range in Kingston? d) What is the mean temperature in Kingston? e) Calculate T(30). and explain what it represents in terms of the situation.
13.5 m? e) What is the displacement of the drive wheel when the large wheel first has a displacement of 2 m? f ) What is the displacement of the point on the large wheel at 5 min?
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. How many pieces of information do you need to know to sketch a sinusoidal
C
function.6.
14. and what does it represent in this 8 16 d(t) Ball A Ball B t 24
situation?
b) What is the equation of the axis of each function.
Ontario. and what does it
represent in this situation?
c) What is the amplitude of each function.2 sin(30(t 2 4. where t represents the number of months. Graph vertical displacement versus time. the month is February. Begin the table at 0 s and end it at 24 s. of the point where the two gears first touched. and so on. to show the vertical displacement.9. graph T(t) and sketch the graph. T(t). Two wrecking balls attached to different cranes swing back and forth. The
distance the balls move to the left and the right of their resting positions in terms of time can be modelled by the graphs shown. for t 5 2. in degrees Celsius in Kingston.
the equation of the axis is y 5 0.
Examples 1 and 2. (A cycle of a sinusoidal function is a portion of the graph that repeats. It can be determined with the formula
y5 (maximum value 1 minimum value) 2
Amplitude The amplitude is the vertical distance from the function's axis to the minimum or maximum value.
Period The period is the change in x corresponding to one cycle.
Equation of the Axis The equation of the axis is the equation of the line halfway between the maximum and minimum values on a sinusoidal function.6
Study
Mid-Chapter Review
FREQUENTLY ASKED Questions
Q:
A:
Aid
• See Lesson 6. as well as any periodic function. Unlike other periodic functions.
EXAMPLE
2 1 0 1 2
y
f(x)
sin x amplitude x
axis
90°
180° 270° 360° period
For the function f (x) 5 sin x.2.
What are sinusoidal functions. sinusoidal functions form smooth symmetrical waves such that any portion of a wave can be horizontally translated onto another portion of the curve. It is always positive. Sinusoidal functions are formed from transformations of the functions y 5 sin x and y 5 cos x . the period is 360°. and what characteristics are often used to describe them?
Sinusoidal functions. • Try Mid-Chapter Review Question 3. and the amplitude is 1. and the amplitude.) One way to determine the period is to look at the change in x between two maxima. The three characteristics of a sinusoidal function. like other periodic functions.
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Chapter 6
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. repeat at regular intervals. the equation of the axis. are the period.
A maximum for y 5 cos x occurs at 0° and at increments of 360° from that point. A minimum for y 5 cos x occurs at 180° and at increments of 360° from that point.3.
The graph of the function y 5 sin x can be changed to a graph of the function y 5 cos x by applying a horizontal translation of 90° to the left. • the motion of objects in a circular orbit • the motion of swinging objects.
• Try Mid-Chapter Review
Questions 5 and 6. The amplitude 5 1. The graph of the function y 5 cos x can be changed to a graph of the function y 5 sin x by applying a horizontal translation of 90° to the right. Q: A:
Why might it be useful to learn about sinusoidal functions? Study
Many real-world phenomena that have a regular repeating pattern can be modelled with sinusoidal functions.
The range is 5 y [ R |21 # y # 16.
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375
.Mid-Chapter Review
Q:
How do the graphs of y 5 sin x and y 5 cos x compare? 2 1 0 1 2 180° y = cos x y y = sin x x 360°
A:
Similarities
Differences
A maximum for y 5 sin x occurs at 90° and at increments of 360° from that point. Examples 1. The period is 360°. A minimum for y 5 sin x occurs at 270° and at increments of 360° from that point. and 3. For example. such as a pendulum • the number of hours of sunlight for a particular latitude • the phase of the Moon • the current for an AC circuit
Aid
• See Lesson 6. The equation of the axis is y 5 0.
2.
in centimetres per second. One mark is made on the
376
Chapter 6
Height of mark (cm) b) c) d) e) f)
a) What is the period of each function. d) Determine the values of x.
and what does it represent in this situation?
d) What is the amplitude of the function.3
5.50 0. the position in terms of the day of the year can be modelled by the function 360 P(d ) 5 28 sin Q d 2 81 R°. how would the graph of this function compare with the other two graphs?
a) Graph the function using a graphing calculator
and adjust the WINDOW settings as required. The position. psi) in the tank of an air compressor at different times. Adjust the WINDOW settings so that 0° # x # 360° and 0 # g(x) # 15.
6. a)
Graph the function g(x) 5 5 cos(2x) 1 7 using a graphing calculator. d. depends on the latitude and the day of the year. equation of the axis. and what
does it represent in this situation?
c) What is the equation of the axis of the function.
60 40 20 0 0. Determine the amplitude. Sketch the graph of a periodic function whose period 2. The two graphs show the relationship between the heights of the white marks above the ground in terms of time as the car moves forward. c) Calculate g(125). Determine the speed of each mark.
Lesson 6. If a third mark were placed on the tire but closer to the centre. amplitude. of the Sun at sunset. and what
does it represent in this situation?
e) Determine the range of the function. 0) from the point (7. and what
does it represent in this situation? What is the equation of the axis of each function. Determine the period. P(d ) .
north or south of due west. for which g(x) 5 12. and range of the function.
square inch. and what does it represent in this situation? Determine the range of each function. b) Explain why the function is sinusoidal. 0).PRACTICE Questions
Lesson 6. in degrees
Lesson 6. 0° # x # 360°.2
3. For a specific latitude.25 0.1
1. Determine the coordinates of the new point after a
rotation of 64° about (0. How do you know that the graph is periodic? Determine the period of the function.00 h(t) Mark 1 Mark 2
Pressure 60 (psi) Time (s) 10
80 100 100 90 80 70 60
11
12
13 90
14 15 16 17 18
19
Pressure 80 100 100 (psi)
80 70 60 60 80 100
a) Create a scatter plot of the data and the curve b) c) d) e) f) g) h)
that best models the data. the other mark is made a few centimetres from the edge.
365
4.75 Time (s) t 1. The following data show the pressure (in pounds per
is 10 and whose range is 5 y [ R | 4 # y # 106.
b) What is the period of the function. and what does it represent in this situation? What is the amplitude of each function. Determine the equation of the axis. f ) What is the angle of sunset on February 15?
NEL
. Two white marks are made on a car tire by a
parking meter inspector.
Time (s) 0 1 60 2 3 4 5 6 7 8 9 60
outer edge of the tire. How fast is the air pressure increasing when the compressor is on? How fast is the air pressure decreasing when the equipment is in operation? Is the container ever empty? Explain.
amplitude.
Part 2 The graphs of y 5 sin x 1 c and y 5 cos x 1 c
F. 0° # x # 720°. and k affect the graphs of f(x) 5 a sin(k(x 2 d) ) 1 c and f(x) 5 a cos(k(x 2 d) ) 1 c
• graphing calculator
EXPLORE the Math
2 1 0 1 2
y y = sin x x 180° y = cos x 360°
They want to know if these same transformations can be applied to y 5 sin x and y 5 cos x. c. Repeat parts A to C using y 5 a cos x. will look like for a 5 1.4
GOAL
Exploring Transformations of Sinusoidal Functions
YOU WILL NEED
Paula and Marcus know how various transformations affect several types of 1 functions.
Predict what the graphs of y 5 a sin x.
Tech
Support
B. and 2. and 2. Sketch the graphs on the same axes. and graph using ZoomFit by pressing
ZOOM
0
. and maximum or minimum values change for each function. and then verify your predictions using a graphing calculator.
NEL
377
. Explain how the value of a affects the graphs of y 5 a sin x and y 5 a cos x. 21. for a 5 21. and f (x) 5 | x |. and do they have the same effect on the graph and the equation?
Part 1 The graphs of y 5 a sin x and y 5 a cos x
A. On a new set of axes. how the equations and graphs of these functions change.
? Can transformations be applied to sinusoidal functions in the same manner. 22. 1. E. C. 21.6. 2. 0° # x # 720°. Y3. Predict what the graphs of y 5 cos x 1 c.
Sinusoidal Functions
G. D. 0° # x # 720°. f (x) 5 x . and so on. will look like for c 5 22.
Predict what the graphs of y 5 sin x 1 c. verify your sketches by graphing the parent function (y = sin x or y = cos x) in Y1 and each transformed function in Y2. f (x) 5 !x. Sketch the graphs on the same axes.
For Parts 1 and 2. Use an Xscl = 90°.
H. repeat part A for the graphs of y 5 a sin x. and 23. How do the graphs in part A compare with those in part B? Discuss how the zeros. 2 4 Verify your sketches using a graphing calculator. Explain how the value of c affects the graphs of y 5 a sin x 1 c and y 5 a cos x 1 c. and then verify your predictions using a graphing calculator.
Determine how changing the values of a. d. 1. will look like for c 5 22. Discuss which features of the graph have changed. 0° # x # 720°. and if so. such as f (x) 5 x 2. and 3 and for a 5 1 and a 5 1. Discuss which features of the graph have changed. Sketch the graphs on the same axes.
O. K.
Tech
Support
Reflecting
Q. T. How could you determine the period of y 5 sin kx and y 5 cos kx knowing that the period of both functions is 360°? Explain how the value of k affects y 5 sin kx and y 5 cos kx. S.
P.Tech
Support
Part 3 The graphs of y 5 sin kx and y 5 cos kx
I.
a) Predict the effect of d on the graph of y 5 sin(x 2 d ). y
i)
x 45 45 135 225 315
y
ii)
x 120 210 300 390 480
a) Predict the effect of d on the graph of y 5 cos(x 2 d ). R. Repeat parts I and J using y 5 cos kx. and discuss which features of the graph have changed.
For Part 4. M. from the graphing calculator before entering another equation. and 4. b) Copy and complete the tables of values at the left.
Explain how the value of d affects the graphs of y 5 sin(x 2 d ) and y 5 cos(x 2 d ).
For Part 3.
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378
Chapter 6
. Repeat part I for k 5 1. i) y 5 sin(x 2 60°) ii) y 5 sin(x 1 120°) c) Use your tables to sketch the graphs of the two sinusoidal functions from
part (b) on the same coordinate system. Clear the previous equation.
J. U. 3. L.
i)
x 60 150 240 330 420
y
ii)
x 120 30 60 150 240
y
Part 4 The graphs of y 5 sin(x 2 d) and y 5 cos(x 2 d)
N.
Predict what the graphs of y 5 sin kx will look like for k 5 2. k 5 1. Verify your sketches with a graphing calculator. verify your sketches by graphing the parent function in Y1 and each transformed function in Y2. Sketch each graph. 0° # x # 720° . Use an Xscl 5 90º and graph using ZoomFit. Include the graph of y 5 sin x.
What transformation affects the period of a sinusoidal function? What transformation affects the equation of the axis of a sinusoidal function? What transformation affects the amplitude of a sinusoidal function? What transformations affect the location of the maximum and minimum values of the sinusoidal function? Summarize how the graphs of y 5 a sin (k(x 2 d) ) 1 c and y 5 a cos(k(x 2 d) ) 1 c compare with the graphs of y 5 sin x and y 5 cos x. and k 5 21. Adjust the WINDOW on the 2 4 graphing calculator so that you can see one complete cycle of each graph. and discuss which features of the graph have changed. verify your sketches using a domain of 0° # x # 360° and an Xscl = 30°. i) y 5 cos(x 1 45°) ii) y 5 cos(x 2 120°) c) Use your tables to sketch the graphs of the two sinusoidal functions from
part (b) on the same coordinate system. and then verify your predictions using a graphing calculator. Graph using ZoomFit. Verify your sketches using a graphing calculator. Include the graph of y 5 cos x. Discuss which features of the graph have changed. b) Copy and complete the tables of values at the left. but not the base equation.
• The values a. domain. amplitude. but has no effect on the amplitude. amplitude. or domain. amplitude. domain. amplitude. Which two of these transformations do not affect the period. and range unless the situation forces a change in the domain or range. a) y 5 3 cos x c) y 5 2cos x b) y 5 sin(x 2 50°) d) y 5 sin(5x)
e) y 5 cos x 2 6 f ) y 5 cos(x 1 20°)
2. • Changing the value of d results in a horizontal translation and slides the graph to the left or right but has no effect on the period. • Changing the value of k results in a horizontal stretch or compression and affects the period. State the transformation to the graph of either y 5 sin x or y 5 cos x that has
occurred to result in each sinusoidal function. In each case. and f(x) 5 |x|. the maximum and minimum values.25 cos x b) y 5 4 sin x d) y 5 sin(2x 1 30°) f ) y 5 sin(0. maximum and minimum values. or equation of the axis. equation of the axis.5x)
3. If k is negative. and d in the functions f(x) 5 a sin(k(x 2 d ) ) 1 c and f(x) 5 a cos(k(x 2 d ) ) 1 c affect the graphs of y 5 sin x and y 5 cos x in the same way that they affect the graphs of y 5 f (k(x 2 d) ) 1 c. a reflection in the x-axis also occurs. where 1 f(x) 5 x 2 .
360°
FURTHER Your Understanding
1. Each sinusoidal function below has undergone one transformation that has
affected either the period. a) y 5 sin x 1 2 c) y 5 cos(8x) e) y 5 0. or
equation of the axis of a sinusoidal function? a) reflection in the x-axis d) horizontal stretch/ b) vertical stretch/vertical compression horizontal compression c) vertical translation e) horizontal translation
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Sinusoidal Functions
379
. and range of the function but has no effect on the period or domain. • Changing the value of a results in a vertical stretch or compression and affects the maximum and minimum values. and the range of the function but has no effect on the period. f(x) 5 x . determine which characteristic has been changed and indicate its value. f(x) 5 ! x . c. equation of the axis. The differences are only in the placement of the graph and how stretched or compressed it is. If a is negative. a reflection in the y-axis also occurs. amplitude.6. or range unless the situation forces a change in the domain or range. changing it to |k| .4
In Summary
Key Ideas
• The graphs of the functions f(x) 5 a sin(k(x 2 d ) ) 1 c and f(x) 5 a cos(k(x 2 d) ) 1 c are periodic in the same way that the graphs of f(x) 5 sin x and f(x) 5 cos x are. k.
Need to Know
• Changing the value of c results in a vertical translation and affects the equation of the axis.
5
YOU WILL NEED
Using Transformations to Sketch the Graphs of Sinusoidal Functions
GOAL
• graph paper
Sketch the graphs of sinusoidal functions using transformations. I graphed y 5 3 sin(2(x 2 60°) ) 1 4 (in black) by
x 60° 120° 180° 240° 300° 360°
applying a vertical translation of 4 to y 5 3 sin(2(x 2 60°) ) .
4
380
Chapter 6
NEL
. This means that the whole graph has slid up 4 units and that the equation of the axis is now y 5 4 because 4 has been added to all the y-coordinates of the points on the previous graph.
Glen's Solution
y Graph A
1 I started by graphing y 5 sin x (in green).
1
1
60° 120° 180° 240° 300° 360°
0 1
2
3 Then I graphed y 5 sin(2(x 2 60°) ) (in blue) by applying 3
a horizontal translation of y 5 sin (2x) 60° to the right because 60° has been added to all the x-coordinates of the points on the previous graph. 2 Then I graphed y 5 sin (2x) (in red). It has a horizontal
x
compression of 1. I graphed y 5 3 sin(2(x 2 60°) ) (in purple) by
applying a vertical stretch of 3 to y 5 sin(2(x 2 60°) ) .
? How can you graph sinusoidal functions using transformations?
EXAMPLE
1
Using transformations to sketch the graph of a sinusoidal function
Sketch the graph of f (x) 5 3 sin(2(x 2 60°)) 1 4. The amplitude is now 3 because all the y-coordinates of the points on the previous graph have been multiplied by 3.6.
LEARN ABOUT the Math
Glen has been asked to graph the sinusoidal function f (x) 5 3 sin(2(x 2 60°) ) 1 4 without using technology.
7 5 3 1 10 3
y
Graph B
4 Next.
5
5 Finally. so the period is 360° 5 180° instead of 2 2 360° because all the x-coordinates of the points on the graph of y 5 sin x have been divided by 2.
The period is 120°.
In what order were the transformations applied to the function y 5 sin x? If the equation of the function y 5 3 sin(2(x 2 60°) ) 1 4 were changed to y 5 3 sin(2(x 2 60°) ) 2 5. period. E. B. I did the vertical
translation. I graphed y 5 22 cos(3x) (in blue) starting at its lowest value due to the reflection. F. how would the graph of the function change? Which transformations affect the range of the function? How? Which transformations affect the period of the function? How? Could Glen graph this function faster by combining transformations? If so. changing its amplitude to 2 due to the vertical stretch.
Steven's Solution
a) 3 2 1 10 2 3 y
3
2
1
1 I started by graphing y 5 cos x
x
(in green). and range of
phase shift the horizontal translation of a sinusoidal function
this sinusoidal function.
60° 120° 180° 240° 300° 360°
2 I dealt with the horizontal
compression first. which ones?
C. phase shift. The phase shift is 0. The equation of the axis is y 5 21. b) State the amplitude.
4
3 I dealt with the vertical stretch
and the reflection in the x-axis.
The range is 5 y [ R |23 # y # 16. I graphed y 5 cos(3x) (in red) using a period of 360° 5 120° instead of 3 360°. I graphed y 5 22 cos(3x) 2 1 (in black) by sliding the previous graph down 1 unit. so the equation of the axis is y 5 21. how would the graph of the function change? How would it stay the same? If the equation of the function y 5 3 sin(2(x 2 60°) ) 1 4 were changed to y 5 3 sin(9(x 2 60°) ) 1 4.5
Reflecting
A.6.
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. D. equation of the axis.
4 Finally.
APPLY the Math
EXAMPLE
2
Connecting transformations to the graph of a sinusoidal function
a) Graph y 5 22 cos(3x) 2 1 using transformations.
b) The amplitude is 2.
(360°. following the order of operations (multiplication and division before addition and subtraction) for all vertical transformations and for all horizontal transformations.
John's Solution
y 5 2sin(0. you need to apply the transformations to the key points of f(x) 5 sin x or f(x) 5 cos x only. respectively.5
3 I applied the phase shift and graphed
y 5 2sin(0. • Key points for f(x) 5 sin x (0°.
1 I started by graphing y 5 sin x (in green). I
dealt with all stretches/compressions and reflections at the same time. 21). 0). The horizontal and vertical transformations can be completed in either order.5(x 1 90°) ) (in black) by shifting all the points on the previous graph 90° to the left. 1) (continued)
382
Chapter 6
NEL
. one at a time.5 1 90° 270° 450° 630° y
1
2
x
2 Rather than graph this one transformation at a time. 1).
3
In Summary
Key Idea
• Functions of the form g(x) 5 a sin(k(x 2 d) ) 1 c and h(x) 5 a cos(k(x 2 d) ) 1 c can be graphed by applying the appropriate transformations to the graphs of f(x) 5 sin x and f(x) 5 cos x. (270°. 0) • Key points for f(x) 5 cos x (0°.5x) (in red) by using a period of 360° 5 720° and reflecting this across the x-axis. (90°. 0).
1 0.5(x 1 90°)4
I factored the expression inside the brackets so that I could see all the transformations. 0). • As with other functions. 1). (360°. 0. I graphed y 5 2sin(0. (180°.5 0 0. you can apply all stretches/compressions and reflections together followed by all translations to graph the transformed function more efficiently.
Need to Know
• To graph g(x) .5x 1 45°) y 5 2sin30. not to every point on f(x) .5 from 0. (180°.5x 1 45°) using transformations.
EXAMPLE
3
Using a factoring strategy to determine the transformations
Graph y 5 2sin(0. (270°.5x and 45. 0). I divided out the common factor 0.You can graph sinusoidal functions more efficiently if you combine and use several transformations at the same time. (90°. 21).
•
CHECK Your Understanding
sinusoidal function. 0. 1.6. and range. State the transformations. • Reflect the graph in the x-axis if a . Use transformations to predict what the graph of
g(x) 5 5 sin(2(x 2 30°) ) 1 4 will look like.
3. amplitude. you end up with a function with • an amplitude of |a| 360° • a period of |k| • an equation of the axis y 5 c • Horizontal and vertical translations of sine and cosine functions can be summarized as follows: Horizontal • Move the graph d units to the right when d .25 cos(x 2 20°) c) g(x) 5 2sin(0. If the function f (x) 5 4 cos 3x 1 6 starts at x 5 0 and completes two
full cycles. 0. |a| . a) f (x) 5 sin(4x) 1 2
b) y 5 0. 1. for each
d) y 5 12 cos(18x) 1 3 1 e) f (x) 5 220 sin c (x 2 40°) d 3
2. 1. 0. • Horizontal and vertical stretches of sine and cosine functions can be summarized as follows: Horizontal 1 • Compress the graph by a factor ` ` when |k| . 0. k 1 • Stretch the graph by a factor ` ` when 0 . • Move the graph |d| units to the left when d . k • Reflect the graph in the y-axis if k . |k| .5x)
1. 0.5
By doing so. domain. determine the period.
NEL
Sinusoidal Functions
383
. Vertical • Move the graph |c| units down when c . • Move the graph c units up when c . • Compress the graph by a factor |a| when 0 . in the order you would apply them. equation of the axis. 0. Verify with a graphing calculator. 1. Vertical • Stretch the graph by a factor |a| when |a| .
b) Determine the equation of the cosine function that results in the same graph as your function in part (a). a)
T
Extending
12. D(t). the function P(t) 5 220 cos(300t)° 1 100 models the blood pressure. The function D(t) 5 4 sin3 365 (t 2 80)4° 1 12 is a model of the number
360
of hours. in millimetres of mercury at time t seconds. But there is a range of blood
A
pressure values that is considered healthy.
Determine the equation of a sine function that would have the range 5 yPR |21 # y # 76 and a period of 720°. For a person at rest. what transformation would map the resulting sine curve onto the resulting cosine curve?
13. a) Explain why a trigonometric function is a reasonable model for predicting the number of hours of daylight. Each person's blood pressure is different.5.5
9. If the functions y 5 sin x and y 5 cos x are subjected to a horizontal
compression of 0. of daylight on a specific day.
10. P(t). at latitude 50° north. a) What is the period of the function? What does the period represent for an individual? b) What is the range of the function? Explain the meaning of the range in terms of a person's blood pressure. t. b) How many hours of daylight do March 21 and September 21 have? What is the significance of each of these days? c) How many hours of daylight do June 21 and December 21 have? What is the significance of each of these days? d) Explain what the number 12 represents in the model. Explain how you would graph the function f (x) 5 2 cos(120x) 1 30 2 C using transformations. 1 11.
NEL
Sinusoidal Functions
385
.6.
equation of the axis. so |k| 5 k
386
Chapter 6
NEL
.
Sasha's Solution
Horizontal compression factor: k period 5 360 |k|
I calculated the period.
Height of a Nail 3 h(t) Height (cm) 2 1 0 1 10 20 Time (s) ? How can you determine the equation of a sinusoidal function from its graph? 30 t
EXAMPLE
1
Representing a sinusoidal graph using the equation of a function
Determine an equation of the given graph. Then I figured out how they are related to different transformations. The horizontal stretch or compression factor k had to be positive because the graph was not reflected in the y-axis. The period is 10 s since the peaks are 10 units apart. The height of the nail in terms of time can be modelled by the graph shown.6
Investigating Models of Sinusoidal Functions
GOAL
Determine the equation of a sinusoidal function from a graph or a table of values. I used the formula relating k to the period.
LEARN ABOUT the Math
A nail located on the circumference of a water wheel is moving as the current pushes on the wheel. k .
The period is 10 s. 0. and amplitude.6.
3. so d 5 0.6.6
10 5 k5 360 k 360 10
k 5 36 The graph was compressed by a 1 factor of 36.
Tanya says that another possible equation of the sinusoidal function created by Sasha is y 5 2 cos(36(x 2 10))° 1 1.5. On this graph. and the amplitude of this graph is 2. Vertical translation: c equation of the axis 5 5 max 1 min 2 3 1 (21) 2
I calculated the amplitude by taking the maximum value. just as this graph does. what would be the equation of the resulting function?
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Sinusoidal Functions
387
. 1. the domain is restricted to x $ 0 because it represents the time elapsed. there is a horizontal translation of 7. This means that for a cosine curve. there isn't any horizontal translation. A sine curve increases from a y-value of 0 at x 5 0. so c 5 7. Is she correct? Why or why not? If the period on the original water wheel graph is changed from 10 to 20. that happens at 7.5. D. I found the equation of the function by substituting the values I calculated into f(x) 5 a cos(k(x 2 d)) 1 c. and subtracting the axis. C. The cosine curve is easier to use for my equation since the graph has its maximum on the y-axis.
5 1 (vertical translation) c51 Vertical stretch: a a 52 Base graph: y 5 cos x As a cosine curve: y 5 2 cos(36x)° 1 1 As a sine curve: y 5 2 sin(36(x 2 7. Since the amplitude of y 5 cos x is 1. 3. what would be the new equation of the sinusoidal function? If the speed of the current increases so that the water wheel spins twice as fast. I could have used the sine function instead. This gave me the vertical translation and the value of c. 21.
Reflecting
A. the vertical stretch is 2.5. and the minimum. what would be the new equation of the sinusoidal function? If the maximum value on the original water wheel graph is changed from 3 to 5. B. for a sine curve. The axis is halfway between the maximum.5) ) ° 1 1 For both functions. This means that.
Since there was no horizontal reflection. unless they've been translated horizontally. The equation of the axis of this cosine curve is y 5 1. and a maximum at (0. A minimum would be halfway between the two maximums. 3). Represent the function with an equation in two different ways. 1) corresponds to the start of the cycle of the sine function. k .
388
Chapter 6
NEL
. so there has been a horizontal compression. I found the equation of the function by substituting the values into f(x) 5 a cos(k(x 2 d )) 1 c. The sine curve with the same period and axis as this cosine curve has the equation y 5 2 sin(2x) 1 1. a period of 180°. 3) and a period of 180°. Since the amplitude is 2. The equation of the axis gave me the vertical translation. so the next maximum would be at (180. but its starting point is (0°. the point (135°. 2 For a cosine curve: No horizontal translation so d50 Equation: y 5 2 cos(2x) 1 1 For a sine curve: horizontal translation 5 135° y 5 2 sin(2(x 2 135°)) 1 1
Cosine curves have a maximum at x 5 0. the minimum would have to be at (90°. The amplitude gives me the vertical stretch.
k52 Compression factor is 1 . so d 5 135°.
Rajiv's Solution
3 2 1 0 1 45° 90° 135° 180° x y
The graph has a maximum at (0. 21 ). and 2 2 3 5 21. 3). Since the equation is y 5 1 instead of y 5 0. This curve starts at its maximum. This means the function y 5 2 sin(2x) 1 1 must be translated horizontally to the right by 135°. I took 360° and divided it by the period.APPLY the Math
EXAMPLE
2
Connecting the equation of a sinusoidal function to its features
A sinusoidal function has an amplitude of 2 units. 0. On this cosine curve. so there would be no horizontal translation with a cosine function as a model. To find k. 1) .
Vertical translation: c 5 1 Vertical stretch: a amplitude 5 3 2 1 5 2 a52 Horizontal compression: k period 5 180° 5 k5 360° |k| 360° k 360° 180°
The period is 180°. there was a vertical translation of 1.
00 0.8 0.00 0.5. How much of the Moon we see depends on where it is in its orbit around Earth.41 0.00
a) Determine the equation of the sinusoidal function that models the
proportion of visible Moon in terms of time. so the period must be 30 days.02 4 7 10 14 20 24 29 34
We see: Last Quarter
Earth
We see: First Quarter
0.12 0. I figured out some of the important features of the sinusoidal function.00 0.0 0.73 0.
Rosalie's Solution
a) Cycle of the Proportion of the Moon Visible Proportion of Moon visible 1. When I drew the curve.22 0.6 0.92
44
48
53
56
59
63
70
74
We see: Waning Crescent We see: Waxing Crescent We (don't) see: New Moon
1.83 1.2 1.86 0. The graph repeats every 30 days.4 0.6.2 0 20 40 60 80 100 120 140 Day of year
I plotted the data.
Day of Year Proportion of Moon Visible Day of Year Proportion of Moon Visible 1 0.34 0. c) Use the equation to determine the proportion of the Moon that is visible
sunlight is coming from this direction
on day 110. and the minimum value 0. The table shows the proportion of the Moon that was visible from Southern Ontario on days 1 to 74 in the year 2006.6
EXAMPLE
3
Connecting data to the algebraic model of a sinusoidal function
We see: Full Moon We see: Waning Gibbous We see: Waxing Gibbous
The Moon is always half illuminated by the Sun.
Vertical translation: c Equation of the axis is y 5 0. c 5 0. the graph looked like a sinusoidal function.
b) Determine the domain and range of the function.55 0.23 0. The axis is halfway between the maximum of 1 and the minimum of 0.28
41 0.88 1. The maximum value was 1.00 0.5
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Sinusoidal Functions
389
.
I chose the x-coordinate of the maximum closest to the origin.5 2 1 5 cos(12(110 2 14)°) 1 0. since y 5 cos x has a maximum at x 5 0. x 5 14. Since x represents the time in days. since they are days of the year. so |k| 5 k 30 5 k5
k 5 12 Horizontal translation: d
b) domain: 5x [ R | 0 # x # 3656 c)
Using a cosine curve: d 5 14
A sine curve or a cosine curve will work. 65% of the Moon is exposed. I substituted 110 for x in the equation to calculate the amount of the Moon visible at that time.Vertical stretch: a amplitude 5 (1 2 0) 2 1 2
The amplitude is the vertical distance between the maximum and the axis.5 Horizontal compression: k period 5 360 |k| 360° k 360° 30
I used the period to get the compression.
k . The horizontal translation is equal to the x-coordinate of a maximum. Then I solved for y. it is 0. The domain is only the non-negative values of x up to 365. I used the cosine curve. 0.3090) 1 0.5 or a 5 0.
range: 5 y [ R | 0 # y # 16 y5
1 cos(12(x 2 14)°)1 0. 2
5 0.
1 y 5 cos(12(x 2 14)°) 1 0.5.5 2 8 1 (0.5 2
I put the information together to get the equation. or 1.5 2 1 5 cos(1152)° 1 0.65 On day 110. In this case.
390
Chapter 6
NEL
.5 2
5 0. The range is 0 to 1.
respectively.
a)
8 6 4 2 2 0 2
y
i
b)
35 25 15
y
i ii iii 1 2 3 4 5 6 x
ii x 2 4 6 iii 8 10
5 1 50 15
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Sinusoidal Functions
391
. 9). • The domain and range of a sinusoidal model may need to be restricted for the situation you are dealing with. The value of d is determined by estimating the required horizontal shift (left or right) compared with the graph of the sine or cosine curve.6. and c.6
In Summary
Key Idea
• If you are given a set of data and the corresponding graph is a sinusoidal function. 2. and equation of the axis. in the equations g(x) 5 a sin(k(x 2 d ) ) 1 c and h(x) 5 a cos(k(x 2 d ) ) 1 c.
PRACTISING
4. Determine the equation for each sinusoidal function. This information will help you determine the values of k. Determine the function that models the data in the table and does not
8 f(x) 6 4 2 0 2 4
a b
involve a horizontal translation. a. then you can determine the equation by calculating the graph's period. Determine the equation of the function. and a
maximum at (0. a period of 120°. A sinusoidal function has an amplitude of 4 units.
x y 0° 9 45° 7 90° 5 135° 7 180° 9 225° 7 270° 5
x 90° 180° 270° 360° c
3. it may be easier to use the cosine function as your model. Determine an equation for each sinusoidal function at the right.
CHECK Your Understanding
1.
Need to Know
• If the graph begins at a maximum value. amplitude.
82 1. (The units of stress are megapascals (MPa). the tire picks
up a nail.5 L/s?
10. graph the data as a scatter plot.75 1.5 0. Matthew is riding a Ferris wheel at a constant speed of 10 km/h.
10 8 6 4 2
Stress on a Motor Shaft f(t)
12.06 0. The
boarding height for the wheel is 1 m.0 0. and Moscow. and the wheel has a radius of 7 m. B-11. c) Explain the differences in the amplitude and the vertical translation for
each city. b) Graph the data.75 0.25 0. e) How many seconds have passed when the velocity is 0.
d) What does this tell you about the cities?
11.45 0. How high above the ground is the nail after the car has travelled 1 km?
14. The table shows the average monthly temperature for three cities: Athens.25 0.143 s? on a given graph. What is the equation of the function that describes Matthew's height in terms of time.6. The table shows the velocity of air of Nicole's breathing while she is at rest. and determine an equation that models the situation.22 0.75 0. While the car is being driven.85 1. c) Using a graphing calculator.02
Extending
13. see Technical Appendix.5 0.83 2 0.6
9.23 3 0
a) Explain why breathing is an example of a periodic function.
For help creating a scatter plot using a graphic calculator.
Lisbon.10 Time (s)
0 0.75 0.43 2. Time (s) Velocity (L/s) 0 0 0.74 2.
Time (month) Athens (8C) Lisbon (8C) Moscow (8C) J 12 13 29 F 13 14 26 M 15 16 0 A 19 18 10 M 24 21 19 J 30 24 21 J 33 26 23 A 32 27 22 S 28 24 16 O 23 21 9 N 18 17 1 D 14 14 24
a) Graph the data to show that temperature is a function of time for each city.25 0.61 2.61 1. b) What do the peaks of the function represent in this situation? c) How much stress was the motor undergoing at 0. The relationship between the stress on the shaft of an electric motor and time
Equivalent stress (MPa)
C
can be modelled with a sinusoidal function. The diameter of a car's tire is 60 cm. Describe a procedure for writing the equation of a sinusoidal function based
T
t 0. b) Write the equations that model each function. d) What is the velocity of Nicole's breathing at 6 s? Justify. Enter your
Tech
Support
equation and graph.) a) Determine an equation of the function that describes the equivalent stress in terms of time. assuming Matthew starts at the highest point on the wheel?
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Sinusoidal Functions
393
.5 0. Comment on the closeness of fit between the scatter plot and the graph.
6. John. That gave me the vertical translation and the value of c. I drew a smooth curve to connect the points to look like a wave.
LEARN ABOUT the Math
A group of students is tracking a friend. They know that John reaches the maximum height of 11 m at 10 s and then reaches the minimum height of 1 m at 55 s.
394
Chapter 6
NEL
. then it would take him 90 s to do one complete revolution and be back to a height of 11 m at 100 s.
? How can you develop the equation of a sinusoidal function that models John's height above the ground to determine his height at 78 s?
EXAMPLE
1
Connecting the equation of a sinusoidal function to the situation
Justine's Solution
John's Height above the Ground f(t) 10 Height (m) 8 6 4 2 0 t 10 20 30 40 50 60 70 80 90 100 110 Time (s)
I found the equation of the axis by adding the maximum and minimum and dividing by 2. Since it takes John 45 s to go from the highest point to the lowest. That gave me the vertical stretch and the value of a. I plotted the two points I knew: (10.7
Solving Problems Using Sinusoidal Models
GOAL
Solve problems related to real-world applications of sinusoidal functions. who is riding a Ferris wheel. 11) and (55. 1).
Vertical translation: c equation of the axis: y5 11 1 1 56 2
c56 Vertical stretch: a amplitude 5 11 2 6 5 5 a55
I found the amplitude by taking the maximum and subtracting the y-value for the equation of the axis.
Once I had the equation.17 m At 78 s. That gave me the value of d.035) 1 6 8 6.
y 5 5 cos(4(78 2 10)° ) 1 6 5 5 cos 272° 1 6 y 8 5(0.6.17 m.7
For the horizontal compression. I substituted x 5 78. so k is positive.
C. how would the sinusoidal function change? State the value and type of transformation associated with this change. So there was a horizontal translation of 10. I used the formula relating the period to k.17 m looks reasonable based on the graph.
Reflecting
A.
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Sinusoidal Functions
395
. If the radius of the Ferris wheel remained the same but the axle of the wheel was 1 m higher.
If it took John 60 s instead of 90 s to complete one revolution.
The answer 6. his height will be about 6. the first maximum is at x 5 0. I got the equation of the sinusoidal function by substituting the values I found into y 5 a cos(k(x 2 d )) 1 c. and solved for the height.
Horizontal compression: k period 5 k . how would the sinusoidal function change? State the value and type of transformation associated with this change. what would be the equation of the sinusoidal function describing John's height above the ground in terms of time?
B. If both characteristics from parts A and B were changed. The curve wasn't reflected. so the period 5 90 5 k5 360 |k| 360 k 360 k 360 90
k 54 Horizontal translation: d d 5 10 y 5 5 cos(4(x 2 10)° ) 1 6
If I use the cosine function. The first maximum of the new function is at x 5 10. 0.
50 0. the pole was momentarily at its resting position. and the lowest will be 210.25
240 60 54
period 5
Vertical translation: c equation of the axis: y 5 0 so c 5 0 Vertical stretch: a amplitude 5 10 so a 5 10
The axis is at y 5 0.25 0. 0) because the pole was at its resting position at t 5 0. The top sways 10 cm to the right (110 cm) and 10 cm to the left (210 cm) of its resting position and moves back and forth 240 times every minute. and the distance the top of the pole moves is the dependent variable.
I took the distance between a peak and the equation of the axis to get the amplitude. Ryan's Solution
a) Distance flag pole sways (cm) 10 5 0 5 10 Time (min) 0. At t 5 0.
396
Chapter 6
NEL
. Since the pole sways back and forth 240 times in 60 s. what is the new equation of the sinusoidal function? Assume that the period remains the same. Then it started moving to the right.75 f(t) t 1.APPLY the Math
EXAMPLE
2
Solving a problem involving a sinusoidal function
The top of a flagpole sways back and forth in high winds. This is the period.0
I drew a graph where time is the independent variable. I started at (0. b) How does the situation affect the domain and range? c) If the wind speed decreases slightly such that the sway of the top of the pole is reduced by 20%.25 s. the time to complete one sway must be 0. a) Determine the equation of a sinusoidal function that describes the distance the top of the pole is from its resting position in terms of time.
Number of sways each second 5 1 4 5 0. This gives the vertical translation. The highest point on my graph will be 10.
I used the x-coordinate of the first maximum of the new function.25
I decided to use the sine function since this graph starts at (0°.
360 |k|
k 5 1440 The sine function: y 5 10 sin(1440x)° Horizontal translation: d d5 1 16 1 ° b b 16
y 5 10 cosa1440ax 2
b) For either function. the horizontal translation is equal to the x-coordinate of any maximum. The amplitude will then change from 10 to 8. the domain is restricted to positive
values because the values represent the time elapsed. 0). Using the values of a and k.
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Sinusoidal Functions
397
.
c) 80% of 10
or y 5 8 sin(1440x)°
y 5 8 cos a1440ax 2 5 0. If the sway is reduced by 20%. The range of each function depends on its amplitudes.0 period 5 0. it's 80% of what it used to be. I put all these transformations together to get the equation of the function.7
Horizontal compression: k period 5 k. That maximum is 1 at t 5 16. i determined the equation For the cosine function.80 3 10 58
1 ° bb 16
If the sway is the only thing that's changing. then the amplitude is going to change on the graph.6.25 5 k5 360 |k| 360 k 360 0. The vertical stretch is 8. I found the horizontal compression from the formula relating the period to the value of k. since the maximum of a cosine function is at 0.
e) Determine the equation of the sinusoidal function. a fictional character in a Spanish novel.
Need to Know
• When you have a description of an event that can be modelled by a sinusoidal graph rather than data. and what does it represent in this situation? d) If Don Quixote remains snagged for seven complete cycles. f ) If the wind speed decreased. and what does it represent in this situation? b) What is the amplitude of the function. At one point. This information will help you determine the values of k. he got snagged by one of the blades and was hoisted into the air.0 1. causing the rear end of the trailer to swing
1
2 3 Time (s)
t 4
left and right. The load on a trailer has shifted. in the equations g(x) 5 a sin(k(x 2 d) ) 1 c and h(x) 5 a cos(k(x 2 d) ) 1 c. a) What is the equation of the axis of the function. and equation of the axis. Don Quixote. a) What is the equation of the axis of the function. attacked windmills
Height (m)
16 12 8 4 0
Height above Ground f(t)
40 20 Time (s)
t 60
because he thought they were giants.5 1. e) What are the domain and range of the function in terms of the situation? f ) How far is the tail light from the curve at t 5 3.
CHECK Your Understanding
Distance (m) 2.In Summary
Key Idea
• Algebraic and graphical models of the sine and cosine functions can be used to solve a variety of real-world problems involving periodic behaviour. The graph models this behaviour. and what does it represent in this situation? c) What is the period of the function. determine the domain and range of the function.2 s?
2.0 0. and c. and what does it represent in this situation? b) What is the amplitude of the function.5 0 Distance Between the Tail Light and the Curb f(t)
1. and what does it represent in this situation? d) Determine the equation and the range of the sinusoidal function. amplitude. The graph shows his height above ground in terms of time. respectively. • You will have to determine the equation of the sinusoidal function by first calculating the period. The distance from one of the tail lights on the trailer to the curb varies sinusoidally with time. it is useful to organize the information presented by drawing a rough sketch of the graph. how would that affect the graph of the sinusoidal function?
398
Chapter 6
NEL
. a. and what does it represent in this situation? c) What is the period of the function.
the top
A
floor of a building swayed 30 cm to the left (230 cm). and at t 5 12.
5. The interior and exterior temperatures of an igloo were recorded over a 48 h
K
period.
determine the equation of each function. 4 m. a) Determine the equation of the function that expresses Milton's distance from the bottom of the pool in terms of time.6. A stopwatch starts timing at this point. 12 m.5 m from the
NEL
Sinusoidal Functions
399
.25 s.
PRACTISING
4.
Interior and Exterior Temperatures T(h) h 12 24 36 48
20 Temperature (°C) 10 0 10 20 30
Time (h) a) How are these curves similar? Explain how each of them might be related
to this situation. at t 5 2 s.
6.
b) Describe the domain and range of each curve. he is on the crest of the wave. are often added to the top floors of skyscrapers to reduce the severity of the sways. a) What is the equation of a sinusoidal function that describes the motion of the building in terms of time? b) Dampers. In one case. Milton is floating in an inner tube in a wave pool. The data were collected and plotted.1 m from the bottom of the pool. Determine an equation of a sinusoidal function that describes Chantelle's distance from the vertical beam in relation to time. What is the equation of the new function that describes the motion of the building in terms of time? bottom of the pool when he is at the trough of a wave. she is the minimum distance from the beam. and two curves were drawn through the appropriate points. she is the maximum distance from the beam. 2. Skyscrapers sway in high-wind conditions. Her distance from a
vertical support beam in terms of time can be modelled by a sinusoidal function. Chantelle is swinging back and forth on a trapeze. the top floor swayed 30 cm to the right (130 cm) of its starting position.7
3. In 1. c) Assuming that the curves can be represented by sinusoidal functions. He is 1. it will reduce the sway (not the period) by 70%. At 1 s. in the forms of large tanks of water. If a damper is added to this building. At 3 s.
the mark moves in a circular motion. The paintball leaves a circular
mark 10 cm from the outer edge of the wheel.b) What is the amplitude of the function. how many complete cycles of the
sinusoidal function will there be? e) If the period of the function changes to 3 s. determine the domain and range of the sinusoidal function. The device records the current in amperes (A) on the vertical axis and the time in seconds on the horizontal axis.6 s.
8. With a little concentration. Determine the equation of the function that expresses the current in terms of time. The first maximum height occurs at 0. the current reads its first minimum value of 24. A paintball is shot at a wheel of radius 40 cm. An oscilloscope hooked up to an alternating current (AC) circuit shows a sine
curve. a) Determine the equation of the sinusoidal function that represents the height of the lead ball in terms of time.5 A.5 A. Candice is holding onto the end of a spring that is attached to a lead ball. what is the equation of this new function?
7. At t 5 120 s. c) What is the height of the mark when the wheel has travelled 120 cm from its initial position?
400
Chapter 6
NEL
. As
she moves her hand slightly up and down. she can repeatedly get the ball to reach a maximum height of 20 cm and a minimum height of 4 cm from the top of a surface. and what does it represent in this
situation?
c) How far above the bottom of the pool is Milton at t 5 4 s? d) If data are collected for only 40 s.2 s. At t 5 0 s. As the wheel rolls. b) If the wheel completes five revolutions before it stops. and the first minimum height occurs at 0. b) Determine the domain and range of the function. the current reads its 1 first maximum value of 4.
40 cm
10 cm
a) Assuming that the paintball mark starts at its lowest point. determine the
equation of the sinusoidal function that describes the height of the mark in terms of the distance the wheel travels.3 s?
9. and what does it represent in this situation? d) What is the height of the lead ball at 1. c) What is the equation of the axis. the ball moves up and down.
Examine the graph of the function f (x). and
Pulley B has a radius of 6 cm. then which of the following is true for x? i) 180° 1 360°k. k [ I d) If f (x) 5 21. model of a situation that could be modelled with a sinusoidal function?
11. The population of rabbits. k [ I ii) 360° 1 180°k. Initially. k [ I iii) 90° 1 360°k. Using graphing technology. in a given
T
region are modelled by the functions R(t) 5 10 000 1 5000 cos(15t)° and F(t) 5 1000 1 500 sin(15t)°. Pulley A has a radius of 3 cm. k [ I iv) 90° 1 180°k.
2 f(x) 1 0 1 2 3 4 a) Determine the equation of the function.6.7
10. and the population of foxes. where t is the time in months. a drop of paint on the circumference of Pulley B rotates around the axle of Pulley B. determine x when f (x) 5 7 for the function
NEL
f (x) 5 4 cos(2x) 1 3 in the domain 5x [ R | 0° # x # 360°6. c) If f (x) 5 2. Two pulleys are connected by a belt. As Pulley A rotates. What information would you need to determine an algebraic or graphical
C
Extending
12. explain how the number of rabbits and the number of foxes are related. k [ I iii) 90° 1 180°k. b) Evaluate f (20). k [ I iv) 270° 1 360°k. R(t) . k [ I ii) 360° 1 90°k.
Sinusoidal Functions
401
. Referring to each graph. then which of the following is true for x? i) 180° 1 360°k. Determine the equation of a sinusoidal function that describes the height of the drop of paint above the ground in terms of the rotation of Pulley A. F(t) . k [ I 180° 360° 540° x 720°
14. the paint drop is 7 cm above the ground.
13.
Which combinations display consonance and which display dissonance? a) CC (C in first octave.o. Graph
this function using your graphing calculator. a string vibrates. Graph the function for this series of notes using your graphing calculator. Is the function for the A major chord periodic. where x is time in seconds and f (x) is the displacement (or movement) of air molecules in micrometres (1 3 1026 m).o.
3. and C in the next octave. When you strike a key. Some combinations of keys produce dissonance. C#. Compare with the C major and the A major graphs. sinusoidal. causing the air to vibrate. The sound waves caused by striking various notes can be described by the functions in the table. Compare the C major graph with the A major graph. or both? 2.
Equations for Notes (n. E. and F. means next octave)
Note A A# B C C# Equation f(x) 5 sin(158 400x)° f(x) 5 sin(167 831x)° f(x) 5 sin(177 806 x)° f(x) 5 sin(188 389x)° f(x) 5 sin(199 584x)° Note D D# E F F# Equation f(x) 5 sin(211 427x)° f(x) 5 sin(224 026x)° f(x) 5 sin(237 348x)° f(x) 5 sin(251 465x)° f(x) 5 sin(266 402x)° Note G G# A n.
4. Equation f(x) 5 sin(282 239x)° f(x) 5 sin(299 015x)° f(x) 5 sin(316 800x)° f(x) 5 sin(355 612x)° f(x) 5 sin(376 777x)°
YOU WILL NEED
• graphing calculator
One combination of notes is the A major chord. E.o. the sound will be dissonance rather
A major
than consonance. The sound can be modelled by graphing the sum of the equations for each note in Y1 using the WINDOW settings shown. The C major chord is made up of C. C#. B n. This vibration of air produces a sound wave that your ear detects.o. Graph and sketch each combination of notes below using your graphing
calculator and the WINDOW settings shown above.
1. which is made up of A. and A in the next octave. or pleasant sounds. If you strike the keys A. G. C in next octave) b) CF c) CD d) CB (B in next octave)
402
Chapter 6
NEL
. B. or unpleasant sounds. Sketch the graph in your notebook.Curious Math
Music
Pressing certain piano keys at the same time produces consonance. C n. Sketch the resulting curve.
Determining this gives you the value of d. then the range would be 5 y [ R | 4 # y # 106. in many questions. Use the formula amplitude 5 maximum 2 axis to determine the
amplitude of the function. and then go above and below that value an amount equivalent to the amplitude.
How do you use transformations to determine the domain and range of a sinusoidal function?
Aid
• See Lesson 6. which is equivalent to the vertical stretch or compression and the value of a.
360° 0k0
5. You then determine the amplitude.
2. Example 2. based on the vertical translation. • Try Chapter Review
Question 12. A restriction in the domain can occur when you consider the real-world situation you are trying to model. which is equivalent to the vertical translation and the value of c.
maximum 1 minimum y5 2 to determine the equation of the axis.
3. based on the vertical stretch or compression. • Try Chapter Review
Question 11.
How do you determine the equation of a sinusoidal function from its graph?
1. Incorporate all the transformations into the equation
y 5 a cos(k(x 2 d )) 1 c or y 5 a sin(k(x 2 d )) 1 c. If you are transforming y 5 cos x. it is easier to identify the coordinates of the peak of the function rather than points on the axis. the horizontal translation is equivalent to the x-coordinate of any maximum. you must determine the equation of the axis.6
Q:
A:
Chapter Review
Study
FREQUENTLY ASKED Questions
The domain of a sinusoidal function is 5x [ R6. If the graph is reflected in the x-axis. 0k0 4. Determine the horizontal translation. For example. 1 . Use the formula
Study
Aid
A:
• See Lesson 6. Example 1.6. It is often easier to transform the function y 5 cos x than to transform y 5 sin x because. if the equation of the axis is y 5 7 and the amplitude is 3.
Q:
To determine the range. Use the formula
period 5
to determine the horizontal stretch or compression.
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Sinusoidal Functions
403
. then a is negative. Determine the equation of the axis.5.
Sketch the graph of a sinusoidal function that has a
period of 6. e) Determine the amplitude. and whose equation of the axis is y 5 22.
4. f ) What is the range of this function?
Graph the function h(x) 5 4 cos(3x) 1 9 using a graphing calculator in DEGREE mode for 0° # x # 360°. a)
a) Plot the data. in
metres per second?
f ) What is the range of the function? g) If the building is 6 m tall. equation of the axis. Colin is on a unique Ferris wheel: it is situated on
the top of a building. an amplitude of 4.4 150 22.3
5. b) Is the function sinusoidal? c) Calculate h(45). b) Is the graph periodic? c) What is the period of the function. If one has.
does it represent in this situation? d) Determine the equation of the axis. A ship is docked in port and rises and falls with the
3. amplitude.4 130 9. a) What is the period of the function. d) Determine the values of x. in metres at t seconds. what was Colin's
boarding height in terms of the building?
Lesson 6.
Time (s) Height (m) Time (s) Height (m) Time (s) Height (m) 0 25 60 16 120 7 10 22. and answer the following questions.
Lesson 6.1
a) What is the period of the function. and what
does it represent in this situation?
d) Was the Ferris wheel already in motion when the
data were recorded? Explain. Graph the function using a graphing calculator. 0° # x # 360°.2
is 20 and whose range is 5 y [ R | 3 # y # 86.7 20 16 80 25 140 16 30 9.
Time (min) Volume (L) Time (min) Volume (L) Time (min) Volume (L) 0 0 8 16 16 0 1 16 9 16 17 16 2 16 10 0 18 16 3 16 11 16 19 16 4 16 12 16 20 0 5 16 13 16 6 0 14 16 7 16 15 16
it represent in this situation?
c) What is the amplitude of the function.PRACTICE Questions
Lesson 6. or equation of the axis of the function. Each sinusoidal function has undergone one
transformation that may have affected the period. and what
2. Use Xscl 5 90°. Sketch a graph of a periodic function whose period
Lesson 6. indicate its new value.4 70 22. what would be the depth of the propeller? c) What is the depth of the propeller at t 5 5.7 90 22. d(t). 0). In each case.7
waves.
e) How fast is Colin travelling around the wheel. and what does it represent in this situation? b) If there were no waves. The function d(t) 5 2 sin(30t)° 1 5 models the depth of the propeller. The table shows the amount of water in the dishwasher at different times.
6. at what times is the propeller at a depth of 3 m?
7. 0) from the point (4. Colin's height above the ground at various times is recorded in the table. Determine the coordinates of the image point after a
rotation of 25° about (0. The automatic dishwasher in a school cafeteria runs
does it represent in this situation?
b) What is the equation of the axis. and what does
constantly through lunch. and draw the resulting graph.5 s? d) What is the range of the function? e) Within the first 10 s.
NEL
404
Chapter 6
. and what
1.4
8. determine which characteristic has been changed. and the range of the function.7 110 9.4 40 7 100 16 160 25 50 9. Determine the period. amplitude. for which h(x) 5 5.
the chair is closest to the wall and d(1) 5 18 cm.5 A 22. Write an equation for the axis of the curve.
b) Draw a curve of good fit. Use transformations to graph each function for
0° # x # 360°.2
13.75) 5 34 cm. The average daily maximum temperature in Kenora.
213. is shown for each month. the chair is farthest from the wall and d(1. between the wall and the rear of the chair varies sinusoidally with time t. Determine the range of each sinusoidal function
x 5 y 10
without graphing. Explain why this type
of data can be expressed as a periodic function.Chapter Review
a) b) c) d)
y 5 sin x 2 3 y 5 sin(4x) y 5 7 cos x y 5 cos(x 2 70°)
12.2 A 8. a) y 5 5 cos(2x) 1 7 b) y 5 20.6 O 9.1 J 24. d) What is the period of the curve? Explain why
e) f) g) h)
this period is appropriate within the context of the question. How can the table be used to confirm this prediction?
d(t) .
Time (months) Temperature (°C) Time (months) Temperature (°C) Time (months) Temperature (°C) J F M 21. a graph. a) y 5 23 sin(4x) 1 2 b) y 5 0. and what does it represent in this situation? f ) What is the equation of the sinusoidal function? g) What is the distance between the wall and the chair at t 5 8 s? 14.9 D 210.0 M 16.3
a) Prepare a scatter plot of the data. a) y
3 2 1 0 b) 6 4 2 0
Lesson 6. At t 5 1. The distance.5 cos(3(x 2 40° ) )
Lesson 6. Use the equation to predict the temperature for month 38. Summarize how you can determine the equation of a sinusoidal function that represents real phenomena from data.7
Lesson 6. explain how each part of the equation relates to the characteristics of the graph. Meagan is sitting in a rocking chair.3 J 21.75 s. and what does it represent in this situation? b) How far is the chair from the wall when no one is rocking in it? c) If Meagan rocks back and forth 40 times only. a) What is the period of the function. Let January
represent month 0. or a description of the situation.5
9.8 S 16.1 29. Determine the sine function y 5 a sin k(u 2 d) 1 c
for each graph.
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Sinusoidal Functions
405
.
Ontario. At t 5 1 s. What is the phase shift if the cosine function acts as the base curve? Use the cosine function to write an equation that models the data. what is the domain of the function? d) What is the range of the function in part (c)? e) What is the amplitude of the function. In your summary.7 N 21.6
x 5 10
11.5 sin(x 2 30°) 2 4
10.
c) State the maximum and minimum values.
Determine the coordinates of the point after a rotation of 65° about (0. and what does it represent in this situation? b) Determine the equation of the axis for this periodic function. d) Determine the range of f (x). what is the domain of the periodic function? f ) Steven states that the stair will be at ground level at t 5 300 s.
5. and an equation of the axis y 5 21. through the hole. and equation of the axis. If she rotates the stick at the same constant rate. then the distance from its short end to the top of the plywood can be modelled by the function in red. what is the domain of the sinusoidal function? g) Determine the equation of each sinusoidal function. Keri has drilled a hole at the 30 cm mark in a metre stick.
3. then the distance from its long end to the top of the plywood can be modelled by the function in blue in the graph shown. h) What is the distance between the short end of the metre stick and the top of the plywood at t 5 19 s?
90 1m
406
Chapter 6
NEL
.5(x 1 90°)) 2 6 using transformations of f (x) 5 cos x.
2. 24) and has a period of
20. Sketch a sinusoidal function that passes through (0. c) Calculate f (135°). 0)
from the point (7. a)
top short end
10 20 30 40 50 60 70 80
Graph f (x) 5 24 cos(0. The height of the step in terms of time can be modelled by the graph shown. and what does it represent in this situation? e) What is the range of each sinusoidal function? f ) If Keri rotates the metre stick five complete revolutions. b) State the amplitude.6
Height of Step 5 h(t) 3 1 t 0 1 10 20 30 40 50 60 70 80 Time (s) Height (m)
Chapter Self-Test
1. e) If the escalator completes only 10 cycles before being shut down. Steven is monitoring the height of one particular step on an escalator that
takes passengers from the ground level to the second floor. an amplitude of 3. If she rotates the stick at a constant rate. a) What do the troughs of the sinusoidal functions represent in this situation? b) How do the periods of the sinusoidal functions compare? Why is this so? c) How far is the nail from the top of the plywood? d) What is the amplitude of each sinusoidal function. 0).
4. Is he correct? Justify your answer. a) What is the period of the function. She then nails the
long end bottom
275 f(t) 225 175 125 75 t 25 0 2 4 6 8 10 12 14 16 Time (s) Distance (cm)
metre stick onto a piece of plywood. c) What do the peaks of the periodic function represent in this situation? d) State the range of the function. period.
Draw a line around the cylinder connecting points D and E and continue back to D. D.
Select one of the cylindrical objects. F. Determine the equation of the resulting sinusoidal function. marking the points D and E in different locations. G. Remove the tape and unroll the paper. C. f ) Another cylinder has a radius of 7 cm. point D at 12 cm high. what would have to be included in the instructions? d) How could you do a similar activity and create a function that was periodic but not sinusoidal? e) If the period of the resulting sinusoidal function was 69. a pop can. Record the equation in the table. Take a sheet of paper. Determine an equation that models the resulting curve. Record the position in the table. complete the table.6
?
Radius of the Cylinder
Chapter Task
YOU WILL NEED
Cylinders and Sinusoidal Functions
Can sinusoidal functions be obtained from cylinders?
• three cylindrically shaped
objects. Make sure that the paper is narrow enough that the top portion of the cylinder is still exposed. a) What is the relationship between the circumference of the cylinder and the resulting sinusoidal function? b) What effect does changing the locations of points D and E have on the resulting sinusoidal function? c) If you wanted to see three complete cycles on the paper. H. Mark a point D along the seam of the paper. and tape it in place. for example. cut along the line you drew. Mark a point E on the opposite side of the cylinder at least 4 cm below the height of point D. a wooden dowel. and record the measurement in the table. Determine its circumference. leaving the tube of paper. somewhere near the top of the paper. Remove the cylindrically shaped object. E.
Circumference of the Cylinder Height of Point D Height of Point E Equation of the Resulting Sinusoidal Function
Cylinder 1 2 3 A. Record the position in the table. and point E at 8 cm high. calculate the radius of the cylinder.
Task
D
B. Repeat this procedure two more times using the other cylindrical objects. wrap it around the cylinder.12 cm. Using scissors.
E
Checklist
Did you show and explain
the steps you used to determine the equations?
Did you support your
choice of data used to determine each equation?
Did you explain your
thinking clearly when answering the questions asked in part H?
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Sinusoidal Functions
407
. and an empty paper towel roll • 216 3 279 mm (letter-size) paper • tape • scissors
As you follow the instructions.
5t)°). The coordinates of her position are (x. or the domain. where t is the time in minutes and x and y are in metres. and the domain and range. the simplified form of the x sin2 u 1 cos2 u expression is cos u sin u x y x y a) b) c) d) y x r r sin x sin x is (1 2 sin x) (1 1 sin x) sin2 x cos x
25. y) 5 (20 cos (7.9 cm c) 85. A circular dining room at the top of a skyscraper
22.5 cm 1 2 !3 2
23. and the equation of the axis. The period of the function y 5 sin 4u in degrees is
a)
2 b) sin x
1 1 b b is equivalent to a b 21 b a 2a a) b) c) a b b 360°
b) 180°
21
1
sin2 x 1 1 sin2 x
c) 90°
d) 1440°
d)
2b a 1 64
20.18. 16. Identify which of the following statements is true
regarding sinusoidal functions of the form y 5 a sin(k(x 2 d)) 1 c. a) Changing the value of a affects the maximum and minimum values. Eighteen minutes later she realizes that her table has moved. radius of 14 cm. c 5 10 cm. b) The amplitude is 1.7 cm b) 56. the amplitude. the equation of the axis.8 cm b) 8. cos u 5 . A woman sits next to the window ledge and places her purse on the ledge as shown.5t)°.0 m d) 62. Which of the following statements is not true about
28. The exact value of cos (2420°) is
a)
c)
!3 2
b) 2
d) 1
24. a a b a
sin x 26. c) Changing the value of c affects the period.9 cm
a)
576
b) 64
c) 16
d)
410 Chapters 4–6
NEL
. d) Changing the value of d affects the period. and the range. Using the definitions sin u 5 . but her purse is on the ledge where she left it.0 cm c) 13.0 cm d) 42.
19. and b 5 15 cm.
A possible height of ^ ABC is a) 10. the amplitude. If 3x 2 5 12. then x is equal to
21. b) Changing the value of k affects the amplitude. y . /A 5 85°.0 m b) 37. c) The equation of the axis is y 5 0.6 cm d) 12.1 m c) 114. The perimeter is a) 32. A regular octagon is inscribed inside a circle with a
the graph of y 5 sin x? a) The period is 360°. 20 sin (7. the amplitude. d) The range is 5 y [ R | 0 . What is the shortest distance she has to walk to retrieve her purse? a) 54. In ^ ABC .9 m
y x r r y and tan u 5 . The simplified form of the expression
a)
c) tan2 x d)
woman's table
woman's purse
27.
rotates in a counterclockwise direction so that diners can see the entire city.
Give a complete solution.94 21:00 1.25 22:00 1.55
Graph the data. they measured the depth of the water every hour over a 24 h period. 125°. In the figure.95 12:00 2.56 4:00 6. Dock Dilemma
A
C
The Arps recently bought a cottage on a small.05 20:00 2.56 15:00 5.25 16:00 6.Cumulative Review
Investigations
29.81 19:00 3.55 18:00 4.94 9:00 1. b) What is the maximum depth of the water at this location? c) The hull of their boat must have a clearance of at least 1 m at all times. To determine the tide's effect at this position.69 2:00 5. If it was possible to keep folding indefinitely. and determine an equation that models this situation over a 24 h period. which is 553 m?
30. On the 7th fold it is about as thick as a notebook. 165°. using paper of any size or shape.1 mm thick.89 23:00 1. AC is 75 cm and AB is 55 cm.62 17:00 5. Determine the positions for the notches on BC that will produce the required angles. A sheet of letter paper is about 0. 145°.45 3:00 5.
31.95 24:00 2. The Paper Folding Problem
The Paper Folding Problem was a well-known challenge to fold paper in half more than seven or eight times. On the third fold it is about as thick as your fingernail.
a)
NEL
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411
. and 175°.
Time Depth (m) Time Depth (m) Time Depth (m) Time Depth (m) 1:00 3. The task was commonly known to be impossible until April 2005.69 13:00 3.05 8:00 2.25 10:00 1. They wish to build a dock on an outcropping of level rocks. Is this location suitable for their dock? Explain. sheltered inlet on Prince Edward Island.81 7:00 3. when Britney Gallivan solved it.25 5:00 5.45 14:00 5. Lawn Chairs
B
The manufacturer of a reclining lawn chair would like to have the chair positioned at the following angles: 105°.89 11:00 1.62 6:00 4. how many folds would be required to end up with a thickness that surpasses the height of the CN Tower.
honeybees don't
always have two biological parents. How can you determine the number of ancestors a bee has over a given number of generations?
NEL
413
.Chapter
7
Discrete Functions: Sequences and Series
GOALS
You will be able to
• • • •
Identify and classify sequences Create functions for describing sequences and use the sequences to make predictions Investigate efficient ways to add the terms of a sequence Model real-life problems using sequences
? Unlike humans. Male drones have only one parent (a female queen). while a queen has two parents (a drone and a queen).
2. The base of the triangle will contain 40 golf balls. A display involves constructing a stack of golf balls within a large equilateral triangle frame on one of the walls. 6. Then determine the 1st differences between the numbers of counters used. 3.
YOU WILL NEED
• counters or coins • graphing calculator • spreadsheet software
(optional)
How many golf balls are needed to construct the triangle?
Use counters or coins to construct a series of equilateral triangles with side lengths 1. 5.
?
A.Getting Started
APPLYING What You Know
Stacking Golf Balls
George's Golf Garage is having a grand-opening celebration. What does this tell you about the type of function that models the number of balls needed to create an equilateral triangle? How is the triangle with side length • 4 related to the triangle with side length 2? • 6 related to the triangle with side length 3? • 2n related to the triangle with side length n? Repeat part C for triangles with side lengths of • 5 and 2 • 7 and 3 • 2n 1 1 and n Use your rules from parts C and D to determine the number of golf balls in a triangle with side length 40.
D. 4. with each stacked row using one less ball than the previous row.
E.
Number of Counters Used 1 3 6
Side Length 1 2 3
Diagram
4
B.
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415
.
Create a scatter plot of number of counters versus side length. respectively. Record the total number of counters used to make each triangle in a table.
C. and 7.
Reflecting
F. then to calculate the 12th term (t12).7. 625. How are the two sequences similar? Different? How does the definition of an arithmetic sequence help you predict the shape of the graph of the sequence? A recursive formula for Chris's sequence is t1 5 1. D. Subscripts are usually used to identify the positions of the terms.
figure 2
figure 3
How many linking cubes are there in the 100th figure?
Create the next three terms of Chris's arithmetic sequence. Use the general term to calculate the 100th term. For example. He wrote the sequence that represents the number of cubes in each shape.1
YOU WILL NEED
Arithmetic Sequences
• linking cubes • graphing calculator or graph
paper
GOAL
Recognize the characteristics of arithmetic sequences. . the common difference. 1.
figure 1
?
A. where n [ N and n . The first three shapes are shown. 630. E. that expresses each term of a sequence as a function of its position.. if the general term is tn 5 2n. C. labelled tn. How is this recursive formula related to the characteristics of Chris's arithmetic sequence?
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Chapter 7
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. How is each term of this recursive sequence related to the previous term? Construct a graph of term (number of cubes) versus figure number.
Chris's sequence is an arithmetic sequence. tn 5 tn21 1 2 . t12 5 2(12) 5 24 recursive formula a formula relating the general term of a sequence to the previous term(s)
INVESTIGATE the Math
Chris used linking cubes to create different shapes. substitute n 5 12.
• spreadsheet software
sequence an ordered list of numbers term a number in a sequence. 620. H.. Another arithmetic sequence is 635. arithmetic sequence a sequence that has the same difference. G. B. between any pair of consecutive terms recursive sequence a sequence for which one term (or more) is given and each successive term is determined from the previous term(s) general term a formula. and express the general terms in a variety of ways. What type of relation is this? Determine a formula for the general term of the sequence.. 615.
.
Wanda's Solution: Using Differences
a) 12 2 3 5 9
I knew that the sequence is arithmetic. a 1 3d. I simplified by collecting like terms. b) State a formula that defines each term of any arithmetic sequence.
tn 5 3 1 (n 2 1) (9) 5 3 1 9n 2 9 5 9n 2 6 The general term is tn 5 9n 2 6. So for the nth term. 3 1 3(9).. The general term of the sequence of multiples of 9 is 9n.. to get each new term.... so I subtracted 6 to get the general term of my sequence. 3 1 9. General term: tn 5 a 1 (n 2 1)d
I wrote an arithmetic sequence using a general first term. and a common difference.
5 a. Each multiple of d that I added was one less than the position number. a 1 d. Each term of my sequence is 6 less than the term in the same position in the sequence of multiples of 9.
Nathan's Solution: Using Multiples of 9
a) 12 5 3 1 9
Since the sequence is arithmetic. So I knew that I had a formula for the general term. I added 9 to the previous term. (a 1 2d ) 1 d. Each multiple of 9 that I added was one less than the position number. I thought about the sequence of multiples of 9 because each term of that sequence goes up by 9s. . a 1 2d. 12. 27. 21.. . a 1 d. so the terms increase by the same amount. 18. a. Since I added 9 each time.1
APPLY the Math
EXAMPLE
1
Representing the general term of an arithmetic sequence
a) Determine a formula that defines the arithmetic sequence 3. . 21. 30. 12....
21 5 12 1 9 30 5 21 1 9 tn 5 9n 9.
b) a. (a 1 d ) 1 d.7. d. . 30. 3..
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. I needed to add (n 2 1) 9s.. . I subtracted t1 from t2 to determine the common difference.. I wrote this sequence as 3.. 36. . The general term is tn 5 9n 2 6. 3 1 2(9).
23. 3d.
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. a 1 2d. But the first term of my sequence was a. 1. I had to add a to.
21 5 12 1 9 30 5 21 1 9 The recursive formula is t1 5 3.. you can use it to determine any term in the sequence. I expressed the general term of the sequence using a recursive formula. 4d. to get each new term. a 1 3d.? David's Solution: Using Differences and the General Term 11 2 18 5 27 4 2 11 5 27 23 2 4 5 27
I subtracted consecutive terms and found that each term is 7 less than the previous term. a 1 d.
b) a. . 2d. 2d 1 (a 2 d). tn 5 tn21 1 9. not d. Since I added 9 each time. where n [ N and n . General term: tn 5 nd 1 (a 2 d) 5 a 1 nd 2 d 5 a 1 (n 2 1)d
Since I added the common difference d each time..
Tina's Solution: Using a Recursive Formula
a) 12 5 3 1 9
Since the sequence is arithmetic. .. To get the terms of any arithmetic sequence. 4d 1 (a 2 d ). 11. 4. . and subtract d from. d 1 (a 2 d).
EXAMPLE
2
Connecting a specific term to the general term of an arithmetic sequence
What is the 33rd term of the sequence 18.. So the sequence is arithmetic. . tn 5 tn – 1 1 d..b) tn 5 nd
d. I thought about the sequence of multiples of d.
Recursive formula: t1 5 a. where a is the first term. I added 9 to the previous term. 1
Once you know the general term of an arithmetic sequence.... I would add d to the previous term each time. where n [ N and n . each term of the sequence of multiples of d. I simplified to get the general term. So to get my sequence. 3d 1 (a 2 d).
I used this information to write the function that describes the line.
Communication
Tip
A dashed line on a graph indicates that the x-coordinates of the points on the line are natural numbers. The y-intercept corresponds to the term t0 but it is not a term of the sequence since x [ N. I substituted x 5 33 into the equation f(x) 5 27x 1 25. Since the terms are decreasing. n 1 2 3 4 tn 18 11 4 23 1st Differences 27 27 27
The 1st differences are constant so these points lie on a line.
The first term of the sequence is 18.
Leila's Solution: Using a Graph and Function Notation
I represented the sequence as a function using ordered pairs with the term number (n) as the x-coordinate and the term (tn ) as the y-coordinate.
f (33) 5 27(33) 1 25 5 2206 The 33rd term is 2206.
To get the 33rd term of the sequence.
f (x) 5 27x 1 25
The slope of the line is m 5 27 and the y-intercept is b 5 25. I let n 5 33.
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. I used a dashed line to connect the points. the common difference is 27.1
a 5 18. I substituted these numbers into the formula for the general term of an arithmetic sequence. To get the 33rd term. d 5 27 tn 5 a 1 (n 2 1)d t33 5 18 1 (33 2 1) (27) 5 2206 The 33rd term is 2206.7. Since n [ N and the terms lie on this line.
1 2 3 4
Year
1 $300.00
I used the spreadsheet to continue the pattern until the amount reached $732.00 $408.00 $678.00 $498. fill down. When will his investment be worth $732? Philip's Solution: Using a Spreadsheet 6% of $300 5 $18
A B
Terry earns 6% of $300.00 $570. B-21.00 $732. I set up a formula to increase the amount by $18 per year.Arithmetic sequences can be used to model problems that involve increases or decreases that occur at a constant rate.00 2 5 B2118
Terry's $
Tech
Support
For help using a spreadsheet to enter values and formulas.00 $462.00 $444. I entered the amount.00 $624. In one column I entered the year number. and in the other column. and fill right.
420
Chapter 7
NEL
.00 $390.00 $516.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
A Year 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
B Terry's $ $300. I set up a spreadsheet.00 $336.00 $588. see Technical Appendix.
From the spreadsheet.00 $642. Terry's investment will be worth $732 at the beginning of the 25th year.00 $372.00 $696.00 $318.00 $660.00 $426. which is $18 interest per year.00 $552.
EXAMPLE
3
Representing an arithmetic sequence
Terry invests $300 in a GIC (guaranteed investment certificate) that pays 6% simple interest per year.00 $534.00 $714.00 $606.00 $480.00 $354.
I divided by the amount of interest paid per year to determine how many interest payments were made. I solved for n. per year. This is an arithmetic sequence. So the amount at the start of each year will form an arithmetic sequence. where a 5 300 and d 5 18. d 5 18
Terry earns 6% of $300 5 $18 interest per year. So his investment increases by $18/year.1
Jamie's Solution: Using the General Term
Terry earns 6% of $300.
732 2 300 5 432
432 4 18 5 24
The investment will be worth $732 at the beginning of the 25th year. or $18 interest. Since interest was paid every year except the first year.
tn 5 a 1 (n 2 1)d tn 5 300 1 (n 2 1)(18)
732 5 300 1 (n 2 1)(18) 732 5 300 1 18n 2 18 732 2 300 1 18 5 18n 450 5 18n 25 5 n Terry's investment will be worth $732 in the 25th year. I needed to determine when tn 5 732.7.
Suzie's Solution: Using Reasoning a 5 300. $732 must occur in the 25th year. I calculated the difference between the starting and ending values to know how much interest was earned.
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421
.
I substituted t7 5 53 and n 5 7 into the general term. you can determine any term in the sequence.
I knew that the sequence is arithmetic.
EXAMPLE
4
Solving an arithmetic sequence problem
The 7th term of an arithmetic sequence is 53 and the 11th term is 97. For the 11th term. a and d are the same in both equations. To solve for d. Tanya's Solution: Using Reasoning t11 2 t7 5 97 2 53 5 44 4d 5 44 d 5 11 t100 5 97 1 89 3 11 5 97 1 979 5 1076 The 100th term is 1076.
97 5 a 1 10d 253 5 2(a 1 6d ) 44 5 4d 11 5 d
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Chapter 7
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.
Deepak's Solution: Using Algebra tn 5 a 1 (n 2 1)d
I knew that the sequence is arithmetic. So I divided 44 by 4 to get the common difference. so the terms increase by the same amount each time. The equations for t7 and t11 represent a linear system.If you know two terms of an arithmetic sequence. I knew that to get the 100th term. There are four differences to go from t7 to t11. Since both equations describe terms of the same arithmetic sequence.
t7 53 5 a 1 (7 2 1)d 53 5 a 1 6d
t11 97 5 a 1 (11 2 1)d 97 5 a 1 10d
For the 7th term. Determine the 100th term. I would have to add it to t11 89 times. so I wrote the formula for the general term. I substituted t11 5 97 and n 5 11. I subtracted the equation for t7 from the equation for t11. Since the common difference is 11.
I substituted a 5 213. tn 5 tn 2 1 1 d. For example.. To get the 100th term.. where b 5 t0 5 a 2 d. and so on). . the domain is the set of natural numbers. 14. is decreasing with a common difference of 23.. 1st term 2nd term
Domain: {1. . 4. 12. 20. . 1.} Range: {4..
14 14 14
t2 2 t1 5 6 2 2 5 4 t3 2 t2 5 10 2 6 5 4 t4 2 t3 5 14 2 10 5 4 ( t2 2 t1 5 6 2 9 5 23 t3 2 t2 5 3 2 6 5 23 t4 2 t3 5 0 2 3 5 23 (
and 9. 6.}.
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. 2. 28. d 5 11. . .. or • by a discrete linear function f (n) 5 dn 1 b..
To solve for a. 3. n [ N. 4.. 3. 2.7.1
53 5 a 1 6(11) 53 5 a 1 66 213 5 a tn 5 a 1 (n 2 1)d t100 5 213 1 (100 2 1) (11) 5 1076 The 100th term is 1076. N 5 {1... is increasing with a common difference of 4. .. 2nd. 12.
In Summary
Key Ideas
• Every sequence is a discrete function.} • An arithmetic sequence is a recursive sequence in which new terms are created by adding the same value (the common difference) each time. 2. Since each term is identified by its position in the list (1st. and d is the common difference. 10. In all cases. 0. • recursively by t1 5 a. The range is the set of all the terms of the sequence.
23 23 23
Need to Know
• An arithmetic sequence can be defined • by the general term tn 5 a 1 (n 2 1)d.. 3. 6. where n . a is the first term. I substituted d 5 11 into the equation for t7. For example. and n 5 100 into the formula for the general term.. 28. 20.
16.. 34 seats in the second row. 27. . 15th. a) How many seats are in the 10th row? b) How many rows of seats are in the opera house? of $0. . . The 50th term of an arithmetic sequence is 238 and the 93rd term is 539. 226. 230. terms of the original sequence as t1. 15. . Create an arithmetic sequence that has t1 . .1
9. of your new sequence. ..7.. The first term of an arithmetic sequence is 13. 86 f ) 28. the terms described by the sequence. 100 e) 233. 10.) Is this new sequence always arithmetic?
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425
. . you would choose the 3rd. An opera house has 27 seats in the first row. for the sequence 3..25/h. . Two terms of an arithmetic sequence are 20 and 50.
15. You are given the 4th and 8th terms of a sequence. t3...
A
11.. How long will he have to leave his money in the account if he wants to have $7800?
13. Her boss promises her a raise
12. t2. Describe how to
determine the 100th term without finding the general term...
a) 7.. Janice gets a job and starts out earning $9. 11. . and so on. 11. 23. from the original sequence. i)
Determine whether each general term defines an arithmetic sequence. The common difference between consecutive terms is an integer. . Two other terms of the
sequence are 37 and 73. 23. . 9.5% simple
interest per year. 1 1 a) tn 5 8 2 2n c) f (n) 5 n 1 2 4 2n 1 5 b) tn 5 n2 2 3n 1 7 d) f (n) 5 7 2 3n 41 seats in the third row. State the general term.
C
Create three different arithmetic sequences given these terms.. 225
T
d) 9. 244
14. 19. . (For example. When will Janice start earning at least twice her starting wage?
10. 219... . b) How are the common differences related to the terms 20 and 50?
a)
Extending
17. 16.15/h after each month of work. 30. The last row has 181 seats. 7th.. 7. 19. 235. Phil invests $5000 in a high-interest savings account and earns 3. state the first five terms and the common difference. Each of the three sequences should have a different first term and a different common difference. Determine the number of terms in each arithmetic sequence. 225. . t4. ii) If the sequence is arithmetic. .. 11th. 212.
18.. Create a new sequence by picking. 13. . 1. 63 b) 220.. 0 and in which each term is
greater than the previous term. 2205 c) 31. Determine all possible values for the 100th term..
how many trees will be planted at the 10th stage?
Create the first five terms of the geometric sequence that represents the number of trees planted at each stage. How is this recursive formula related to the characteristics of this geometric sequence?
NEL
G.
426
Chapter 7
. How are the two sequences similar? Different? How is the general term of a geometric sequence related to the equation of its graph? A recursive formula for the tree-planting sequence is t1 52. 1. What type of relation is this? Determine a formula for the general term of the sequence.
Reflecting
F. D.7.tn 57tn 2 1. 500 000. 250 000. Denise and Lise both initially accepted the challenge. Use the general term to calculate the 10th term. the common ratio.. E.
stage 1
stage 2
?
geometric sequence a sequence that has the same ratio. Another geometric sequence is 1 000 000.
The tree-planting sequence is a geometric sequence. C. where n [ N and n ... between any pair of consecutive terms A.
If the pattern continues. B.2
YOU WILL NEED
Geometric Sequences
• graphing calculator • graph paper
GOAL
Recognize the characteristics of geometric sequences and express the general terms in a variety of ways. .
INVESTIGATE the Math
A local conservation group set up a challenge to get trees planted in a community. The challenge involves each person planting a tree and signing up seven other people to each do the same. H. 125 000. How is each term of this recursive sequence related to the previous term? Use a graphing calculator to graph the term (number of trees planted) versus stage number.
ar 2. I recognized this as an exponential function. . I would multiply the previous term by r each time. I made a table starting with the first term. To get the 13th term..
The 13th term is 36 864. I wrote a geometric sequence using a general first term. I simplified the terms.. (ar)r.
Leo's Solution: Using a Recursive Formula
a)
n tn n tn 1 9 8 1152 2 18 9 2304 3 36 10 4608 4 72 11 9216 5 144 12 18 432 6 288 13 36 864 7 576 I knew that the sequence is geometric so the terms increase by the same multiple each time. so I knew that I had a formula for the general term. where a is the first term. the result was one less than the position number.
Discrete Functions: Sequences and Series
NEL
427
. and I multiplied each term by 2 to get the next term until I got the 13th term..7. ar. where n [ N and n .
b) a.
5 a. ar. and a common ratio. (ar 2 )r.. . Then I multiplied the common ratio 12 times.
b) State a formula that defines each term of any geometric sequence.
b) a. r. 1
To get the terms of any geometric sequence. .
Tamara's Solution: Using Powers of r
a)
a59 r52 t13 5 9 3 212 5 36 864 The 13th term is 36 864. a. ar. (ar 2)r. tn 5 rtn21. ar 3. Each time I multiplied by r..
Recursive formula: t1 5 a.
I knew that the sequence is geometric with first term 9 and common ratio 2.2
APPLY the Math
EXAMPLE
1
Connecting a specific term to the general term of a geometric sequence
a) Determine the 13th term of a geometric sequence if the first term is 9
and the common ratio is 2. I started with the first term. (ar)r. General term: tn 5 ar n21 or f (n) 5 ar n21
Geometric sequences can be used to model problems that involve increases or decreases that change exponentially..
3 3 0. r 5 0.9510021 533 0. I wrote the formula for the general term. I determined the value of n when f (n) 5 11 by substituting a 5 52 612 659.95. 3 3 0.95 f (n) 5 ar n21 f (100) 5 3 3 0. . How much radioactive material will be left after 100 years?
Jacob's Solution
3.. there will be about 19 g of radioactive material left. I calculated the common ratio by dividing t2 by t1. 11?
Suzie's Solution
a 5 52 612 659 r5 17 537 553 1 5 52 612 659 3
I knew that the sequence is geometric with first term 52 612 659. I needed to determine the value of f (n) when n 5 100.
2 3
Every year.952 ) 3 0.019 After the 100th year.95. so its position number will be equal to the number of terms in the sequence..95) 3 0. 5 3.EXAMPLE
2
Solving a problem by using a geometric sequence
A company has 3 kg of radioactive material that must be stored until it becomes safe to the environment.
1
f (n) 5 ar n21
1 n21 11 5 52 612 659 3 a b 3
428
Chapter 7
NEL
. After one year. and n 5 100 into the formula.95. and f(n) 5 11 into the formula. 17 537 553.95.. .
EXAMPLE
3
Selecting a strategy to determine the number of terms in a geometric sequence
How many terms are in the geometric sequence 52 612 659. (3 3 0.. .9599 8 0. The last term of the sequence is 11. r 5 3.. a53 r 5 0. 3 3 0. 95% of the radioactive material remains.95.95.95 . So I substituted a 5 3. 95% of the radioactive material remains. The sequence is geometric with first term 3 and common ratio 0. . The terms show the amounts in each year. (3 3 0. I wrote the formula for the general term of a geometric sequence. 3 3 0..95 . I represented the amount of radioactive material as a sequence.
the sequence has terms that alternate from positive to negative. 220. Then I found the point of intersection. 36.
In Summary
Key Idea
• A geometric sequence is a recursive sequence in which new terms are created by multiplying the previous term by the same value (the common ratio) each time.
3 33 3 3 3
t2 6 5 53 t1 2 18 t3 5 53 t2 6 t4 54 5 53 t3 18 o
1 and 144. 2
3 1 1 1 3 3 2 2 2
72 1 t2 5 5 t1 144 2 36 1 t3 5 5 t2 72 2 t4 1 18 5 5 t3 36 2 o
If the common ratio is negative. 6. t 5 rt 1 n n21...
3 (24) 3 (24) 3 (24)
Need to Know
• A geometric sequence can be defined • by the general term t 5 ar n21. For example.. 80. 54. has a common ratio of 24. 72. I graphed the functions Y1 5 52 612 659(1/3)^(X21) and Y2 5 11 using my graphing calculator. 2. is increasing with a common ratio of 3. see Technical Appendix. n • recursively by t 5 a.. 1. or • by a discrete exponential function f (n) 5 ar n21.. The x-coordinate represents the number of terms in the sequence. where n . In all cases. n [ N. 2320. 5.
There are 15 terms in the geometric sequence.
Tech
Support
For help using a graphing calculator to determine the point of intersection of two functions. a is the first term. 18. . B-12.7. 18. . is decreasing with a common ratio of .
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. and r is the common ratio.. . For example.2
Instead of using guess and check to determine n.
How many doses must be administered to reduce the bacterial population to the desired level?
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. what percent of the original bacterial population is left? b) Biologists have determined that when a person has a bacterial infection. If the pattern continues. 1 3 b) t1 5 28. 2 n21 t1 5 18. i)
Determine whether each recursive formula defines a geometric sequence. The 5th term of a geometric sequence is 45 and the 8th term is 360.
a)
tn 5 4n
d) tn 5 7 3 (25) n24 e) f (n) 5
b) tn 5 3n 1 5 c)
f (n) 5 n2 2 13n 1 8
2 3n 1 1 11 f ) f (n) 5 n 13
11. ii) If the sequence is geometric. state the first five terms and the common ratio. a) If four doses of the antibiotic are taken. A certain antibiotic reduces the number of bacteria in your body by 10%
each dose. Sam invested $5000 in a GIC earning 8% compound interest per year. ii) If the sequence is geometric. where n [ N. as well as on the original amount. 2
a)
10. 1 3 d) t1 5 10. tn 5 4tn22. 1 t c) t1 5 123. where n . so the next year Sam gets interest on the interest already earned. where n [ N. where n . how many bacteria will be present at the 9th observation?
Observation 1 2 3 4 Number of Bacteria 5 120 7 680 11 520 17 280
13. t2 5 20. state the first five terms and the common ratio. if the bacterial level can be reduced to 5% of its initial population.2
9. where n . the person can fight off the infection. The
interest gets added to the amount invested.
Determine the 20th term. tn 5 23tn21. How much will Sam's investment be worth at the end of 10 years?
14. The
A
table below shows his first four observations.
12. where n . tn 5 n21 . i)
Determine whether each general term defines a geometric sequence.7. tn 5 a b tn21. A doctor makes observations of a bacterial culture at fixed time intervals.
provide an example.
stage 1 a)
stage 2
stage 3
If the process continues indefinitely. . How many shaded triangles would be present in the sixth stage? b) If the triangle in the first stage has an area of 80 cm2. Determine the 10th term of the sequence 3. Is it possible for the first three terms of an arithmetic sequence to be equal
to the first three terms of a geometric sequence? If so. The midpoints of the square are joined
creating a smaller square and four triangles. 3..
21. Is it possible to
determine the 29th term without finding the general term? If so.
20. and 4 terms. describe how you would do it. the stages get closer to the Sierpinski gasket. Create an arithmetic sequence such that some of its terms form a
geometric sequence. 2.. what is the area of the shaded portion of the 20th stage?
17. what will be the total area of the shaded region in stage 6?
stage 1
stage 2
stage 3
stage 4
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. The first three stages are shown. At each
T
stage. Given the geometric sequence with t1 5 1 and r 5 .
16.
State the general term. 176. calculate the 2
sum of the first 1. In what ways are arithmetic and geometric sequences similar? Different?
C
Extending
1 18. How is the geometric sequence related to the arithmetic sequence?
22. You are given the 5th and 7th terms of a geometric sequence. The Sierpinski gasket is a fractal created from an equilateral triangle.. If you continue this process. the "middle" is cut out of each remaining equilateral triangle. A square has a side length of 12 cm.15. 10. 72. 28. What would happen to the sum if you added more and more terms?
19.
with the smallest on top.7.
start
end ?
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.
• circles (or squares) of paper
of increasing size
LEARN ABOUT the Math
The Tower of Hanoi is a game played with three pegs and 10 discs of increasing size.3
GOAL
Creating Rules to Define Sequences
YOU WILL NEED
Create rules for generating sequences that are neither arithmetic nor geometric. The object of the game is to stack all the discs on a different peg in the same order of size as you started with in the fewest number of moves. The rules for moving discs are: • You may move only one disc at a time. • You may place a disc only on an open peg. • You may move only a disc that is alone on a peg. or one that is on top of a pile. all of the discs are arranged in order of size on one of the pegs. At the start of the game. or on top of another disc that is larger than it.
I noticed that each term was double the previous term plus 1.
Reflecting
A. C. tn 5 2tn21 1 1 t2 5 3. 2. B. I noticed that for 3 discs. and 3 discs. and it worked for the first three cases. t2 5 2 3 1 1 1 t3 5 7. I then used the formula to calculate the number of moves needed for 10 discs. and finally move the top two discs on top of the third disc. I assumed that this pattern was correct.
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. respectively.
t1 5 1.
Mario's Solution
Number of Discs 1 2 3 Number of Moves 1 3 7 I started with a simpler problem by counting the moves needed with 1. Use these new numbers to help you determine the general term of the sequence. I first had to move the top two discs to another peg. Then I used my formula to calculate the number of moves needed for 4 discs. then move the third disc to the open peg. I wrote my pattern rule as a recursive formula. t3 5 2 3 3 1 1 t4 5 2 3 7 1 1 5 15
t5 5 2 3 15 1 1 5 31 t6 5 2 3 31 1 1 5 63 t7 5 2 3 63 1 1 5 127 t8 5 2 3 127 1 1 5 255 t9 5 2 3 255 1 1 5 511 t10 5 2 3 511 1 1 5 1023 To move 10 discs to a new peg requires 1023 moves.
How is Mario's recursive formula useful for understanding this sequence? Why would it be difficult to use a recursive formula to figure out the number of moves if there were 1000 discs? Add 1 to each term in the sequence.EXAMPLE
1
Using a pattern to represent the moves
Determine the minimum number of moves needed to move 10 discs to another peg.
There was no common ratio so the sequence is not geometric. Explain your reasoning.. 39. I worked backward again to calculate the next two terms. determine the next three terms. .. I could determine the next terms of the sequence. so the pattern rule seemed to be valid.
Term 1 8 16 26 39 39 1 17 5 56 56 1 22 5 78
1st Difference 7 8 10 13 13 1 4 5 17 17 1 5 5 22
2nd Difference
1 2 3 31154 41155
I calculated the next two 2nd differences and worked backward to get the next two 1st differences.
Tina's Solution
Term 1 8 16 26 39 1st Difference 7 8 10 13 I calculated the 1st differences of the first five terms to determine whether the sequence is arithmetic. 26.
t2 8 5 58 t1 1 t3 16 5 52 t2 8
Term 1 8 16 26 39 1st Difference 7 8 10 13 1 2 3 2nd Difference
I checked to see if the sequence is geometric. 56.
I calculated the 2nd differences. 16. My values of t6 and t7 matched those in the given sequence.
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. The 2nd differences go up by 1.
EXAMPLE
2
Using reasoning to determine the next terms of a sequence
Given the sequence 1.7. identify the type of pattern (if one exists) that relates the terms to each other to get the general term. The sequence is not arithmetic since the 1st differences were not constant. . 78. If this pattern continues. 8. I first checked whether this pattern was valid by calculating the next two terms.3
APPLY the Math
If a sequence is neither arithmetic nor geometric.
365. the recursive formula for the sequence 5. . 1 Assuming that the pattern continues.
t1 5 5.. I created a table to compare each term tn in the given sequence with 3 times the previous term... . Since the ratios were almost 3. is t1 5 5. where n [ N and n . 4 9 16 25 36 49 64 Explain your reasoning. so the numerators formed an arithmetic sequence. 41. 13. . There was a common difference of 2 between terms. determine the general term. 3281. . where n [ N and n . tn 5 3tn21 2 1. 1094. . 7. .3
n 1 2 3 4 5 6 7 tn 5 14 41 122 365 1094 3281 3tn21 — 15 42 123 366 1095 3282 I wrote a recursive formula for the sequence based on the pattern in my table. .
I looked at just the numerators to see if they formed a sequence.
Monica's Solution
3.. 15. 1. I wrote the numerator in terms of n by substituting a 5 3 and d 5 2 into the general formula for an arithmetic sequence. 5. . 9. 3tn21. .. you can sometimes find a pattern between terms if you look at the numerators and the denominators on their own. I pretended that they were actually 3.7. I noticed that the value of tn was one less than the value of 3tn21.
Nn 5 a 1 (n 2 1)d 5 3 1 (n 2 1) (2) 5 2n 1 1
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. tn 5 3tn21 2 1. 11.
EXAMPLE
4
Using reasoning to determine the general term of a sequence
3 5 7 9 11 13 15
Given the sequence . 122.
If the terms of a sequence are rational numbers. 14. ..
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. 82. 3.
Need to Know
• A sequence has a general term if an algebraic rule using the term number. I looked at the denominators to see if they formed a sequence. 62.. n. The terms were perfect squares but of the next position's value. 49. • If a sequence is arithmetic or geometric. 7. a general term can always be found because arithmetic and geometric sequences follow a predictable pattern. 36. For example.
Next. 52. the sequence of primes. The general term of the given sequence is Nn tn 5 . . not the position value for the current term. 22....
Dn 5 (n 1 1) 2
tn 5
Nn Dn 2n 1 1 (n 1 1) 2
5
Assuming that the pattern continues. it is not always possible to find a general term. 64. (n 1 1) 2
In Summary
Key Idea
• A sequence is an ordered list of numbers that may or may not follow a predictable pattern.4. I wrote the denominator in terms of n by using a general expression for perfect squares. 16. the general term of the given sequence is tn 5
2n 1 1 . For any other type of sequence. 11. 5. but no function or recursive formula has ever been discovered to generate them.. 2. 25. . . 32. is well understood. 72. 42. Dn I substituted the expressions for Nn and Dn into tn. .. 9. can be found to generate each term. where n [ N.
Explain your reasoning. b) How will your rule change if the row of triangles is replaced with a row of squares? Explain your reasoning. You are given the sequence 0.
1 2 3 2. 2. .
Determine a rule for calculating the general term. .
Determine a rule for calculating tn . . 1. 3. Explain your reasoning. Leila used toothpicks to make a row of triangles.. Sam wrote a solution to determine the 10th term of the sequence 1. . Did you have to modify your rule to do this? If so. 23. Determine a rule for calculating the terms of the sequence . 24. 2 3 4 4 5 6 .... .. .. y 1 1 2x 1 2. 3x 1 3 .. . y y
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. 21. the number of toothpicks needed for n triangles. b) Compare your rule with that of a classmate... what is your new rule?
a)
1 5. 22. Explain your reasoning. Did you come up with the same rule? Which rule is "better"? Why? c) Determine t12345.
figure 1
figure 2
figure 3
d) Show that both your rules work for n 5 4.. .
4. Determine an expression for the general term of the sequence x 1 . the number of toothpicks needed to create an n 3 n grid of squares.
21.3
CHECK Your Understanding
1.. Sam's Solution
tn = tn–1 – tn–2 ∴t10 = – 1
Do you think Sam is right? Explain. 4. 5 6 7
PRACTISING
3.
a)
c)
Determine a rule for calculating tn .7.. 5. Explain your reasoning. 25.
nicknamed Fibonacci. French mathematician Edouard Lucas (1842291) named the sequence in the rabbit problem "the Fibonacci sequence.
What relationships can you determine in the Fibonacci sequence?
The first five terms of the Fibonacci sequence are 1. 3. Explain how these terms are related and generate the next five terms.4
GOAL
Exploring Recursive Sequences
YOU WILL NEED
Explore patterns in sequences in which a term is related to the previous two terms. and 5. 3. 2. 4. in the sequence. If the cycle continues..
B. which eventually mate. Determine an expression for generating any term. how many pairs of rabbits are there every month?
The sequence that represents the number of pairs of rabbits each month is called the Fibonacci sequence in Pisano's honour. . described a situation like this: A man put a pair of newborn rabbits (one male and one female) in an area surrounded on all sides by a wall..
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. Fn.
• graph paper
EXPLORE the Math
In his book Liber Abaci (The Book of Calculation). they produce a new pair of rabbits every month (one male and one female). . Italian mathematician Leonardo Pisano (117021250).
?
A.7. When the rabbits are in their second month of life. whose terms are generated in the same way as the Fibonacci sequence. 1." He studied the related sequence 1. Generate the next five terms of the Lucas sequence.
How are these two sequences related? What happens if you repeat this process with the Lucas sequence?
F. how are the Fibonacci and Lucas sequences related to a geometric sequence?
E. 1 F2 5 5 1.
Repeat this process with the Lucas sequence. How is this new sequence related to the Fibonacci sequence?
Reflecting
G. create a new sequence by adding terms that are two apart. The second is the products of Fibonacci terms that are two apart.
Starting with the Fibonacci sequence.
The first new sequence is the squares of the Fibonacci terms. F2 1 F4 3 5 5 1. How are these new sequences related to the Fibonacci and Lucas sequences?
D.
How are the Fibonacci and Lucas sequences similar? different? Although the Fibonacci and Lucas sequences have different starting values.C. The first three ratios are shown. The first four terms are shown. they share the same relationship between consecutive terms. F1 1 2 F3 5 5 2.
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. create two new sequences as shown. and they have many similar properties. H.5 F3 2
What happens to the ratios if you continue the process? What happens if you repeat this process with the Lucas sequence? Based on your answers. How are these sequences similar? different?
I. What properties do you think different sequences with the same relationship between consecutive terms have? How would you check your conjecture? From part D. the Fibonacci and Lucas sequences are closely related to a geometric sequence.
Starting with the Fibonacci sequence.
Create a new sequence by multiplying a Fibonacci number by a Lucas number from the same position.
Determine the ratios of consecutive terms in the Fibonacci sequence.
B-16. tn 5 ar n21. tn 5 tn21 1 tn22 . the terms depend on one or more of the previous terms. and has many of the properties of the Fibonacci sequence. these sequences are similar to a geometric sequence. where n [ N and n . where n [ N and n . and the number of spirals of seeds on a sunflower head. Pick any two numbers and use the same relationship between consecutive
terms as the Fibonacci and Lucas sequences to generate a new sequence. A sequence is defined by the recursive formula t1 5 1. t2 5 1. What happens to the ratios? c) Develop a formula for the general term. 2. • The Lucas sequence is defined by the recursive formula t1 5 1.
Tech
Support
For help using a graphing calculator to generate sequences using recursive formulas. where n [ N and n . among other naturally occurring phenomena. 2. the number of spirals on a pineapple. This sequence models the number of petals on many kinds of flowers. t2 5 3. Substitute the general term for a geometric sequence. a) Generate the first 10 terms. What properties does this new sequence share with the Fibonacci and Lucas sequences?
2. into the recursive formulas for the Fibonacci and Lucas sequences.
FURTHER Your Understanding
1. and solve for r.4
In Summary
Key Ideas
• The Fibonacci sequence is defined by the recursive formula t1 5 1. tn 5 tn21 1 tn22 . 2. • Two different sequences with the same relationship between consecutive terms have similar properties. see Technical Appendix. b) Calculate the ratios of consecutive terms. Since the ratios of consecutive terms of the Fibonacci and Lucas sequences
are almost constant.
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.
tn 5 tn21 1 2tn22. How does this value of r relate to what you found in part D?
3. t2 5 5.7.
Need to Know
• In a recursive sequence.
down. Artists have been known to incorporate the golden ratio into their works. like the Fibonacci sequence. The ancient Greeks thought that it defined the most pleasing ratio to the eye. and so on. The start of the spiral is shown at the right. draw a 3 3 3 square touching one of the 1 3 1 squares and the 2 3 2 square.
Human works aren't the only places where the golden ratio occurs. so they used it in their architecture. and spirals in seed heads of flowers can be expressed using the golden ratio. How is this spiral related to the Fibonacci sequence and the
golden ratio?
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. draw a 5 3 5 square touching both 1 3 1 squares and the 3 3 3 square.Curious Math
The Golden Ratio
The golden ratio (symbolized by f. trace a 1 3 1 square. seems to pop up in unexpected places. • • • • On a piece of graph paper. Greek letter phi) was known to the ancient Greeks. draw a 2 3 2 square. On the right side of your picture.
A C B
AC AB
The golden ratio. • Repeat this process of adding squares around the picture. Euclid defined the golden ratio by a point C on a line segment AB such that f 5 CB 5 CB. The ratio of certain proportions in the human body are close to the golden ratio. up. On top of these two squares. alternating directions left. • Below your picture.
1. Draw another 1 3 1 square touching the left side of the first square. as well as in the movie and book The Da Vinci Code. It has even received some exposure in an episode of the crime series NUMB3RS. right.
. d.
EXAMPLE
If the sequence 13. state the recursive formula and the general term. f(n)
1
2
3 4 5 6 7 8 Position number. t1 t3 5 r.2. 27. t4 5r t3
• See Lesson 7. t4 2 t3 5 d
• See Lesson 7. Examples 1. . Its general term is defined by a discrete exponential function since the graph of term versus position number gives an exponential curve.1.
The recursive formula of an arithmetic sequence is based on adding the same value to the previous term. 34.7
Q:
A:
Mid-Chapter Review
Study
FREQUENTLY ASKED Questions
How do you know if a sequence is arithmetic?
Aid
A sequence is arithmetic if consecutive terms differ by a constant called the common difference. and 3. . t2 . where n [ N and n . 1. t2 2 t1 5 d . r. 20. So the recursive formula is t1 5 13.
Solution
Each term is 7 more than the previous term. 41. The recursive formula of a geometric sequence is based on multiplying the previous term by the same value. n
9 10
Q:
A:
How do you know if a sequence is geometric?
Study
Aid
A sequence is geometric if the ratio of consecutive terms is a constant called the common ratio. t3 2 t2 5 d . Its general term is defined by a discrete linear function since the graph of term versus position number gives a straight line. t2 5 r. The general term is tn 5 13 1 (n 2 1) (7) 5 7n 1 6 and its graph is a discrete linear function. is arithmetic.. 2. • Try Mid-Chapter Review Questions 4 to 6.
.
80 70 60 50 40 30 20 10 0
f(n) = 7n + 6
Term.. tn 5 tn21 1 7.
Examples 1 to 4
• Try Mid-Chapter Review
Questions 1 to 5.
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. . .
It is also useful to look at the 1st.. So the recursive formula is t1 5 16. 6. work backward using the 3rd. and 9.
Solution
Each term is 5 times the previous term.
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. Once you find a pattern.
Solution
Start by looking at the 1st. . 11. and then the 1st differences.
Look for a pattern among the terms. 41. and possibly higher. is geometric. and 70. To determine the next three terms. Since the 2nd differences seem to be going up by a constant. the next three terms are 22. 80.
Determine the next three terms of the sequence 1. 6.
How do you determine terms of a sequence that is neither arithmetic nor geometric?
Q:
A:
Study
Aid
• See Lesson 7.
EXAMPLE
• Try Mid-Chapter Review
Questions 7. differences.. 10 000. where n [ N and n . 8. f(n) 25 000 000 20 000 000 15 000 000 10 000 000 5 000 000 0 1 2 3 4 5 6 7 8 9 10 Position number. 1. 5. the 3rd differences are the same.3. 2nd. 2000.35 000 000 30 000 000 Term. and 3rd differences.. 2nd..
Examples 1 to 4. you can use it to generate terms of the sequence. The general term is tn 5 16 3 5n21. Assuming that the pattern continues. 7. . n f(n) = 16(5)n – 1
EXAMPLE
If the sequence 16. 6.. 2nd. 3rd.
Term 1 6 7 6 5 6 11 22 41 70 1st Difference 5 1 21 21 1 5 11 19 29 2nd Difference 3rd Difference
24 22 0 2 4 6 8 10
2 2 2 2 2 2 2
The 1st differences are not constant. tn 5 5tn21. state the recursive formula and the general term. 400. and its graph is a discrete exponential function.
. ." What are the next three terms? Explain your reasoning.75. the recursive
formula. 21. An IQ test has the question "Determine the next
an arithmetic sequence.. .. b) Determine the next three terms of the sequence. where n .
b) 28.25... c) State the general term of the sequence. ___.4
10. i)
d) 3000. 6. . 45. a)
Determine the 15th term of the sequence 3. if it hasn't sold. and t6. d) How many unit cubes does Sarah need to build the 15th cube?
a)
23.Mid-Chapter Review
PRACTICE Questions
Lesson 7. tn 5 2tn21 2 tn22. The number of seats in the rows of a stadium form
Lesson 7. . . 1 tn21 e) t1 5 8. 2
b) tn 5
tn 5 5n
Lesson 7.
d) 3. determine
i) the recursive formula ii) the general term iii) t10 a)
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. 15. For each arithmetic sequence. 3x 1 3y . where n . 900.2
three numbers in the sequence 1. 13. A work of art is priced at $10 000. or neither). If the art is not sold. 29. how many weeks will you have to wait before being able to afford it?
2. How many seats are in the 55th row?
Lesson 7. 5 22 . ___. where n [ N. geometric..
ii) Determine the general term. 12. Your mother really likes the art and you would like to purchase it for her.2.
9. 2. Determine the general term of the sequence
x 1 y. . 2 3 6 f ) x... . . 21. Sarah built a sequence of large cubes using unit cubes. 67. but you have only $100.
8. x 3 1 3y. e) c)
217.
a) 15. 1 tn d) t1 5 22. 246. where n .. tn 5 tn21 2 12.
1 1 2 . Two employees of the stadium determine that the 13th row has 189 seats and the 25th row has 225 seats. . Explain your reasoning.. After one week. 5. . f)
State the sequence of the number of unit cubes in each larger cube.... Determine the recursive formula and the general term
for the arithmetic sequence in which a) the first term is 17 and the common difference is 11 b) t1 5 38 and d 5 27 c) the first term is 55 and the second term is 73 d) t3 5 234 and d 5 238 e) the fifth term is 91 and the seventh term is 57
3. 9. .. 30..1
6. .. 2 3 9
Determine the type of each sequence (arithmetic. .
i) Determine whether each sequence is arithmetic
or geometric...
a)
3 4n 1 3 c) t1 5 5.. e) 3. its price is reduced by 10%... ii) State the first five terms. . t2 5 11. b) Write the recursive formula for the sequence in part (a).... if
1. .. 92.3
7. 5. c)
5. b) 640. 270.
1 2 5 . 129. 9.. 221.
4. Each week after that. Explain your reasoning. 160.8. 7.5. x 2 1 2y.. . 216. 29.. 224. its price is reduced by another 10%. ___.
the art isn't sold. .. 320.. 5x 1 6y . .
When German mathematician Karl Friedrich Gauss (177721855) was a child.
D.
INVESTIGATE the Math
Marian goes to a party where there are 23 people present. He then paired terms from the two lists to solve the problem. including her. but include plus signs between terms. they shake hands with the host and everyone who is already there. Each person shakes hands with every other person once and only once. Use this method to determine the sum of your arithmetic series. his teacher asked him to calculate the sum of the numbers from 1 to 100. C. + 98 + 99 + 100 100 + 99 + 98 + .. What type of sequence is this? Write your sequence from part A.
1 + 2 + 3 + . Gauss wrote the list of numbers twice. once forward and once backward.. + 3 + 2 + 1 101 + 101
How can Marian determine the total number of handshakes that take place?
A. Solve the handshake problem without using Gauss's method. Create a sequence representing the number of handshakes each person will make.7. This expression is a series and represents the total number of handshakes. When they enter..
?
series the sum of the terms of a sequence arithmetic series the sum of the terms of an arithmetic sequence B.
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.5
YOU WILL NEED
Arithmetic Series
• linking cubes
GOAL
Calculate the sum of the terms of an arithmetic sequence..
Suppose the people join the party one at a time.
. What would you notice about the 1st and 2nd differences? Why is Gauss's method for determining the sum of an arithmetic series efficient? Consider the arithmetic series 1 1 6 1 11 1 16 1 21 1 26 1 31 1 36. of the first n terms of a sequence
F. Do you think this method will work for any arithmetic series? Justify your answer.
The sum of the first n terms of an arithmetic series is Sn 5 5 5 5 n 3a 1 (a 1 (n 2 1)d )4 2 n(t1 1 tn ) 2
n 32a 1 (n 2 1)d 4 2
n 3a 1 a 1 (n 2 1)d 4 2
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. I noticed that a is the first term of the series and a 1 (n 2 1)d is the nth term. The nth term of the series corresponds to the general term of an arithmetic sequence.. 1 (a 1 d )
1 .. I regrouped the terms in the numerator.. Next.. first forward and then backward.7. 1 32a 1 (n 2 1)d 4 1 32a 1 (n 2 1)d 4
Sn 5
n 32a 1 (n 2 1)d 4 2
1 (a 1 d )
1 3a 1 (n 2 2)d 4
1 ... Since the terms in the top row go up by d and the terms in the bottom row go down by d. I added each column. I knew that 2a 5 a 1 a. tn 5 a 1 (n 2 1)d.. To find Sn. Using Gauss's method.
partial sum the sum. 1 tn
˛
1 Sn 5 3a 1 (n 2 1)d 4
2Sn 5 n 3 32a 1 (n 2 1)d 4
Sn 5 a
2Sn 5 32a 1 (n 2 1)d 4 1 32a 1 (n 2 1)d 4 1 . so I divided by 2. I wrote the sum out twice.. 1 3a 1 (n 2 2)d 4 1 3a 1 (n 2 1)d 4
1a
The series is arithmetic. each pair of terms has the same sum.5
Reflecting
E.. but that represents 2Sn.
APPLY the Math
EXAMPLE
1
Representing the sum of an arithmetic series
Determine the sum of the first n terms of the arithmetic series a 1 (a 1 d ) 1 (a 1 2d ) 1 (a 1 3d ) 1 .
˛ ˛ ˛
Barbara's Solution
Sn 5 a 1 (a 1 d ) 1 (a 1 2d ) 1 (a 1 3d ) 1 . Use Gauss's method to determine the sum of this series. There are n pairs that add up to 2a 1 (n 2 1)d.. so I wrote this formula another way.
Suppose the partial sums of an arithmetic series are the terms of an arithmetic sequence. I added all terms up to tn. Sn. G.
..
Jasmine's Solution
t2 2 t1 5 235 2 (231) 5 24 t3 2 t2 5 239 2 (235) 5 24
I checked to see if the series was arithmetic. seats are arranged in 50 semicircular rows facing a domed stage.
EXAMPLE
3
Selecting a strategy to calculate the sum of a series when the number of terms is unknown
Determine the sum of 231 2 35 2 39 2 . The differences were the same.
EXAMPLE
2
Solving a problem by using an arithmetic series
In an amphitheatre. The first row contains 23 seats.. the number of seats in each row forms an arithmetic sequence.
In order to determine the sum of any arithmetic series.. How many seats are in the amphitheatre?
Kew's Solution
a 5 23. 2403. and each row contains 4 more seats than the previous row. you can use the formula for the sum of an arithmetic series. I used the formula for the sum of an arithmetic series in terms of a and d. 1 t50 (50) 32(23) 1 (50 2 1) (4) 4 2
n 32a 1 (n 2 1)d 4 Sn 5 2
S50 5
5 6050
There are 6050 seats in the amphitheatre.
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.If a problem involves adding the terms of an arithmetic sequence.
23 1 27 1 31 1 . you need to know the number of terms in the series. Since I knew the first term and the common difference. I wrote an arithmetic series to represent the total number of seats in the amphitheatre. so the series is arithmetic. d 5 4
Since each row has 4 more seats than the previous row. I substituted n 5 50 since there are 50 rows of seats. So I calculated a few 1st differences.
5 220 398
The sum of the series 231 2 35 2 39 2 . Since I knew the first and last terms of the series. S4 5 t1 1 t2 1 t3 1 t4 5 2 1 10 1 18 1 26 5 56
Need to Know
• The sum of the first n terms of an arithmetic sequence can be calculated using n32a 1 (n 2 1)d 4 • Sn 5 or 2 n(t1 1 tn ) • Sn 5 .. . For example.
In Summary
Key Idea
• An arithmetic series is created by adding the terms of an arithmetic sequence. Sn 5 t1 1 t2 1 t3 1 . . and t 94 5 2403.. 1 tn. t1 5 231. . 18. For the sequence 2. 18. • You can use either formula. 26. 10.. So I substituted a 5 31. a is the first term. 2 In both cases.. I needed to determine the value of n when tn 5 2403. d 5 4. the related arithmetic series is 2 1 10 1 18 1 26 1 . n [ N. Sn. I substituted n 5 94.. . use the formula in terms of t1 and tn. If you know the last term.. 2403 is 220 398.. • The partial sum. but you need to know the number of terms in the series and the first term.. and d is the common difference. and tn into the formula for the general term of an arithmetic sequence and solved for n. 26. I used the formula for the sum of an arithmetic series in terms of t1 and tn.. If you can calculate the common difference.5
tn 5 a 1 (n 2 1)d 2403 5 231 1 (n 2 1) (24) 2403 1 31 5 (n 2 1) (24) 2372 5 (n 2 1) (24) 2372 (n 2 1) (24) 5 24 24 93 5 n 2 1 93 1 1 5 n 94 5 n Sn 5 S94 5 n(t1 1 tn ) 2 943231 1 (2403)4 2
There are 94 terms in this sequence... of a series is the sum of a finite number of terms from the series.
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. for the sequence 2. use the formula in terms of a and d.7. 10.
In level 1. The arithmetic series 1 1 4 1 7 1 . she continues to fall 9. what total distance will Sara run in a 10 week training session? uses linking cubes to represent the sum of the series 2 1 5 1 8 1 11 1 14 two ways. and the sum of the first 20 terms
is 710. you are given 3 min 20 s to complete the task. Chandra jumps out of a plane and falls 4. he assembled 137 toys. each day she runs 2 km farther than she ran the days of the previous week. After 15 s. How far did Chandra fall before she opened her parachute?
11. He noticed that since he started. The first week she runs 5 km each day. 1 tn has a sum of 1001. Jamal got a job working on an assembly line in a toy factory. How many toys did Jamal assemble altogether during his first 20 days?
12.8 m more than the previous second.. The
T
next week. The 10th term of an arithmetic series is 34. On the 20th day
of work. These representations are shown at the right.5
9. she runs 7 km each day.3%
simple interest on his investments. at level 20. a fixed number of seconds are removed from the time until. Determine the 25th term." a number of shapes have to be
arranged into a predefined form. she opens her parachute. During each successive week. Explain how the linking-cube representations can be used to explain the formulas for an arithmetic series. What would be the total amount of time given if you were to complete the first 20 levels?
13.9 m
during the first second. In the video game "Geometric Constructors.
14. If she runs for five days each week.7. During a skydiving lesson. How many
terms does the series have?
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. Joan is helping a friend understand the formulas for an arithmetic series. Joe invests $1000 at the start of each year for five years and earns 6. 1 min 45 s are given. every day he assembled 3 more toys than the day before. Sara is training to run a marathon. For each second afterward.
16. How much will all his investments be worth at the start of the fifth year?
10. At each level afterward.. She
C
Extending
15.
How do you know that this sequence is geometric? Based on your sequence. rSn and Sn. determine how many people are in John's ancestor tree.6
YOU WILL NEED
Geometric Series
• spreadsheet software
GOAL
Calculate the sum of the terms of a geometric sequence. create a geometric series to represent the total number of people in John's ancestor tree.
INVESTIGATE the Math
An ancestor tree is a family tree that shows only the parents in each generation. Subtract Sn from rSn.
How many people are in John's ancestor tree?
Create a sequence to represent the number of people in each generation for the first six generations. His complete ancestor tree includes 13 generations. B. C.
3rd generation
2nd generation
1st generation
?
A.7. Based on your calculation in part C. John started to draw his ancestor tree. starting with his own parents. Write both series. but this time multiply each term by the common ratio. Write the series again.
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. so that equal terms are aligned one above the other.
geometric series the sum of the terms of a geometric sequence
D.
r21
ar n 2 a Sn 5 r21 Sn 5 tn11 2 t1 . If I multiplied every term by the common ratio.6
Reflecting
E. Only the first term of Sn and the last term of rSn would remain. I wrote this new series above the original series and lined up equal terms. r21
I wrote this formula another way by expanding the numerator. so I would get zero for most of the terms when I subtracted... 1 ar n22 1 ar n21 ) (r 2 1)Sn 5 2a 1 0 1 (r 2 1)Sn 5 2a 1 ar n r21 a r 2 1 b Sn 5
1 1
I wrote the sum out.
(r 2 1)Sn 5 a(r n 2 1) a(r n 2 1) r21 a(r n 2 1) r21
Sn 5
The sum of the first n terms of a geometric series is Sn 5 a(r n 2 1) . I solved for Sn by dividing both sides by r 2 1.. The nth term of the series corresponds to the general term of a geometric sequence.. To find Sn. 1 01 01 ar n
2Sn 5 2(a 1 ar 1 ar 2 1 ar 3 1 .
How is the sum of a geometric series related to an exponential function? Why did lining up equal terms make the subtraction easier?
APPLY the Math
EXAMPLE
1
Representing the sum of a geometric series
Determine the sum of the first n terms of the geometric series. most of the terms would be repeated.
rSn 5
ar 1 ar 2 1 ar 3 1 . where r 2 1. I added all terms up to tn.7..
Tara's Solution
tn 5 ar n21
The series is geometric. where r 2 1..
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. F. 1 ar n22 1 ar n21 1 ar n 01 0 1 . I noticed that a is the first term in the series and ar n is the (n 1 1)th term.
So I calculated the ratio of consecutive terms. 10. and n. 50.
5 4 882 812 A total of 4 882 812 fish hatched during the first 10 days. respectively. The first term is 2 and there are 10 terms. fish hatch at different times even though the eggs were all fertilized at the same time. Since I knew the first term. Since all the ratios are the same. you can use the formula for the sum of geometric series. 250. If the pattern continues. r 5 5. calculate the total number of fish hatched during the first 10 days.
EXAMPLE
2
Solving a problem by using a geometric series
At a fish hatchery. and 250... the common ratio. and the number of terms. and n 5 10.
Joel's Solution
10 t2 5 t1 2 55 6r55 a52 n 5 10 Sn 5 S10 5 a(r 2 1) r21 2(510 2 1) 521
n
50 t3 5 t2 10 55
t4 250 5 t3 50 55
I checked to see if the sequence 2. the sequence is geometric.If a problem involves adding together the terms of a geometric sequence.
456
Chapter 7
NEL
. I substituted a 5 2. I used the formula for the sum of a geometric series in terms of a. 50. r. is geometric. . The number of fish that hatched on each of the first four days after fertilization was 2. 10.
. I used the formula for the sum of a geometric series in terms of t1 and tn11. 1 92 160. 61 440 corresponds to the (n 1 1)th term.7.6
EXAMPLE
3
Selecting a strategy to calculate the sum of a geometric series when the number of terms is unknown
Calculate the sum of the geometric series 7 971 615 1 5 314 410 1 3 542 940 1 . I saw that the 12th term is 92 160..
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. So I calculated the common ratio.
Jasmine's Solution: Using a Spreadsheet
t2 5 314 410 5 t1 7 971 615 5 6r5 2 3 2 3
A n tn B
I knew that the series is geometric..
1 2 3 4 5 6 7 8 9 10 11 12 13 14
1 2 3 4 5 6 7 8 9 10 11 12 13
7971615 5314410 3542940 2361960 1574640 1049760 699840 466560 311040 207360 138240 92160 61440
I needed to determine the number of terms.
5 23 730 525 The sum of the series 7 971 615 1 5 314 410 1 3 542 940 1 .
Sn 5 S12 5
tn11 2 t1 r21 61 440 2 7 971 615 2 21 3
From the spreadsheet. n.. Since I knew the first term and the (n 1 1)th term. to get to tn 5 92 160. So I set up a spreadsheet to generate the terms of the series. 1 92 160 is 23 730 525.
to get to tn 5 92 160.
Tech
Support
Since I knew that a 5 7 971 615. n. Then I used the table function to determine the term number whose value was 92 160. I 2 substituted a 5 7 971 615 and r 5 3 into the formula.Mario's Solution: Using a Graphing Calculator
t2 5 314 410 5 t1 7 971 615 5 6r5 2 3 2 3
I wrote the formula for the general term of a geometric sequence.
2
The sum of the series 7 971 615 1 5 314 410 1 3 542 940 1 . B-6. So I calculated the common ratio.
2 n21 92 160 5 7 971 615 3 a b 3 tn 5 ar n21
2 n21 5 7 971 615 3 a b 3
I needed to determine the number of terms. I knew that the series is geometric.. and n 5 12.
I entered the function Y1 5 7 971 615(2/3)^(X21) into my graphing calculator. r. see Technical Appendix. It was n 5 12. 1 92 160 is 23 730 525. and n..
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. I used my calculator to evaluate.
For help using the table on a graphing calculator.
a(r n 2 1) Sn 5 r21
r 5 3. I substituted these values into the formula for the sum of a geometric series in terms of a.
the ball reaches 60% of its previous height..5.
1 1 6 1 36 1 .8225 26 1 24 2 96 1 . 2 334 611 24 000 1 3600 1 540 1 . At each new stage.
At the top of each bounce. A simple fractal tree grows in stages. an isosceles right triangle and two squares are attached to the last square(s) drawn. 1 279 936 960 1 480 1 240 1 .. A ball is dropped from a height of 3 m and bounces on the ground. 1 15 17 2 51 1 153 2 . Calculate the sum of each geometric series. The formula for the sum of a geometric series is Sn 5
tn112t1 . 1 98 304 1 f ) 4 1 2 1 1 1 .. two new line
A
a(r n 2 1) or r21
segments branch out from each segment at the top of the tree.
stage 1
stage 2
stage 3
NEL
460
Chapter 7
. Calculate the area of the tree at the 10th stage. Sn 5
9..
8. Calculate the total distance travelled by the ball when it hits the ground for the fifth time. each of which is valid only if r 2 1. A Pythagorean fractal tree starts at stage 1 with a square of side length 1 m. How many line segments need to be drawn to create stage 20?
stage 1
stage 2
stage 3
stage 4
stage 5
10... Determine the sum of the first seven terms of the geometric series in which
a) b) c) d) e) f) a) b) c) d) e)
t1 5 13 and r 5 5 the first term is 11 and the seventh term is 704 t1 5 120 and t2 5 30 the third term is 18 and the terms increase by a factor of 3 t8 5 1024 and the terms decrease by a factor of 3 t5 5 5 and t8 5 240
2
6. The first five stages are shown.. 1 1. Explain how you would r21 determine the sum of a geometric series if r 5 1..... 1 1024
7. The first three stages are shown.
At every consecutive stage..
How many terms are in the series?
17. Justify your reasoning. Factor x15 2 1.. how many employees does the company have?
12. John solved the 3 problem by considering another geometric series with common ratio 5.6
11.
13. In a geometric series. Represent the next three terms on the diagram. who each call three other employees.
Use John's method to calculate the sum. John wants to calculate the sum of a geometric series with 10 terms.
16.
a)
1 2
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Discrete Functions: Sequences and Series
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.. t3 5 92. t1 5 23. Each of the five senior managers calls three employees.7. What is the greatest possible
value for t5? Justify your answer. 5 where the 10th term is 5 859 375 and the common ratio is 3. The company wants to come up with a number of prizes that satisfy all of these conditions: a) The total of the prizes is at most $2 000 000.. Suppose you want to calculate the sum of the infinite geometric series
1 1 1 1 1 1 1 1 . each prize is a constant integral multiple of the next smaller prize and is • more than double the next smaller prize • less than 10 times the next smaller prize Determine a set of prizes that satisfies these conditions. Describe several methods for calculating the partial sums of an arithmetic and a
C
Extending
15. 18. In a geometric series. 2 4 8 16 The diagram shown illustrates the first term of this series. geometric series. If the tree consists of seven levels. t1 5 12 and S3 5 372. b) How can the formula for the sum of a geometric series be used in this case? c) Does it make sense to talk about adding together an infinite number of terms? Justify your reasoning. How are the methods similar? different?
14. and the sum of all of the terms of the
series is 62 813. b) Each prize is in whole dollars (no cents). c) When the prizes are arranged from least to greatest. and so on. A large company has a phone tree to contact its employees in case of an
emergency factory shutdown. A cereal company attempts to promote its product by placing certificates for a
T
cash prize in selected boxes.
7. D.
Blaise Pascal
462
Chapter 7
. it has the same chance of going either left or right. A version of the toy with nine levels is shown at the right.
Reflecting
E. Use your pattern to calculate the number of paths to each bin in a toy with nine levels. How is the number of paths for a toy with three levels related to the number of paths for a toy with two levels? Why is this so? Use the pattern to predict how many paths lead to each bin in a toy with four levels.
How is the number of paths for each bin in a given level related to the number of paths in the level above it? The triangular pattern of numbers in the "Rockin' Rollers" toy is known as Pascal's triangle. and use one of these patterns to expand binomials efficiently. Calculate the number of paths to each bin at the bottom.
INVESTIGATE the Math
A child's toy called "Rockin' Rollers" involves dropping a marble into its top. who explored many of its properties.
B.
level 0 level 1 level 2 level 3 level 4 level 5 level 6 level 7 level 8 level 9
?
A. C. What other pattern(s) can you find in Pascal's triangle?
NEL
F.
How many paths are there to each of the bins at the bottom of this version of "Rockin' Rollers"?
Consider a "Rockin' Rollers" toy that has only one level.7
Pascal's Triangle and Binomial Expansions
GOAL
Investigate patterns in Pascal's triangle. When the marble hits a pin. named after French mathematician Blaise Pascal (1623 62). Check your prediction by counting the number of paths. Repeat the calculation with a toy having two and three levels.
I used the 5th row of Pascal's triangle to determine the coefficients.
(x 2 2)5 51(x)5 1 5(x)4(22)1 1 10(x)3(22)2 1 10(x)2(22)3 1 5(x)1(22)4 1 1(22)5
5 x 5 1 5(x 4)(22) 1 10(x 3)(4) 1 10(x 2)(28) 1 5(x)(16) 1 (232) 5 x 5 2 10x 4 1 40x 3 2 80x 2 1 80x 2 32
EXAMPLE
3
Selecting a strategy to expand a binomial power involving a variable in each term
Expand and simplify (5x 1 2y)3. I used the terms x and 22 and applied the pattern for expanding a binomial. As the x exponents decrease by 1 each time.
Tanya's Solution
1 1 1 1 1 1 5 4 10 3 6 10 2 3 4 5 1 1 1 1 1
Since the exponent of the binomial is 5.
464
Chapter 7
NEL
.
Jason's Solution
1 1 1 1 3 2 3 1 1 1
Since the exponent of the binomial is 3. I simplified each term. the exponents of 2 increase by 1.
EXAMPLE
2
Selecting a strategy to expand a binomial power involving a variable in one term
Expand and simplify (x 2 2)5. The exponents in each term always add up to 5.Any binomial can be expanded by using Pascal's triangle to help determine the coefficients of each term. I wrote out the 3rd row of Pascal's triangle.
. 1. . tn 5 tn 2 1 1 n. which can be defined by the recursive formula t1 5 1... • In each term. Each row is generated by calculating the sum of pairs of consecutive terms in the previous row. 2. • The diagonal beside that is the counting numbers. while the exponents of b start at zero and increase by 1 up to n. • In the expansion. 1. 1.
1 1 1 1 1 1 5 4 10 3 6 10 2 3 4 5 1 1 1 1 1
0th row 1st row 2nd row 3rd row 4th row 5th row
• The numbers in Pascal's triangle correspond to the coefficients in the expansion of binomials raised to whole-number exponents. Some of these relationships are recursive. 6. where n [ N and n . . a power of a... the exponents of a start at n and decrease by 1 down to zero. • The next diagonal is the triangular numbers. . • The coefficients in the expansion correspond to the numbers in the nth row in Pascal's triangle. • For example. I simplified each term.7.
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. down the sides are constant sequences: 1..
Need to Know
• Pascal's triangle has many interesting relationships among its numbers.. which form an arithmetic sequence. 1.. • There are patterns in the expansions of a binomial (a 1 b)n: • Each term in the expansion is the product of a number from Pascal's triangle. . and a power of b.
5 1(125x 3) 1 3(25x 2)(2y) 1 3(5x)(4y 2) 1 1(8y 3) 5 125x 3 1 150x 2y 1 60xy 2 1 8y 3
In Summary
Key Ideas
• The arrangement of numbers shown is called Pascal's triangle. the sum of the exponents of a and b is always n. 3.7
(5x 1 2y)3 5 1(5x)3 1 3(5x)2(2y)1 1 3(5x)1(2y)2 1 1(2y)3
I used the terms 5x and 2y and applied the pattern for expanding a binomial. 1. 10. 3.
expand each binomial power to
describe a pattern in Pascal's triangle. and 220. 3.CHECK Your Understanding
1. without calculating. 2 2
are also constant.
PRACTISING
4. Expand and simplify each binomial power. Using the diagram at the left. a) 2n 5 (1 1 1)n b) 0 5 (1 2 1)n. Expand and simplify the first three terms of each binomial power. the third differences are constant. the 1st differences are constant. Use the pattern for expanding a binomial to demonstrate that if a relation is cubic. The first four entries of the 12th row of Pascal's triangle are 1. Expand and simplify each binomial power.
a) a)
(x 1 2)5 (x 1 5)10
b) (x 2 1)6 b) (x 2 2)8
c) (2x 2 3)3 c) (2x 2 7)9
6 e) Q !2x 1 !3 R f ) (2z 3 2 3y 2)5
3.)
How are the terms related to tossing the coin 10 times?
466
Chapter 7
NEL
. 66. If a relation is linear.
2. Using the pattern for expanding a binomial.
Extending
12.
Determine the first four entries of the 13th row of the triangle. !5 2 2 How are the terms related?
d) Q !a 1 !5 R 10 c) (z5 2 z 3)11
f ) (5x 3 1 3y 2)8
e) a3b2 2 b
2 b
14
could walk to school from her house if she always travels either north or east. where n 5 1. determine the number of different ways that Joan
A
1 1 2 !5 n 1 1 !5 n ca b 2a b d . binomial to expand (x 1 y 1 z)10. 2. When a fair coin is tossed. Expand and simplify (3x 2 5y)6.
9. 11. Using the pattern for expanding a binomial. Expand and simplify the first three terms of each binomial power. 12. (Hint: You may want to look at x 3 and (x 1 1)3. where n $ 1
7. how you can use the pattern for expanding a
T
10. the probability of getting heads or tails is 2. If the 2nd differences
1 13. and 4. the relation is quadratic. 1 1 Expand and simplify the first three terms in the expression Q 1 R10. expand and simplify the
expression
school N E Joan's house
8.
K
a) (k 1 3)4 b) ( y 2 5)6
c) (3q 2 4)4 d) (2x 1 7y)3
5.
a)
(x 2 2)13
b) (3y 1 5)9
6. Summarize the methods of expanding a binomial power and determining a
C
term in an expansion. Explain.
a is the first term. 2.
Examples 1. n [ N. r21 r21
Study
Aid
• See Lesson 7.
• Try Chapter Review
Question 23. 2.
• Try Chapter Review
Questions 14 to 17. You can use either of the formulas Sn 5
n(t1 1 tn ) n32a 1 (n 2 1)d 4 2
• See Lesson 7. use the formula in terms of t1 and tn11. For either formula. one above the other. use the formula in terms of a. where n is a whole number.
How do you expand a binomial power?
Study
Aid
Use the pattern for expanding a binomial. Each term in the expansion is a product of a number from Pascal's triangle. and d is the 2 common difference. n [ N. 2. and 3. Q:
A:
How do you determine the sum of a geometric series?
You can use either of the formulas Sn 5
where r 2 1.5. and 3. the sum of the exponents of a and b is always n. Choose the nth row of Pascal's triangle for the coefficients. If you can calculate the number of terms. Suppose you have the binomial (a 1 b)n.
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Discrete Functions: Sequences and Series
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. you need to know the common ratio and the first term. a power of a.
A2:
or
Sn 5 .7. and 3.
Examples 1. The exponents of a start at n and decrease by 1 down to zero. r.6. and n. and r is the common ratio. you need to know the number of terms in the series and the first term. use the formula in terms of a and d. In both cases. use the formula in terms of t1 and tn.
• Try Chapter Review
Questions 18 to 22. For either formula. once forward and once backward. In each term of the expansion. When the terms of the two series are paired together. If you know the (n 1 1)th term.
• See Lesson 7. while the exponents of b start at zero and increase by 1 up to n. they have the same sum. a is the first term. Q:
A:
tn11 2 t1 a(r n 2 1) or Sn 5 . and a power of b.
Examples 1.7
Q:
A1:
Chapter Review
Study
FREQUENTLY ASKED Questions
What strategies can you use to determine the sum of an arithmetic sequence?
Aid
Write the series out twice. In both cases. This method works for calculating the sum of any arithmetic series. If you can calculate the common difference. If you know the last term.
A student plants a seed.3
12... For each geometric sequence. How can you determine whether a sequence is
Determine the type of each sequence (arithmetic. 3.. c) 81..4.m. there were 23 000 bacteria present. a) At 1 p. . .. 1
arithmetic?
3. Explain your reasoning.. d) 29 2 14 2 19 2 .
11. 384.PRACTICE Questions
Lesson 7.. . . c) 31 1 52 1 73 1 . i) the recursive formula iii) the first five terms a) d) 10.. and 33 mm. or neither. b) 21 1 17 1 13 1 . 45. the number of toothpicks needed for a stack of squares n high. Guy purchased a rare stamp for $820 in 2001. state
i)
the general term
ii) the recursive formula
10. ii) If a sequence is arithmetic or geometric.. 69. 75.2
value of the stamp increases by 10% per year.5 1 18. . 20. f ) 512... where n . ii) the general term
13. b) 0. .. 90. . the
bacteria doubles every hour. f ) 239 2 31 2 23 2 .
4.. How many bacteria will be present at midnight? b) Can this model be used to determine the bacterial population at any time? Explain. where n [ N. For each arithmetic sequence.. Explain your reasoning. 14.
6.. After the seed sprouts. 8. 231.. the count of a certain
a) 58..9 1 20. For each arithmetic series.. .72. 15.
a) 5. . Determine a rule for calculating tn.. geometric.5
8.... tn 5 4n 1 5 1 b) tn 5 7n 2 3
a) c) tn 5 n 2 2 1 d) t1 5 217.. or neither). i)
Determine whether each sequence is arithmetic. . 73. 10. Determine the 100th term of the arithmetic
sequence with t 7 5 465 and t13 5 219. calculate the sum of the
the first term is 7 and the common ratio is 23 1 b) a 5 12 and r 5 2 c) the second term is 36 and the third term is 144
first 50 terms. 288. 0. 1.. e) 17. 50. a) 1 1 9 1 17 1 . 240.
5. If the
student monitors the growth of the plant by measuring its height every week.. c) 288. Determine the 100th term of the sequence
1 2 3 4 ..
468
Chapter 7
NEL
..... tn 5 tn21 1 n 2 1. In a laboratory experiment. respectively. determine t 6. Represent the sequence 2. .. geometric.. The height after each of the first three weeks was 7 mm.. .. determine
14.1
9. How can you determine whether a sequence is
geometric?
7. Toothpicks are used to make a sequence of stacked
squares as shown. ii) State the first five terms. 14. how much will the stamp be worth in 2010?
Lesson 7. in what week will the plant be more than 100 mm tall?
Lesson 7. If this pattern of growth continues.3 1 .. i)
1. .
a) in words b) algebraically c) graphically
2. 2 5 8 11
Lesson 7. 20 mm. 88. 8. .. e) 19. b) 249.
Expand and simplify. The 1st. how far did it descend?
21.
a) b) c)
(a 1 6) 4 (b 2 3) 5 (2c 1 5) 3
d) (4 2 3d) 6 e) (5e 2 2f ) 4 f)
2
19.7
23... calculate t6 and S6..
15 625 5 25 1 1 . 1 3584 b) 23 2 6 2 12 2 24 2 . 0.6
18. Calculate the sum of each series. 1 4068 c) 123 1 118 1 113 1 ..06) 1 1000(1.5 1 38..111 111 1 1. For each month afterward.... 9 1 3 f) 1 1 1 . 2 10 50
a) b) c) d) e)
Lesson 7. 32 805 1 21 870 1 14 580 1 ..Chapter Review 15...
a) 7 1 14 1 28 1 . A spacecraft leaves an orbiting space station to
descend to the planet below. 1 1000(1. Determine the sum of the first eight terms of the
a3f
2 4 2 b f
geometric series in which a) the first term is 26 and the common ratio is 4 b) t1 5 42 and t9 5 2112
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Discrete Functions: Sequences and Series
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. 1 2 4 64 d) 96 000 2 48 000 1 24 000 2 .. 6 2 12 1 24 2 .. 5th. For each geometric series... A catering company has 15 customer orders during
its first month. 1 145 b) 9 1 42 1 75 1 ..06) 2 1 . The spacecraft descends 64 m during the first second and then engages its reverse thrusters to slow down its descent.111 11 1 11. It travels 7 m less during each second afterward. If the spacecraft lands after 10 s..1111 1 . If the 21st term of the arithmetic sequence is 72.. 17 2 25. 1 375 e) 1000 1 1000(1.. 2 768
Lesson 7.
c) the first term is 320 and the second term is 80 d) the third term is 35 and the terms increase by a
factor of 5
20. 2 122
17.25 2 ..... How many orders in total did the company fill at the end of its first year?
a) 1 1 13 1 25 1 . and 13th terms of an arithmetic sequence
are the first three terms of a geometric sequence with common ratio 2.
22. the company has double the number of orders than the previous month... Calculate the sum of each series. calculate the sum of the first 10 terms of the geometric sequence. Determine the sum of the first 25 terms of an
arithmetic series in which a) the first term is 24 and the common difference is 11 b) t1 5 91 and t25 5 374 c) t1 5 84 and t2 5 57 d) the third term is 42 and the terms decrease by 11 e) the 12th term is 19 and the terms decrease by 4 f ) t5 5 142 and t15 5 12
16..06) 12
c)
11
11 1 33 1 99 1 ...
• Each pill contains 30 mg of the active ingredient. nausea) can occur when more than 180 mg are in the body. which schedule is best for Tom? Is another schedule more appropriate?
Task
Checklist
Did you justify your
"reasonable" conditions?
B. as well as times of the day when you would take the medication.
Did you show your work? Did you support your
choice of medication schedule?
C. Tom decides to try a certain brand of antihistamine. Side effects (sleepiness.
How many hay fever pills should Tom take. It is unhealthy to ingest more than 180 mg within a 24 h period.7
Chapter Task
Allergy Medicine
It is estimated that 1 in 7 Canadians suffers from seasonal allergies such as hay fever. • For his size. The label says: • The half-life of the antihistamine in the body is 16 h.
?
A. and how often should he take them?
What are some conditions that would be reasonable when taking medication? For example. maximum relief is felt when there are 150 mg to 180 mg in the body. D. think about dosages.
Did you explain your
thinking clearly?
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Discrete Functions: Sequences and Series
471
. A typical treatment for hay fever is over-the-counter antihistamines. Based on your calculations in part C. headaches. determine the amount of the medication present in Tom's body for the first few days. For each of your schedules. Determine three different schedules for taking the pills considering the appropriate amounts of the medication to ingest and your conditions in part A.
472
NEL
.
your
parents deposit $1000 into a bank account that pays 3% interest per year on the balance. $1500. $1100. How can you determine which amount will be closest to what will be in the account when you are ready to go to college or university.Chapter
8
Discrete Functions: Financial Applications
GOALS
You will be able to
• • • •
Determine how interest is earned and charged Use the difference between future value and present value to solve problems Solve problems about money invested at regular intervals over a period of time Calculate payments that must be made when a purchase is financed over a period of time
? On your first birthday. or $2500?
NEL
473
. $2000.
2525.. 9724. 80.. . 15. The 4th... 19. and 6th terms of a sequence are 9261. 2. respectively.. 2500.2 7.. . 20. 223. .
2.004512x 5 400
8.07x 5 30 000 b) 5 3 3x 5 228 d) 250 3 1.5 and 7.1 and 7. 31. Complete the chart shown by writing what you know about exponential
functions.. a) 2x 5 1 000 000 c) 14 000 3 1. The fourth term of an arithmetic sequence is 46 and the sixth term is 248.. Round your answers to two decimal places.. Determine the sum of the first 10 terms of each series. d) 1000.
Example: Visual representation:
Definition in your own words:
Personal association:
474
Chapter 8
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.... 320. b) What will be the expected population 10 years from now?
6. . Determine the value of x that makes the equation 2x 5 4096 true. 4. 7. b) 58. a) What type of sequence is this? Justify your reasoning.
c) 5. Solve each equation by graphing the corresponding functions on a graphing
calculator.7
i)
the next two terms
ii) the general term
iii) the recursive formula
a) 7. 6 Lesson 7. 5th.
5. determine
Aid
For help. b) Determine the recursive formula. c) State the general term.
Determine a) the 5th term b) the common difference
c) the 1st term d) the 100th term
3. see the following lessons in Chapters 4 and 7. and
10 210.8
Study
Getting Started
SKILLS AND CONCEPTS You Need
1. The population of a city is 200 000 and increases by 5% per year.
Determine the expected population at the end of each of the next 3 years. 3 4 5.6 4. Question 1. 8 192 000 2 4 096 000 1 2 048 000 2 . 250.05. 11.. 227 2 21 2 15 2 . 48 1 31 1 14 1 ... For each sequence..
a) b) c) d) a)
3 1 5 1 7 1 .
4. d) Determine the 10th term.
or continue the table in part D to determine how long it will take Mark to save for the trip.
Assuming that the cost of the trip stays the same.35% on the minimum monthly balance means? How will earning interest affect the amount of time he will have to save? Determine how much Mark will have in his account at the end of the 1st.
Deposits
Use a spreadsheet or the lists on a graphing calculator. He already has $200 in his account at the start of the month. D. B.Getting Started
APPLYING What You Know
Saving for a Trip
Mark is saving for an overseas trip that costs $1895.35% on the minimum monthly balance.
Discrete Functions: Financial Applications
NEL
475
.
Starting Balance Interest Earned Final Balance
C. Determine the additional costs he will incur. Each week. Record your values in a table as shown. 2nd. What do you think earning 0. and 3rd months. how long will it take Mark to save enough money to pay for it?
How do you know it will take less than three years for Mark to meet his goal? When Mark pays for the trip. Calculate the total price of the trip.
YOU WILL NEED
• graphing calculator • spreadsheet software
(optional)
?
A. he deposits $50 into a savings account that pays him 0.
Month 1 2 3 E. he will have to pay the Goods and Services Tax (GST).
and I is the interest. D. Record your results in a table as shown.1
YOU WILL NEED
Simple Interest
• graphing calculator • spreadsheet software
GOAL
Calculate simple interest.
?
A.
F. see Technical Appendix.
Calculate the interest earned and the amount of the investment at the end of the second and third years. Determine the equation of the function that models the amount of Amanda's investment over time. C.8. using Year as the independent variable. What type of function best models the growth of Amanda's money? You may need to calculate more data points before you decide. P is the principal. Enter your data for Year and Amount into either lists on a graphing calculator or columns in a spreadsheet. B-11. where A is the amount. Year 0 1 2 3 Interest Earned — Amount $2000
What function can be used to model the growth of Amanda's money?
Calculate the interest earned and the amount of the investment at the end of the first year.
Tech
Support
For help using a graphing calculator to enter lists and to create scatter plots. B. principal a sum of money that is borrowed or invested simple interest interest earned or paid only on the original sum of money that was invested or borrowed interest the money earned from an investment or the cost of borrowing money amount the total value of an investment or loan. Create a scatter plot from your two lists or columns. Record your results in your table. E.
476
Chapter 8
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. The amount is given by A 5 P 1 I.
INVESTIGATE the Math
Amanda wants to invest $2000. Explain your reasoning. Her bank will pay 6% of the principal per year each year the money is kept in a savings account that earns simple interest.
52 + 77.52 3395. for t years at r%/a simple interest. To calculate the interest earned each year.4% as a decimal is 0. I.04
I set up a table to calculate the amount at the end of each year. These rates are sometimes abbreviated to 5%/a. c) Determine the total amount. Jasmine's Solution
a) I 5 0.76 77.024 3 $3240
Each year. A. Then I entered the year and amount into separate lists on a graphing calculator. or year. H. and the interest.1
Reflecting
G. Allen earns 2.
NEL
Discrete Functions: Financial Applications
477
.76
b)
Year Interest ($) 0 1 2 3 4 0 77. $P. I multiplied the principal by the interest rate.
5 $77.76 3240
Amount ($) 3240 + 77.
a) Calculate the interest earned each year.024.76 3317. I added the interest earned each year to the previous amount.
What type of sequence could you use to represent the amount of Amanda's investment for successive years? How do you know? How does the recursive formula for this sequence relate to her investment? How do the principal.76 = 3473.8.76 + 77.76 = 3395. which means 5% per annum. and amount of Amanda's investment relate to • the sequence from part G that represents the amount of the investment over time? • the function from part F that represents the amount of the investment over time?
APPLY the Math
EXAMPLE
1
Representing any situation earning simple interest as a function
Communication
Tip
Allen invests $3240 at 2. earned if he invested a principal.
Interest rates are often advertised as a certain percent per year.28 + 77.76 = 3317. interest.76 = 3551.76 77.28 3473.76 77. 2. b) Calculate the amount and the total interest
earned after 20 years.4% of the principal as interest. So a rate of 5% means that 5% interest is earned each year. I.4%/a simple interest.
Allen will have $4795.76(20) 1 3240 5 4795. The increase would be the interest earned in one year.
f (t) 5 77. The graph is linear. I subtracted the principal from the amount.20 2 3240 5 1555. the time in years.
478
Chapter 8
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. P 3 r. That is the rate of change of the amount.20 and will have earned $1555.20 After 20 years. I substituted t 5 20.
A 5 P(1 1 rt) The total amount of an investment of $P for t years at r%/a simple interest is A 5 P(1 1 rt). the amount. The y-intercept is Allen's principal of $3240.I used the calculator to create a scatter plot of amount versus time. I used two points to calculate the slope and I found that it was $77. I wrote a formula for interest in terms of t.20 in interest.76. Since interest depends on time. The amount is the sum of the principal and the interest. the time in years. I wrote the formula for the amount by factoring out the P. To determine the total interest earned. the total interest earned would go up by the same amount. and the total interest earned is I 5 Prt.20 I 5 4795. and f(t).
f (20) 5 77. Allen would earn r% of his original investment of $P.
c) end of 1st year:
I1 5 Pr end of 2nd year: I2 5 Pr 1 Pr 5 2Pr end of 3rd year: I3 5 Pr 1 Pr 1 Pr 5 3Pr end of tth year: I 5 Prt A5P1I A 5 P 1 Prt
After one year. Since the interest earned at the end of each year depends on time. To determine the amount after 20 years. the amount must also depend on time.76t 1 3240
I used the slope and y-intercept to create a linear function in terms of t. Each year.
will be charged in interest.26 3 b 365 8 ($540) (1.00 $1 020. $32.
NEL
479
. B-21. How much will he need to pay back at the end of the loan period. see Technical Appendix.00
C Total Amount of Loan $15 000. 6. Discrete Functions: Financial Applications
P 5 $540
r 5 26% 5 0. I subtracted the principal to get the interest.00 $25 200.
EXAMPLE
3
Selecting a strategy to calculate the amount owed after less than a year
Philip borrows $540 for 85 days by taking a cash advance on his credit card.00 $22 140. The interest rate is 26%/a simple interest.00 $1 020.00 $21 120.70
Of the $572.00 $19 080.1
EXAMPLE
2
Using a spreadsheet to represent the amount owed
Tina borrows $15 000 at 6. and how much interest will he have paid? Lara's Solution t5 85 365
Philip isn't borrowing the money for a full year. or $1020. I substituted the values of P.00 "= C2 + B3" "= C3 + B4"
1 2 3 4
1 2 3 4 5 6 7 8 9 10 11 12
A Time (Years) 1 2 3 4 5 6 7 8 9 10
B Total Interest Charged $1 020.00 $17 040. Each year. and t into the formula.00 $1 020.
I used the spreadsheet to calculate the amount Tina would need to pay back at the end of each year during the 10-year period.00 $16 020. Calculate how much she will owe at the end of each year during this period.8.8/100)" 2 "= C2* (6.70 that Philip has to pay back.8%/a simple interest. as a decimal. and to fill down. so I expressed the time as a fraction of 365 days in a year.00 $1 020. P.061) 5 $572.00
I set up a spreadsheet to calculate the interest charged every year and the loan amount.00 $24 180.00 $18 060. r.00 $1 020.00 $1 020.00 $20 100.8/100)" C Total Amount of Loan $15 000.70 is interest. and I wrote the interest rate.00 $1 020. I knew the principal.
Tech
Support
For help using a spreadsheet to enter values and formulas.26 A 5 P(1 1 rt)
85 5 ($540) a1 1 0. She plans to pay back the loan in 10 years. r.00 $1 020. Tom's Solution
A Time (Years) B Total Interest Charged 1 "= C2* (6.8% of $15 000. To calculate the amount at the end of the loan period.00 $1 020. I rounded to the nearest cent.00 $23 160.
I got a value greater than 8.6% 5 0.00 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Year Total Interest Earned Total Amount
(continued)
480
Chapter 8
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.2 days Tanya would have to invest her money for 8 years and 200 days to get $8000.6%/a simple interest. I substituted the values of P.00 $400. are linear functions in terms of time.546 3 365 days 8 199. and A into the formula and solved for t.076 A 5 $8000 A 5 P 1 Prt 8000 5 4850 1 (4850) (0.076)t 8000 5 4850 1 368. so their graphs are straight lines (see graph below). r.EXAMPLE
4
Calculating the time needed to earn a specific amount on an investment
Tanya invests $4850 at 7. I.546 of a year was.00 Dollars $300.6
When I solved for t.00 $200. • The total amount.
In Summary
Key Ideas
• Simple interest is calculated only on the principal.00 $0.00 $500.6t 3150 5 368.6t t5 3150 368. how long will she need to invest her money? Josh's Solution P 5 $4850 r 5 7.
$300 Earning 5%/a Simple Interest $600. A. and interest earned. The 8 meant 8 years. the interest rate. and the total amount. If she wants the money to increase to $8000. so I had to figure out what 0.
8 8.546 0.00 $100. I knew the principal. The values of A and I at the end of each interest period form the terms of two arithmetic sequences.
How long will he
have to leave his investment in the bank before earning $200 in interest?
4. Each situation represents an investment earning simple interest. where I is the interest.1
Need to Know
• Simple interest can be calculated using the formula I 5 Prt.8. • The total amount. For each investment. and 3rd years ii) 15 years
a) $500 at 6.
CHECK Your Understanding
1. expressed as a decimal. calculate the interest earned and the total amount. 2nd. a) What is the principal? b) How much interest is earned in 5 years? c) What interest rate is being applied? d) State the equation that represents the amount as a function of time.9% 18% 24% 12%
Time 8 years 12 years 16 months 5 months 17 weeks 100 days Discrete Functions: Financial Applications
a) b) c) d) e) f)
NEL
$500 $3 200 $5 000 $128 $50 000 $4 500
481
. • Unless otherwise stated. an interest rate is assumed to be per year.19 interest for the 12 days her payment is late?
PRACTISING
5.00 $500. Calculate the
total amount at the end of each period. usually per year.00 $1 500.00 $2 500. The graph at the right represents the total amount of an
investment of principal $P earning a fixed rate of simple interest over a period of 5 years. Sally has a balance of $2845 on her credit card. r is the interest rate.00 1 2 3 Year 4 5
2.00 $1 000. Michel invests $850 at 7%/a simple interest. and t is time.00 Total Amount $2 000. What rate of simple interest is
she being charged if she must pay $26. i) 1st.
3. expressed in the same period as the interest rate. P is the principal.
K
Principal
Rate of Simple Interest per Year 4.3%/a
An Investment Earning Simple Interest $3 000.8% 3.1%/a c) $25 000 at 5%/a d) $1700 at 2. A.00 $0.8% 9. of an investment earning simple interest can be calculated using the formula A 5 P 1 Prt or A 5 P(1 1 rt).4%/a b) $1250 at 4.
96
Dec. Lotti invests some money at a fixed rate of simple interest. Determine a formula for the doubling time.25 $2312. The amount she
owes at the end of each of the first five years is shown at the left. Sara's parents decide to invest $500 on each of her birthdays from the day
she is born until she becomes 25. 1. His bank sends him statements after each quarter. d) Graph this sequence. Len invests $5200 at 3%/a simple interest. c) State the total amount as the general term of a sequence.75t to calculate how much her investment will be worth after t years. Each investment earns 6. How long will it take for Dave's investment to be worth more than Len's? function A(t) 5 750 1 27.50 $2543.5 years at a fixed rate of simple interest. Ahmad deposits an amount on September 1. 2005 Mar. The doubling time of an investment is the length of time it takes for the total
amount invested to become double the original amount invested.32 $4248.25
T A
Balance $3994. 1. 1. What will be the total amount of the investments when Sara is 25?
482
Chapter 8
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. while his friend Dave invests
12.28
10. a) How much did Ahmad invest? b) What rate of simple interest is he earning?
Statement Date 1 2 3 4 Year 1 2 3 4 5 Amount Owed $2081. a) By how much does the amount in her account increase each year? b) Determine the amount in her account at the end of each of the first 5 years. 2006 Jun. 2006
Sept. a) How much did Anita borrow? b) State the total amount as the general term of a sequence. 1.
14.4%/a simple interest. Anita borrows some money at a fixed rate of simple interest. She uses the
C
Extending
13.
11. At the
end of that time.64 $4502. Nina deposits $3500 into a savings account.6. Mario borrows $4800 for 8.5%/a. D. he owes $8000.75 $2775. 2006 $4757. The rate of simple interest
is 5.
9. What interest rate is he being charged?
7.3%/a simple interest to earn $250
in interest each month?
8. into an account that
earns simple interest quarterly. c) How much time will have passed before Anita owes $7500? $3600 at 5%/a simple interest. The amounts for the first four quarters are shown.00 $3006. 2005. How much did she invest and what interest rate is she earning? Explain your reasoning. How much money must be invested at 6. of an investment of principal $P earning a rate of simple interest of r%/a.
C.8. What is the advantage of earning compound interest over simple interest? How are compound interest. annually 1 1 time per year semi-annually 1 2 times per year quarterly 1 4 times per year monthly 1 12 times per year
Complete the table for the 2nd to 5th years. using Year as the independent variable. F. compounding period the intervals at which interest is calculated.
Compare the total amount of Mena's investment with that based on the same principal earning simple interest. for example. The savings account is called the "Accumulator" and pays compound interest. Determine the function that models the amount of her investment over time.
Interest Earned —
compound interest interest that is added to the principal before new interest earned is calculated.
NEL
483
. Create a scatter plot. Interest is paid at regular time intervals called the compounding period. E.
• graphing calculator • spreadsheet software
(optional)
LEARN ABOUT the Math
Mena invests $2000 in a bank account that pays 6%/a compounded annually.
Reflecting
G. Record your answers in a table as shown. What type of function best models the growth of Mena's money? You may need to calculate more data points before you decide. Explain how you know. D. exponential functions. Enter your data for Year and Balance at End of Year into either lists on a graphing calculator or columns in a spreadsheet. and geometric sequences related?
Discrete Functions: Financial Applications
H. So interest is calculated on the principal and on interest already earned.2
GOAL
Compound Interest: Future Value
YOU WILL NEED
Determine the future value of a principal being charged or earning compound interest.
Balance at Start of Year — $2000 Balance at End of Year $2000
Year 0 1 2 3 4 5 B.
?
A.
What type of function will model the growth of Mena's money?
Calculate the interest earned and amount at the end of the first year.
80.84 after 10 years.56 ( tn 5 5300 3 1.APPLY the Math
EXAMPLE
1
Representing any situation earning compound interest as a function
future value the total amount. The general term of the sequence is tn 5 5300 3 1. Each time.56 t1 5 5300 3 1. and I got the same numbers as in my previous calculations. I.046(1) 4
end of 3rd year: A 5 5798.046) 5 $5543. A.81
5 530031 1 0.6%/a compounded annually.0462 8 $5798.0463 8 $6065.80(1.046 each time.04610 8 $8309. This is a geometric sequence with common ratio 1. a) How much will he have to pay back if he borrows the money for 10 years? b) Determine the future value.046.
NEL
I drew a timeline for the situation. I substituted n 5 10 into the formula for the general term.84 Tim would have to pay back $8309.0461 5 $5543. Tim gets charged 4.6% of his original $5300 loan.046n.046) 8 $5798.80 5 5543. $5543.81(1. of an investment after a certain length of time
Tim borrows $5300 at 4. A. I rounded to the nearest cent.
Shelley's Solution
a) Time now
1 2 3 8 9 10 A=? At the end of the first year. At the end of the second year.
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Chapter 8
. I used this formula to calculate the first three terms. if he invested a principal of $P for n years at i%/a compounded annually.8031 1 0. To determine how much Tim would owe after 10 years.046(1)4 5 5798.8131 1 0. I noticed that I was multiplying by 1.046) 5 $6065. I used the same method to calculate the amount at the end of the third year.
P = $5300 i = 4. he gets charged 4. So I calculated the amount he would owe at the end of that year. So I calculated the amount he would owe at the end of that year.80 t2 5 5300 3 1.6%/a compounded annually
end of 1st year:
A 5 P(1 1 rt)
end of 2nd year:
A 5 5543.046(1) 4 5 5300(1.81 t3 5 5300 3 1.6% of the amount he owed at the end of the first year.046n t10 5 5300 3 1. and interest earned.
To determine the interest earned over a period of n years.8%/a compounded quarterly
NEL
Discrete Functions: Financial Applications
485
.8%/a compounded quarterly when she was born. I subtracted the principal from the total amount.b) end of 1st year:
A2 5 3P(1 1 i)4 (1 1 i) 5 P(1 1 i) end of 2nd year: 5 P(1 1 i) 2 end of 3rd year: 5 P(1 1 i)
A1 5 P 31 1 i(1)4 A 5 P(1 1 in)
8.
EXAMPLE
5 P 3(1 1 i) n 2 14
2
Selecting a strategy to determine the amount when the compounding period is less than a year
Lara's grandparents invested $5000 at 4. and 3rd years. as it did for simple interest. I substituted i for r into the formula for the total amount and calculated the future value for the 1st. 2nd. Since the interest rate is i%/a. so I wrote the general term.2
For compound interest. which gave me the amount after n years. I factored out the common factor P. The amount is an exponential function in terms of time.
P = $5000 i = 4.
I5A2P
5 P(1 1 i) n 2 P
The future value of an investment of $P for n years at i%/a compounded annually will be A 5 P(1 1 i) n. the amount or future value depends on time. How much will the investment be worth on her 21st birthday?
Herman's Solution
Time now 1 2 20 21 A=? I drew a timeline for the situation.
A3 5 3P(1 1 i) 24 (1 1 i)
3
end of nth year: A 5 P(1 1 i) n
This is a geometric sequence with first term P(1 1 i) and common ratio 1 1 i. and the total interest earned will be I 5 P 3(1 1 i) n 2 14.
62 on Lara's 21st birthday.012 n 5 21 3 4 5 84
Since interest is paid quarterly for each compounding period.94
Trudy's investment will be worth $538 781.
486
Chapter 8
NEL
.P 5 $5000 i 5 0. where i is the interest rate per compounding period and n is the number of compounding periods.048 4 4 5 0. Interest is paid 4 times per year. When Lina turns 45. I divided the annual interest rate by 4 to get the interest rate.
A 5 P(1 1 i) n
A 5 5000(1 1 0. I first determined the interest rate per month as a fraction.012) 84 8 $13 618. If both women leave their investments until they are 65. so I calculated the number of compounding periods.2%/a compounded annually for 84 years. she invests $10 000 at 8%/a compounded monthly. I used the formula A 5 P(1 1 i) n.
EXAMPLE
3
Calculating the difference of the amounts of two different investments
On her 15th birthday. This is the total number of times that the interest would be calculated over the 21 years. there will be 50 3 12 5 600 compounding periods. Since interest is compounded monthly and she is investing for 50 years. I noticed that solving this problem is the same as solving a problem in which the money earns 1.94 when she turns 65.
n 5 (65 2 15) 3 12 5 600 A 5 10 000a1 1 8 $538 781.08 12 0.62 The $5000 investment will be worth $13 618. how much more will Trudy's investment be worth?
Henry's Solution
P 5 $10 000 i5 0.08 600 b 12
To calculate how much Trudy's investment will be worth when she turns 65. Trudy invests $10 000 at 8%/a compounded monthly.
91 Trudy's investment will be worth $489 513. To solve for t.
EXAMPLE
4
Comparing simple interest and compound interest
Nicolas invests $1000. I substituted the values of P.03 5 $489 513.08 240 b 12
Lina's investment has the same principal and interest rate per month as Trudy's. the interest earned must be the same as the principal.03
Lina's investment will be worth $49 268.05) 5a bt 1000(0.05). Since Lina invested for 20 years.05 I 5 $1000 I 5 Prt 1000 1000(0. I divided both sides of the equation by 1000(0.
I subtracted Lina's amount from Trudy's amount. but fewer compounding periods.05)t 20 5 t It will take 20 years for Nicolas's investment to double at 5%/a simple interest. r. there will be 20 3 12 5 240 compounding periods.
n 5 (65 2 45) 3 12 5 240 A 5 10 000a1 1 8 $49 268. Since Nicolas's investment will double and he is earning simple interest.91 more than Lina's.94 2 $49 268.08 12 0.05) 1000(0. $538 781.8. How long would it take for his investment to double for each type of interest earned? a) 5%/a simple interest b) 5%/a compounded semi-annually
Jesse's Solution
a)
P 5 $1000 r 5 5% 5 0.03 when she turns 65.
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Discrete Functions: Financial Applications
487
.05) 1000 5 1000(0. and I into the formula for the interest earned.
I knew the principal and the interest rate.2
P 5 $10 000 i5 0.
P is the principal.00 $0.00 $2 000.00 Simple Interest Compound Interest
0
5
10 Year
15
20
25
It will take about 14 years for Nicolas's investment to double at 5%/a compounded semi-annually. interest is calculated and added to the principal. From the graph. interest is calculated on the new balance (principal plus interest earned from the previous year). and n is the number of compounding periods.
I graphed the amount of the investment for both cases.00 $1 000. i is the interest rate per compounding period. This pattern continues every year the investment is kept.
Total Amount
In Summary
Key Ideas
• Compound interest is calculated by applying the interest rate to the principal and any interest already earned. So compound interest results in exponential growth. At the end of the second year. • The future value of an investment earning compound interest can be calculated using the formula A 5 P(1 1 i) n. where I is the total interest. of an investment after a certain period is called the future value of the investment.
Tech
Support
For help using a spreadsheet to graph functions. while the compound-interest situation is modelled by an exponential function growing at a changing rate. • The most common compounding periods are: annually semi-annually quarterly 1 time per year 2 times per year 4 times per year i 5 annual interest rate i 5 annual interest rate ÷ 2 i 5 annual interest rate ÷ 4 n 5 number of years n 5 number of years 3 2 n 5 number of years 3 4
• Compound interest can be calculated using the formula I 5 A 2 P or I 5 P 3(1 1 i) n 2 14.00 $500. where A is the future value.00 $1 500. A. monthly 12 times per year i 5 annual interest rate ÷ 12
NEL
n 5 number of years 3 12
Discrete Functions: Financial Applications
489
. the simpleinterest situation is modelled by a linear function growing at a constant rate. see Technical Appendix. expressed as a decimal. B-21.2
Comparing Simple and Compound Interest $3 000.00 $2 500. If interest is compounded annually. then at the end of the first year.
Need to Know
• Banks pay or charge compound interest at regular intervals called the compounding period. • The total amount. • The total amounts at the end of each interest period form a geometric sequence.8.
Determine the future value of the bond. Serena wants to borrow $15 000 and pay it back in 10 years. Calculate the value of his investment two years after this change.
11. Create a problem for this situation and solve it. Sima invests some money in an account that earns a fixed rate of interest
compounded annually. Cliff has some money he wants to invest for his retirement. of an investment. and the compounding period.2%/a compounded monthly. b) How much did Sima invest?
6. Noreen used her graphing calculator to investigate two sequences.8. A. the interest rate will be 6%/a compounded quarterly. how much will Serena save by choosing the second option? the future value. Dieter deposits $9000 into an account that pays 10%/a compounded
quarterly. a) Determine the annual rate of compound interest earned. • Option 2: Borrow the money at 12%/a compounded quarterly for 5 years and then renegotiate the loan based on the new balance for the last 5 years. Three
T
screenshots from her investigation are shown. He is offered
two options: • 10%/a simple interest • 5%/a compounded annually Under what conditions should he choose the first option?
13.40 $4764. the interest rate changes to 9%/a compounded semi-annually. in dollars.
12. Interest rates are
A
high.00 $4494. Margaret can finance the purchase of a $949. the annual interest rate. Explain your reasoning.007512n to represent
9.99 refrigerator one of two ways:
• Plan A: 10%/a simple interest for 2 years • Plan B: 5%/a compounded quarterly for 2 years Which plan should she choose? Justify your answer. After three years.06
for his investment to grow to $25 000?
7. Eric can redeem the bond in 7 years. in 5 years.
8. Ted used the exponential function A(n) 5 5000 3 1.
10. Eric bought a $1000 Canada Savings Bond that earns 5%/a compounded
annually. so the bank makes her two offers: • Option 1: Borrow the money at 10%/a compounded quarterly for the full term. If. The amounts of the investment at the end of the first three years are shown at the right. Chris invests $10 000 at 7. How long will it take
Year 1 2 3
Total Amount $4240.2
5. Determine the principal.
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Discrete Functions: Financial Applications
491
.
and you find these rates: • 6. The interest rate on this account is then increased by 0.45/a compounded monthly Rank the rates from most to least return on your investment.2% each year. On July 1. her grandparents deposited $500 into a savings
account that earns 4.55%/a compounded semi-annually • 6. Liz decides to save money to buy an electric car. What total amount of money will she have at the end of the 10th year?
20.14. 1996.5%/a compounded quarterly • 6. she moved the total amount to a new account that paid 8%/a compounded quarterly. Show how the calculations of
C
simple and compound interest are related to functions and sequences. They deposited the same amount on her 5th.3% 4.6%/a compounded annually • 6. 2008. 2001. You are searching different banks for the best interest rate on an investment. Create a mind map for the concept of interest.
Rate of Compound Interest per Year Compounding Period semi-annually monthly quarterly
a) b) c)
6. Anna invested $2000 in an account that earned 6%/a
compounded monthly.2%
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Chapter 8
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.2% 3. Round to two decimal places. and 15th birthdays.8%/a compounded semi-annually. On the day Rachel was born. Determine the balance in her account on January 1. assuming that compounding occurs annually. Bernie deposited $4000 into an account that pays 4%/a compounded
quarterly during the first year.
16. Determine the balance in the account on Rachel's 18th birthday.8%/a compounded monthly. Calculate the balance in Bernie's account after three years. Calculate the effective annual interest rate for each loan. She invests $500 every
6 months at 6. An effective annual interest rate is the interest rate that is equivalent to the
given one.
18. 10th.
17.
Extending
19. On July 1.
15.
06(1.06(1.06P 1 0.
Determine the present value of Anton's parents' investment if it must be worth $15 000 ten years from now.
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. They can earn 6%/a compounded annually on their investment.06P(1.062P
The interest rate is 6%/a compounded annually.
Tina's Solution: Working Backward
Time now P=? 1 2 3 8 9 10 I drew a timeline for the situation.06) 5 1.
• graphing calculator • spreadsheet software
LEARN ABOUT the Math
Anton's parents would like to put some money away so that he will have $15 000 to study music professionally in 10 years. I calculated the interest earned and the amount at the end of the first year.3
GOAL
Compound Interest: Present Value
YOU WILL NEED
Determine the present value of an amount being charged or earning compound interest. since P is used for principal. I calculated the interest earned and the amount at the end of the second year.06P(1 1 0.06P A 5 P 1 0. PV is used for present value instead of P.06P end of 2nd year: I 5 0.
? How much money should Anton's parents invest now so that it will grow to $15 000 in 10 years at 6%/a compounded annually?
EXAMPLE
1
Selecting a strategy to determine the principal for a given amount
present value the principal that would have to be invested now to get a specific future value in a certain amount of time.
For the second year.06P) A 5 1.06) 5 1.8. interest is earned on the amount at the end of the first year.06P) 5 1.
A = $15 000 i = 6%/a compounded annually
end of 1st year: I 5 0.06P 5 1.
92 now to get $15 000 in 10 years. The amounts at the end of each year form a geometric sequence with common ratio 1..
end of 10th year: 15 000 5 1. Since Anton's parents want $15 000 at the end of 10 years.063P.92 Anton's parents would need to invest $8375. I got 106% of what I started with. To calculate the principal that Anton's parents would have to invest.
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.. 1. . and solved for PV. 1. I rearranged the formula. I set the 10th term of the sequence equal to $15 000 and solved for P.0610P
Since 6% is added at the end of each year. so I wrote the amount as an exponential formula.06 gives the next term of the sequence.062P.92 now to get $15 000 in 10 years.
Jamie's Solution: Representing the Formula for the Amount in a Different Way
A 5 $15 000 i 5 6% 5 0. The amount at the end of the 10th year has an exponent of 10.1.92 Anton's parents would need to invest $8375.0610P P5 15 000 1.06P. So multiplying by 1.06. An investment earning compound interest grows like an exponential function.06 n 5 10 A 5 PV(1 1 i) n
Anton's parents invest a certain amount and let it grow to $15 000 at 6%/a compounded annually. substituted.0610
8 $8375. . 1.
PV 5 5
A (1 1 i) n 15 000 (1 1 0.06) 10
8 $8375.
Next I calculated the present value of the $20 000 at the given interest rate.3
Reflecting
A.4%/a compounded quarterly I calculated the interest rate Monica would be charged each compounding period and how many periods the loan would last.
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.4%/a compounded quarterly. B.81 5 $5440.19 The most Monica can borrow is $14 559.016) 20
8 $14 559.81.19 in interest. she will pay $5440.
I determined the interest by subtracting the present value from the amount.
How are the problems of determining the present value and the amount of an investment related? Based on Example 1.016 4 n 5 5 3 4 5 20 PV 5 5 A (1 1 i) n 20 000 (1 1 0.064 i5 5 0.81 I 5 A 2 PV 5 $20 000 2 $14 559.
0. What is the maximum amount that she can borrow and how much interest will she pay if she doesn't pay anything back until the end of the 5 years?
Kwok's Solution
Time now P=? 1 2 4 5 I drew a timeline for the situation. Monica wants to repay the loan in 5 years.8. but doesn't want the amount she pays back to be more than $20 000. Her bank will charge her 6. which method do you prefer to use to calculate the present value? Why?
APPLY the Math
EXAMPLE
2
Solving a problem involving present value
Monica wants to start a business and needs to borrow some money.
A = $20 000 i = 6.
What annual interest rate. and the number of years. To calculate i.
I graphed both functions on the same graph and used the calculator to find the point of intersection. see Technical Appendix. B-12. I thought of the intersection of two functions: 50 000 Y1 5 and Y2 5 5000. X (1 1 100 ) 40 I entered these into my graphing calculator.
Philip's Solution: Using a Graphing Calculator
PV 5 $5000 n 5 40 A 5 50 000 PV 5 5000 5 A (1 1 i) n 50 000 (1 1 i) 40
I knew the principal.EXAMPLE
3
Selecting a strategy to determine the interest rate
Tony is investing $5000 that he would like to grow to at least $50 000 by the time he retires in 40 years.
Tony would need to get at least 5. I used Y1 and Y2 for the present value and X for the interest rate. the amount (or future value). will provide this? Round your answer to two decimal places. compounded annually. I wrote the formula for the present value and substituted the given information.93%/a compounded annually to reach his goal.
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.
Tech
Support
For help using a graphing calculator to find the point of intersection of two functions.
I used the inverse operation of raising something to the 40th power. i is the interest rate per compounding period. where PV is the calculated using the formula PV 5 n or PV 5 A(1 1 i)
(1 1 i)
present value. A. PV.0593 i 5 0. then I divided both sides by 5000. and n is the number of compounding periods.
Need to Know
• The present value of an investment earning compound interest can be A 2n .93% Tony would need to get at least 5. is called the present value.
5000(1 1 i) 40 5 50 000 (1 1 i) 40 5 10
40
1 1 i 8 1.
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. I wrote the formula for the present value.0593 i 5 5.8. the amount (or future value). and rearranged to solve for i.3
Derek's Solution: Using the Formula
PV 5 $5000 n 5 40 A 5 $50 000 PV 5 5000 5 A (1 1 i) n 50 000 (1 1 i) 40
I knew the principal. or future value. then substituted the given information. To calculate i. which is determining the 40th root. A is the total amount. expressed as a decimal.
1 1 i 5 "10
In Summary
Key Idea
• The principal. that must be invested now to grow to a specific future value. I multiplied both sides of the equation by (1 1 i) 40. and the number of years.93%/a compounded annually to reach his goal.
K
Rate of Compound Interest per Year
Compounding Period annually semi-annually quarterly monthly
Time 4 years 3 years 15 years 9 years
Future Value $10 000 $6 200 $20 000 $12 800
a) b) c) d)
6% 8.2%
4.8%/a compounded monthly. After another five years.6% 4.CHECK Your Understanding
1. Rate of Compound Interest per Year
Compounding Period annually semi-annually quarterly monthly
Time 10 years 5 years 15 years 100 years
Future Value $10 000 $100 000 $23 000 $2 500
a) b) c) d)
4% 6. he paid $5000 toward the principal and the interest. she repays $12 033. He borrowed the rest at an interest
rate of 18%/a compounded monthly. He would like
$15 000 in 10 years.52 for the principal and the interest. Two years later. he paid the remainder of the principal and the interest. For each investment. which totalled $5000. After
5 years. Kevin can invest at
5%/a compounded annually and Lui can invest at 4. Chandra borrows some money at 7. Calculate the present value of each investment. he paid another $5000. determine the present value and the interest earned. Who has to invest more money to reach his goal?
PRACTISING
3. Rico can invest money at 10%/a compounded quarterly.2%/a compounded annually. How much does he need to invest now?
7. Three years
later. Nazir saved $900 to buy a plasma TV. How much money did Chandra borrow?
5. Kevin and Lui both want to have $10 000 in 20 years.16%/a compounded quarterly. How much did the TV originally cost?
6. he paid $1429. How much money did he originally borrow?
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.2% 5.50 for the principal and the interest. Colin borrowed some money at 7.2% 5.6%
2. After another two years.2% 6.
After 30 years. Who made the greater original investment and by how much? takes the principal and interest and reinvests it all at 7. he will probably get 5%/a compounded quarterly for the remainder of the term.
9. He can get only 3. each investment is worth $25 000. How much interest will you have paid on your purchase?
16.2%/a interest
12. must Tia invest her money? Round your answer to two decimal places. She would like
the money to grow to $12 000. compounded quarterly.9%/a compounded annually and David
A
invests some money at 6. • In 5 years. How long will she have to wait?
14. At the end of this time. he will probably get 4%/a compounded quarterly for the remainder of the term.4%/a compounded semi-annually. What annual interest rate.
15. Louise invests $5000 at 5.
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.3
8. Tia is investing $2500 that she would like to grow to $6000 in 10 years. the 5-year one or the 8-year one? b) How much does he need to invest?
10.
Extending
13.8.17 for 21 years and are charged 19.2%/a interest 2 compounded monthly. Describe how determining the present value of an investment is similar to
C
solving a radioactive decay problem. • In 8 years. would cause an investment
to triple every 10 years? Round your answer to two decimal places. Determine a formula for the present value of an investment with future value. Steve wants to have $25 000 in 25 years. You make monthly
payments of $268. a) Which guarantee should Steve choose. After 5 years. her investment is worth $14 784. she
T
11.56. Franco invests some money at 6. How much did Sally originally invest? compounded quarterly. Sally invests some money at 6%/a compounded annually. You buy a home entertainment system on credit.
A. compounded quarterly.9%/a compounded monthly. At
what annual interest rate. earning simple interest at a rate of i% per interest period for n interest periods.2%/a compounded quarterly for 6 more years. His bank will guarantee the rate for either 5 years or 8 years.
08) 9 8 $1999.00 10. How close is your estimate to the actual amount after 45 years?
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Chapter 8
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. then it will take 72 4 r years to double in value.24 9.00 14. the rule indicates that it will take 72 4 8 5 9 years for the money to double in value.21 11.67 14. predicted by the rule of 72 is very close to the actual doubling time.29 9.00 which is very close to double your money. How could you use the rule of 72 to determine how much a $1000
investment earning 8%/a compounded annually would be worth after 45 years?
2.00 18.00 23.01 8.00 7.66 35.
A
1 2 3 4 5 6 7 8 9 10 11 12 13
Interest Rate
1 2 3 4 5 6 7 8 9 10 11 12
B Double Time Using the Rule of 72 72.00 36.45 17. Suppose you invested $1000. Here is how the rule works: If an investment is earning r%/a compounded annually.04 7.55 6. So if you are earning 8%/a compounded annually. For years. banks. in years. investors.90 10. The spreadsheet below shows how the time.00
C Actual Double Time 69.20 6.40 12. Then the formula for the future value gives A 5 P(1 1 i) n 5 1000(1.00 24.12
1.00 8.64 6. and the general public have used "the rule of 72" to help approximate calculations involving compound interest.Curious Math
The Rule of 72
Working with compound interest can be difficult if you don't have a calculator handy.27 6.
grow at a constant rate.00 $80. r is the interest rate.8
Q:
A1:
Mid-Chapter Review
Study
FREQUENTLY ASKED Questions
What strategies can you use to solve problems involving simple interest?
Aid
Since simple interest is paid only on the principal.00 $180. A.00 $200. expressed as a decimal.00 $540.00 $640. where I is the interest. or future value.00 $700.
Examples 1 to 4.
• Try Mid-Chapter Review
Questions 1 to 3.
EXAMPLE
• See Lesson 8.00 $120.00 $740.00 $400.00 $320.00 $280. A.
Examples 1 to 4.00 $800.00 $560. where A is the total amount.00 $300.00 $700.00 $380. The formulas for I and A are linear in terms of time.00 $580.00 $100.00 $200.2.00 0 5 10 Time 15 20 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Dollars
Interest Amount
A2:
To calculate the interest.00 $340.
How can you solve problems involving compound interest? Study
Q:
A1:
Aid
Compound interest is paid at regular intervals.00 $300. the total amount.00 $0.00 $780. So A can be modelled by an exponential function in terms of time or a geometric sequence. where I and A are in dollars and t is time in years. use the formula A 5 P 1 Prt or A 5 P(1 1 rt). and the amount.
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. So interest is calculated on the principal and interest already earned.00 $860. or $20. You can model the interest and the amount by the functions I(t) 5 20t and A(t) 5 500 1 20t.00 $900.00 $800. I.00 $620.
• Try Mid-Chapter Review
Questions 4 to 6.00 $220. use the formula I 5 Prt.00 $500 Earning 4%/a Simple Interest $1 000.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 A Year B Interest $20. P is the principal. expressed in the same period as the interest rate.00 $40.00 $400.
• See Lesson 8.00 $240. To calculate the amount.00 $140. After each compounding period.00 $160.00 $360. grows by a fixed rate.00 $840.00 $680. both the interest.00 $900.00 $60. and t is time.00 $500.1.
If $500 is invested at 4%/a simple interest.00 C Amount $520. called compounding periods.00 $600.00 $760.00 $100.00 $820.00 $660. interest is earned each year. So I and A can be modelled by linear functions in terms of time or arithmetic sequences.00 $600.00 $720.00 $260. then 4% of $500.00 $880. and is added to the principal.
is the amount you start with (the principal). and n is the number of compounding periods.3.00 $973.00 $684.73 $720.00 $608. then 4% of the total amount is earned as interest each year.00 $1 000.00 $584.80 $540. These two values are related through the formulas A 5 P(1 1 i) n and A or PV 5 A(1 1 i) 2n PV 5 (1 1 i) n which are just rearrangements of each other.
Present value.84 $780. PV.42 $880. what is the difference between present value and future value?
Study
Aid
Q:
A:
• See Lesson 8. is the amount you end up with after the last compounding period.00 $769. i is the interest rate per compounding period. expressed as a decimal.97 $640. • Try Mid-Chapter Review Questions 7 to 9.93 $580. The amount grows exponentially. while future value.28 $660.
Examples 1 to 3.00 $540. where A is the future value in dollars and t is time in years.43 $560.EXAMPLE
If $500 is invested at 4%/a compounded annually. use the formula A 5 P(1 1 i) n.
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.00 $632.
For problems involving compound interest.00 $562.33 $600.00 $800.12 $700.00 $520. This formula is exponential in terms of the number of compounding periods.95 $840.56 $900.00 Dollars $600.00 $400.47 $800.00 $832. or future value.00 $200.00 $740. P is the principal.00 $800.04t.66 $620.00 Comparing Simple and Compound Interest $1 200.54 $760.00 $1 095.66 $680.00 $936.00 $900.00 $865. where A is the future value.00 $711.00 $657.00 0 5 10 Time 15 20 25 Simple Interest Compound Interest
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A2:
To calculate the total amount.91 $860.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 A Year B C Simple Interest Compound Interest $520.00 $1 053. you are determining the principal that should be invested to yield the desired amount. So the amount can be modelled by A(t) 5 500 3 1. faster than the linear growth of simple interest.52 $740.49 $820.00 $0. A. When you solve for PV.00 $1012.
Rate of Simple Interest per Year 6.8%
What interest rate is she being charged? Round your answer to two decimal places. was Sara being charged? Round your answer to two decimal places.78 when they took it out 65 years ago. How much money does Maria need to invest at
2.6% 14.
Lesson 8.9% 8.80
made at 7.2%/a compounded
1. The balances from his statements for the first three months are shown. A year and a half later. For each investment. compounded semi-annually.91 $9125.4% 22. Iris borrowed some money at a fixed rate of
How much interest is he being charged each month? b) How much did Tom borrow? c) What interest rate is he being charged?
a)
Lesson 8.8% 4. Clive inherits an investment that his grandparents
simple interest to earn $1200 interest?
3.
Rate of Compound Interest per Year 4.2%/a compounded quarterly in order to have $25 000 after 25 years?
8. The investment was worth $39 382.1
5.
Statement 1 2 3 Balance $1014. How long will it take for his investment to grow to $34 000?
6.20 $1143. but she forgot what the interest rate was.4%/a compounded semi-annually.3% 27. she pays off the loan.56
4. calculate the future value and
the total interest earned.2%/a
9.3
7. She pays
Principal
Time 15 years 16 months 80 weeks 150 days
a) b) c) d)
$5 400 $400 $15 000 $2 500
$500 and borrows the remaining amount. Calculate the interest being charged and the total amount. Tom borrows some money and is charged simple
interest on the principal.2
compound interest.Mid-Chapter Review
PRACTICE Questions
Lesson 8. which amounted to $2112.
Statement 1 2 Time 6 months 1 year Balance $8715. How long would you have to invest $5300 at 7. She knew that the interest was compounded semi-annually. Sara buys a washer and dryer for $2112. George invests $15 000 at 7. What annual interest rate.60 $1079.5 years
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Discrete Functions: Financial Applications
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. b) How much did Iris borrow?
a)
a) b) c) d)
Principal $6 300 $14 000 $120 000 $298
Compounding Period annually semi-annually quarterly monthly
Time 7 years 10.7% 9.1%
monthly. Each situation represents a loan being charged simple
interest.5 years 44 years 1. The balances of her first two statements are shown. How much did Clive's grandparents invest?
9.
Complete the entries under Date Invested up to Dec. 1. 31.8. 1. 31. Copy the data already entered.
?
A.00 $1 000. 2014.
How much will her investments be worth 10 years later. 2017. 2015. 31.5%/a compounded annually. 2007. How is the value in cell D3 (FV9) related to the value in cell D2 (FV10)? How is the value in cell D4 (FV8) related to the value in cell D3 (FV9)? Use the pattern from part D to complete the rest of the entries under Value on Jan.
Fill in cells D3 and D4 to show what the investments made on Dec. 31. 31. 2014 B Amount Invested $1 000. 31. on January 1. What type of sequence do the values on Jan. 1. and Dec. 2007. How would you calculate each of the future values FV1 to FV10?
i = 6. Her first deposit is on December 31.00 C Number of Years Invested 0 1 2 D Value on Jan. 1. F. E. 2015 Dec.
A 1 2 3 4 Date Invested Dec.
Set up a spreadsheet with columns as shown. 2017 form?
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504
Chapter 8
.
INVESTIGATE the Math
Christine decides to invest $1000 at the end of each year in a Canada Savings Bond earning 6. 2017 $1 000. 2017?
Copy the timeline shown. respectively will be worth on Jan.4
YOU WILL NEED
Annuities: Future Value
• graphing calculator • spreadsheet software
GOAL
Determine the future value of an annuity earning compound interest. 2016 Dec. D.00
C. 2017.5%/a compounded annually
Compounding 0 Period Payment $0
1
2
3
8
9
10
$1000 $1000 $1000 $1000 $1000 $1000
Amount of each payment at the end of the term FV10 FV9 FV8 FV3 FV2 FV1
B.00 $1 000.
024)37 1000(1 + 0. FV. ordinary annuity. A simple annuity is an annuity in which the payments coincide with the compounding period.024)36 1000(1 + 0.
The values of all of the investments at the end of each year for 10 years formed a specific type of sequence.024)38 1000(1 + 0.024)39
Payment $0
$1000 $1000 $1000 $1000
$1000 $1000 $1000
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Discrete Functions: Financial Applications
505
.
Reflecting
H.024)2 1000(1 + 0.8%/a compounded semi-annually Compounding Period 0 1 2 19 20 Amount of each payment at the end of the term 1000 1000(1 + 0. each annuity in this chapter is a simple. What will be the future value.
i = 4. Unless otherwise stated.4
G. and I represented the amount of each investment.024)1 1000(1 + 0. or conversion period. It didn't earn interest during the first compounding period because it was deposited at the end of that period. How is the total value of the annuity at the end of 10 years related to a series? How could you use the related series to solve problems involving annuities?
I.
Calculate the total amount of all the investments at the end of 10 years for this annuity.024 2
n 5 20 3 2 5 40
Since the interest is paid semi-annually.
annuity a series of payments or investments made at regular intervals.8%/a compounded semi-annually for the next 20 years.8. I calculated the interest rate per compounding period and the number of compounding periods. What will be the future value of his annuity? b) You plan to invest $R at regular intervals in an annuity that earns i% compounded at the end of each interval.048 5 0. The last $1000 investment earned no interest because it was deposited at the end of the term. An ordinary annuity is an annuity in which the payments are made at the end of each interval. The first $1000 investment earned interest over 39 periods. I drew a timeline of the investments for each compounding period. of your annuity after n intervals?
Barbara's Solution
a) i 5
0.
APPLY the Math
EXAMPLE
1
Representing the future value of an annuity earning compound interest as a series
a) Hans plans to invest $1000 at the end of each 6-month period in an annuity
that earns 4.
1 R(1 1 i) n22 1 R(1 1 i) n21 a(r n 2 1) r21 R 3(1 1 i) n 2 14 (1 1 i) 2 1
Sn 5
5
The future value of an annuity in which $R is invested at the end of each of n regular intervals earning i% of compound interest per interval is (1 1 i) n 2 1 b.. The last $R investment earned no interest.. I rounded to the nearest cent. . R(1 1 i) 2..024) 2 1 ..
R.08 The future value of Hans's annuity at the end of 20 years is $65 927.024) 1 1000(1.
S40 5 1000 1 1000(1.024) 2. 1000(1. . R(1 1 i) n22..024) 38..
b)
Compounding Period 0 1 $R 2 $R 3 $R n –2 $R i% per compounding period n –1 $R n $R
Payment $0
Amount of each payment at the end of the term R R(1 + i )1 R(1 + i )2 R(1 + i )n–3 R(1 + i )n–2 R(1 + i )n–1
I drew a timeline of the investments for each compounding period to show the amount of each investment. . R(1 1 i) n21 Sn 5 R 1 R(1 1 i) 1 R(1 1 i) 2 1 .024 2 1
8 $65 927. 1000(1.024) 38 1 1000(1. The total amount of all these investments is the first 40 terms of a geometric series.024) 39
The future values of all of the investments form a geometric sequence with first term $1000 and common ratio 1.4
1000. The total amount of all these investments is the first n terms of a geometric series.02440 2 1) 1. FV 5 R 3 a i
5R3a
(1 1 i) n 2 1 b i
506
Chapter 8
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.08. where i is expressed as a decimal. . 1000(1. 1 1000(1.. R(1 1 i).024). I used the formula for the sum of a geometric series to calculate the total amount of all the investments.024) 39 Sn 5 S40 5 a(r n 2 1) r21 1000(1.024.8.. The first $R investment earned interest n 2 1 times. 1000(1. I used the formula for the sum of a geometric series to calculate the total amount of all of Hans's investments. The values of all of the investments form a geometric sequence with first term R and common ratio 1 1 i.
Sn 5 S100 5
a(r n 2 1) r21 500(1.013)99
I drew a timeline of the investments for each quarter to show the amounts of each investment..013) 99
The values form a geometric sequence with first term $500 and common ratio 1.052 5 0.013100 2 1) 1. 1 500(1. .013)96 500(1 + 0. I used the formula for the sum of a geometric series to calculate the total amount of all of Chie's investments. How much will her annuity be worth in 25 years?
Kew's Solution: Using a Geometric Series
i5 0.
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Discrete Functions: Financial Applications
507
.
500. I rounded to the nearest cent. 500(1.013) 2 1 .013)98 500(1 + 0.013)97 500(1 + 0.2%/a compounded quarterly.013)1 500(1 + 0.013) 1 500(1.013) 98.013).013.
i = 5.013 4
First I calculated the interest rate per compounding period and the number of compounding periods. .8.. 500(1. 500(1. I calculated the value of each investment at the end of 25 years.2%/a compounded quarterly Compounding Period 0 1 24 25
n 5 25 3 4 5 100
Payment $0 $500 $500 $500 $500
$500 $500 $500 $500 $500
Amount of each payment at the end of the term 500 500(1 + 0.013) 99 S100 5 500 1 500(1.013)2 500(1 + 0.91 The total amount of all of Chie's investments at the end of 25 years will be $101 487..013) 98 1 500(1..013) 2.013)3 500(1 + 0. 500(1.91.4
EXAMPLE
2
Selecting a strategy to determine the future value of an annuity
Chie puts away $500 every 3 months at 5.013)4 500(1 + 0. The total amount of all these investments is the first 100 terms of a geometric series.013 2 1
8 $101 487.
ordinary annuity.6%/a compounded monthly. and n into the formula for the future value of an annuity.008 12 (1 1 i) n 2 1 b i
I calculated the interest rate per compounding period and the number of compounding periods.013 4 n 5 25 3 4 5 100 5 500 3 a 8 $101 487.008) 360 2 1 b 0.013
I rounded to the nearest cent. How much should he deposit each month?
Chantal's Solution
i5 0. The future value of the annuity is $500 000. and n into the formula for the future value of a simple.008
I substituted the values of FV.91.013) 100 2 1 b 0.052 5 0. He would like to have $500 000 in the account at the end of 30 years.
508
Chapter 8
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. i.Tina's Solution: Using the Formula for the Future Value of an Annuity
R 5 $500 (1 1 i) n 2 1 b FV 5 R 3 a i i5 0.096 5 0. I substituted the values of R.
(1 1 0. i.
EXAMPLE
3
Selecting a strategy to determine the regular payment of an annuity
Sam wants to make monthly deposits into an account that guarantees 9.91
I calculated the interest rate per compounding period and the number of compounding periods.
The future value of Chie's annuity will be $101 487.
FV 5 $500 000 500 000 5 R 3 a
FV 5 R 3 a
n 5 30 3 12 5 360
(1 1 0.
00 $750.00 $427.
To solve for R.80 Sam would have to deposit $240.94 $6.413 2076.72 –$22. She agrees to repay the loan by making equal monthly payments of $750 until the balance is paid off.55 $94 028.50 $323. The part of the principal that is paid off with each payment is $750.00 $750. which would take about 15 years and 8 months.59 $423.95 $325.
1
1 2 3 4 5 6 7 188 187 188 189 190
Payment Number 1 2 3 4 5 184 185 186 187 188
A
Payment $750.55 $13.34 $733.50 $426. The 15 meant 15 years. and to fill down.00 $94 677.66 $16. If Nahid is being charged 5. see Technical Appendix.75 $740. B-21.50 $322.
After 188 payments. less the interest.26
C
Principal Paid $322.00 $750.00 $750. The new balance is the old balance.413.00 $750.4
500 000 8 R 3 2076.14 $93 701. I calculated the number of years needed to make 188 payments by dividing by 12.92 $2 944.13 $421. less the part of the principal that is paid.00 "=E3*0.17 $1 468. The interest is always charged
1 2 3 4 Payment Number 1 2
A
Payment
B
Interest Paid Principal Paid Balance $95 000.00 $750.4% since it is compounded monthly. the balance is close to zero.00 $750.50 $94 677.667 0.00 $750. how long will it take her to pay off the loan?
Zak's Solution
I set up a spreadsheet to calculate the balance after every payment.02
E
I used the FILL DOWN command to complete the spreadsheet until the balance was close to zero.25 $9.
Tech
Support
For help using a spreadsheet to enter values and formulas.667 3 12 months 8 8 months Nahid can pay off the loan after 188 payments.27 $93 372.00
B
Interest Paid $427.06 $743.667 of a year was.45 $736.
EXAMPLE
4
Selecting a strategy to determine the term of an annuity
Nahid borrows $95 000 to buy a cottage.4%/a compounded monthly. I rounded to the nearest cent.00 $750.413 2076.054/12" "=B4-C4" "=E3-D4"
C
D
E
on the balance and is 12 of 5.413 R 5 $240.00 $750.61 $3. so I had to figure out what 0.87 $328.
t5
188 12
8 15.
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.39 $746.41 $326.80 into the account each month in order to have $500 000 at the end of 30 years. I got a value greater than 15.74
D
Balance $95 000.05 $424.92 $2 208.8.11 $724.413 500 000 5R3 2076.50 $94 353.50 $750. since there are 12 payments each year. I divided both sides of the equation by 2076.
In Summary
Key Ideas
• The future value of an annuity is the sum of all regular payments and interest earned. i is the interest rate per compounding period. • The payment interval of an annuity is the time between successive payments. • The formula for the sum of a geometric series can be used to determine the future value of an annuity. and n is the number of compounding periods.
3 2 1
Need to Know
• A variety of technological tools (spreadsheets.
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. i% per compounding period Compounding 0 Period Payment $0 1 $R 2 $R 3 $R n 2 $R n 1 $R n $R Amount of each payment at the end of the term R R(1 + i)1 R(1 + i)2 R(1 + i)n R(1 + i)n R(1 + i)n • The future value can be written as the geometric series FV 5 R 1 R(1 1 i) 1 R(1 1 i) 2 1 . expressed as a decimal. • The term of an annuity is the time from the first payment to the last payment. i is the interest rate per compounding period.. 1 R(1 1 i) n22 1 R(1 1 i) n21 where FV is the future value. graphing calculators) can be used to solve problems involving annuities. and n is the number of compounding periods. R is the regular payment. • The formula for the future value of an annuity is FV 5 R 3 a (1 1 i) n 2 1 b i
where FV is the future value. R is the regular payment each compounding period.. expressed as a decimal.
Mike wants to invest money every month for 40 years. b) What type of sequence do the values form? c) Determine the total amount of all of Eric's investments.5%
Regular Payment
Compounding Period monthly quarterly semi-annually annually
Time 50 years 15 years 8 years 10 years
a) b) c) d)
$100 per month $1500 per quarter $500 every 6 months $4000 per year
3. Josh borrows some money on which he makes monthly payments of $125.3% 3.
a)
2. Lois invests $650 every 6 months at 4.6% 4.1%/a compounded monthly
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.2%/a compounded monthly b) 5. For each investment option.4
CHECK Your Understanding
1.2% 5.8% 8%
Compounding Period annually semi-annually quarterly monthly
Time 10 years 3 years 7 years 35 years
a) b) c) d)
$1500 per year $250 every 6 months $2400 per quarter $25 per month
6. Rate of Compound Interest per Year 3.43
for 3 years. Eric invests $2500 at 8. He would like to have
A
$1 000 000 at the end of the 40 years. If the interest rate is 5. Calculate the future value of each annuity. Calculate the future value of each annuity.4%/a compounded monthly.
K
Regular Payment
Rate of Compound Interest per Year 6. Each year. how much does he need to invest each month? a) 10. what will be the total amount of all of the payments at the end of the 3 years?
PRACTISING
5. How much interest will she earn after the 25th year?
4.6% 6.6%/a compounded semi-annually for
25 years.
Calculate the value of each of the first four investments at the end of 25 years.8.2%/a compounded annually for 25 years.6% 4.
Sonja and Anita want to make equal monthly payments for the next 35 years. compounded monthly. Greg borrows $123 000 for the purchase of a house.8%/a compounded monthly. At the end of that time.6%/a compounded monthly.2% 7.8%/a compounded monthly. related to interest. How many equal monthly payments would you have to make to get
100 times the amount you are investing each month if you are earning 8.7. At what annual interest rate. What monthly payments will Greg have to make?
14. Sonja's bank will give her 6. He would like to have at least $6500 at the end of his investment.2%/a compounded
quarterly. The bank is charging Greg 6.
At the end of that time. each person would like to have $500 000. He plans to make
regular monthly payments over the next 20 years to pay off the loan.2% 7. She decides to
make monthly payments of $250. sequences. how much more money will she have at the end of 35 years?
10. does Jamal need to invest to reach his goal? Round your answer to two decimal places. Kiki has several options for investing $1200 per year: Rate of Compound Interest per Year 7.
8. Draw a mind map for the concept of future value of annuities. Carmen borrows $10 000 at 4.
T
he would like to have $25 000. Jamal wants to invest $150 every month for 10 years. Show how it is
C
Extending
12. a) How long will it take her to pay off the loan? b) How much interest will she pay over the term of the loan?
13.2% 7.6%/a compounded monthly. How long will he need to make regular payments?
9.4%/a compounded monthly?
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. which investment would be best? Justify your reasoning. and series.2%
Regular Payment
Compounding Period monthly quarterly semi-annually annually
a) b) c) d)
$100 per month $300 per quarter $600 every 6 months $1200 per year
Without doing any calculations.
11. a) How much more per month does Sonja have to invest? b) If Anita decides to invest the same monthly amount as Sonja. Kenny wants to invest $250 every three months at 5. Anita can invest through her work and earn 10.
?
A. D.00 $1 000.5%/a compounded annually Compounding 0 Period Payment $0 Present value of each payment PV1 PV2 PV3 PV23 PV24 PV25 1 2 3 23 24 25
$1000 $1000 $1000 $1000 $1000 $1000
B. Use the relationship among the present values to complete the rest of the entries under Present Value.8.
How much would Kew need to invest now if he wanted to provide $1000 at the end of the 1st year? How much would Kew need to invest now if he wanted to provide $1000 at the end of the 2nd. C.00 $1 000. Use the values in the Present Value column to determine how much Kew would need to invest now in order to provide the scholarships for the next 25 years. 3rd.00
C Present Value
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.00 $1 000.00 $1 000.00 $1 000. He would like the investment to provide $1000 for scholarships at his old high school at the end of each year for the next 25 years.5
GOAL
Annuities: Present Value
YOU WILL NEED
Determine the present value of an annuity earning compound interest. How would you calculate each of the present values PV1 to PV25?
i = 5. Enter your values of PV1 and PV2 in the Present Value column.00 $1 000.
• graphing calculator • spreadsheet software
INVESTIGATE the Math
Kew wants to invest some money at 5. respectively? How is the present value after 2 years (PV2) related to the present value after 1 year (PV1)? Set up a spreadsheet with columns as shown at the right. G.
How much should Kew invest now?
Copy the timeline shown.5%/a compounded annually.
1 2 3 4 5 6 7 8 9 10 11
A Year
1 2 3 4 5 6 7 8 9 10
B Scholarship Payment
$1 000.00 $1 000.00 $1 000.00 $1 000. E. F. and 4th years.
I.083)2 500 (1 + 0.
Payment $0 $500 Present value of each payment 500 (1 + 0.Reflecting
H.083)10
$500 $500
$500 $500 $500
PV 5 PV1 5 PV2 5 PV3 5 (
A (1 1 i)n 500 (1.083)2 500 (1.083)9 500 (1 + 0.083) 500 (1.
What type of sequence do the present values in part F form? Describe a method that you could have used to solve this problem without using a spreadsheet.
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.083)8 500 (1 + 0.3%/a compounded annually 2 3 8 9 10 I drew a timeline showing the $500 payments for the next 10 years.
APPLY the Math
EXAMPLE
1
Representing the present value of an annuity earning compound interest as a series
a) How much would you need to invest now at 8.083)3
I considered each payment separately and used the present-value formula to determine how much would need to be invested now to provide each $500 payment.083)3 500 (1 + 0.083)1 500 (1 + 0.3%/a compounded annually
to provide $500 per year for the next 10 years?
b) How much would you need to invest now to provide n regular payments
of $R if the money is invested at a rate of i% per compounding period?
Tara's Solution
a)
Compounding Period 0 1 i = 8.
072 133.072
1 2 (1 1 0.0055 12
I calculated the interest rate per compounding period and the number of compounding periods.072. he will take 20 years to pay off the loan.0055
133.60 in interest. Len will have paid $160 706. I rounded to the nearest cent.066 5 0.60 I 5 A 2 PV 5 $ 360 706. I substituted the values of PV.
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.EXAMPLE
3
Selecting a strategy to determine the regular payment and total interest of an annuity
Len borrowed $200 000 from the bank to purchase a yacht. i.60 Over the 20-year term of the loan.
200 000 8 R 3 133.60 2 $ 200 000 5 $ 160 706.072 R 8 1502. If the bank charges 6.
I calculated the total amount that Len will have paid over the 20-year term.0055)2240 b 0.6%/a compounded monthly. I determined the interest by subtracting the present value from the total amount that Len will have paid.
a) How much will each monthly payment be? b) How much interest will he have paid over the term of the loan?
Jasmine's Solution: Using the Formula
a)
200 000 5 R 3 a
1 2 (1 1 i)2n b PV 5 R 3 a i n 5 20 3 12 5 240 PV 5 $200 000
i5
0. I divided both sides of the equation by 133.072 200 000 5R3 133.
5 $ 360 706. and n into the formula for the present value of an annuity.
b) A 5 1502.94 per month for 20 years to pay off the loan.94 Len will have to pay $1502.94 3 240
To solve for R.
expressed as a decimal.8. and n is the number of compounding periods. 1 $R 2 $R 3 $R n 2 $R n 1 $R n $R
Need to Know
• The formula for the present value of an annuity is
where PV is the present value. It is the sum of all present values of the payments and can be written as the geometric series PV 5 R 3 (1 1 i)21 1 R 3 (1 1 i)22 1 R 3 (1 1 i)23 1 ...
PV 5 R 3 a
1 2 (1 1 i)2n b i
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. i is the interest rate per compounding period. 1 R 3 (1 1 i)2n where PV is the present value. R is the regular payment. expressed as a decimal.5
In Summary
Key Ideas
• The present value of an annuity is the value of the annuity at the beginning of the term. and n is the number of compounding periods. i is the interest rate per compounding period. i% per compounding period Compounding 0 Period Payment $0 Present value of each payment R (1 + i)1 R (1 + i)2 R (1 + i)3 R (1 + i)n 2 R (1 + i)n 1 R (1 + i)n • The formula for the sum of a geometric series can be used to determine the present value of an annuity. R is the regular payment each compounding period.
iii) Calculate the present value of the annuity. Calculate the present value of each annuity. Each situation represents a simple. Calculate the amount of the original loan.
Rate of Compound Interest per Year 9% 8% 8%
Regular Payment
Compounding Period annually semi-annually quarterly
Time 7 years 3.2% 23.4%
Compounding Period annually semi-annually weekly monthly
Time 5 years 12 years 100 weeks 1 2 years 2
a) b) c) d)
$5000 per year $250 every 6 months $25.6% 6.50 per week $48.73 per quarter $183.
Rate of Compound Interest per Year 3.
K
Regular Payment
Rate of Compound Interest per Year 7.17 per month
10 years
2.6%
Regular Payment
Compounding Period annually semi-annually quarterly monthly
Time 5 years 9 years 3 1 years 2
a) b) c) d)
$650 per year $1200 every 6 months $84.7% 9. Write the series that represents the amount of the original loan. Calculate the interest paid.2% 4.
i) ii) iii) iv)
Draw a timeline to represent the amount of the original loan. ii) Write the present values of the payments as a series.
i) Calculate the present value of each payment.4% 3.50 per month
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. ordinary annuity.CHECK Your Understanding
1.8% 5. Each situation represents a loan.5 years 2 years
a) b) c)
$8000 per year $300 every 6 months $750 per quarter
PRACTISING
3.
8.5
4. You want to buy a $1300 stereo on credit and make monthly payments over
2 years. If the store is charging you 18%/a compounded monthly, what will be your monthly payments?
5. Lily wants to buy a snowmobile. She can borrow $7500 at 10%/a
compounded quarterly if she repays the loan by making equal quarterly payments for 4 years. a) Draw a timeline to represent the annuity. b) Write the series that represents the present value of the annuity. c) Calculate the quarterly payment that Lily must make.
6. Rocco pays $50 for a DVD/CD player and borrows the remaining amount.
He plans to make 10 monthly payments of $40 each. The first payment is due next month. a) The interest rate is 18%/a compounded monthly. What was the selling price of the player? b) How much interest will he have paid over the term of the loan?
7. Emily is investing $128 000 at 7.8%/a compounded monthly. She wants to
withdraw an equal amount from this investment each month for the next 25 years as spending money. What is the most she can take out each month?
8. The Peca family wants to buy a cottage for $69 000. The Pecas can pay
$5000 and finance the remaining amount with a loan at 9%/a compounded monthly. The loan payments are monthly, and they may choose either a 7-year or a 10-year term. a) Calculate the monthly payment for each term. b) How much would they save in interest by choosing the shorter term? c) What other factors should the Pecas consider before making their financing decision?
9. Charles would like to buy a new car that costs $32 000. The dealership offers
A
to finance the car at 2.4%/a compounded monthly for five years with monthly payments. The dealer will reduce the selling price by $3000 if Charles pays cash. Charles can get a loan from his bank at 5.4%/a compounded monthly. Which is the best way to buy the car? Justify your answer with calculations. compounded monthly. She has a choice between a 5-, 10-, or 15-year term. a) Determine the monthly payment for each term. b) Calculate how much interest Nina would pay in each case.
10. To pay off $35 000 in loans, Nina's bank offers her a rate of 8.4%/a
11. Pedro pays $45 for a portable stereo and borrows the remaining amount. The
loan payments are $25 per month for 1 year. The interest rate is 18.6%/a compounded monthly. a) What was the selling price of the stereo? b) How much interest will Pedro have paid over the term of the loan?
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12. Suzie buys a new computer for $2500. She pays $700 and finances the rest at
$75.84 per month for 21 years. What annual interest rate, compounded 2 monthly, is Suzie being charged? Round your answer to two decimal places.
Leo's financial advisor tells him that he should take out a regular amount quarterly when he retires. If Leo has 20 years until he retires and wants to use the investment for recreation for the first 10 years of retirement, what is the maximum quarterly withdrawal he can make?
14. Charmaine calculates that she will require about $2500 per month for the
first 15 years of her retirement. If she has 25 years until she retires, how much should she invest each month at 9%/a compounded monthly for the next 25 years if she plans to withdraw $2500 per month for the 15 years after that?
15. A lottery has two options for winners collecting their prize:
T
• Option A: $1000 each week for life • Option B: $660 000 in one lump sum The current interest rate is 6.76%/a compounded weekly. a) Which option would you suggest to a winner who expects to live for another 25 years? b) When is option A better than option B? terms. Give examples. a) a lump sum or an annuity b) future value or present value
16. Classify situations and factors that show the differences between each pair of
C
Extending
17. Stefan claims that he has found a different method for calculating the present
value of an annuity. Instead of calculating the present value of each payment, he calculates the future value of each payment. Then he calculates the sum of the future values of the payments. Finally, he calculates the present value of this total sum. a) Use Stefan's method to solve Example 1 (a). b) Create another example to show that his claim is true. Include timelines. c) Use the formula for present value to prove that Stefan's claim works for all annuities.
18. Kyla must repay student loans that total $17 000. She can afford to make
$325 monthly payments. The bank is charging an interest rate of 7.2%/a compounded monthly. How long will it take Kyla to repay her loans?
19. In question 14, Charmaine invested a fixed amount per month so that her
annuity would provide her with another monthly amount in her retirement. Derive a formula for the regular payment $R that must be made for m payments at an interest rate of i% per compounding period to provide for a regular withdrawal of $W after all the payments are made for n withdrawals.
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8.6
GOAL
Using Technology to Investigate Financial Problems
YOU WILL NEED
Use technology to investigate the effects of changing the conditions in financial problems.
• graphing calculator • spreadsheet software
INVESTIGATE the Math
Tina wants to buy a stereo that costs $566.47 after taxes. The store allows her to buy the stereo by making payments of $29.95 per month for 2 years.
?
A. B.
What annual interest rate, compounded monthly, is the store charging?
Draw a timeline for this situation. Will you be calculating present values or future values? Use a spreadsheet to set up an amortization schedule as shown.
1 Interest Rate Payment Number 2 3 0.01 1 4 "=B3+1" A B C
Payment
amortization schedule
F
D
Interest Paid
E
Principal Paid
Balance
$29.95 "=F2*A3" $29.95 "=F3*A3"
"=C3-D3" "=C4-D4"
$566.47 "=F2-E3" "=F3-E4"
a record of payments showing the interest paid, the principal, and the current balance on a loan or investment
C.
Use the COPY DOWN command to complete the spreadsheet so that 24 payments are showing. The spreadsheet shown here is set up with an interest rate of 1% per compounding period. Adjust the value of the interest rate to solve the problem. Enter the formula for the present value of the annuity into a graphing calculator, where Y is the (unknown) present value and X is the annual interest rate compounded monthly. Graph the equation in part D, as well as y 5 566.47. Use these graphs to solve the problem. On your graphing calculator, activate the TVM Solver. Enter the corresponding values and then solve the problem.
Tech
Support
For help creating an amortization schedule using a spreadsheet, see Technical Appendix, B-22.
D.
E. F. G.
Tech
Support
For help using the TVM Solver on a graphing calculator, see Technical Appendix, B-19.
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Reflecting
H. I.
Why could you not solve this problem easily with pencil and paper? Which of the three methods (the spreadsheet in parts B and C, the graphs in parts D and E, or the TVM Solver in parts F and G) used to solve the problem do you prefer? Why?
APPLY the Math
EXAMPLE
1
Selecting a tool to investigate the effects of varying the interest rate
Jamal has $10 000 to invest. Bank of North America offers an interest rate of 4.2%/a compounded monthly. Key Bank offers an interest rate of 5%/a compounded quarterly. How much longer will it take the money invested to grow to $50 000 if Jamal chooses Bank of North America?
Lina's Solution: Using Guess-and-Check
i5 0.042 5 0.0035 12
I first looked at the Bank of North America. Since interest is paid monthly, I divided the annual interest rate by 12 to get the interest rate per month. I substituted the values of i and P into the compoundinterest formula. I thought 10 years might be a good guess. That would give n 5 120 compounding periods. My guess was way too small, so I tried 40 years, which gives n 5 480 compounding periods. That guess was much closer. Eventually, I tried 460 months. It was a little low, so I tried 461 months.
I determined how long 461 months is in terms of years. First I divided 461 by 12 to get 38 years. Then I multiplied 0.417 by 12 to get 5 months. Next, I looked at Key Bank. Since interest is paid quarterly, I divided the annual interest rate by 4 to get the interest rate per quarter.
I substituted the values of i and P into the compoundinterest formula. Since it took Bank of North America 38 years to grow to $50 000, I used 35 years as my first guess for Key Bank because the interest rate is higher. 35 years is 35 3 4 5 140 quarters.
NEL
A 5 10 000(1.0125) 140
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Try n 5 129: 8 49 654.56 130 5 32.5 4 n 5 32 years and 6 months It will take 38 years and 5 months to get $50 000 if Jamal chooses Bank of North America. It will take 32 years and 6 months if he chooses Key Bank. So it will take almost 6 years longer to reach his goal if Jamal chooses Bank of North America. Try n 5 130: A 5 10 000(1.0125) 130 8 50 275.24
I determined how long 130 quarters is in terms of years. I divided 130 by 4 to get 32 years. I knew that 0.5 years is 6 months. This result was close, but a bit high. Eventually, I tried 129 quarters and then 130 quarters.
A 5 10 000(1.0125) 129
George's Solution: Using a Graphing Calculator
A 5 P(1 1 i) n Bank of North America: i5 0.042 5 0.0035 12 Key Bank: i5 0.05 5 0.0125 4
At Bank of North America, interest is compounded monthly. At Key Bank, interest is compounded quarterly. I calculated the interest rate per compounding period at each bank. Then I used the compound-interest formula to calculate the amounts. I entered the equations for the amounts into my graphing calculator, using Y1 and Y2 for the amounts for Bank of North America and Key Bank, respectively, and X for the number of compounding periods. I entered Y3 5 50 000.
A 5 10 000(1.0035) n
A 5 10 000(1.0125) n
I graphed the three equations and used the calculator to find the point of intersection of each exponential function with the horizontal line, which indicated when the amount of the investment had reached $50 000.
It will take about 460 months, or 38 years and 5 months, to get $50 000 if Jamal chooses Bank of North America. It will take about 129 quarters, or 32 years and 6 months, if he chooses Key Bank. So it will take almost 6 years longer to reach his goal if Jamal chooses Bank of North America.
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Coco's Solution: Using the TVM Solver
I entered the information on the investment with Bank of North America into the TVM Solver. Jamal pays into the account at the start, so the present value is 2 $10 000. Also, no regular payments are being made. This is a lump-sum investment, so I set PMT on my calculator to 0. I determined that it would take a bit more than 460 months, or 38 years and 5 months, to reach his goal with Bank of North America.
I entered the information on the investment with Key Bank into the TVM Solver. I determined that it would take a bit more than 129 quarters, or 32 years and 6 months, to reach his goal with Key Bank.
If Jamal chooses Bank of North America, it will take about 6 years longer to reach his goal.
EXAMPLE
2
Selecting a tool to investigate the effects of increasing the monthly payment
Lia borrows $180 000 to open a restaurant. She can afford to make monthly payments between $1000 and $1500 at 4.8%/a compounded monthly. How much sooner can she pay off the loan if she makes the maximum monthly payment?
Teresa's Solution: Using a Spreadsheet
I set up a spreadsheet to solve the problem. Since the interest is compounded monthly, I divided 4.8% by 12 to get the interest rate per month. For the $1000 minimum payment, I calculated the proportion of the principal paid for each payment. Then I subtracted that proportion from the previous balance to get the balance at the end of the next month. Next, I used the FILL DOWN command to complete the other rows.
I continued until the balance became negative, indicating that the loan was paid offMike's Solution: Using the TVM Solver
I entered the information on the loan into the TVM Solver on a graphing calculator. I entered the minimum monthly payment of $1000 and then used the calculator to determine the number of payments needed. I then changed the monthly payment to the maximum amount of $1500, and used the calculator to determine the number of payments neededNEL
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Communication
Tip
EXAMPLE
3
Sometimes you can make a large purchase by paying a small portion of the cost right away and financing the rest with a loan. The portion paid right away is called a down payment.
Selecting a tool to investigate the effects of paying more frequently
Sarah and John are both saving for a down payment on their first home. Both plan to save $2400 each year by depositing into an account that earns 4%/a. • John makes monthly deposits of $200 into an account on which the interest is compounded monthly. • Sarah makes annual payments of $2400 into an account on which the interest is compounded annually. Determine the difference in their account balances at the end of 5 years.
Jason's Solution
I used the TVM Solver on my graphing calculator and entered the information on John. I found that his balance would be $13 259.80 at the end of 5 years.
I repeated the same type of calculation, but this time with the information on Sarah. I found that her balance would be $12 999.17 at the end of 5 years.
$13 259.80 2 $12 999.17 5 $260.63 John's account will have $260.63 more than Sarah's after 5 years.
I subtracted to calculate the difference in the amounts.
EXAMPLE
4
Connecting the interest paid on a loan with time
You borrow $100 000 at 8.4%/a compounded monthly. You make monthly payments of $861.50 to pay off the loan after 20 years. How long will it take to pay off a) the first $25 000? b) the next $25 000? c) the next $25 000? d) the last $25 000? e) Why are the answers to parts (a) through (d) all different?
I used a spreadsheet to create an amortization schedule. I then used the FILL DOWN feature to complete the spreadsheet. I noticed that the balance is reduced to $74 743.48 after 106 months, so it took 8 years and 10 months to pay off the first $25 000. The balance is reduced to $49 625.69 after 166 months, so it took 60 months, or 5 years, to pay off the next $25 000. The balance is reduced to $24 624.00 after 208 months, so it took 42 months, or 3 years and 6 months, to pay off the next $25 000. The loan is paid off after 240 months, or 20 years. It takes 208 months to pay about $75 000, so I subtracted 208 from 240 to determine how long it takes to pay the last $25 000 of the loan. The last $25 000 takes 32 months, or 2 years and 8 months, to pay off.
b)
164 165 166 167 168
162 163 164 165 166
$861.50 $861.50 $861.50 $861.50 $861.50
$365.00 $361.53 $358.03 $354.50 $350.95
$496.50 $499.97 $503.47 $507.00 $510.55
$51 646.68 $51 146.71 $50 643.23 $50 136.24 $49 625.69
c)
206 207 208 209 210
204 205 206 207 208
$861.50 $861.50 $861.50 $861.50 $861.50
$195.99 $191.33 $186.64 $181.92 $177.16
$665.51 $670.17 $674.86 $679.58 $684.34
$27 332.96 $26 662.79 $25 987.93 $25 308.35 $24 624.00
d)
239 240 241 242 243
237 238 239 240 241
$861.50 $861.50 $861.50 $861.50 $861.50
$23.72 $17.86 $11.95 $6.01 $0.02
$837.78 $843.64 $849.55 $855.49 $861.48
$2 551.46 $1 707.82 $858.28 $2.78 –$858.70
e) It takes different lengths of time to pay off the same amount of
money because the interest paid is greater when the balance owed is greater. Less of the payment goes toward the principal.
In Summary
Key Idea
• Spreadsheets and graphing calculators are just two of the technological tools that can be used to investigate and solve financial problems involving interest, annuities, and amortization schedules.
Need to Know
• The advantage of an amortization schedule is that it provides the history of all payments, interest paid, and balances on a loan. • More interest can be earned if • the interest rate is higher • there are more compounding periods per year • If you increase the amount of the regular payment of a loan, you can pay it off sooner and save a significant amount in interest charges. • Early in the term of a loan, the major proportion of each regular payment is interest, with only a small amount going toward paying off the principal. As time progresses, a larger proportion of each regular payment goes toward the principal.
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CHECK Your Understanding
1. Use technology to determine how long it will take to reach each investment goal. Rate of Compound Interest per Year 8.3% 6.8% 12.4% 3.6%
Principal
Compounding Period annually semi-annually quarterly monthly
Future Value $13 000 $4 000 $4 500 $20 000
a) b) c) d)
$5 000 $2 500 $450 $15 000
2. Use technology to determine the annual interest rate, to two decimal places,
being charged in each loan. The compounding period corresponds to when the payments are made.
Principal Regular Payment $357.59 per year $1497.95 every 6 months $374.56 per quarter $29.62 per month Time 10 years 6 years 3 years 18 months
a) b) c) d)
$2 500 $15 000 $3 500 $450
PRACTISING
3. Trevor wants to save $3500. How much will he have to put away each month 4. Nadia borrows $120 000 to buy a house. The current interest rate is 6.6%/a
A
at 12.6%/a compounded monthly in order to have enough money in 2 1 years? 2 compounded monthly, and Nadia negotiates the term of the loan to be 25 years. a) What will be each monthly payment? b) After paying for 3 years, Nadia receives an inheritance and makes a onetime payment of $15 000 against the outstanding balance of the loan. How much earlier can she pay off the loan because of this payment? c) How much will she save in interest charges by making the $15 000 payment? • Bank A has offered them 6.6%/a compounded monthly. • Bank B has offered them 7.8%/a compounded monthly. How much more will they end up with by choosing the second offer? off a $25 000 loan. After 2 years, he is making a bit more money and decides to increase the monthly payment. If he pays $50 extra per month at the end of each 2-year period, how long will it take him to pay off the loan?
5. Lisa and Karl are deciding to invest $750 per month for the next 7 years.
K
8.6
7. Natalie borrows $150 000 at 4.2%/a compounded monthly for a period of
20 years to start a business. She is guaranteed that interest rate for 5 years and makes monthly payments of $924.86. After 5 years, she renegotiates her loan, but interest rates have gone up to 7.5%/a compounded monthly. a) If Natalie would like to have the loan paid off after the original 20-year period, what should her new monthly payment be? b) If she keeps her payments the same, how much extra time will it take her to pay off the loan?
8. Peter buys a ski vacation package priced at $2754. He pays $350 down and
finances the balance at $147 per month for 11 years. Determine the annual 2 interest rate, compounded monthly, being charged. Round your answer to two decimal places.
9. a)
T
Suppose you have a loan where the interest rate doubles. If you want to keep the same amortization period, should you double the payment? Justify your reasoning with examples. b) Suppose you are borrowing money. If you decide to double the amount borrowed, should you double the payment if you want to keep the same amortization period? Justify your reasoning with examples.
10. Laurie borrows $50 000 for 10 years at 6.6%/a compounded monthly. How
much sooner can she pay off the loan if she doubles the monthly payment after 4 years?
11. What are the advantages and disadvantages of using each technology to solve
C
financial problems? • a spreadsheet • a graphing calculator
Extending
12. A music store will finance the purchase of a rare guitar at 3.6%/a compounded
monthly over 5 years, but offers a $250 reduction if the payment is cash. If you can get a loan from a bank at 4.8%/a compounded annually, how much would the guitar have to sell for to make it worthwhile to take out the loan?
13. The interest on all mortgages is charged semi-annually. You are given a choice
of monthly, semi-monthly, bi-weekly, and weekly payments. Suppose you have a mortgage at 8%/a, the monthly payments are $1000, and the amortization period is 20 years. Investigate the effect on the time to pay off the mortgage if you made each of these payments. a) $500 semi-monthly b) $500 bi-weekly c) $250 weekly
14. Steve decides to pay $150 per month to pay off a $6800 loan. In the
beginning, the interest rate is 13%/a compounded monthly. The bank guarantees the interest rate for one year at a time. The rate for the next year is determined by the going rate at the time. Assuming that each year the rate drops by 0.5%/a, how long will it take Steve to pay off his loan?
NEL
Discrete Functions: Financial Applications
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8
Study
Chapter Review
FREQUENTLY ASKED Questions
Q:
A1:
Aid
How do you determine the future value of an annuity?
• See Lesson 8.4,
Examples 1 to 4.
• Try Chapter Review
Questions 11 and 12.
An annuity is a series of payments or investments made at regular intervals. The future value of an annuity is the sum of all regular payments and interest earned. You can determine the future value of each payment or investment by using the formula A 5 P(1 1 i) n. Since an annuity consists of regular payments, the future values of the investments, starting from the last, will be P, P(1 1 i), P(1 1 i) 2, .... These form a geometric sequence with common ratio 1 1 i. So the future value of all of the investments is the geometric series P 1 P(1 1 i) 1 P(1 1 i) 2 1 ... , which can be calculated with the formula for the sum of a geometric series.
A2:
You can use technology such as a spreadsheet or the TVM Solver on a graphing calculator to calculate the future value of an annuity.
EXAMPLE
The spreadsheet below is set up for an annuity in which 40 regular investments of $250 are made at the end of each compounding period. The annuity earns 2% interest per compounding period. Since the last $250 investment was deposited at the end of the term, it earned no interest. The first $250 investment earned interest 39 times, but didn't earn interest during the first compounding period because it was deposited at the end of that period.
A Number of Compounding Periods Invested "=A2+1" "=A3+1" 0 B C
Use the formula for the future value of an annuity, FV 5 R 3 a The future value of this annuity is $15 100.50
where FV is the future value; R is the regular payment each compounding period; i is the interest rate per compounding period, expressed as a decimal; and n is the number of compounding periods.
(1 1 i) n 2 1 b, i
532
Chapter 8
NEL
Chapter Review
Q:
A1:
How can you determine the present value of an annuity?
Study
Aid
The present value of an annuity is the amount of money you have to invest to get a specific amount some time in the future. You can determine the present value of each investment by using the formula PV 5 A(1 1 i) 2n. Since an annuity consists of regular payments, the present values of the investments, starting from the first, will be A(1 1 i) 21, A(1 1 i) 22, A(1 1 i) 23, .... These form a geometric sequence with common ratio (1 1 i) 21. So the present value of all of the investments is the geometric series A(1 1 i) 21 1 A(1 1 i) 22 1 A(1 1 i) 23 1 ... , which can be calculated with the formula for the sum of a geometric series.
• See Lesson 8.5,
Examples 1, 2, and 3.
• Try Chapter Review
Questions 13 to 17.
A2:
You can use technology such as a spreadsheet or the TVM Solver on a graphing calculator to calculate the present value of an annuity.
EXAMPLE
The spreadsheet below is set up for an annuity earning 0.5% interest per compounding period and providing 20 regular payments of $50.
A Number of Compounding Periods Invested "=A2+1" "=A3+1" B Payment 1 C Present Value $50.00 "=B2/1.005" $50.00 "=B3/(1.005)^A3" $50.00 "=B4/(1.005)^A4"
The present value of all of the investements in this annuity is $949.37.
A3:
Use the formula for the present value of an annuity,
2n
PV 5 R 3
payment each compounding period; i is the interest rate per compounding period, expressed as a decimal; and n is the number of compounding periods.
Q 1 2 (1 1 i)
i
R,
where PV is the present value; R is the regular
NEL
Discrete Functions: Financial Applications
533
a) What annual rate of simple interest is Pia earning? Round your answer to two decimal places.6% 24. compounded monthly.00 $2 000.1
5.5 years
Future Value $8 000 $1 280 $100 000 $850
a) a) How much did Karl borrow? b) What annual interest rate is he being charged? c) How long will it take before he owes $20 000?
Lesson 8. she earns $400 in interest.4% 27. How long will it take for her investment to grow to $10 000?
monthly for 2.25 in interest. is Deana earning? Round your answer to two decimal places.
Karl's Loan being Charged Simple Interest $12 000. Calculate the total amount and the interest being charged.2% 29% 7. she earns $432 in interest.
Rate of Compound Interest per Year 9. Pia invests $2500 in an account that earns simple
6. b) How much did she invest?
7. Vlad purchased some furniture for his apartment.5 years. At the end of the first year. How much money did Roberto borrow?
NEL
534
Chapter 8
.
Lesson 8. Calculate the present value of each investment.48.6%/a compounded
9.1% 10.2
6. Roberto financed a purchase at 9.6%
b) c) d)
4.00 Total Amount $8 000.00 $4 000. He paid $850 down and financed the rest for 18 months.00 $10 000.7% 8. The
interest.
Principal Rate of Simple Interest per Year 6% 11% 3. What annual interest rate.3% Time 10 years 3 years 34 months 100 weeks 42 days
compound interest. Isabelle invests $4350 at 7. Deana invests some money that earns interest
interest. a) What interest rate.53. Each situation represents a loan being charged
1. she earns $11. At the end of the financing period.6%/a compounded
quarterly.6%
a) b) c) d) e)
$3 500 $15 000 $280 $850 $21 000
Principal
Compounding Period annually semi-annually quarterly monthly
Time 8 years 11. b) How much money will be in her account after 7 years? c) How long will it take for her money to double?
3. Karl borrows some money and is charged simple
compounded annually. The graph below shows how the amount he owes grows over time. At the end of the second year.00 $0.PRACTICE Questions
Lesson 8.00 $6 000. At the end of each month. Vlad owed $2147.5 years 3 years 2. Rate of Compound Interest per Year
Compounding Period annually semi-annually quarterly monthly
Time 5 years 2.4% 6.5 years
a) b) d)
$4 300 $500
c) $25 000
$307
2. Each situation represents an investment earning
simple interest. At the end of the finance period. was he being charged? Round your answer to two decimal places.37.8% 5. he still owed $847.5 years 8 years 1.00 1 2 3 4 5 6 Year 7 8 9 10 11
total cost was $2942.3
8. compounded annually. Calculate the interest earned and the total amount.
4
11.2% 4.
Rate of Compound Interest Compounding per Year Period Time 7. How much sooner can she pay off the loan if she pays the maximum amount compared with the minimum amount?
20. At the end of that time. so she makes
Lesson 8. How much more money will Adam need to invest each month if he wants his investment to be worth the same as Ken's by the time they are 55 years old?
19. He agrees to make monthly
investment is worth $2262.2%/a compounded monthly. $650 down. Eden finances a purchase of $611.
Rate of Compound Interest Compounding per Year Period Time 5. would yield the same results? Round your answer to two decimal places. Starting at age 20. Starting at age 37. calculate the future value and the
payments for the next 20 years.6%/a compounded monthly over 4 years. a) How much will Paul have to pay each month? b) How much interest is he being charged over the term of the loan?
16.5
13. She finances the balance at 6. Chantal purchases a moped for $1875. For each loan.17 for 21 years.73
per quarter
d) $105. How much money does he have to put away each month?
14.8% 19. Paul borrows $136 000.47 with
b) $500 every
6 months
c) $2500
per quarter 12.5 years
4.6
18. compounded annually. For each annuity. he would like to have $25 000. starts saving money in an account that pays 7. What interest rate. Marisa invests $1650 for 3 years.4%/a compounded quarterly.2% annually semi-annually quarterly monthly 6 years 4.6%/a compounded monthly. How much will Chantal have to pay each month?
a) $2500
per year
17.85 per week for 21 years at 13%/a compounded 2 weekly.4%/a compounded monthly. Kevin purchases a guitar on a payment plan of
Regular Payment
$17.
Lesson 8. What was the selling price of the guitar?
a) $450
per year
b) $2375 every
6 months
c) $185.3% annually semi-annually quarterly 12 years 9.27
per month
NEL
Discrete Functions: Financial Applications
535
. his twin brother. Adam.2%/a compounded monthly. calculate the amount of the loan and
the interest being charged. compounded monthly.5 years 3. at which time her 15. Naomi wants to save $100 000.1% 9. Ken invests $100 per month in
quarterly payments of $1500 into an account that earns 4. Jenny starts a business and borrows $100 000 at
9%/a compounded monthly for 6 years. She can afford to make payments between $1000 and $1500 per month. is she being charged? Round your answer to two decimal places.70. What 2 annual interest rate.6% 7.5 years 3 years
Regular Payment
monthly payments of $26.03 by making
interest earned.5 years 1. How long will it take her to reach her goal?
Lesson 8.Chapter Review 10. The interest rate being charged is 6.2% 12. Ernie wants to invest some money each month at
an account that earns 5.
4%/a compounded monthly for the next 10 years?
536
Chapter 8
NEL
.35
payment on a house in 8 years. determine whether simple interest or compound interest is being charged. If she would like to have $25 000 for the down payment.18 $1098.2%/a compounded quarterly for the first 17 years and 8. Simone wants to save money for her retirement. What annual interest rate.8% compounded monthly Time 6 years 13 years 6 1 years 2
a) b) c)
Loan #1 Year 1 2 3 Amount Owed $3796 $3942 $4088
$850 $5460 $230 per month
2. Which option should she choose? Why?
6. compounded monthly.2%/a compounded monthly. The amounts owed for two different loans are shown at the left.
For each loan. Derek invests $250 per month for 62 years at 4. are Anand's parents earning? Round your answer to two decimal places. Her two best options are
5.
a)
3. c) How much was each loan originally? d) Determine the future value of each loan after 10 years. she plans to reinvest the money at 7. The fund is now worth $450 000. she wants to earn 8. For each investment.8%/a compounded monthly.4% compounded semi-annually 4. • She wants to withdraw $5000 per month for the 10 years after the initial 17 years. Principal Rate of Interest per Year 9% simple interest 8.
7. How much more would she have to invest per quarter if she earned 7. Justify your answer.88%/a compounded monthly or 6%/a compounded annually.8
Chapter Self-Test
1. Betsy inherits $15 000 and would like to put some of it away for a down
Loan #2 Year 2 3 4 Amount Owed $977. determine the total amount and the interest earned.
1
How much will his investment be worth at the end of the 61 years? 2
5.53 $1036. Yvette wants to invest some money under these conditions:
• Each quarter for the next 17 years. Anand's parents have been paying $450 per month into a retirement fund for
the last 30 years.4%/a compounded quarterly. • After 17 years. b) What annual interest rate is each loan being charged? Round your answer to two decimal places. how much of her inheritance must she invest at 9.2%/a compounded quarterly?
4.
B. Carol. When they retire. Carol. Carol wants hers to last 15 years. and Lisa get their first full-time jobs and talk about saving for retirement. Carol.
?
A. Steve starts investing immediately and puts aside $150 per month. Carol and Lisa want to accumulate the same amount as Steve upon retirement. F. Assume that Steve. Carol wants to enjoy life a bit and decides to start contributing when she is 30. and Lisa be able to withdraw monthly upon retirement?
What strategies will you use to solve this problem? Justify your strategies.
How much will Steve. They are each 22 years old and plan to work until they are 55. C. Steve wants his investment to last 10 years. How much money will Steve have accumulated by the time he is 55? For how many months will Carol and Lisa be making payments? How much will Carol and Lisa have to put away each month to meet their goals? For how many months will each person withdraw money? How much will each person be able to withdraw from his or her nest egg each month?
Task
Checklist
Did you explain and justify
your strategies?
Did you show your work? Did you support your
calculations with appropriate reasoning?
Did you explain your
thinking clearly?
NEL
Discrete Functions: Financial Applications
537
. and Lisa are each earning 9%/a compounded monthly.8
Chapter Task
Saving for Retirement
Steve. D. Lisa thinks that they are both starting too early and decides to wait until she is 42 before starting to save. and Lisa wants hers to last 20 years. E.
Which option. 32% of the medication remains in his body. Financial Planner
Marcus has a bacterial infection and must take 350 mg of medication every 6 h. Medicine Dosage 18. As his financial planner. Use it to verify your answer in part (b). a) the average of all interest payments b) the fixed periodic payment made up of interest and principal c) the average of all principal payments d) the payment of principal only
Investigations
17. b) What will the amount of medication in his body level off to? c) How long will it take for the medication to reach this level?
You are a financial planner with a new client. The fund earns interest at 6%/a. Mr. would allow you to repay a
to deposit $300 at the end of each month into an account that earns 6%/a interest. c) Create a spreadsheet to represent the monthly balance in the fund. by depositing $25 at the end of each month until Bart turns 18. By the time he takes his next dose. if any. compounded monthly. he uses the balance in the account as a down payment on a $56 000 sailboat. If interest rates remain constant. Cowan. is celebrating his son's fourth birthday. a) Show why the sequence of the monthly amounts in the fund is a geometric sequence. Los is planning to buy a sailboat. compounded monthly. how long will it take him to repay the loan? a) 10 years and 6 months b) 12 years and 9 months c) 9 years d) 8 years and 10 months
loan in less time? a) decrease the regular payment b) increase the regular payment and decrease the interest rate c) decrease the regular payment and increase the interest rate d) none of the above
16. He gets financing for the balance at a rate of 8%/a. b) How many payments will take place by the time Bart turns 18? Determine the balance in the fund on Bart's 18th birthday.Cumulative Review
14. Mr. He can afford payments of $525 per month. Bart. Mr. d) How much would there be in the fund if Mr. who just turned 37. he asks you to develop a financial plan for him. Cowan wants to set up an education fund for his son. compounded monthly. Cowan deposits $50 per month instead of $25?
NEL
Cumulative Review
539
. a) Determine a recursive formula that models this situation. the value of the first deposit after n months. Determine an expression for tn . He decides
15. At the end of four years. Determine which best describes the regular payment
on an amortized loan.
c2 a2 hypotenuse b2 a2 b2 c2
Appendix B
a
c
b
Practising
1.
Glossary
a)
6 cm x
c)
9m
y
8 cm 5m
Answers
b)
d)
3. For each right triangle. Every right triangle has a longest side. One of the important relationships in mathematics is known as the Pythagorean theorem. write the equation for the Pythagorean theorem. which is always opposite the right angle. called the hypotenuse.A–4
The Pythagorean Theorem
Appendix A
The three sides of a right triangle are related to each other in a unique way.2 cm 13 cm c
a 8.5 cm
Index
6 cm
NEL
Appendix A: Review of Essential Skills and Knowledge
547
. It states that the area of the square of the hypotenuse is equal to the sum of the areas of the squares of the other two sides.
10 m
5.72 5 c 2 c)
5m 5.32 1 4. An apartment building casts a shadow.
3. The tip of the shadow is 72 m from the base of the building.8 m
3 cm
5.
6. An isosceles triangle has a hypotenuse 15 cm long.2 cm
b)
6 cm
d) 1.
a) b) c) d) a)
a 2 5 52 1 132 102 5 82 1 m2 262 5 b 2 1 122 2.2 cm
4.
Round all answers to one decimal place. Determine the length of
the two equal sides. Determine the length of the diagonals of each rectangle to the nearest tenth. How tall is the building?
548
Appendix A: Review of Essential Skills and Knowledge
NEL
. Find the value of each unknown measure to the nearest hundredth.2 m
4.2. From the tip of the shadow to the top
of the building is 100 m. Calculate the length of the unknown side of each triangle in question 1.
Verify this result algebraically by substituting (5. y) is the intersection point of the linear equations in the system.
8 6 4 2 0 2 4 6 8 10 2 4 (5. 5 R. although this method does not always yield an exact solution. 5 R.S. 2x 2 8 5 2(5) 2 8 52
In equation 2 . L. 2) into equations 1 and 2 .S. Point P(x. y52 R. 2)
x y
Glossary
y = 2x
8
6
8 10
In equation 1 . they are called a linear system of equations.S. 2) appears to be the point of intersection.S. L.
Appendix B
EXAMPLE
1
Solve the system graphically.S.
x+y=7
Answers
L.S. Point P is called the solution of the linear system and satisfies all equations in the system. 7
Therefore. y 5 2x 1 8 1 x1y57 2
Solution
Draw both graphs on the same axes and locate the point of intersection.S.
Solving a Linear System Graphically
Linear systems can be solved graphically. Point (5. x1y 5512 57
R.S.A–6
Solving Linear Systems
Appendix A
Many kinds of situations can be modelled with linear equations.
Index
551
NEL
Appendix A: Review of Essential Skills and Knowledge
. When two or more linear equations are used to model a problem. L.
Therefore.
The points in a scatter plot often show a general pattern. with the remaining points grouped equally above and below the line. or trend.4
1996 12. then the correlation is weak. Ages 15–19
Answers
1981
1985
1989 Year
1993
1997
1981
1985
1989 Year
1993
1997
The scatter plot shows a negative correlation.4
1989 12.1
1986 15.2
Glossary
Solution
a)
Number of cigarettes smoked/day 18 16 14 12 10 8 6 4 2 0
Average Number of Cigarettes Smoked/Day. Data that have a positive correlation have a pattern that slopes up and to the right. If the points are dispersed. b) Draw the line of best fit.
EXAMPLE
Appendix A Appendix B
1
a) Make a scatter plot of the data and describe the kind of correlation the scatter plot shows.A–11 Creating Scatter Plots and Lines or Curves of Good Fit
A scatter plot is a graph that shows the relationship between two sets of numeric data. Ages 15–19
b)
Number of cigarettes smoked/day 18 16 14 12 10 8 6 4 2 0
Average Number of Cigarettes Smoked/Day. but still form some linear pattern.6
1985 15. If the points nearly form a line. Data that have a negative correlation have a pattern that slopes down and to the right.0
1983 16.6
1995 11. A line of best fit passes through as many points as possible.8
1994 12.5
1991 14.
Index
559
NEL
Appendix A: Review of Essential Skills and Knowledge
. then the correlation is strong. A line that approximates a trend for the data in a scatter plot is called a line of best fit.9
1990 13. Long-Term Trends in Average Number of Cigarettes Smoked per Day by Smokers Aged 15–19
Year Number Per Day
1981 16.
sketch a curve of good fit. Therefore. Here are the results:
Speed (m/s) Distance (m)
10 10
16 25
19 47
22 43
38 142
43 182
50 244
54 280
Draw the line or curve of good fit. The final horizontal distance of the ball from the tee is also recorded. Each time she drives the ball from the tee.EXAMPLE
2
A professional golfer is taking part in a scientific investigation. a motion sensor records the initial speed of the ball.
Solution
The scatter plot shows that a line of best fit does not fit the data as well as an upward-sloping curve does.
Distance (m)
Horizontal Distance of a Golf Ball 300 250 200 150 100 50 0 10 20 30 40 50 60 Speed (m/s)
560
Appendix A: Review of Essential Skills and Knowledge
NEL
.
86
4. the motion sensor may not be measuring accurately.51
a) Draw a curve of good fit for the data.4
1981 20.3 0. A motion
sensor detects the speed of the marble at the start of the ramp. However. Here are the data:
Speed (m/s) Final Height (m)
1.3
1996 12.Practising
1. In an experiment for a physics project.4
2.07
2. For each set of data.2 0.8 0.1 1.5 1.49
4.6
Glossary
1.1 0.7
1986 17.21
2.
Answers Index
561
NEL
Appendix A: Review of Essential Skills and Knowledge
.7
1991 14. marbles are rolled up a ramp.
Appendix A
i) create a scatter plot and draw the line of best fit ii) describe the type of correlation the trend in the data displays a) Population of the Hamilton–Wentworth. Region
Year Population
1966 449 116
1976 529 371
1986 557 029
1996 624 360
1998 618 658
b) Percent of Canadians with Less than Grade 9 Education
Appendix B
Year Percent of the Population
1976 25. b) How consistent are the motion sensor's measurements? Explain.02
5.38
3. and the final height of the marble is recorded. Ontario.36
5.0 0.
A c B a b C
Appendix A
In any right triangle. there are three primary trigonometric ratios that associate the measure of an angle with the ratio of two sides. where c is the length of the hypotenuse and a and b are the lengths of the other two sides. in Figure 2. for /ABC. a 2 1 b 2 5 c 2 for any right triangle. in Figure 1.
A hypotenuse B side adjacent to angle A C For sin A cos A tan A A opposite —————— hypotenuse adjacent —————— hypotenuse opposite ————— adjacent
Answers
side opposite to angle A
Figure 2
Note how the opposite and adjacent sides change in Figures 1 and 2 with angles A and B.
Index
569
NEL
Appendix A: Review of Essential Skills and Knowledge
.
A hypotenuse side opposite to angle B For sin B cos B tan B B opposite —————— hypotenuse adjacent —————— hypotenuse opposite ————— adjacent
Appendix B
B side adjacent C to angle B
Figure 1
Glossary
Similarly. For example.A–16 Trigonometry of Right Triangles
By the Pythagorean relationship.
and the hypotenuse.EXAMPLE
1
C 5 A
State the primary trigonometric ratios of /A. run 8
A 1 C
B
8
570
Appendix A: Review of Essential Skills and Knowledge
NEL
. Draw a labelled sketch. What is the angle of inclination of the ramp?
Solution
The slope of the ramp is
1 rise 5 . Then label the opposite side. the adjacent side.
4 3 B
Solution
Sketch the triangle. sin A 5 opposite hypotenuse 4 5 5 adjacent hypotenuse 3 5 5 opposite adjacent 4 5 3
C hypotenuse 5 opposite 4
cos A 5
A adjacent B 3
tan A 5
EXAMPLE
2
A ramp must have a rise of one unit for every eight units of run.
Determine x to one decimal place. How tall is the other building?
a) sin A 5 b) cos A 5
5 8 13 22
c) tan B 5 d) cos B 5
19 22 3 7
572
Appendix A: Review of Essential Skills and Knowledge
NEL
.
a)
15. How tall is the tree?
8. The angle of elevation from her eyes to the top of the tower is 70°.2 cm
a)
C
b)
13 12
A
5
B
c)
9. State the primary trigonometric ratios for /A.
x 7 x b) cos 65° 5 16 a) sin 39° 5
c) tan 15° 5
x 22 31 d) tan 49° 5 x
communications tower.5 cm
c)
6. A tree casts a shadow 9. In ^ ABC. Solve for x to one decimal place.0 m from the base of a
3. /B 5 90° and AC 5 13 cm. How long is the 5. A rectangular lot is 15 m by 22 m. Determine
a) BC if /C 5 17° b) AB if /C 5 26° c) /A if BC 5 6 cm d) /C if BC 5 9 cm
7.
diagonal. Solve for /A to the nearest degree.3 m long when the angle of
d)
B
the sun is 43°.
building that is 80 m tall.Practising
1.8 m above the ground?
9. Janine stands 30. How high is the tower if her eyes are 1.2 cm x 35 x 17 21. to the nearest metre?
2.2 cm x
7. The angle of elevation from the camera to the top of another building is 42°. A surveillance camera is mounted on the top of a
4.2 cm
b)
A 9 B 15 12 C B 15 A 17 20 12 C 16 8 C A
25 x
d)
18. The angle of depression from the camera to the same building is 32°.
Index
573
NEL
Appendix A: Review of Essential Skills and Knowledge
. for ^ ABC. /Z 527°.
Solution
In the triangle shown. and z 54 cm. two angles and one side (AAS) or two sides and an opposite angle (SSA) must be given. Two angles and one side are known.2 Then XZ is about 8.
Glossary
EXAMPLE
1
Determine the length of XZ to one decimal place. /X 584°. sin Y sin Z 5 y z sin 69° sin 27° 5 y 4 4 sin 69° y5 sin 27° 8 8. so the sine law can be used.
• a2 5 b2 1 c2 – 2bc cos A or • b2 5 a2 1 c2 – 2ac cos B To use the cosine law. two sides and the contained angle (SAS) or three sides (SSS) must be given.
B c A b a C
The cosine law states that.
B c A b a
Appendix B
C
•
sin A sin C sin B 5 5 c a b
or
• c2 5 a2 1 b2 – 2ab cos C
or
a b c • 5 5 sin A sin B sin C To use the sine law.
Sine Law Cosine Law
Appendix A
The sine law states that. for ^ ABC. so the primary trigonometric ratios do not apply. /Y 5 180° 2 (84° 1 27°) 5 180° 2 111° 5 69°
X 4 cm 84 Y 27 Z
Answers
This is not a right triangle.A–17 Trigonometry of Acute Triangles: The Sine Law and the Cosine Law
An acute triangle contains three angles less than 90°.2 cm.
Clear all equations in the equation editor. enter 7 2 2x 1 3y 5 7 in the form y 5 2 x 1 .
3. 3 3
3.
1.
If m or b are fractions.
Press
ZOOM
6
. The calculator will display the graph. ENTER TRACE Press ZOOM to trace using integer 8 intervals. be sure to clear any information left on the calculator from the last time it was used.
Appendix B
Press 2nd
Y=
4
ENTER
. Graph. Enter all linear equations in the form y 5 mx 1 b.
Answers
2. to find the coordinates of any point on a graph. Turn off all stat plots. Press 4. as shown.
Press 2nd
1
4
ENTER
. T.
2. The graph will be displayed as shown. enter them between brackets. then press
CLEAR
for each equation.
Press
Y= .
To graph y 5 2x 1 8. Press
GRAPH TRACE
to view the graph. press
GRAPH
Y=
2
X.
4. For example. Set the window so that the axes range from 210 to 10. n
1
8
.
Glossary
B–2
Entering and Graphing Functions
Enter the equation of the function into the equation editor. You should always do the following:
1.
Use the left and right arrow keys to cursor along the graph. Clear all data in the lists.
Index
577
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Appendix B: Review of Technical Skills
. use the up and down arrow to scroll between graphs.PART 1 USING THE TI-83 PLUS AND TI-84 GRAPHING CALCULATORS
B–1 Preparing the Calculator
Before you graph any function. Press WINDOW to verify. If you are working with several graphs at the same time. U.
U.
Glossary
B–6
Using the TABLE Feature
A function such as y 5 20. stands for "change in.
To enter y 5 20.
Press 2nd
WINDOW . press
1
ENTER
. press 2nd and adjust the settings as shown.1x3 1 2x 1 3 can be displayed in a table of values. to see the table start at x 5 0 and increase in increments of 0. Press
ENTER
and
to select this. Enter the function into the equation editor. Enter y 5 x 2 in Y1 of the equation editor. press cursor to Horiz. n
. To view the table. U. the Greek capital letter delta. T.
2. The cursor is now alongside
"DTbl5" (D.
Appendix B
It is possible to view the table with different increments. press Y=
X.
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Appendix B: Review of Technical Skills
579
.
1. then press
. then press
MODE
to return to the home screen. n
^
(2) X. Press
GRAPH WINDOW
. press (2)
. press
3.1x 3 1 2x 1 3.") To
Index
increase the x-value in steps of 1. Set the start point and step size for the table. The cursor is alongside "TblStart5.
2nd
GRAPH
.5." To start at 5 ENTER
x 5 25. 1
1 3
Answers
3
1
2
. Notice that you can look
at higher or lower x-values than the original range.
Use
and
to move up and down the table. Press
ENTER
and cursor to G–T
GRAPH
to select this. T.
MODE
To see a graph and a table at the same time: press (Graph-Table). For example. then press
GRAPH
.B–5
Using the Split Screen
MODE 2nd
To see a graph and the equation editor at the same time.
Scroll right and up to select L2.
)
. use x-values from 22 to 4. use the STAT lists.
to see the first differences displayed in L3.
4. Enter the function f (x). Find the first differences.
2.
For the function f(x) 5 3x 2 2 4x 1 1. Start by entering y 5 2(x 1 3)(x 2 5) in the equation editor.
to display the values of the function in L2. Repeat step 4.
Scroll right and up to select L4.
5. Find the second differences.
B–8
Finding the Zeros of a Function
To find the zeros of a function.
Press 2nd
TRACE
2
.
2.
1. Then press 2nd Scroll right to OPS and press Enter L2 by pressing 2nd Press
ENTER 7 2
STAT . then press
GRAPH
ZOOM
6
.
to choose DList(. Press
ALPHA 1 1
1 1
3
2nd ALPHA
1 1
x2
2
2nd ENTER
. Press
ENTER
to see the second differences displayed in L4. Access the zero operation. using L3 in place of L2. use the zero operation. using L1 as the variable x.B–7
Making a Table of Differences
To create a table with the first and second differences for a function. Press
4
3.
1. Press STAT
1
and enter the x-values into L1. Enter the function.
Scroll right and up to select L3.
580
Appendix B: Review of Technical Skills
NEL
.
3. Press
ENTER
to set the left bound. so it has a maximum. Press
ENTER ENTER
to set the right bound. again to display the coordinates of the zero
(the x-intercept). again to display the coordinates of the optimal
Index
value.
NEL
Appendix B: Review of Technical Skills
581
. Press
ENTER
to set the left bound.
1. Cursor along the curve to any point right of the maximum value. Enter y 5 22x 2 2 12x 1 30. Repeat to find the second zero.
4.
3.
Answers
Press 2nd
2nd
TRACE 3
4
.
2.
Press
5. Use the maximum operation. Press
ENTER ENTER
to set the right bound.
Press
5. Cursor along the curve to any point to the right of the zero.
Graph it and adjust the window as shown. For parabolas that open upward. Use the left and right arrow keys to cursor along the curve to any point
to the left of the maximum value.
6. Use the left and right arrow keys to cursor along the curve to any point
to the left of the zero.
Appendix B
B–9
Finding the Maximum or Minimum Values of a Function
Glossary
The least or greatest value can be found using the minimum operation or the maximum operation. press
TRACE
to use the minimum operation.
4. This graph opens downward.
2. At Y1T=. Use window settings such as the ones shown to display the graph.
y 5 2 2 t 2 to x 5 2 2 t 2. Press
GRAPH
to display the inverse function. enter X. Change the setting to the parametric mode by scrolling down to the fourth line and to the right to Par. U. Replace x with t. Press ENTER . is also the domain of t. For a parametric equation. For example. n . Then x 5 t and y 5 2 2 t 2. t. as shown on the screen. The inverse of this function can now be graphed. Enter the inverse function by swapping the parametric equations x 5 t. Press WINDOW . Press
Y= . U. the function y 5 2 2 x 2 with domain x $ 0 can be graphed using parametric mode.B–10 Graphing the Inverse of a Function
Parametric equations allow you to graph any function and its inverse. T.
4. Clear the calculator and press
MODE . At X1T=. n
x2
ENTER
.
The original domain. enter 2
2
X.
3. both x and y must be expressed in terms of a parameter.
1. x $ 0. T. y 5 t.
582
Appendix B: Review of Technical Skills
NEL
.
and L1 and L2 appear after Xlist and Ylist.
Index
583
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Appendix B: Review of Technical Skills
. Display the graph. to enable ExpReg.
Glossary
4. Repeat this for height in L2. Scroll over to
twice. Press
ENTER . Turn on and 1 Plot 1 by making sure the cursor is over On.
VARS .
Answers
Press 2nd Y-VARS. Enter the data into lists. Display and analyze the results.
2nd Y=
ENTER
2. Create a scatter plot.B–11 Creating Scatter Plots and Determining Lines and Curves of Best Fit Using Regression
This table gives the height of a baseball above ground. b.
Time (s) Height (m) 0 2 1 27 2 42 3 48 4 43 5 29 6 5
Create a scatter plot of the data.
C 1 1
ALPHA
to enable SinReg. the equation is about y 5 24. This action stores the equation of the line or
curve of best fit into Y1 of the equation editor. In this case.
Press
ENTER
.
2nd
2
. the letters a.
5. Move the cursor over to the first position in L1 and enter the values for time.
Appendix B
1. Press
.93x 1 1. ZOOM 9
3. from the time it was hit to the time it touched the ground.98. and c are the coefficients of
the general quadratic equation y 5 ax 2 1 bx 1 c for the curve of best fit.90x 2 1 29. To start press STAT
. Press:
4 5 0
to enable LinReg(ax1b) to enable QuadReg. In this case. To determine the equation of
the line or curve of best fit press STAT and scroll over to CALC. Press
to activate ZoomStat. R 2 is the percent of data variation represented by the model. the Type is set to the graph type you prefer. Press ENTER after each value. Apply the appropriate regression analysis.
Use SIN21.
6
584
Appendix B: Review of Technical Skills
NEL
.
5. Press 2nd
TRACE 5
. Press
ENTER
. Use the intersect operation. Put the calculator in degree mode. press SIN
.
3. Determine any additional points of intersection. Repeat steps 4 to 6. Use the
MODE
.
6.
SIN .
COS
.
Press
2. COS21. Press
TRACE
and
move the cursor close to the other point you wish to identify. 5 4 ) ENTER
To find the value of sin 54°.
twice.
GRAPH
.
ENTER
4.Note: In the case of linear regression. The point of intersection is exactly (2. if r is not displayed. Adjust the window settings until
the point(s) of intersection are displayed. Graph both functions. Press 3. Press 2nd Press
ENTER 0
and scroll down to DiagnosticOn. press 2nd ) ENTER . Scroll down and across to Degree. or TAN key to calculate trigonometric ratios. or TAN21 to calculate angles. Repeat step 4. In this case we will use
y 5 5x 1 4 and y 5 22x 1 18. COS .
To find the angle whose cosine is 0.
Press
GRAPH
B–12 Finding the Points of Intersection of Two Functions
1. You will be asked to verify the two curves
and enter a guess (optional) for the point of intersection.6. Enter both functions into the equation editor.
B–13 Evaluating Trigonometric Ratios and Finding Angles
1. turn on the diagnostics function. 14).
2. Plot the curve. Press after each screen appears. Determine a point of intersection.
n 2
step 2
586
Appendix B: Review of Technical Skills
NEL
. U. Change the graphing mode from function to sequence.
The graphing modes are listed on the fourth line of the MODE menu.
You will need to enter the following: • the expression of the general term • the variable n — let X.
5
. the first term is generated. because the difference between each pair of consecutive natural numbers is always 1 Press X.
Press
step 3
ENTER
. Scroll down to sequence and press
ENTER
. U. You can change the minimum value of n (nMin). Enter the information for sequence.
2. Scroll down to u(n) and position the cursor to the right of the 1 .
1. You will need the general term. n
.
1
.
1. and w(n) represent the general terms
of sequences.
B–17 Graphing Sequences
Part 1: Creating a Table and Graphing — The General Term
Using the TI-83 or TI-84 Plus calculator. Press 3 X. n represent n • the first position number • the last position number • the increment — the increment is 1. n
x2 )
step 2
. step 1
Press
Y= . v(n). U. Press ENTER . Press MODE and scroll down and across to Seq. Enter the general term into the sequence editor.
Press 2nd
step 1
STAT
.
1
. you will not need to change the value 1.
X. T. u(n). In this editor. In most cases. T. Select sequence from the List OPS menu. T.
2. when n 5 1.B–16 Generating the Terms of a Sequence
Generate or list the first five terms of the sequence defined by tn 5 n2. Generate the first five terms of the sequence. because. equal sign.
3. T. U. Create and graph the sequence defined by tn 5 3n 2 1. you can first create a table for a sequence and then a graph for the sequence.
Trace along the graph to identify specific terms of the sequence. Use
and
to move from point to point.
Press WINDOW .
6.
step 4
5.
Appendix B
step 3 4. or position. then scroll down and across to G–T (on the last line of
ENTER GRAPH
Index
the MODE menu). The y-coordinate represents the value of the term. The setting nMin indicates the smallest n-value for the calculator to evaluate.3.
Press
GRAPH
.and y-coordinates of each term are displayed at each point. Adjust the window to display the required number of terms.
Press
TRACE
. Set TblStart to 1 and DTbl to 1. Press
. The
Glossary
n-value.
a) Press 2nd b) Press 2nd
WINDOW . while nMax indicates the largest n-value for the calculator to evaluate.
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Appendix B: Review of Technical Skills
587
. PlotStep=1 means that each consecutive term will be plotted. GRAPH
to display the table. but use these window settings for this example. Use the cursor keys to
scroll through the table. You can change these settings. PlotStart=1 means that the graph starts at the first term. use split-screen mode. Graph the sequence. and the x.
step 5
Answers
step 6 a)
step 6 b)
Note: To see the graph and the table at the same time. View the terms of the sequence in a table. Press
MODE
.
Enter the recursive formula in the sequence editor and set the initial
value. tn 5 500 1 0.
1. Then.26tn–1.
Press WINDOW and enter the values shown. or the first term.
588
Appendix B: Review of Technical Skills
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. t 1 5 0 and t 2 5 1.
0
2nd
)
ENTER
. Graph the sequence t1 5 500. if you were to
(
enter. T. ) 2 6 2nd 7
5 (
0
0 X. for the sequence u(n).
Set the initial value to 500. However. n
. Position the cursor to the right of the equal sign for u(nMin) and press 2nd
) ENTER ( 5 0 0 2nd
. then you would press 2nd
1
.
2. press
. Set the window. you can graph recursive sequences in the same way.Part 2: Creating a Table and Graphing — The Recursive Formula
Using the TI-83 Plus or TI-84 calculator.
Note: You do not have to enter the braces ( 2nd
)
(
and 2nd
) around the initial value. U. Press
1 2 Y= 0 1
. for example. with one exception: you must specify an initial value or values for u(nMin) in the sequence editor.
Set TblStart to 1 and DTbl to 1.
Appendix B
Press 2nd Press 2nd
WINDOW . Use the cursor keys to
scroll through the table.
Press
GRAPH
.
Glossary Answers Index
589
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Appendix B: Review of Technical Skills
.
4. Draw the graph.3. GRAPH
to display the table. View the terms of the sequence in a table.
step 3
590
Appendix B: Review of Technical Skills
NEL
.
b) Enter the expression of the general term. Find the sum of the series. and the increment.
Part 1: Finding the Sum of the Terms of a Series
1.
2nd
2
.
Follow step 1 of Part 1.
3.
3. 1.
Use Ans.
Press 2nd Press
step 2
STAT )
and scroll down to 6:cumSum.
ENTER
. Find the cumulative sums of the terms of the sequence.
Press 2nd Press
step 2
STAT
and scroll down to 5:sum. Press
ENTER
to generate the first five terms. the ending value of the variable.
2.
a) Select sequence from the List OPS menu. the starting value of the variable. Press
ENTER
ENTER
. Generate the terms of the sequence again.
2. ) 2 ENTER . The cumulative sums are 151 55114 14 5 1 1 4 1 9 30 5 1 1 4 1 9 1 16 55 5 1 1 4 1 9 1 16 1 25
displayed.B–18 Finding the Sum and Cumulative Sum of a Series
Find the sum and the cumulative sum of the first five terms of the sequence defined by tn 5 n 2. Press 2nd Scroll down to 5:seq and press
step 1
STAT
. The sum is displayed. Select sum from the List MATH menu. Select cumulative sum from the LIST OPS menu. as shown. Press 2nd
Part 2: Finding the Cumulative Sum of the Terms of a Series
The cumulative sum displays the progression of sums of the terms of a series.
step 3 1.
ENTER
. the variable. Generate the first five terms of the sequence. last answer. to insert the terms in sum.
move the cursor to that variable and press
ENTER ALPHA
Glossary Answers
. and the calculator will calculate this value. From the Finance CALC menu. across to 2. and press ENTER . you will see that the calculator adds the decimal and two zeros. You may enter different values for the variables. A small
Index
shaded box to the left of the line containing the calculated value will appear. Press APPS and then select 1:Finance. The screen that appears should be similar to the second one shown. select 1:TVM Solver. or the number of interest conversion periods for simple annuities I% annual interest rate as a percent. N total number of payment periods.B–19 Analyzing Financial Situations Using the TVM Solver
Part 1: Introducing the TVM Solver
Press MODE and change the fixed decimal mode to 2. When you enter a whole number. To solve for a variable. enter the value of money that is received as a positive number.
Appendix B
You will notice eight variables on the screen. since the investment is a cash outflow. because most of the values that you are working with here represent dollars and cents.
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Appendix B: Review of Technical Skills
591
. Enter the value for money that is paid as a negative number. not as a decimal PV present or discounted value PMT regular payment amount FV future or accumulated value P/Y number of payment periods per year C/Y number of interest conversion periods per year PMT Choose BEGIN if the payments are made at the beginning of the payment intervals. since the money is a cash inflow. Choose END if the payments are made at the end of the payment intervals. but the values may be different. Scroll down to Float.
C/Y. because the future value of the investment will be "paid" to Maeve. except the value for PV. Scroll to the line containing ENTER to get 2$2242. and PV 5 27500. I% 5 8. the future value.48 after nine years. Open the TVM Solver and enter these values.
592
Appendix B: Review of Technical Skills
NEL
. and the compounding periods per year. compounded quarterly. PMT 5 0 and FV 5 0.
Solution
The number of interest conversion periods. is 9 3 12 5 108. because the investment represents a cash outflow. Scroll to the line containing FV. How much money should she deposit now in a savings account that pays 9%/a. because Maeve must pay this money and the payment is a cash outflow. representing a cash inflow. P/Y. The payments per year. The value for FV is positive. is negative. and press ALPHA ENTER .
EXAMPLE
2
Maeve would like to have $3500 at the end of five years.Part 2: Determining Future Value and Present Value
EXAMPLE
1
Find the future value or amount of $7500 invested for nine years at 8%/a. to finance her trip?
Solution
Open the TVM Solver and enter the values shown in the screen. compounded monthly. N.86. so she can visit Europe. are both 12. The solution for PV is PV and press ALPHA negative.
The investment will be worth $15 371. The value for present value. PV.
The payment appears as 2473. Scroll to the line containing FV and press ALPHA
Appendix B Glossary
The balance in Celia's account at the end of the year will be $8206. is positive. is negative.24 at the end of each month for 15 years. Scroll to the line containing PMT and press
ALPHA ENTER
. which is a cash outflow. compounded monthly?
Solution
Answers
Open the TVM Solver and enter the values shown.
Index
Mr.
NEL
Appendix B: Review of Technical Skills
593
. PMT 5 21500. compounded annually. Note that N 5 12 3 15 5 180. because it is a cash outflow. Open the TVM Solver and enter these values.Part 3: Determining the Future or Accumulated Value of an Ordinary Simple Annuity
EXAMPLE
3
Celia deposits $1500 at the end of each year in a savings account that pays 4.
EXAMPLE
4
Mr. The payment. PV 5 0. What will be the balance in the account after five years?
Solution
N 5 5 and I% 5 4. because Celia makes a payment. Bartolluci would like to have $150 000 in his account when he retires in 15 years. since he will receive the money at some future time. Because there is no money in the account at the beginning of the term. except for PMT. PMT. How much should he deposit at the end of each month in an account that pays 7%/a. P/Y 5 1 and C/Y 5 1.06.5%/a. and the future value.5. FV. Bartolluci must deposit $473. ENTER .24.
The value for PMT is positive. since the fund will be depleted at the end of the term. The total cash price is PV 1 $150 5 $706.
Solution
Open the TVM Solver and enter the values as shown in the screen.5 years.
594
Appendix B: Review of Technical Skills
NEL
. 9 for I%.82. compounded monthly. because it represents what Monica would have to pay now if she were to pay cash. The payment. because the school must pay this money to establish the fund. because the payments are a cash inflow for the snowboard's seller. If the finance charge is 16%/a. The payment is a cash outflow. The selling price is the sum of the positive present value and the down payment. PMT.41 now for the scholarship fund. except the value for PV. Under this finance plan. find the selling price of the snowboard. what does the school need to invest now to pay for the fund?
Solution
Open the TVM Solver and enter 8 for N.82. Scroll to the line containing PV and press ALPHA The school must invest $2767. is positive. Enter 0 for FV. Since the down payment is also a payment.
EXAMPLE
6
Monica buys a snowboard for $150 down and pays $35 at the end of each month for 1. C/Y. and 500 for PMT. Monica will pay $35 3 18 1 $150 5 $780. ALPHA The present value is $556. because someone will receive $500 each year. A $500 scholarship will be awarded at the end of each school year for the next eight years. Enter 1 for both P/Y and ENTER . Scroll to the line containing PV and press ENTER .Part 4: Determining Present or Discounted Value of an Ordinary Simple Annuity
EXAMPLE
5
Northern Lights High School wishes to establish a scholarship fund. The present value appears as 22767. If the fund earns 9%/a. compounded annually. add both numbers.41. The present value appears as a negative value on the screen.
B–20 Creating Repayment Schedules Using the TVM Solver
In this section. The greater roundvalue is. roundvalue) calculates the sum of the interest paid from period A to period B calculates the sum of the principal paid from period A to period B calculates the balance owing after period x
Appendix B
The calculator rounds as it calculates. You will need to tell the calculator the value for rounding. a) How much is the monthly payment? b) How much will she pay in interest? c) How much will she still owe on the loan after the 30th payment. P/Y 5 12.
EXAMPLE
1
Eleanor finances the purchase of a new pickup truck by borrowing $18 000. FV 5 0. and C/Y 5 12. She will repay the loan with monthly payments. Enter I% 5 14. at the halfway point in repaying the loan? d) What portion of the 30th payment reduces the principal?
Solution
Answers
a) Press APPS and select 1:Finance. Here are three other functions: SInt(A. by applying some of the financial functions from the Finance CALC menu.
ALPHA ENTER
Index
. which is also the value that banks use. The calculator can use the information that you have entered into the TVM Solver to perform other functions. future value and present value. that is. because Eleanor pays this amount each month. PV 5 18 000. is a positive number because Eleanor receives (a cash inflow) $18 000 from the bank. the greater the accuracy of the calculations.
Part 1: Introducing Other Finance CALC Menu Functions
You have used the TVM Solver to find.83. The interest rate is 14%/a.
NEL
Appendix B: Review of Technical Skills
595
.83. for example. B. roundvalue. because 12 3 5 5 60. roundvalue) SPrn(A. The
The payment appears as a negative value. compounded monthly. you will create a repayment schedule. Then press ENTER to select 1:TVM Solver from the Finance CALC menu. B. Scroll to the line containing PMT and press monthly payment is $418. roundvalue) bal(x.
Part 2: Using the TVM Solver and Other Finance CALC Menu Functions
Press
MODE
Glossary
and change the fixed decimal mode to 2. the roundvalue is 6. Notice that the present value. In this section. Enter N 5 60.828 515 3. The actual value is 2418. which the calculator rounded to 2418. because most of the
values in this section represent dollars and cents. The term of the loan is five years. PV.
(Why is this amount not $9000?) d) Find the portion of the 30th payment that reduces the principal by calculating the sum of the principal paid from the 30th payment to the 30th payment.27 after the 30th payment.
3
0
. In the words. Press 2nd
MODE
to return to the home screen. and the total number of payments. select SPrn by scrolling down or by pressing
0
. Eleanor will have paid $7129. Eleanor will have paid $7129.80.71 in interest and principal for the truck.
. Note that the product of the payment. Press APPS and
select 1:Finance from the Finance CALC menu. that is.83. The difference of $0.828 515 3 was rounded to 418. From the Finance CALC menu.
6
)
ENTER
. 6. roundvalue must be consistent. you are calculating the sum of only one item.83. c) Find the balance on the loan after the 30th payment.71 1 $18 000 5 $25 129.
Eleanor still owes $10 550. roundvalue is again 6. Press . select bal by scrolling or by pressing .71 in total interest. The other portion of this payment.09 is due to rounding.b) Use SInt(A. By the end of the loan. 0 Press 3 6
9
. $418. is $25 129. $126. because 418. Select SInt by scrolling down or by pressing Press Press
ENTER 1 ALPHA MATH
.
3
0
.50. From the Finance CALC menu. the 30th payment.
.33. roundvalue) to calculate the total interest that Eleanor will pay. B. is interest. 60.
596
Appendix B: Review of Technical Skills
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. ) ENTER .
6
)
ENTER
The portion of the 30th payment that reduces the principal is $292.
The sum of the interest paid from the first period to the 60th period is calculated.
6
0
.
T. press 2nd Then press APPS Press X. You will combine sequence (List OPS menu) and bal. She will repay the loan with monthly payments.
. beginning with the
first month. n
.
Glossary
to select bal. The increment for this sequence is 1.
to calculate the sequence of balances. n
ENTER 9 ) ) STAT 5
to select sequence. 7.
Answers Index
597
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Appendix B: Review of Technical Skills
.
Solution
Appendix B
a) Find the outstanding balance after each payment for the first seven months. c) Use the repayment schedule to verify that the loan is completely paid after five years or 60 payments. The term of the loan is five years.
Press
7 ENTER
.
EXAMPLE
2
Recall that Eleanor borrows $18 000 to purchase a pickup truck. 1. U. and ending with the last month. U. From the home screen. . a) What will be the monthly outstanding balance on the loan after each of the first seven months? b) Create a repayment schedule for the first seven months of the loan.
1
.Part 3: Using the Finance Functions to Create Repayment Schedules
Use the functions described in parts 1 and 2 to create repayment schedules or amortization tables. Recall that the increment is the change from payment number to payment number.
X.
1
6
. The interest rate is 14%/a compounded monthly. T. Scroll right ( ) to see the other balances. which is the last value entered.
83 payment is interest.
. Notice that a substantial portion of the $418. Each payment. The outstanding balance after seven payments is $16 486. T. c) Scroll or reset the tables's start value to see other entries in the amortization table. Press X. The final outstanding balance is 2. or an amortization.
To store the principal portion of each payment in Y2. T. The table will start with payment 0 and the payment number will
Press 2nd
GRAPH
to see the amortization table. T.
WINDOW . this value is $0. Therefore. U. n
. none of the payment is applied to the principal. Set TblStart to 1 and
Before viewing the table. which is a combination of interest and principal.
6
)
.
X. the outstanding balance.
ENTER ENTER
. n
to select SInt. U.b) Create a repayment. Scroll up to the beginning of the table. n
ENTER
. Press X.03. T. The loan will be paid completely after five years or 60 payments. n
ENTER
. schedule for the first seven months by comparing the interest. Clear Y1 to Y3. press APPS
9
to select bal. and balance in a table. n
. At the end of the amortization. the amortization period of 60 payments is correct. Press X.
DTbl to 1. Notice that the
interest portion and the principal portion of each payment appear as negative values. Move the cursor to the right of Y15. U. Scroll down the table. Scroll right to see the values for Y3.
. T. As a decimal.
598
Appendix B: Review of Technical Skills
NEL
. Store the interest portion of each payment in Y1. U. is a cash outflow for Eleanor. Press Y= . if necessary. press 2nd increase by 1 at each step.5E25. U. Press APPS
MATH 6 ) ENTER ALPHA X. press APPS
ENTER 0 6 )
to select SPrn. principal.000 25. Begin by opening the equation editor.
To store the outsanding balance after each payment in Y3.
Use the Fill Down command. 1955.94 A 1 Number of Quarters 2 0 3 1 4 2 . and cell C1 as Amount ($).
A 1 Number of Quarters 2 0 3 1 4 2 5 3 6 4 B Time (years) 0 0. a number.PART 2 USING A SPREADSHEET
B–21 Introduction to Spreadsheets
A spreadsheet is a computer program that can be used to create a table of values and then graph the values.. . A cell can hold a label.50 1025.8328 1980. . .. the computer automatically calculates and enters the values in each cell.75 1 C Amount ($) 1000 1012.0125)^A3 in C3 to generate the next values in the table.2807 2005. label cell A1 as Number of Quarters. 0 in B2. and a caret (^) is used for exponents. 13. or a formula. and 1000 in C2.1563 . Cell B1 as Time (years).97 1050. as shown below in the screen on the left. This command inserts the appropriate formula into each selected cell.25 C Amount ($) 1000
Answers
When the Fill Down command is used.
Creating a Table
Use spreadsheets to solve problems like this: How long will it take to double your money if you invest $1000 at 5%/a compounded quarterly? To create a spreadsheet.0342
Index
599
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Appendix B: Review of Technical Skills
. A3/4 in B3.
A 1 Number of Quarters 2 0 3 1 4 B Time (years) 0 0..75 14 C Amount ($) 1000 1012.25 0.. 56 54 57 55 58 56 B Time (years) 0 0.16 1037.5 0. Next. Enter the initial values of 0 in A2. Continue to select the cells in the last row of the table and use the Fill Down command to generate more values until the solution appears. and 1000*(1.50 1025. such as A2 or B5.5 .25 0.
A 1 Number of Quarters 2 0 3 =A2+1 4 B Time (years) 0 =A3/4 C Amount ($) 1000 -1000*(1. use the cursor to select cell A3 to C3 and several rows of cells below them. Enter the formulas A2 1 in A3. . as shown below in the screen on the right. .0125^A3)
Appendix B Glossary
Notice that an equal sign is in front of each formula. an asterisk (*) is used for multiplication.5 13. It is made up of cells that are identified by column letter and row number.
. In this case. 1955.5 13. This graph appears above on the right. Use the cursor to highlight the portion of the table you would like to graph..
Value of $1000 Over Time
A 1 Number of Quarters 2 0 3 1 4 2 .25 0. .. Time versus Amount. .8328 1980.75 14 C Amount ($) 1000 1012.
Determining the Equation of the Curve of Best Fit
Different spreadsheets have different commands for finding the equation of the curve of best fit using regression.0342
2200 2000 Amount ($) 1800 1600 1400 1200 1000 1 11 21 31 41 Time (years) 51
Different spreadsheets have different graphing commands. .
600
Appendix B: Review of Technical Skills
NEL
. .1563 . Check your spreadsheet's instructions to find the proper command..5 . . 56 54 57 55 58 56 B Time (years) 0 0.2807 2005.Creating a Graph
Use the spreadsheet's graphing command to graph the results.50 1025. 13. Check your spreadsheet's instructions to find the proper command for the type of regression that suits the data.
Graph the function y
x2 3x 2. The graph of y x2 3x 2 should appear on the grid. Turn on the grid.3 * x Then press the "OK" button on the calculator keypad.
3. choose Plot New Function.
The function calculator should appear.PART 3 USING THE GEOMETER'S SKETCHPAD
B–23 Graphing Functions
1.
2". choose Show Grid.
2.
From the Graph menu. From the Graph menu. Enter the function. Use either the calculator keypad or the keyboard to enter "x ^ 2 .
602
Appendix B: Review of Technical Skills
NEL
.
Left-click and drag this point to change the scale. Adjust the origin and/or scale. then left-click on the point at (1. 0) to select it. left-click in blank space to deselect.
Appendix B Glossary Answers Index
603
NEL
Appendix B: Review of Technical Skills
. To adjust the scale.4. left-click on the point at the origin to select it. Then left-click and drag the origin as desired.
To adjust the origin.
• Click on "No" in the pop-up panel to keep degrees as the angle unit.
From the Graph menu. Adjust the origin and/or scale." use the pull-down "Functions" menu on the calculator keypad.
3. Turn on the grid. 0). choose Plot New Function. To enter "sin. • From the Graph menu. The function calculator should appear.B–24 Graphing Trigonometric Functions
1. The graph of y 2 sin (30x) 3 should appear on the grid.
2. • Click on the "OK" button on the calculator keypad. Graph the function y
2 sin (30x) 3.
Left-click on and drag either the origin or the point (1.
604
Appendix B: Review of Technical Skills
NEL
. choose Show Grid. • Use either the calculator keypad or the keyboard to enter "2 * sin (30 * x) 3".
Left-click and drag around the three
Appendix B Glossary
3. Choose the Point tool
2.
. Select the vertices. Draw the sides of the figure.
Answers Index
605
NEL
Appendix B: Review of Technical Skills
. choose Segments. Plot the vertices of the figure.
From the Construct menu.B–25 Creating Geometric Figures
To draw a triangle. The sides of the triangle should appear.
. 1. follow these steps.
Choose the Arrow Selection tool vertices to select them.
3. choose Length. then C. then A. From the Measure menu. Left-click on a side to select it.
606
Appendix B: Review of Technical Skills
NEL
. 1. choose Angle. To measure BAC. Repeat for the other two angles.
2. Repeat for the other two sides. Find the measure of each side. See "Creating Geometric Figures" (B-25).
• • • • •
Choose the Arrow Selection tool.B–26 Measuring Sides and Angles. Draw a triangle. select the vertices in that order: first B. Find the measure of each angle. Left-click on any blank area to deselect.
• • • •
Left-click on any blank area to deselect. draw a triangle and verify that the sine law holds for it. From the Measure menu. and Using the Calculator
As an example.
Left-click on the measure "mAB" to select it.
Appendix B Glossary Answers Index
607
NEL
Appendix B: Review of Technical Skills
. Select the measure "m/ACB". Repeat step 4 for the other two pairs of corresponding sides and angles. choose Calculate…. For side AB and the opposite ACB.
5. Use the "Functions" menu on the keypad to select "sin". Use the calculator keypad to enter "÷". Press the "OK" button on the calculator keypad. sin ACB From the Measure menu. The ratio should appear as a new measure.4. The calculator keypad should appear. calculate the ratio
• • • • • •
AB .
) When you name your first attribute. Deleting the collection deletes your data. The collection is where your data are actually stored.
2.
608
Appendix B: Review of Technical Skills
NEL
. repeat this step to add more attributes. and drop it in a blank space in your document. Drag a new graph from the object shelf at the top of the
Fathom window. empty box). Drag a case table from the object shelf. (Press Tab to move from cell to cell. type a name for the new variable
or attribute and press Enter. and drop it in
the document. Graph the data.)
3. pressing Tab instead of Enter moves you to the next column. When you add cases by typing values. and drop it on the prompt below and/or to the left of the appropriate axis in the graph. Drag an attribute from the case table. Create a case table. Click <new>. the collection icon fills with gold balls. Enter the Variables and Data. To enter the data click in the blank cell under the attribute name and begin typing values. (If necessary. Fathom creates an empty collection to hold your data (a little.PART 4 USING FATHOM
B–27 Creating a Scatter Plot and Determining the Equation of a Line or Curve of Good Fit
1.
Enter
your function using a parameter that can be adjusted to fit the curve to the scatter plot.2). and drop it in a blank space below your graph. Drag a new slider
from the object shelf at the top of the Fathom window. Create a slider for the parameter(s) in your equation. Create a function. Click on the number then adjust the value of the slider until you are satisfied with the fit.
Index
609
NEL
Appendix B: Review of Technical Skills
. (a was used in this case). Right click the graph and select Plot Function.4.
Appendix B
5. Type in the letter of the parameter used in your function in step 4 over V1.8(x
0.2)(x
6.
Glossary Answers
The equation of a curve of good fit is y
4.
7.5 and 5. tell why you think your answer is correct
M
measure: Use a tool to describe an object or determine an amount (e. Show something in picture form (e.) create: Make your own example explore: Investigate a problem by questioning. judge 2. Use a ruler and a protractor to construct an angle.. Determine if something makes sense.. Model addition of integers using red and blue counters. Calculate the value as a number explain: Tell what you did...)
D
describe: Tell. drawings.g. organize into categories compare: Look at two or more objects or numbers and identify how they are the same and how they are different (e. Pull or select an object (e.. Use a ruler to measure the height or distance around something. show how you know
S
show (your work): Record all calculations.) model: Show or demonstrate an idea using objects and/or pictures (e.g. drawings. and trying new ideas extend: 1. Draw a card from the deck. Use a measuring cup to measure capacity. 2. tell about a process in a step-by-step way determine: Decide with certainty as a result of calculation.g. show your mathematical thinking at every stage.) conclude: Judge or decide after reflection or after considering data construct: Make or build a model.. numbers. make a reasonable guess (e.g. brainstorming. continue the pattern 2. or symbols that make up the solution
610
Glossary
NEL
... Estimate how many leaves are on a tree. Predict the next number in the pattern 1..)
P
predict: Use what you know to work out what is going to happen (e.. words. Use balance scales to measure mass. or a solution. 4.g. Draw a diagram. Use a protractor to measure an angle. Compare the size of the students' feet. analyze relevant information to show understanding relate: Describe how two or more objects. Draw a tile from the bag. What is your estimate of 3210 1 789?) evaluate: 1. In patterning. or numbers are similar represent: Show information or an idea in a different way that makes it easier to understand (e. ideas.g. Use a stopwatch to measure the time in seconds or minutes. Compare the numbers 6.)
E
estimate: Use your knowledge to make a sensible decision about an amount.g.6.g. experiment.. create a new problem that takes the idea of the original problem further
J
justify: Give convincing reasons for a prediction. Draw a graph.)
R
reason: Develop ideas and relate them to the purpose of the task and to each other. draw an accurate geometric shape (e.g. . or exploration draw: 1. or write about what something is or what something looks like. compute clarify: Make a statement easier to understand. Make a model. Estimate how long it takes to cycle from your home to school.) 2. an estimate. In problem solving..Glossary
Instructional Words
C
calculate: Figure out the number that answers a question. provide an example classify: Put things into groups according to a rule and label the groups. draw. Compare two shapes.
1. This occurs when you know two side lengths and an angle opposite one of the sides rather than between them (an SSA triangle).g. A simple annuity is an annuity in which the payments coincide with the compounding period. For example. A bearing is a clockwise angle from magnetic north. but never meets.sketch: Make a rough drawing (e. equals x when x $ 0 or 2x when x .g. 0. If the given angle is obtuse. or conversion period. on some portion of its domain
y
the ambiguous case of the sine law: A situation in which 0. you can add or multiply in any order: (a 1 b) 1 c 5 a 1 (b 1 c) and (ab)c 5 a(bc) asymptote: A line that the graph of a relation or function gets closer and closer to.) solve: Develop and carry out a process for finding a solution to a problem sort: Separate a set of objects. describes the distance of x from 0.. and I is the interest amplitude: Half the difference between the maximum and minimum values. 0. the bearing of the lighthouse shown is 335°
N
25
W 335 S
E
Glossary
611
. 1. P is the principal. for example.) absolute value: Written as 0 x 0 . or 2 triangles can be drawn given the information in a problem. an ordinary annuity is an annuity in which the payments are made at the end of each interval arithmetic sequence: A sequence that has the same difference. the principal. 0 or 1 triangle is possible amortization schedule: A record of payments showing the interest paid. if a perpendicular is constructed. 0 3 0 5 3 and 0 23 0 5 2 (23) 5 3
V
validate: Check an idea by showing that it works verify: Work out an answer or solution again. ideas. The amount is given by A 5 P 1 I. where A is the amount. and the current balance on a loan or investment amount: The total value of an investment or loan. or numbers according to an attribute (e. usually in another way. it is also the vertical distance from the function's axis to the maximum or minimum value angle of depression (declination): The angle between a line below the horizontal and the horizontal angle of elevation: The angle formed by the horizontal and the line of sight (to an object above the horizontal) annuity: A series of payments or investments made at regular intervals. Sketch a picture of the field with dimensions. imagine
Mathematical Words
A
associative property: With addition and multiplication. show evidence of visualize: Form a picture in your head of what something is like. Sort 2-D shapes by the number of sides. the common difference. If the given angle is acute. drawings.. between any pair of consecutive terms arithmetic series: The sum of the terms of an arithmetic sequence
NEL
Glossary
asymptotes
x
axis of symmetry: A line in a two-dimensional figure such that. any two points lying on the perpendicular and the figure are at equal distances from this line
B
bearing: The direction in which you have to move in order to reach an object. or 2 triangles are possible.
which is the square of x 1 3 compound interest: Interest that is added to the principal before new interest earned is calculated. A. each of its terms is multiplied or divided by the number or term outside of the brackets
2
t12 5 2(12) 5 24 geometric sequence: A sequence that has the same ratio. and c. between any pair of consecutive terms geometric series: The sum of the terms of a geometric sequence
NEL
612
Glossary
. annually 1 1 time per year semi-annually 1 2 times per year quarterly 1 4 times per year monthly 1 12 times per year cosine law: The relationship. Interest is paid at regular time intervals called the compounding period compounding period: The intervals at which interest is calculated. of an investment after a certain length of time
G
general term: A formula. substitute n 5 12
˛
D
direction of opening: The direction in which a parabola opens. used to represent the value of the dependent variable—the output—for a given value of the independent variable. then a 2 5 b 2 1 c 2 2 2bc cos A curve of best fit: The curve that best describes the distribution of points in a scatter plot. if the general term of a sequence is tn 5 2n. used to determine unknown sides and angles in triangles. for any triangle. for example "12
E
equation of the axis: The equation of the horizontal line halfway between the maximum and the minimum. then to calculate the 12th term (t12 ). when a polynomial is expanded. labelled tn. Typically found using an informal process
domain: The set of all values of the independent variable of a relation down payment: The partial amount of a purchase paid at the time of purchase entire radical: A radical with coefficient 1. and if the angle A is opposite side a. For example. b. a 1 b 5 b 1 a and a 3 b 5 b 3 a completing the square: The process of adding a constant to a given quadratic expression to form a perfect trinomial square. but if 7 is added it becomes x 2 1 6x 1 9. If a triangle has sides a. x—the input future value: The total amount. The result is the same.C
common difference: The constant difference between two consecutive terms in an arithmetic sequence or series common ratio: The constant ratio (quotient) between two consecutive terms in a geometric sequence or series commutative property: The order in which you add or multiply numbers does not matter. For example. x 2 1 6x 1 2 is not a perfect square. such as f (x). for example. up or down discriminant: The expression b 2 4ac in the quadratic formula distributive property: The principal that says that. it is determined by y5 maximum value 1 minimum value 2
exponential function: A function of the form y 5 a(bx )
F
factored form: A quadratic relation in the form f (x) 5 a(x 2 r) (x 2 s) family: A collection of functions (or lines or curves) sharing common characteristics family of parabolas: A group of parabolas that all share a common characteristic function: A relation where each value of the independent variable corresponds with only one value of the dependent variable function notation: Notation. the common ratio. involving the cosine of one of the angles and the lengths of the three sides. that expresses each term of a sequence as a function of its position. Typically found using regression analysis curve of good fit: A curve that reasonably describes the distribution of points in a scatter plot. So interest is calculated on the principal and on the interest already earned.
undoes what the original function has done parabola: The graph of a quadratic relation of the form y 5 ax 2 1 bx 1 c (a 2 0). which resembles the letter "U. for example. Any restrictions on a variable must be stated independent variable: In an algebraic relation. Typically found using linear regression analysis line of good fit: The straight line that reasonably describes the distribution of points in a scatter plot. the side that is opposite the right angle
I
identity: A mathematical statement that is true for all values of the given variables.. It tells which root is indicated: 3 for cube root. a variable whose values may be freely chosen and upon which the values of the other variables depend. If there is no number. involving the reciprocals of a list of numbers n H5 1 1 1 . the other decreases or vice versa nonperiodic function: Any function that does not repeat at regular intervals
P
inverse of a function: The reverse of the original function. Often represented by x index (plural indices): The number at the left of the radical sign. The graph. where 1 a2 1 . If the identity involves fractions. an are positive real numbers
˛
hypotenuse: The longest side of a right triangle.. the denominators cannot be zero.. 1 an a1
L
like radicals: Radicals that have the same number under the radical symbol. such as 3"6 and 22"6 linear relation: A relation between two variables that appears as a straight line when graphed on a coordinate system.. . Typically found using an informal process lowest common denominator (LCD): The smallest multiple shared by two or more denominators
a1. 0) for this graph and transformation y
4 y = x2 4 4 2 2 0 2 y= 4 x2 2 4 4 x
M
mixed radical: A radical with coefficient other than 1. denoted by H.. for example. etc.H
half life: The time required for a quantity to decay to half of its initial value harmonic mean: A type of average. May also be referred to as a linear function line of best fit: The straight line that best describes the distribution of points in a scatter plot." is symmetrical
NEL
Glossary
613
. 0) and (2. (22. the square root is intended interest: The cost of borrowing money or the money earned from an investment invariant point: A point on a graph (or figure) that is unchanged by a transformation. 4 for fourth root. 2"3
N
negative angle: An angle measured clockwise from the positive x-axis
y
Glossary
x 0 210
negative correlation: This indicates that as one variable in a linear relationship increases.
.) principal: A sum of money that is borrowed or invested principal angle: The counterclockwise angle between the initial arm and the terminal arm of an angle in standard position. Sn. of the first n terms of a sequence peak: The maximum point on a graph
y peak trough
principal angle initial arm quadrant 4
x
x
period: The change in the independent variable (typically x) corresponding to one cycle. cube. the y-values in the table of values show a repetitive pattern when the x-values change by the same increment phase shift: The horizontal translation of a sinusoidal function is also called a phase shift positive correlation: This indicates that both variables in a linear relationship increase or decrease together present value: The principal that would have to be invested now to get a specific future value in a certain amount of time. or higher root.x20 3x
614
Glossary
NEL
. 1 and itself (e. The formula is phrased in terms of the coefficients of the quadratic equation: x5 2b 6 "b2 2 4ac 2a
periodic function: A function whose graph repeats at regular intervals. for example..parent function: The simplest. such as 23 or 2 3 rational expression: A quotient of polynomials.g. a cycle of a periodic function is a portion of the graph that repeats
y 1 cycle x 0 period
Pythagorean theorem: The conclusion that. or base. y 5 x 2 1 7x 1 10 or " 27 5 3. function in a family
y 4 2 4 2 0 2 4 2 4 x
prime number: A number with only two factors. the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides
Q
quadratic formula: A formula for determining the roots of a quadratic equation of the form ax 2 1 bx 1 c 5 0. PV is used for present value instead of P. for 2x 2 1 example. " is called the radical symbol
R
3
range: The set of all values of the dependent variable of a relation rational form: A number written as an integer or a 2 fraction. 17 is a prime number since its only factors are 1 and 17. Its value is between 0° and 360°
quadrant 2 terminal arm related acute angle vertex 0 quadrant 3 y quadrant 1
partial sum: The sum. in a right triangle. such as "4 5 2 quadratic relation: A relation whose equation is in quadratic form. since P is used for principal
radical: A square.
b. B. used to determine unknown sides and angles in triangles.rational function: Any function whose output can be given by an expression that is the ratio of two polynomials. equivalently. for any triangle. where R and S are polynomials and S 2 0—for example. 3. fractions. x 2 2 2x 1 3 1 f (x) 5 . respectively. values of the independent variable are paired with values of the dependent variable restrictions: The values of the variable(s) in a rational function or rational expression that cause the function to be undefined. graphs of sinusoidal functions can be created by transforming the graph of the function y 5 sin x or y 5 cos x
4 2 0 y x 2 4 6 8 10
1 adjacent cot u 5 5 tan u opposite cot u is the short form for the cotangent of angle u. and C. and c. then a b c 5 5 sin A sin B sin C sinusoidal function: A periodic function whose graph looks like smooth symmetrical waves. where any portion of the wave can be horizontally translated onto another portion of the curve. sec u is the short form for the secant of angle u.x2 4x 2 1 4 real numbers: Numbers that are either rational or irrational. and irrational numbers such as "2 and p reciprocal trigonometric ratios: The reciprocal ratios are defined by dividing 1 by each of the primary trigonometric ratios csc u 5 sec u 5 1 hypotenuse 5 sin u opposite hypotenuse 1 5 cos u adjacent
R(x)
relation: A set of ordered pairs. these include positive and negative integers. zero. and csc u is the short form for the cosecant of angle u recursive formula: A formula relating the general term of a sequence to the previous term or terms recursive sequence: A sequence for which one term (or more) is given and each successive term is determined from the previous term(s) related acute angle: The acute angle between the terminal arm of an angle in standard position and the x-axis when the terminal arm lies in quadrants 2. or 4
quadrant 2 terminal arm related acute angle vertex 0 quadrant 3 y quadrant 1
Glossary
principal angle initial arm quadrant 4
2 x
standard form: A quadratic relation in the form f (x) 5 ax 2 1 bx 1 c
NEL
Glossary
615
. and if the angles opposite each side are A. If a triangle has sides a. A rational function can be expressed as f (x) 5 S(x) . These are the zeros of the denominator or. the numbers that are not in the domain of the function
S
scatter plot: A graph that attempts to show a relationship between two variables by means of points plotted on a coordinate grid sequence: An ordered list of numbers series: The sum of the terms of a sequence simple interest: Interest earned or paid only on the original sum of money that was invested or borrowed sine law: The relationships. involving the sines of two of the angles and the lengths of the opposite sides.
g. then the relation is not a function
T
term: A number in a sequence. A cube has eight vertices.) vertex form: A quadratic function written in the form f (x) 5 a(x 2 h) 2 1 k is in vertex form. Angle u is measured from initial arm to terminal arm (the arm that rotates)
y
m l ar ina m ter
trough: The minimum point on a graph
y peak trough
x
0
initial arm
x
vertex
V
vertex (plural vertices): The point at the corner of an angle or shape (e. k) vertical line test: If any vertical line intersects the graph of a relation more than once.standard position: An angle in the Cartesian plane whose vertex lies at the origin and whose initial arm (the arm that is fixed) lies on the positive x-axis. An angle has one vertex. Subscripts are usually used to identify the position of the terms transformation: A geometric operation such as a translation. rotation. or reflection trend: A relationship between two variables for which the independent variable is time
616
Glossary
NEL
.. the vertex is (h. A triangle has three vertices. dilation. | 677.169 | 1 |
A Simple Plan: Sales
Types of Calculators and Their Uses There are many types of calculators in the market and while some of them are complex, some are simple. While simple calculators can only perform basic arithmetic, specialized ones do perform many tasks including creating graphs and finding points on a parabola. Given the functionalities of these calculators, you have to ensure that you get the right one for the specific task you intend to use it. The prices of calculators increase with the increase in functionality hence finding the right calculator can save you money. With the basic calculator, you have a device that can only do simple arithmetic. They have fewer buttons with the necessary signs including addition, multiplication, division, and subtraction are clearly indicated. Some of these calculators are capable of saving a couple of results from previous calculations. The calculators also have a small solar panel to complement the batteries in powering it. If you are in a finance field, especially bookkeeping and accounting, the printing calculator is what you need. It has additional keys to perform functions such as the grand total, tax rate calculation, among others. In addition, it allows checking for common errors during entry such as the transposed and skipped digits. Some printing calculators even have the ability to produce receipts and round off decimals. Lessons Learned About Calculators
The scientific calculator is mostly used in schools. These calculators are designed to help students solve problems in various subjects such as chemistry, mathematics, and physics. An extra function button enables you to use the multifunctional features of these calculators. Some of the functions that are possible with the scientific calculator include calculating powers, roots, exponential functions, logarithmic functions, and trigonometric functions, among others. The Path To Finding Better Calculators
The business or financial calculator is also another type of calculator commonly found in the market. With this calculator, you can calculate a variety of business functions. They are suitable for use by those who work in the business field as well as those in the business school. With the financial calculator, you can do a couple of calculations such as APR, mortgage repayments, and compound interests. The business functions and some math functions are clearly indicated. The graphing calculator is essential in calculating mathematics and business functions that involve graphs and it is mostly used in colleges. These calculators are highly programmable and have a huge storage to store more data than the normal ones. Despite lacking the print capability, graphing calculators can do almost every other arithmetic. They are designed to plot graphs and solve equations. | 677.169 | 1 |
Fundamentals of Math and Physics for Game Programmers
Fundamentals of Math and Physics for Game Programmers teaches the fundamental math and physics concepts, principles, and formulas that are crucial for developing successful games. Covers topics such as trigonometry snippets, vector and matrix operations, transformations, momentum and collisions, and 1D/2D/3D motion. Concepts are taught in a step-by-step format in order to improve the level of game development. Includes case studies and hands-on projects allowing students to experience the application of essential concepts. End-or-chapter review exercises are provided for additional content reinforcement
"synopsis" may belong to another edition of this title.
From the Inside Flap:
"The book is very well written; the author does an excellent job of explaining the concepts and walking the reader through the various terms and applications of math and physics for the 3-D world." Bill Galbreath Program Manager for Digital Media, Full Sail Real World Education
"There are several other books covering game-specific math and physics on the shelves. Those books focus on intermediate to advanced issues that might be too complex for those learning the basics. This book focuses on topics that are foundational in the area of math and physics that the other books tend to assume the reader already understands." David Astle Executive Producer, GameDev.net
"I could see myself using this book with my students when I'm doing introductory game design and programming. The content is well organized, lucidly presented, and written at a good level for students who lack the basic math and/or physics background." Harvey Duff Department Head[md]Computer Technology, Jasper Place High School, Edmonton, Alberta, Canada
"New developers and students in math/physics will find the topics covered a good start. The topics are covered quite comprehensively to make the book easy to pick up and read. The exercises and examples are also helpful tools." Gianfranco Berardi Computer Science Student, DePaul University
"This book is an excellent practical introduction to game math and physics. The author has an interesting visual approach to trigonometry and matrix arithmetic, which makes it an entertaining read. There's a good mix of theoretical insight (described in an unusually useful visual style) and practical advice." Richard Jones Director, Merjis Ltd.
From the Back Cover:
Whether you're a hobbyist or a budding game design pro, your objective is probably the same: To create the coolest games possible using today's increasingly sophisticated technology. To do that, however, you need to understand some basic math and physics concepts. Not to worry: You don't need to go to night school if you get this handy guide! Through clear, step-by-step instructions, author Wendy Stahler covers the trigonometry snippets, vector operations, and 1D/2D/3D motion you need to improve your level of game development. Each chapter includes exercises to make the learning stick, and Visualization Experience sections are sprinkled throughout that walk you through a demo of the chapter's content. By the end of the volume, you'll have a thorough understanding of all of the math and physics concepts, principles, and formulas you need to control and enhance your user's gaming experience. | 677.169 | 1 |
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Computer Science and Applied Mathematics
How can computational tools solve specific bioinformatics problems like the prediction of protein structures, the identification of protein functional sites, and the discovery of genes? How can mathematical modeling augment our understanding of biological phenomena? Bioinformatics-based activities bridge the divide between computer science, math, and biology classrooms as students learn powerful ways to seek the answers to complex questions. | 677.169 | 1 |
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Show More the market, the authors have incorporated the use of color to highlight different methods of solving a problem using the calculator. By providing students with a centralized resource where they can find information for how to use their specific model of calculator to perform the operations shown in the text, this calculator manual enables professors to focus on teaching how to interpret information from the graphing calculator without wasting time showing students how to perform operations on every calculator model used in their class | 677.169 | 1 |
AoPS Holidays
There are no classes July 4, September 4, November 18 - 26, and December 21 - January 3.
Who Should Take?
Students are ready for this course if they have mastered arithmetic with fractions, decimals, percents,
negative numbers, and integer exponents. Most students who have completed a typical Prealgebra course are ready for this course.
Students finishing our Beast Academy curriculum should take Prealgebra 1 and
Prealgebra 2 prior to taking this course.
Please note: Our Algebra sequence will likely not match up exactly with your school's.
Most students who have completed a Prealgebra course are ready for this course.
This course covers much of the algebra of a typical honors Algebra 1 course, some of the content of an honors
Algebra 2 course,
and some additional material not taught in most standard curricula.
Visit our Recommendations page for help in choosing the right class.
This is the first AoPS course I've ever taken, and I absolutely loved it. Definitely will be taking more classes here. This is the best math program I've ever had. It's very challenging but also very fun. | 677.169 | 1 |
Math
Department Mission Statement
We live in a world where engineering makes possible the conviences of life. Comprehensive knowledge in mathmatics is extremely important to ensure that we are able to maintain the infrastructure on which we have become dependent, and to develop the infrastructure of tomorrow. Our primary goal is to share our love for mathmatics with our students. When students see our excitement for mathmatics they will appreciate the opportunity they have to learn. We strive to create critical thinkers who are able to take the lessons we teach in the classroom and apply them to our daily lives to solve real-world problems. We understand that there is enough formality in mathmatics to allow for freedom of thought and opportunity to try new apporaches, and understand that we need to allow students the opportunity to make mistakes. Our department promotes metacognitive skills that will create more complex thinkers who are capable of success in mathmatics beyond high school.
Courses
Mathematics
It is very difficult to outline a precise math sequence for every student since the math sequence will depend upon the large number of available math courses and the following three influential variables: student career plans, student math ability, and teacher recommendation. Students are urged to seek help from members of the math staff before attempting to select next year's courses. The Milpitas High School Mathematics department strongly recommends that students with a grade of "D" in a particular course repeat that course before progressing with the math continuum of courses. Courses repeated to improve a "D" grade do not earn additional required math units. Students or parents desiring further guidance should contact their present teacher, counselor, or call the Mathematics Department Chairperson.
The table below will represent what will happen in the future
9th
10th
11th
12th
Math 1
Math 2
Math 3
Pre-Calculus OR
AP Calc AB OR
AP Statistics
Math 2
Math 3
AP Calculus AB
AP Statistics OR
AP Calc BC
Math 2A
Math 3A
AP Calculus BC
AP Statistics
Math 1
Description: Math 1 is the first course in the Integrated Common Core Pathway at Milpitas High School. Math 1 is to formalize and extend fundamental math concepts in middle school math. Topics will include number & quantity, algebra, geometry, sequences, and statistics. The eight Mathematical Practice Standards will be implemented in the course as well. It is intended for students to be active learners and use critical thinking to solve problems and formulate certain properties and formulas on their own.
Description: Math 2 is the second year course in the Integrated Common Core Pathway at Milpitas High School. Math 2 builds on understanding of functions from Math 1 by extending characteristic and behavior of linear & exponential relations to quadratic functions. Topics will include exponents, quadratics, probability, complex numbers, and geometry regarding triangles properties, circles, and proofs. The eight Mathematical Practice Standards will be implemented in the course as well. It is intended for students to be active learners and use critical thinking to solve problems and formulate certain properties and formulas on their own.
Math 2 Advanced
#200---
Grade Level: 9th – 11th
Length: 1 year
Pre-requisites: Completion of Math 1 with a grade of "A" with strong math skills & study habits AND teacher recommendation.
Description: Math 2 Advanced is the second year course in the Integrated Common Core Pathway at Milpitas High School with additional Common Core Honors Standards than Math 2. It will align with Math 2 to build on understanding of functions from Math 1 by extending characteristic and behavior of linear & exponential relations to quadratic functions. Topics will include exponents, quadratics, probability, complex numbers, and geometry regarding triangles properties, circles, and proofs. In addition, the eight Mathematical Practice Standards will be implemented in the course as well. It is intended for students to be active learners and use critical thinking to solve problems and formulate certain properties and formulas on their own.
Math 3
#200----
Grade Level: 11th – 12th
Length: 1 year
Pre-requisites: Completion of Math 2 with a grade of "C" or better in Math 2; teacher recommendation.
Description: Math 3 is the third year course in the Integrated Common Core Pathway at Milpitas High
School. It will align with Math 2 to build on understanding of functions from Math 1 by extending characteristic and behavior of linear & exponential relations to quadratic functions. Topics will include polynomial and rational functions, sequences and series, exponential and logarithmic functions, trigonometry and analyzing trigonometric functions, probability & statistics, complex numbers, optimization and geometric modeling. In addition, the eight Mathematical Practice Standards will be implemented in the course as well. It is intended for students to be active learners and use critical thinking to solve problems and formulate certain properties and formulas on their own.
Algebra II/Trigonometry (phased out after 2015-16)
#201445 [M] {UC} (CSU)
Grade level: 10-12
Length: 1 year
Prerequisites: Grades of a "B" or better in Geometry A or an "A" or better in Geometry I and Teacher recommendation Description: All of the concepts covered in Algebra I, Geometry I (or Geometry A) and Algebra II will be expanded upon and presented with a strong emphasis on the underlying structure mathematics. In addition students will study concepts of quadratic functions, logarithmic functions, circular functions, and trigonometry. This course moves at a rapid pace and covers two chapters on the use of rigorous mathematical proof.
Pre-calculus
#203505 [M] {UC} (CSU)
Grade level: 11-12
Length: 1 year
Prerequisites: Algebra II with a grade of "C" or better. Description: This course is designed to be taken after Algebra II. Topics will include trigonometric functions treated as circular functions, trigonometric identities and equations, inverse functions, properties of vectors, complex numbers and analytic geometry. Analytic geometry topics will include lines and vectors in the plane and in space, conic sections, transformation of coordinates, surfaces in space, matrixes and determinants.
Honors Pre-calculus
#203555 [M] {UC} (CSU)
Grade level: 11-12
Length: 1 year
Prerequisites: Algebra II/Trig. with a grade of "B" or better
AND teacher recommendation
Description: Honors Pre-calculus consists of a sound introduction to Calculus. Most of the basic ideas of advanced mathematics and calculus are considered. Mathematical induction sequences, limits, vectors, polynomials, functions and other fundamental mathematical concepts are explored in considerable depth. All the topics prerequisite for a course in complex numbers are treated in much detail. A concentrated effort is made to show the unity and structure of the real number system. The problems in the course are designed to encourage independent and creative thinking in mathematics.
Advanced Placement Calculus AB
#204065 $ for Test [M] {UC} (CSU)
Grade level: 11-12
Length: 1 year
Prerequisites: Pre-calculus with a grade of "C" or better and a teacher recommendation.
Description: AP Calculus AB is a RIGOROUS COLLEGE LEVEL COURSE that will prepare students to take the AP Calculus AB exam in May. The course consists of an introduction, review of inequalities, absolute value, and analytic geometry as well as a thorough study of the following topics: limits of algebraic functions and continuity of algebraic functions, the derivative and differentiation of algebraic functions, applications of the derivative, the definite integral, applications of the definite integral, logarithmic and exponential functions, trigonometric functions, techniques of integration and approximate integration.
Advanced Placement Calculus BC
#204265 $ for Test [M] {UC} (CSU)
Grade level: 11-12
Length: 1 year
Prerequisites: Honors Pre-calculus and a teacher recommendation.
Description: AB Calculus BC is a RIGOROUS COLLEGE LEVEL COURSE, which will prepare students to take the AP Calculus BC exam in May. This course consists of a review of limits using the epsilon-delta approach that will lead to the derivative and its applications. Integration and its applications, techniques of integration, indeterminate forms, polar curves, vector calculus, series, parametric equations, and elementary differential equations are among the major topics covered in this course.
AP Statistics
#204565 $ for Test [M] {UC} (CSU)
Grade level: 11-12
Length: 1 year Prerequisites: Pre-Calculus
Description: Advanced Placement Statistics is a college level course, which upon conclusion students participate in the AP Exam to earn college credit. It is an activity based course in which students actively construct their own understanding of the concepts and techniques of statistics. Topics include Exploring and Analyzing Data, Planning, Designing, and Conducting a Study, Anticipating Patterns, Statistical Inference with Applications, Mean and Variance, Normal distribution and Regression, and Testing of Significance for Means and Proportions.
CAHSEE Math
#205345
Grade level: 11-12
Length: 1 year Prerequisites: None
Description: This Course is strictly for 11th and 12th grade
students that have not passed the math section of the California High School Exit Exam. Students review skills found on the CAHSEE, take regular practice exams and attend CAHSEE lab session to build Math proficiency.
Consumer Math
#205645 [M]
Grade level: 11-12
Length: 1 year (may be taken one or both semesters) Prerequisite: Algebra I with final grade of "C" or better Description: This course is designed to offer the student a practical approach to basic math concepts by using common consumer problems. It satisfies one year of the mathematics graduation requirement. This course is structured so the student will strengthen the basic math skills necessary in everyday consumerism. Topics covered may include:
computing salary and deductions, making change, estimating monthly bills for groceries, entertainment, utility, transportation, medical expenses and insurance. This course will be useful to any student and especially to those who prefer to learn math by using practical everyday life situations. | 677.169 | 1 |
Algebra project
Why U animated videos are designed for mathematics courses on the K-12 and college levels, and as a resource for informal independent study. Rather than focusing on. [SourceForge Summary Page] The ATLAS (Automatically Tuned Linear Algebra Software) project is an ongoing research effort focusing on applying empirical techniques in. A free mathematics software system licensed under the GPL. It combines the power of many existing open-source packages into a common Python-based interface. Details. Software for math teachers that creates exactly the worksheets you need in a matter of minutes. Try for free. Available for Pre-Algebra, Algebra 1, Geometry, Algebra.
CME Project is a four-year, NSF-funded, comprehensive high school mathematics program that is problem-based, student-centered, and organized around the familiar. Downloads. Files can be downloaded from the Files section in the Sourceforge site. That section contains links to various different versions and releases of. Fun math practice! Improve your skills with free problems in 'Complete a function table: quadratic functions' and thousands of other practice lessons. Design a Brochure Algebra 2 Project Linear Equations and Inequalities Purpose: For this project, you will be working as a group of two to design a tri-fold brochure or.
Algebra project
Why U animated videos are designed for mathematics courses on the K-12 and college levels, and as a resource for informal independent study. Rather than focusing on. The Algebra Project, Inc. is a 501 (c) (3) national, nonprofit organization that uses mathematics as an organizing tool to ensure quality public school education for. Pearson Prentice Hall and our other respected imprints provide educational materials, technologies, assessments and related services across the secondary curriculum.
[SourceForge Summary Page] The ATLAS (Automatically Tuned Linear Algebra Software) project is an ongoing research effort focusing on applying empirical techniques in. Downloads. Files can be downloaded from the Files section in the Sourceforge site. That section contains links to various different versions and releases of. Texas: Algebra Ready (TXAR) The Texas: Algebra Ready (TXAR) initiative exists to increase the preparedness of students to meet standards and pass assessments in.
Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. JMAP offers New York teachers and other users of the Common Core State Standards free resources that simplify the integration of Regents exam questions into their. Texas: Algebra Ready (TXAR) The Texas: Algebra Ready (TXAR) initiative exists to increase the preparedness of students to meet standards and pass assessments in. A First Course in Linear Algebra is an introductory textbook designed for university sophomores and juniors. Typically such a student will have taken calculus, but. Pearson Prentice Hall and our other respected imprints provide educational materials, technologies, assessments and related services across the secondary curriculum.
Bored with Algebra? Confused by Algebra? Hate Algebra? We can fix that. Coolmath Algebra has hundreds of really easy to follow lessons and examples. Algebra 1.
LAPACK routines are written so that as much as possible of the computation is performed by calls to the Basic Linear Algebra Subprograms (BLAS).
LAPACK routines are written so that as much as possible of the computation is performed by calls to the Basic Linear Algebra Subprograms (BLAS).
A First Course in Linear Algebra is an introductory textbook designed for university sophomores and juniors. Typically such a student will have taken calculus, but. Maxima, a Computer Algebra System. Maxima is a system for the manipulation of symbolic and numerical expressions, including differentiation, integration, Taylor. By leveraging the existing network of math teachers, nationwide, Bootstrap is built to scale. We work with school districts across the country, reaching hundreds of. By leveraging the existing network of math teachers, nationwide, Bootstrap is built to scale. We work with school districts across the country, reaching hundreds of. | 677.169 | 1 |
(1) Within a well-balanced mathematics curriculum, the primary focal points at Grade 8 are using basic principles of algebra to analyze and represent both proportional and non-proportional linear relationships and using probability to describe data and make predictions.
(2) Throughout mathematics in Grades 6-8, students build a foundation of basic understandings in number, operation, and quantitative reasoning; patterns, relationships, and algebraic thinking; geometry and spatial reasoning; measurement; and probability and statistics. Students use concepts, algorithms, and properties of rational numbers to explore mathematical relationships and to describe increasingly complex situations. Students use algebraic thinking to describe how a change in one quantity in a relationship results in a change in the other; and they connect verbal, numeric, graphic, and symbolic representations of relationships. Students use geometric properties and relationships, as well as spatial reasoning, to model and analyze situations and solve problems. Students communicate information about geometric figures or situations by quantifying attributes, generalize procedures from measurement experiences, and use the procedures to solve problems. Students use appropriate statistics, representations of data, reasoning, and concepts of probability to draw conclusions, evaluate arguments, and make recommendations.
(3) Problem solving in meaningful contexts, language and communication, connections within and outside mathematics, and formal and informal reasoning underlie all content areas in mathematics. Throughout mathematics in Grades 6-8, students use these processes together with graphing technology and other mathematical tools such as manipulative materials to develop conceptual understanding and solve problems as they do mathematics.
(b) Knowledge and skills.
(1) Number, operation, and quantitative reasoning. The student understands that different forms of numbers are appropriate for different situations. The student is expected to:
(A) compare and order rational numbers in various forms including integers, percents, and positive and negative fractions and decimals;
(B) select and use appropriate forms of rational numbers to solve real-life problems including those involving proportional relationships;
(C) approximate (mentally and with calculators) the value of irrational numbers as they arise from problem situations (such as p, Ö2);
(A) compare and contrast proportional and non-proportional linear relationships; and
(B) estimate and find solutions to application problems involving percents and other proportional relationships such as similarity and rates.
(4) Patterns, relationships, and algebraic thinking. The student makes connections among various representations of a numerical relationship. The student is expected to generate a different representation of data given another representation of data (such as a table, graph, equation, or verbal description).
(5) Patterns, relationships, and algebraic thinking. The student uses graphs, tables, and algebraic representations to make predictions and solve problems. The student is expected to:
(A) use variability (range, including interquartile range (IQR)) and select the appropriate measure of central tendency to describe a set of data and justify the choice for a particular situation;
(B) draw conclusions and make predictions by analyzing trends in scatterplots; and
(C) select and use an appropriate representation for presenting and displaying relationships among collected data, including line plots, line graphs, stem and leaf plots, circle graphs, bar graphs, box and whisker plots, histograms, and Venn diagrams, with and without the use of technology.
(13) Probability and statistics. The student evaluates predictions and conclusions based on statistical data. The student is expected to:
(A) evaluate methods of sampling to determine validity of an inference made from a set of data; and
(B) recognize misuses of graphical or numerical information and evaluate predictions and conclusions based on data analysis.
(14) Underlying processes and mathematical tools. The student applies Grade 8 mathematics to solve problems connected to everyday experiences, investigations in other disciplines, and activities in and outside of school. The student is expected to:
(A) identify and apply mathematics to everyday experiences, to activities in and outside of school, with other disciplines, and with other mathematical topics;
(B) use a problem-solving model that incorporates understanding the problem, making a plan, carrying out the plan, and evaluating the solution for reasonableness;
(C) select or develop an appropriate problem-solving strategy from a variety of different types, including drawing a picture, looking for a pattern, systematic guessing and checking, acting it out, making a table, working a simpler problem, or working backwards to solve a problem; and
(D) select tools such as real objects, manipulatives, paper/pencil, and technology or techniques such as mental math, estimation, and number sense to solve problems.
(15) Underlying processes and mathematical tools. The student communicates about Grade 8 mathematics through informal and mathematical language, representations, and models. The student is expected to: | 677.169 | 1 |
Description
Over 200 practice questions with detailed explanations covering all topics and 4 complete practice papers make this an invaluable revision tool at your desk or on the move. The practice questions are suitable for any Higher Tier GCSE. The complete papers are written for the 2-paper Linear courses from Edexcel (1MAO), AQA (4365) and OCR (J567).
The topic questions come with solutions and detailed explanations that are written in student-friendly language. Work through each topic to revise and gain confidence, then try the practice papers to test your knowledge and understanding.
The content of this App is of excellent quality from the author of practice GCSE and A level papers used in hundreds of UK schools. There is no overlap of questions with our current papers sold to | 677.169 | 1 |
Goals: The main goal of the Mid-Cities Math Circle seminar is to provide a stimulating environment for local area middle- and high-school students to learn mathematics. Regular attendance of the (MC)^2 seminar will help students not only improve their individual problem-solving skills, but also enjoy and understand mathematics better. More specifically the Mid-Cities Math Circle seminar will
* attract students' attention to mathematics and motivate them to excel in the subject;
* prepare students for mathematical contests;
* introduce them to the beauty of advanced mathematical theories;
* encourage them to pursue careers related to mathematics such as scientists, educators, engineers, economists, and business leaders.
General Information: The seminar runs every semester and is designed for students in 8th-12th grades. Students in lower grades are welcome as well but should be advised that a solid mathematical background is required in order to follow the discussions. | 677.169 | 1 |
Practical Linear Algebra introduces students in mathematics, science, engineering, and computer science to Linear Algebra from an intuitive and geometric viewpoint, creating a level of understanding that goes far beyond mere matrix manipulations. Practical aspects, such as computer graphics topics and numerical strategies, are covered throughout, and thus students can build a "Geometry Toolbox," based on a geometric understanding of the key concepts." This book covers all the standard linear algebra material for a first-year course; the authors teach by motivation, illustration, and example rather than by using a theorem/proof style. | 677.169 | 1 |
15 Images of Advanced Algebra 2 Practice Worksheets
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Useful Worksheet Designing Tips:
Having all the necessary details before you start to write or create. Research, read, research, resources.
Simple tips: Let the background determine how we align our text.
Do not use a distinctive icon and the symbol that you see everywhere to represent your topic.
We are just like you, some humans which are highly honor creativity from every one, with no exception. That's why we make sure to keep the original images without any editing including the copyright mark. Every pictures gallery we include are always carrying website or blog link where we found it below each photos. Common thing is people ask about their right connected with the photos on our gallery. When you need to ensure your right, please contact the website on each photos, actually we are not able to decide your right. We notice you, if you don't see watermark does not mean the photos can be freely used without permission. | 677.169 | 1 |
Check out the PREVIEW for a sneak peek of this Toolbox Builder!
This functions foldable is designed as an extensive graphic organizer to compare linear functions and exponential functions!
This summary activity focuses on comparing the two types of functions in different representatOVER 50 PAGES! These are the best strategies for teaching linear functions to all students. Tested in grades 4 through high school, this unit helps students develop competency with both the concepts and computations involved in functions. The multi-representational activities cover slope, y-intercept
ASSESSING EVERY STANDARD - FUNCTIONS, Linear, Quadratic and Exponential Models
Have you tried to find quick assessments for Common Core algebra one?
So have I!
After giving up the search, I made my own. My goal is to have a quick resource to ASSESS EVERY STANDARD.
This packet of six quick assess
This product is a two-page foldable, in the style of a mini flip-book, that summarizes various representations of Rate of Change. Students will get practice finding the rate of change of linear functions using graphs, tables of values, equations, ordered pairs and contextual situations. The file formIn the context of birthday present options, students explore linear and exponential functions--their rates of change, their graphs, and their values. The teachers who created and used this task used it as an introduction to the unit and before instruction over the topics. part 2 in a five-part series on developing an understanding of and proficiency with functions. Part 1 offered an introduction helping students to understand slope and y-intercept when integrated across a multi-representational approach. In part 2, the student will develop even greate
This 13 lesson unit explores one of the most important types of functions, linear. It builds from proportional relationships and equations of lines to linear modeling. The fundamental concept of a graph showing solutions to equations is emphasized. Finally, arithmetic sequences are introduced as a ty
Fundamental exponent laws are developed early in this 9 lesson unit and give rise to an exploration of basic exponential functions. Percent calculations are then used to develop increasing and decreasing exponential models. A comparison between linear and exponential models is extensively done. The u
In "Linear Equations of Flags," the students will be drawing different flags with linear equations and researching the meaning behind each one. They will be representing their flags with linear equations in both slope-intercept form and point-slope form. This project will open up the students minds
Algebra that Functions
Quadratic, Linear, and Exponential Functions Properties
Do you want your students to look at patterns on models of tables of values to identify whether the function is linear or quadratic or exponential? In this exploration, your students will explore equati
Algebra that Functions
Exponential Functions and Graphs Activity
Do you want your students to understand the relationships between equations for exponential functions and graphs? In this activity your students will evaluate functions and create tables of values for exponential fun
Algebra that Functions
Linear Functions Real-World Application Math Project
Are you looking for a way to assess your students' understanding of linear functions? My Linear Functions Real-World Application Math Project is just what you are looking for. Your students will create a
At this point in any students' experience of learning to graph linear equations, they are beginning to notice that there are multiple variations available to them in order to graph a linear equation.
* They learn to graph a linear equation by finding the x and y intercepts.
* They learnAlgebra that Functions
Linear and Exponential Functions Math Project: Which Salary Would You Choose?
Are you looking for a way for you students to show you they can model linear and exponential functions with a real world application? My product, Linear and Exponential Functions M
This lesson is a follow-up to my Patterns & Sequences Lesson Plan on TpT. In this lesson, students will distinguish between arithmetic and geometric sequences. They will use the functions for arithmetic and geometric sequences to find any term in a sequence.
There is also a reading activity thatThe Town of Gilbert, since 2002, has seen increases in revenue generated from its recycling program. With an increase in population, Gilbert has also seen an increase in the amount of recyclables collected. Furthermore, a recent newspaper article indicated that Gilbert has gone from getting $5 per to only for
This handout teaches students about the history of the Quadratic Formula and how its used today in various applications including sports and business. The handout aligns perfectly to the Common Core Standards for Reading and helps math teachers provide opportunities for reading in their content area
This activity allows students to get to know each other better, and practice using the slope formula! Students are given a point to start with. They then pair with another student and use the slope formula and their two points to calculate the slope between those two points. They then fine a new par
This is the sixth (in a series of 9) worksheet I created to teach the ideas between linear and exponential functions.
Topics covered in this worksheet are: comparing geometric and arithmetic sequences, comparing geometric and arithmetic tables of values, comparing geometric and arithmetic graphs, f
This is the second lesson (in a series of 6) that I created to introduce my students to quadratic functions. This lesson covers: identifying the different coefficients of a quadratic function (a, b, and c), finding a c-value for a quadratic that will give different numbers of x-intercepts, graphing a
This is the third lesson (in a series of 6) that I created to introduce my students to quadratic functions. This lesson covers: finding the x and y-intercepts of a linear equation in standard form and graphing, multiplying two binomials and simplifying to a trinomial, completing the square of a quadr
This is the fourth lesson (in a series of 6) that I created to introduce my students to quadratic functions. This lesson covers: factoring a trinomial quadratic into two binomials, and simplifying radical expressions that have numbers and variables.
This is the fifth lesson (in a series of 6) that I created to introduce my students to quadratic functions. This lesson covers: multiplying two binomials using a 2-way table, and given a factored quadratic and a quadratic in vertex form--finding the vertex, the x and y-intercepts, and the stretch
This is the sixth lesson (in a series of 6) that I created to introduce my students to quadratic functions. This lesson covers: identifying quadratic, linear and exponential expressions, given one form of a quadratic (graph, table of values, vertex form, or standard form) and then coming up with theI designed these lessons to teach my students about quadratic sequences (the 2nd in a series of 7). This lesson can be purchased as a complete bundled unit at a discounted price under the listing Quadratic Sequences Complete Bundled Unit.
This lesson focuses on: adding and subtracting binomials, mul
I designed these lessons to teach my students about quadratic sequences (the 3rd in a series of 7). This lesson can be purchased as a complete bundled unit at a discounted price under the listing Quadratic Sequences Complete Bundled Unit.
This lesson focuses on: reviewing prime factorization, multip | 677.169 | 1 |
Welcome to 8th Grade Algebra!
Please note that all homework assignments are listed in blue font on the right side of this page. Each homework assignment can be downloaded from the folder below labeled Algebra Homework Worksheets.
Remember that the purpose of algebra homework is to deepen understanding of the concepts that have been discussed in class. Don't just "get it done"; apply yourself, think, and use homework as a tool to reinforce and enhance what we are learning in class! Algebra homework is designed to take 20-30 minutes per night, 4 nights per week. If you have worked diligently for 30 minutes (with TV, Facebook, etc, all off!), stop, and write me a note on the assignment - you will still receive full credit. I do not want students working on algebra homework for more than 30 minutes on any given evening.
Homework assignments are updated on this site daily. Grades are posted on Power School and updated every two weeks. | 677.169 | 1 |
Graph Theory: Modeling, Applications, and Algorithms
Once considered an "unimportant" branch of topology, graph theory has come into its own through many important contributions to a wide range of fields – and is now one of the fastest-growing areas in discrete mathematics and computer science. This practical, intuitive book introduces basic concepts, definitions, theorems, and examples from graph theory. Presents a collection of interesting results from mathematics that involve key concepts and proof techniques. Covers design and analysis of computer algorithms for solving problems in graph theory. Discusses applications of graph theory to the sciences. Includes a collection of graph algorithms, written in Java, that are ready for compiling and running. For anyone interested in learning graph theory, discrete structures, or algorithmic design for graph problems1423848
Book Description Pearson. Book Condition: New. 013142384323823848
Book Description Pearson. PAPERBACK. Book Condition: New. 0131423843142387374 | 677.169 | 1 |
This is a digital textbook for a first course (sequence of courses) in Abstract Algebra covering the essentials of groups, rings and fields. The book is not an electronic version of a traditional print textbook but rather makes use of the digital environment to enhance student learning. One way this is achieved is by spiraling through the material, periodically returning to previous concepts to reinforce students' understanding. Specifically, every section begins with a subsection entitled: What do I Need to Know, where all the concepts that will be used are listed with links to their definitions. Also, throughout the text, all fundamental concepts are linked to their definitions so that students can return over and over again as needed to review them. Also, all citations to lemmas, propositions, theorems and corollaries throughout the text are linked to their statements and proofs.
Bruce Cooperstein:
Bruce Cooperstein received his Ph.D. in mathematics from the University of Michigan in 1975. He has been on the faculty of the University of California, Santa Cruz continuously since 1975, obtaining the rank of full professor in 1989. He has won two prestigious awards, a W.K.Kellogg National Fellowship (1982-85) and a Pew National Scholarship for Carnegie Scholars (1999-2000). Bruce's research areas include finite groups, groups of Lie Type, Lie geometries, incidence and Galois geometry. He is author of one of the first on-line course portfolios, Learning to Think Mathematically. and is the author of over 50 papers that have appeared in referred journals and proceedings of conferences. Bruce was a visiting Fellow of the Carnegie Foundation in Spring, 2007, and has also been involved in mathematics teacher professional development and mathematics education for over two decades. | 677.169 | 1 |
You're wasting time listening to the standard advice about learning math. You can't continue with "random acts of learning" as you study simple math, calculus and statistics formulas - at least not for long. The truth is that learning math and remembering numbers can be incredibly simple. You just need to know how.
Barron's EZ-101 Study Keys: Psychology
Designed to be compatible with virtually every standard textbook in its subject field, Barron's EZ-101 Study Keys gives you a valuable overview of your college-level course. Classroom-style notes emphasize important facts, remind you what you need to remember for term papers and exams, and help guide you through the complexities of lectures and textbooksM. Clarke says:"Dr. Jung's Life Would Make A Good Movie"
Publisher's Summary
"Case Studies" documents personal tragedies over the course of the 20th century often pertaining to uncontrollable health ailments. At other times these tragedies may or may not be at the hands of another. In either case making the transition from healthy to unhealthy can be a difficult endeavor. This can take the form of physical damage to the brain structure, autism, Parkinson's disease, spinal injury, multiple sclerosis, Alzheimer's disease, or brain death. Read about these individuals' personal tragedies and how they would learn to cope in dealing with these health concerns in his or her life. Many may take their health for granted, and many others as described here may have deserved a second chance.
"Algebra, Trigonometry, and Statistics" helps in explaining different theorems and formulas within the three branches of mathematics. Use this guide in helping one better understand the properties and rules within "Algebra, Trigonometry, and Statistics".
"Freud, Jung, Adler, Calkins, James" provides brief biographies of each historic individual within the field. In addition to a biography are theoretical positions of each individual providing input into their philosophical identities.
"Motivation Theories, Theorists, and Emotion" profiles theories of motivation and goes into detail pertaining to personal constructs and incentives. Sources of motivation can be both internal or external as the relationship among motivation and behavior is explained by the sequence of motivation. Enjoy a brief description of both theories of emotion and motivation. | 677.169 | 1 |
Course Description: Math topics to be covered include algebraic equations and inequalities, proportions, exponents, geometry, measurement, data analysis, and statistics. Students will be using the Singapore Math strategy to solve word problems. More information on the strategy can be found on my Educational Website Page.
Course Goals
To develop responsibility and personal advocacy
To develop problem-solving strategies
To develop abstract thinking
To have fun and be successful!
Textbooks: Students are allowed to write in these textbooks. These textbooks are for them to keep. Pages are perforated for the students to hand in for credit.
Go Math 8th Grade
Supply List
Single subject notebook or composition for journal
5 subject binder for note taking
Pencil with eraser or blue/black pen.
Grading Policy:
20% Classwork and Quizzes
80% Projects and Tests
Problem of the Day (P.O.D.):Posted on smartboard when students walk into classroom. The P.O.D is a problem or question related to the previous days' lesson for students to respond to. Students are to complete the P.O.D. in their composition workbook.
Late Work: For each day an assignment is late, one letter grade will be taken from the assignment grade. After one (1) week, late work will not be accepted and will result in an automatic "F." If the student has an excused absence, student will receive two (2) days for every day the student is absent before it is considered late. For example, if a student is absent for one (1) day, student have two (2) days to complete the assignment. After two (2) days, the assignment is late. All work will be posted on my webpage.
BEHAVIOR CODE:
Prepared: Come prepared to class every day and be on time. Wear the proper uniform.
Respectful: Respect others' space, property, and right to learn. No talking during instruction and keep speaking volume appropriate to task.
Inspire: Encourage others to succeed and teach others. Clean up after messes even if they are not yours. Encourage others to show P.R.I.D.E. | 677.169 | 1 |
Jie Xiao
This is a well-done but very limited introduction to elementary number theory, containing just enough material for a one-semester course. The material is severely classical, with everything except RSA encryption being known to Gauss in 1801. A Very Good Feature is the inclusion of many numerical examples. The exercises cover both numerical work and proofs, and all even-numbered problems are answered in the back of the book.
Being so elementary, the book deals primarily with divisibility: primes, unique factorization, solution of linear indeterminate equations, Chinese Remainder Theorem, and the like. But it also covers the irrationality of the square root of 2, quadratic reciprocity, Pythagorean triples, the proof of Fermat's Last Theorem for exponents 3 and 4, and applications of Gaussian integers. | 677.169 | 1 |
Algebra Introduction
Why
teach Algebra?
There
are a number of reasons why algebra holds an important place in the curriculum. We
list some of these below.
It gives us a powerful way to communicate.
Using letters and symbols provides a succinct and very precise way to write
mathematical statements;
It provides us with a method for solving
equations for unknown quantities. We can let x be an unknown, construct an
equation provided by a problem, and then solve the equation for the unknown. In
many situations algebra, if not the only way to solve for an unknown, is
certainly the most efficient;
It provides a way to express formulae and
processes that encapsulate an infinite number of variations and arrangements. By
so doing we don�t have to repeat similar work again and again. One
manipulation with a given quantity will settle a situation for an infinite range
of quantities once and for all.
The
Skeleton
The Algebra
Strand of this site is built around the two threads of
patterns;
and
equations.
These
two aspects of the Algebra curriculum are intimately related and we believe are best
developed together.Consequently most of the units address both of these aspects.
Pattern
covers pattern and relationships. A great deal of the Algebra from Levels 1 to 4
concerns patterns in one way or another, from making and describing patterns, to
making rules and finding general rules to making a rule to predict the results of
patterns. Implicit in this is the notion of relationships. These relationships are
expressed in words and in graphs.
Equations
consists of knowing and using the symbols +, <, and >. It moves on to look at
the fundamentals of solving equations with unknowns. | 677.169 | 1 |
Algebra II, Grades 5 - 8
Item# IF-Algebra-2-Grades-5-8
$12.99
Product Description
This revised edition of Algebra II links all the activities to the NCTM Standards. The activities were designed to provide students with practice in the skill areas necessary to master the concepts presented in a second-level course in algebra. Reviewing concepts presented in beginning algebra plus exercises involving slope, intercepts, graphing linear inequalities, domain and range, graphing exponential functions, matrix operations, quadratic equations, trigonometry, radicals, roots, and vectors are all part of this new edition. Examples of solution methods are presented at the top of each page. New puzzles and riddles have been added to gauge the success of concepts learned. Contains complete answer key. | 677.169 | 1 |
Description - Actuaries Survival Guide by Fred Szabo
This unique book is a guide for students and graduates of mathematics, statistics, economics, finance, and other number-based disciplines contemplating a career in actuarial science. Given the comprehensive range of the cases that are analyzed in the book, the Actuaries' Survival Guide can serve as a companion to existing study material for all courses designed to prepare students for actuarial examinations.
Author Biography - Fred Szabo
Author of: The Linear Algebra Survival Guide, 1st Edition Actuaries' Survival Guide, 2nd Edition Actuaries' Survival Guide, 1st Edition Linear Algebra: An Introduction using Maple, 1st Edition Linear Algebra: An Introduction using Mathematica, 1st Edition Fred E. Szabo is professor in the Department of Mathematics and Statistics at Concordia University in Canada. He completed his undergraduate studies at Oxford University under the guidance of Sir Michael Dummett and received a Ph.D. in mathematics from McGill University under the supervision of Joachim Lambek. After postdoctoral studies at Oxford University and visiting professorships at several European universities, he returned to Concordia University as a faculty member and dean of graduate studies. For more than twenty years, he developed methods for the teaching of mathematics with technology. In 2012 he was honored at the annual Wolfram Technology Conference for his work on "A New Kind of Learning" with a Wolfram Innovator Award. He is currently professor and Provost Fellow at Concordia University. | 677.169 | 1 |
Unit IV: Sequences/Series & Trigonometry 25%
-analyze arithmetic/geometric sequences to solve problems
-demonstrate an understanding of angles in standard position
-solve problems using the three primary trigonometric ratios (principle angles)
-solve problems using the cosine/sine laws, including the ambiguous case
Our semester is approximately 90 days in length; the pacing in the course is determined by that and the relative emphasis of each unit. Students can expect to have some homework throughout; this shouldn't be a huge burden provided we can stay on pace within each unit.
Summative evaluation for the course work will consist of Quizzes/Projects (20%) and Unit Exams (80%); this will determine the student's in-class mark. Formative evaluation will occur within each class; this can be intrinsic (through their own reflection on completed tasks) AND extrinsic (feedback from peers or myself). To arrive at the FINAL mark in Math 20-1, this in-class mark will have a 60% weighting; the Final Exam will make up the remaining 40%.
Saturday, February 5, 2011
Some students WANT to know WHY. I use methods similar to the example below to meet that requirement........
Find where the following functions intersect: y=2x^2 and y=7x-5
The values of the two functions above are equal where they intersect; to find this point of intersection, we set the functions equal to one another. The result is 2x^2=7x-5.
We now set this equation equal to zero: 2x^2-7x+5=0.
This next part takes care in explaining. Since we want to solve for x^1 (which is linear), our goal is to now express the left side of this equation as the PRODUCT of 2 linear functions. We want the product of these two linear functions to equal 0; this only happens if one OR the other is equal to "0″. THIS is why we must first set the equation equal to "0" and then factor the quadratic; break it into the product of linear functions. Since this product must equal zero, one of those linear functions must equal zero to satisfy the given condition given by (2x-5)(x-1)=0. The solution to this equation is therefore x=5/2 or x=1.
Once this first example has been completed, I like making a small change in the original functions that ties in with my notes on transformations. These can be viewed by clicking on the link below.
TRANSFORMATIONS
Here is what I like to do next; replace "x" in each of the two original functions with "(x+3)". This, as you should know, will move everything to the left by 3 units (including the solutions). Those solutions should therefore be x=-1/2 and x=-2.
Now we can verify these solutions. This is a valuable exercise to work through with students as it allows them to visualize the geometric representation AND use REASON to find solutions (or at the very least, to predict what should happen). We need to teach transformations anyways; why not incorporate some of that here while we're at it.
Let's solve the new system:
y=2(x+3)^2 and y=7(x+3)-5
2x^2+12x+18=7x+16 ==> 2x^2+5x+2=0 ==> (2x+1)(x+2)=0 ==>x=-1/2 & x=-2
Represent everything graphically and connect the algebraic representation of each to the corresponding geometric version; I believe this empowers students AND motivates them to learn more.
I don't spend weeks on this type of thing; in fact, this goes very quickly. Once these basic notions are understood by students, I DO use graphing technology; they do NOT have to be mutually exclusive. I'm certain we all want the same for our students and there will always be more than one effective approach. I think the approach I've shown here and in my notes on facebook have value in helping kids have a better UNDERSTANDING of math.
FUTURE APPLICATIONS
The logic behind this process is identical for higher degree polynomial equations; in fact, the same thought process can easily be transfered to exponential, logarthmic, trigonometric equations and others. This becomes very powerful when adopted and my students surprise themselves at how easy these so-called "complicated" tasks become. Representing higher order polynomials as the product of linear "functions" will be carried out at much higher levels of mathematics as well. An example of this can be seen by clicking on the link directly below; the notes contained within relate to the process of Integration by Partial Fractions, which is part of my Introduction to Calculus.
JAMES TANTON James Tanton has made available his view on how QUADRATICS should be taught; a link to this is found below. It would benefit everyone studying quadratic functions to have a good look at the pages contained within. | 677.169 | 1 |
Graphing Calculator by Mathlab Pro APK 4.11.141
Graphing Calculator by Mathlab Pro v4.11.141 APK : In case you're searching out a graphing calculator app that works smoothly and seamlessly, you've discovered it! Graphing Calculator with the aid of Mathlab is a systematic graphing calculator integrated with algebra and is an integral mathematical devicefor college students in standardschool to the ones in university or graduate school, or simplyevery person who wishesmore than what a simple calculator offers. it's miles designed to updatecumbersome and costlyhandheld graphing calculators and works on virtually any Android smartphone or tablet.
furthermore, Graphing Calculator with the aid of Mathlab displays calculations as itplays them on the 86f68e4d402306ad3cd330d005134dac show of the Android device, making it easier for the person to apprehend the calculations and notice them truly. This app has fantastic strengths. First, it acts as a greatclinical calculator, however, more than that, it shows the intermediate steps of the calculations as you type. It permitsthe scholars to each watch and learn the way the calculations are made and a way todiscover the finalanswer. 2d, the graphing ability is certainlystunning! notbest does the calculator superblyshow the graphs, however it automatically generates the x- and y- values and presentations them as properly.
helpweb site with instructions and examples: when you have a query, shipelectronic mail to calc@mathlab.us | 677.169 | 1 |
You will often see problems related to matrices on the ACT. In this video, I will walk you through questions related to multiplying matrices and finding the determinant of a matrix. We will also look at a question of adding 2 matrices.
Knowing the definition of cosine and how to use it in calculations are skills that you should review in your math ACT prep. I will show you several examples of calculating the cosine of an angle given different triangles.
You should be familiar with the general trigonometry (trig) identities because questions using them often appear on the ACT. I will go over several example questions that make use of trig identities to simplify expressions.
A good understanding of what the midpoint of a line is will be useful on the ACT as there can be several different type of questions related to this. In this video, I will walk through 2 questions that make use of midpoint. | 677.169 | 1 |
خلاصه کتاب و اطلاعات بیشتر
Signals and Systems: A Primer with MATLAB® provides clear, interesting, and easy-to-understand coverage of continuous-time and discrete-time signals and systems. Each chapter opens with a historical profile or career talk, followed by an introduction that states the chapter objectives and links the chapter to the previous ones. All principles are presented in a lucid, logical, step-by-step approach. As much as possible, the authors avoid wordiness and detail overload that could hide concepts and impede understanding.In recognition of the requirements by the Accreditation Board for Engineering and Technology (ABET) on integrating computer tools, the use of MATLAB® is encouraged in a student-friendly manner. MATLAB is introduced in Appendix B and applied gradually throughout the book.Each illustrative example is immediately followed by a practice problem along with its answer. Students can follow the example step by step to solve the practice problem without flipping pages or looking at the end of the book for answers. These practice problems test students' comprehension and reinforce key concepts before moving on to the next section.Toward the end of each chapter, the authors discuss some application aspects of the concepts covered in the chapter. The material covered in the chapter is applied to at least one or two practical problems or devices. This helps students see how the concepts are applied to real-life situations.In addition, thoroughly worked examples are given liberally at the end of every section. These examples give students a solid grasp of the solutions as well as the confidence to solve similar problems themselves. Some of the problems are solved in two or three ways to facilitate a deeper understanding and comparison of different approaches.Ten review questions in the form of multiple-choice objective items are provided at the end of each chapter with answers. The review questions are intended to cover the "little tricks" that the examples and end-of-chapter problems may not cover. They serve as a self-test device and help students determine chapter mastery. Each chapter also ends with a summary of key points and formulas.Designed for a three-hour semester course on signals and systems, Signals and Systems: A Primer with MATLAB® is intended as a textbook for junior-level undergraduate students in electrical and computer engineering. The prerequisites for a course based on this book are knowledge of standard mathematics (including calculus and differential equations) and electric circuit analysis.InstructorsWe provide complimentary e-inspection copies of primary textbooks to instructors considering our books for course adoption.Request an e-inspection copyShare this TitleRelated Titles1 of 3 Numerical Techniques in Electromagnetics with MATLAB, Third EditionTable of ContentsBasic ConceptsGlobal Positioning SystemIntroductionBasic DefinitionsClassifications of SignalsBasic Continuous-Time SignalsBasic Discrete-Time SignalsBasic Operations on SignalsClassifications of SystemsApplicationsComputing with MATLAB®SummaryReview QuestionsProblemsConvolutionEnhancing Your Communication SkillsIntroductionImpulse ResponseConvolution IntegralGraphical ConvolutionBlock Diagram RepresentationDiscrete-Time ConvolutionBlock Diagram RealizationDeconvolutionComputing with MATLAB®ApplicationsSummaryReview QuestionsProblemsThe Laplace TransformHistorical ProfileIntroductionDefinition of the Laplace TransformProperties of the Laplace TransformThe Inverse Laplace TransformTransfer FunctionApplicationsComputing with MATLAB®SummaryReview QuestionsProblemsFourier SeriesHistorical ProfileIntroductionTrigonometric Fourier SeriesExponential Fourier SeriesProperties of Fourier SeriesTruncated Complex Fourier SeriesApplicationsComputing with MATLAB®SummaryReview QuestionsProblemsFourier TransformCareer in Control SystemsIntroductionDefinition of the Fourier TransformProperties of Fourier TransformInverse Fourier TransformApplicationsParseval's TheoremComparing the Fourier and Laplace TransformsComputing with MATLAB®SummaryReview QuestionsProblemsDiscrete Fourier TransformCareer in Communications SystemsIntroductionDiscrete-Time Fourier TransformProperties of DTFTDiscrete Fourier TransformFast Fourier TransformComputing with MATLAB®ApplicationsSummaryReview QuestionsProblemsz-TransformCodes of EthicsIntroductionDefinition of the z-TransformRegion of ConvergenceProperties of the z-TransformInverse z-TransformApplicationsComputing with MATLAB®SummaryReview QuestionsProblemsAppendix A: Mathematical FormulasAppendix B: Complex NumbersAppendix C: Introduction to MATLAB® Appendix D: Answers to Odd-Numbered Problems | 677.169 | 1 |
A Published Work from Math 1 On 1, LLC
...an entertaining, motivating, and educating work meant to address problem areas of math in a unique manner...
This book is a reference tool that describes time saving techniques, addresses areas of math that students find most difficult, and shares different ways of explaining problems that many students find challenging.
Hopefully, parents and students can relate to the problem areas presented in this book.
Lessons are presented with real world examples to demonstrate how math is used in every day life. | 677.169 | 1 |
Showing 1 to 22 of 22
Physical chemistry III
PROBLEMS
SS 2007
Exercise II
These exercises refer to chapter 3 of the script, which recapitulates central concepts in physics. Make sure that you know them well before you do your exercises 1. Wave equation Consider a guitar string
Order your Inspection Copy now
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Exercise 7
M/M/ Queueing- and Loss-Systems
1. A pure loss system with negativ exponentially distributed inter-arrival times and service times has exactly n service units. Determine the probability of exactly k = 0, . . . ,n service units being occupied. W
Solving Differential Equations using Laplace and Inverse Laplace Transforms
Introduction to the methodology Response of a system to particular inputs
unit impulse input unit step input unit ramp input
Analyse system in the Laplace domain
without initi
CHAPTER - 3 3. THE Z TRANSFORM
3.1. INTRODUCTION
I
n this chapter, we review the z transform and its properties. We then use the z transform to define the transfer function of a discrete-time system. At the end of the chapter we give two sets of problems;
Hanoi University of Technology
Faculty of Applied mathematics and informatics
Advanced Training Program
Lecture on
Algebra
Dr. Nguyen Thieu Huy
Hanoi 2008
Nguyen Thieu Huy, Lecture on Algebra
Preface
This Lecture on Algebra is written for students of Adva
AN INQUIRY INTO THE NATURE AND CAUSES OF
THE WEALTH OF NATIONS
by
Adam Smith
A PENN STATE ELECTRONIC CLASSICS SERIES PUBLICATION
An Inquiry into the Nature and Causes of the Wealth of Nations by Adam Smith is a publication of the Pennsylvania State Univer
INQUIRY
N C
W
N
Books I, II, III, IV and V
Adam Smith
Libri
Copyright 2007 Libri
this digital edition A LL RIGHTS RESERVED. No part of this digital edition may be reproduced, stored in a retrieval system, or transmitted, in any form, or by any means, el
Introduction to Laplace Transforms for Engineers
C.T.J. Dodson, School of Mathematics, Manchester University
1
What are Laplace Transforms, and Why?
This is much easier to state than to motivate! We state the denition in two ways, rst in words to explain | 677.169 | 1 |
Algebra 2
Algebra 2
0.00
Algebra 2 extends and builds upon concepts learned in Algebra 1 in order to improve problem-solving skills. Algebra 2 continues the study of algebra with a focus on the analysis of functions, including polynomial, rational, exponential, and logarithmic functions. Emphasis will be made on solving, graphing, transforming, and modeling these functions. | 677.169 | 1 |
Maths data handling coursework
Statistics coursework gcse maths statistics coursework gcse maths. A secondary school revision resource for GCSE Maths about foundation and higher level data handling. Free statistics coursework. Maths Statistics Coursework. Height and Weight Data on High School Children - Missing Some Graphs Handling Data Coursework. Many thanks to Pixi_Maths for the fantastic Task on Entry. Histograms - Drawing and Interpreting (GCSE. Data Handling activity based on children's. Year 5 Maths Homework Sheets. Video Card Term Paper Junior High School Life Essay.mayfield high data handling coursework. Graduate Literature Review Sample. Maths. Home » Courses ». GCSE covers Data Handling, Number and Algebra A Level is 7% coursework and 93% exam (the coursework forms 3.5% of the full A. Faculty of Maths, Science and Sports GCSE Mathematics. Data Handling Scheme of. Your final GCSE grade is assessed by external exams and coursework but we. Data Handling all questions from previous SATs paper Level. Data handling by topic and level I created this code breaking maths lesson for an Ofsted.
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Maths data handling coursework
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Marking GCSE Statistics coursework. The marks for GCSE Statistics coursework are. and may normally be a high mark in the GCSE Mathematics data handling project.. If you face such task as maths data handling coursework and you are at loss how to create an interesting piece of work do not panic. Our article can cast light on. Year 9 Data Handling Project 1. SMARTIES – which colour is most popular? 1 In this coursework I am going to find out which _____ smartie is most common. Maths Gcse Statistics Coursework Help. removed from GCSEA secondary school revision resource for GCSE Maths about foundation and higher level data handling Maths. | 677.169 | 1 |
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Tuesday, January 11, 2011
"Microsoft Mathematics provides a graphing calculator that plots in 2D and 3D, step-by-step equation solving, and useful tools to help students with math and science studies.
MSetup_x64.exe: 18.9MB
MSetup_x86.exe: 17.6MB
Version: 4.0
Date Published: 1/11/2011
Language: English
Microsoft Mathematics provides a set of mathematical tools that help students get school work done quickly and easily. With Microsoft Mathematics, students can learn to solve equations step-by-step, while gaining a better understanding of fundamental concepts in pre-algebra, algebra, trigonometry, physics, chemistry, and calculus.Microsoft Mathematics includes a full-featured graphing calculator that's designed to work just like a handheld calculator. Additional math tools help you evaluate triangles, convert from one system of units to another, and solve systems of equations
…"
Since I indeed have a student in the house, I'm going to see if he's interested (or if I get the "zomg, you are SUCH a parent" eye-roll… lol )
Here's a snap of it. It looks pretty cool, doesn't it? How it "shows the work" of solving an equation made my chuckle (man, I hated having to do that…)
2 comments:
I work at a school that has all students 7-12 with tablet laptops. I thought this was really cool so I sent it to the Math and Science department. At first they thought is was useful but then quickly switched to fear of students using this to complete homework and not do it themselves. When used properly this can be a very powerful tool, there is some room for abuse however by students. | 677.169 | 1 |
1 This resource was developed by CSMC faculty and doctoral students with support from the National Science Foundation under Grant No. ESI-0333879. The.
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Presentation on theme: "1 This resource was developed by CSMC faculty and doctoral students with support from the National Science Foundation under Grant No. ESI-0333879. The."— Presentation transcript:
1
1 This resource was developed by CSMC faculty and doctoral students with support from the National Science Foundation under Grant No. ESI-0333879. The opinions and information provided do not necessarily reflect the views of the National Science Foundation. 12-6-04
2
2 Committees and Reports that Have Influenced the Changing Mathematics Curriculum This set of PowerPoint slides is one of a series of resources produced by the Center for the Study of Mathematics Curriculum. These materials are provided to facilitate greater understanding of mathematics curriculum change and permission is granted for their educational use. Report of the National Committee of Fifteen on the Geometry Syllabus Report 1912
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3 Final Report of the National Committee of Fifteen on the Geometry Syllabus National Education Association American Federation of Teachers of the Mathematical and Natural Sciences Mathematics Teacher, December 1912
4
4 Committee of Fifteen on the Geometry Syllabus Appointed 1909 Final Report 1912
5
5 Forces at Work Increasing enrollments placed pressure on high schools to provide education to a broader student population Geometry courses varied widely between two extremes –Formal proof approach –Utilitarian approach There were rising failure rates in algebra and geometry Psychology of "Mental Discipline" was losing favor and both mathematicians and teachers were calling for reform A common syllabus and direction for geometry was needed
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7 Final report had 5 general sections Historical overview of school geometry Logical considerations Special courses Exercises and problems Syllabus of geometry
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8 Historical Overview of School Geometry Reviewed the nature of school geometry in several European countries. Provided a brief review of popular geometry textbooks being used in the United States. Identified books and resources that illustrated attempts to reform the teaching of geometry.
9
9 Logical Considerations Recommended axioms, postulates, and definitions that should be included in the syllabus for a geometry course. Addressed the strengths and dangers associated with informal proofs, and argued that nearly 100 of the propositions (theorems) "must receive formal proof in any well-regulated course in geometry." Recommended algebra as a ninth-grade course, geometry in the tenth grade, and algebra and geometry in the eleventh. Recommended that algebraic approaches and notations be utilized in geometry. Recommended incorporation of some solid geometry and trigonometry in geometry.
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10 Special Courses Rather than special courses, a single syllabus for high school geometry was recommended. The syllabus could be altered by omission of some theorems and by increased emphasis on theorems supporting practical applications. Guidelines were provided for the study of geometry across the grades, including informal geometry in the elementary grades. Definitions should be "stated formally only after the concept is clearly forms in the student's mind." Theorems need not all be proved; for some, heuristic arguments to convince students of their truth would be sufficient. Making scale drawings and models that apply important theorems on measurement was recommended.
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11 Exercises and Problems Expressed concern about the lengthy sets of exercises and their lack of application as found in prevailing textbooks. Recommended distributing exercises by difficulty so that each theorem was accompanied by immediate concrete exercises and applications and more difficult exercises were delayed to a later part of the course. Urged a balance of formal reasoning exercises and application problems to make geometry more appealing to the average student.
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12 Geometry Syllabus Proposed a list of theorems, often stated algebraically, that would be a basic set for a geometry course. For plane geometry, 106 theorems were listed; and for solid geometry, 75 theorems were listed. Used different fonts to convey to teachers and examiners the relative importance of listed theorems. Identified some theorems that could be developed with informal arguments.
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13 Significance of the Report of the Committee of 15 Called for a balance of informal and formal work in geometry and more concrete examples. Emphasized the need for good pedagogy to support the learning of geometry. Recommended a full-year of plane geometry in 10 th grade to be followed by more geometry and algebra in 11 th grade. Geometry textbooks published after 1910 acknowledged the Committee of Fifteen and noted their alignment with the Committee's report. Theorems listed by the Committee of Fifteen were included in forthcoming high school geometry textbooks. College Entrance Examination Board limited testing of propositions to the Committee's list of basic theorems. | 677.169 | 1 |
Description
Computational Modeling, by Jay Wang introduces computational modeling and visualization of physical systems that are commonly found in physics and related areas. The authors begin with a framework that integrates model building, algorithm development, and data visualization for problem solving via scientific computing. Through carefully selected problems, methods, and projects, the reader is guided to learning and discovery by actively doing rather than just knowing physics.
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Engineers around the world depend on MATLAB for its power, usability, and outstanding graphics capabilities. Yet too often, engineering students are either left on their own to acquire the background they need to use MATLAB, or they must learn the program concurrently within an advanced course. Both of these options delay students from solving realistic design problems, especially when they do not have a text focused on applications relevant to their field and written at the appropriate level of mathematics.
Ideal for use as a short-course textbook and for self-study Elementary Mathematical and Computational Tools for Electrical and Computer Engineers Using MATLAB fills that gap. Accessible after just one semester of calculus, it introduces the many practical analytical and numerical tools that are essential to success both in future studies and in professional life. Sharply focused on the needs of the electrical and computer engineering communities, the text provides a wealth of relevant exercises and design problems. Changes in MATLAB's version 6.0 are included in a special addendum.
The lack of skills in fundamental quantitative tools can seriously impede progress in one's engineering studies or career. By working through this text, either in a lecture/lab environment or by themselves, readers will not only begin mastering MATLAB, but they will also hone their analytical and computational skills to a level that will help them to enjoy and succeed in subsequent electrical and computer engineering pursuits | 677.169 | 1 |
Discovering Math: Algebra and Functions 4-Pack: Discovering Math -
Discovering Math: Algebra and Functions 4-Pack: Discovering Math
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$ 159.95$ 169.95
This 4-pack considers fundamental and advanced concepts of functions and their properties, including polynomial and trigonometric functions. It also considers algebraic representations using variables and graphs, inverse functions, systems of equations and inequalities, and sequences and series. | 677.169 | 1 |
Quadratic Acrobatics: Wildcats Explore What the Path of a Home Run Has in Common with the Invention of GPS
Freshmen students at Roberto Clemente Community Academy have begun a study of the quadratic function, an equation that explains architecture, technology, and movement of objects all around us in the natural world. By using real world representations of the quadratic function in the form of the parabolic curve—found in archways, bridges, satellite dishes and even the McDonald's arches—its properties explain the rules that govern the movement of objects through space. TTS algebraic calculation creates the foundation for GPS, cell phones, and many of the technological breakthroughs of today, and students are becoming aware of how pervasive this function is in everyday life.
Freshmen began their investigation by understanding the basic properties of the function's graph, its key elements, and their definitions. They continued by connecting the visual to the algebraic expressions. This week they moved their exploration to understand the mathematical connections and truths behind the shape of the curve and its algebraic equation. By identifying the parts of the curve as they exist in graphic form and relating those to the function's equation using formulas to predict the curves dimensions, our mathematicians are beginning to explore some of the basic foundations of graphic design and engineering.
After their initial analysis, the students will engage in some problem solving where they will be tasked with finding the properties and predicting the path of projectile scenarios and exploring the problem-solving strength of the quadratic model in real-world engineering situations. As they gain confidence in moving between the graphic representation and its algebraic equation, they will begin to use computer programs like Desmos to help them flesh out their calculation and give them a greater understanding of the application they are investigating.
Finally, students will be given an opportunity to develop their own short video or slide presentation of an example of the quadratic function in a real-world setting, applying all they know to showcase their mastery of understanding the application of the parabolic shape | 677.169 | 1 |
Electricity and Electronics,book was written as a math refresher to help your students with the required math in the dc/ac course. The material is presented in a concise, relevant form beginning with the fundamentals of applied mathematics, algebra, trigonometry, and logarithms.
As each topic is introduced, examples are given to help students master the manipulative skills and provide them with a better understanding of the mathematical concepts. Hundreds of drill problems are presented sing a unique three column format so students can see the math problem worked out traditionally and how the same problem is solved using a calculator. | 677.169 | 1 |
Summer Packet Remediation
Directions:
1. Check your results .. Make a list of the topic letters that correspond to the problems you got wrong on the diagnostic from class. The topic letter is on the top of the spreadsheet. You know you got a problem wrong because it is marked with an x. If you have a problem marked witha "/", it means you got the problem wrong but your mistake was so small that you DO NOT NEED TO DO THE REMEDIATION.
2. Complete appropriate remediation module using the material on this page.
3. A remediation cycle consists of the following.
Watch the Remediation Video to get some re-instruction. Taking notes is recommended. (It will be mandatory on Flipped Classroom assignments in the future.)
Practice on the Remediation Worksheet posted here.
Use the Remediation Worksheet Answers Key to check your answers.
If you do not print the worksheet, you must copy the problems onto a clean sheet of paper. (Do not use a scrap or the back side of notes.)
Take the Remediation Re-Evaluation (It is a google form. Follow the directions there about formats for entering your answers.)
You should make sure to copy the problems from the Remediation Re-Evaluation onto the back of the Remediation Worksheet. Both must be submitted for full credit.
If you do not print the Remediation Worksheet and copy the problems from the screen, you need to copy the problems for the Remediation Re-Evaluation onto the back of the sheet of paper onto which you copied the Remediation Worksheet Problems.
PLEASE NOTE THAT IF THERE IS PART OF A MODULE MISSING (a video, a worksheet, or a re-evaluation quiz), SKIP THAT ENTIRE MODULE FOR NOW AND WORK ON ANOTHER ONE. I AM HOPING TO MAKE MORE MATERIALS AS THE COURSE PROGRESSES. | 677.169 | 1 |
Algebra Basics Overview Part 1
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Product Description
An overview of important points that pupils need to understand when starting out on basic algebra. It can also be used as a revision tool and has been used as a refresher for adult support staff (who have not seen algebra since they were at school) prior to them assisting pupils in the classroom. | 677.169 | 1 |
Description
This edition features the exact same content as the traditional text in a convenient, three-hole- punched, loose-leaf version. Books a la Carte also offer a great value—this format costs significantly less than a new textbook.
Mathematical Reasoning for Elementary Teachers presents the mathematical knowledge needed for teaching, with an emphasis on why future teachers are learning the content as well as when and how they will use it in the classroom. The Sixth Edition has been streamlined to make it easier to focus on the most important concepts. The authors continue to make the course relevant for future teachers, including the new features like Examining School Book Pages, as well as the hallmark features like Into the Classroom discussions and Responding to Students questions. | 677.169 | 1 |
Posts filed under: Q and A
Formatting. There's so much that's not simple text in the book (e.g., fractions, graphs, geometrical figures). None of that stuff plays nicely with a Kindle. However, I do have a digital version of the book available through Google Play. That version reads exactly like a physical copy of the book.
Sure. For me, the trick to solving questions where the function isn't defined the simplest way (e.g., ) is thinking a little abstractly about what a function is. Basically, when you see , you know that some combination of mathematical operations must happen to to transform it into .
What could those operations be? Well, there'll have to be some multiplication—that's the only way we're getting the to turn into . So let's just start with that as an experiment; let's say this function is just and see what we'd get by plugging in as the argument:
Of course, that's not exactly what we want, but it is close. All we need to do now is add to turn into . That tells us the function should really be . Let's just double-check:
Yep, that works! Once we know our function is , all we need to do is drop in there:
I used your book(not the hard copy) for the old test and raised my score from a shaky 650 to a solid 750, I was wondering if the google play edition that you had on sale for the old test is also available for the new test. I am looking into buying that for my brother who is outside the United States.
Sidenote: I am reading your blog posts and thinking about trying the new test just for fun.
I don't know how to navigate your site yet, so please forgive me if you have already answered this question. Regarding question number 8 in your book, will you further explain why y=180-x is the same angle degree as the unmarked angle? I know y=2x because of geometry, but I do not understand how y can also equal x?
You're talking about this question from the triangles chapter, right?
You've actually hit on a totally legit way to solve the question: If due to the exterior angle theorem, and , then you can substitute and solve for :
Once you know that , you can determine that the triangle is equilateral, so you need a perimeter that's a multiple of 3.
To answer your question more directly, look at just the top part of the figure:
See how you've got a straight line there? That means the angle that's NOT marked (i.e., the angle inside the triangle) has a measure of . Since the question tells you that , you can substitute:
Hello Mike! Could you please tell me, what is the answer of Question 16, Section 4, Test 3? Thank you veery much for your help and time.
Katerina
The answer is , and here's why. They're asking you to find a value of for which . That can only happen when . So you're looking for a place on the graph where curve and the curve are the exact same distance from the -axis.
The vertices of the graphs of and are and , respectively. In other words, when , the graphs of and are the same distances from the -axis. In other words, , , and therefore
Are you still doing the daily PWN practice question. It looks like question #110 was the last one I received back on April 2nd.
For now, #110 is the last one, although I might start adding more over the summer. To be clear, the way the Daily PWN works is that everyone who signs up starts at the beginning, so someone who signs up today will get a question a day for 110 days—more than 3 and a half months. As it turns out, the data suggests that most people unsubscribe or just stop answering questions before then.
I do love writing the questions, though. If and when I do add more, assuming you're still subscribed, you'll be one of the first to know.
Can you explain #15 in Section 3 of test 4? I understand how to do the problem but was wondering if there is a shortcut or trick to seeing the correct answer?
Great question! First, I don't know if this is really a shortcut but for me this question becomes much easier to tackle when I multiply by 2 to eliminate the fraction. So the quadratic I'm working with after multiplying and rearranging is:
So, for quadratic formula purposes, , , and .
The real shortcut, for this question and ones like it, is to focus on the differences in the answer choices rather than doing the whole quadratic formula process and trying to match a choice.
In this question, for example, you might notice that you only have two choices for the first term: or . If you start by focusing only on that term, you can eliminate half the choices right away!
The quadratic formula is . That means the first term is . In this question, then, the first term is . So we know the answer must be A or B.
From there, we know the bit under the radical must either be or .
Since the radical bit equals , we can plainly see that is the way to go. Choice B is the answer.
Yep! The key to getting this one is recognizing that when they tell you that , they're telling you that . That's just a little trig fact you should keep in mind when you're taking the SAT. It can be derived pretty easily—just picture any right triangle with acute angles measuring and :
Of course, since it's a right triangle, . And if you do your SOH-CAH-TOA, you'll see that and .
Anyway, back to the question. Once we recognize that , the rest is just substitution.
Yeah, this one is trickier than it looks. I think most people get turned around on it because they try to get out of fractional exponent notation too early, and then they get stuck. It's a LOT easier, in my opinion, to do a little manipulation in exponent form before you jump to radicals. In this case, that means making a substitution, which will allow you to reduce the fraction.
Remember than when you raise a power to a power, you multiply those powers. So:
Now you can convert way more comfortably to radical notation. Remember, with fractional exponents, the numerator (top number in the fraction) is the power and the denominator (bottom number in the fraction) is the root.
Would you please explain the solution to Problem #31 in the calculator-allowed section of Practice Test #6?
This one is all about translating carefully between words and math. You can write two equations. Let's use for the original number of friends. If all friends go on the trip, the cost per person is $20 less than if friends go on the trip. Cost per Person with Friends minus Cost per Person with Friends is $20.
To solve, multiply by :
That tells you that the number of friends originally in the group is either 10 or –8. Since the answer can't be negative, it must be 10.
BONUS: The other way to go on this one is the way I actually did it the first time: just dividing 800 by a bunch of numbers in a row until I saw a difference of 20:
This is a fun question! First, note that the polynomials provided have a little something in common:
Do you see it? How about now:
Takeaway lesson from that: when a question at first looks like it might take a very long time to do (in this case, when it looks at first like you're going to have to do long division possibly four times) slow down for a minute and think about whether there's a shortcut built into the problem.
You're looking for a way to combine and so that they're divisible by . Once you see that you can pull a (as in, the same that's part of ) out of and get , you know to use that as your base of operations. Let's just see what happens if we multiply by :
Ready for the last step? All we need to do is substitute because we know :
That matches choice B, and we know that it's divisible by because we made it by multiplying by !
The key to this one is to eliminate every choice that makes an unjustified (i.e., not directly supported by the question) assumption.
You can eliminate choice A because we don't know anything about the preferences of the viewers who didn't vote. Generally speaking, don't draw conclusions about what could have happened, draw conclusions about what DID happen.
You can eliminate choice B because the question mentions nothing about the ages of voters. Don't let your own assumptions about who texts and who uses social media cloud your judgment!
You can eliminate choice C because it's mathematically false (more on this below) but you should also lean towards eliminating it just because it's using something that didn't happen as a basis for its conclusion. As before, resist the urge to make conclusions about what could have happened when you only know what DID happen.
Choice D is totally supported by the data: 70% of social media voters preferred Contestant 2, and only 40% of text message voters preferred Contestant 2. Therefore, social media voters were more likely to prefer Contestant 2 than text message voters.
Now, as for why C isn't mathematically true…let's plug in! Say there were 100 voters. We know 30% of the votes came in via social media, so that's 30 votes. The other 70 votes must have come in via text message.
Of the 30 social media votes, Contestant 2 got 70% of them, or 21 votes. Of the 70 text message votes, Contestant 2 got 40% of them, or 28 votes.
So of the 100 votes, Contestant 2 only got 21 + 28 = 49 votes! That's not enough to win the contest.
I just bought your book and with it in tow, I am learning a lot of helpful, useful ways to do SAT math. Thank you.
Here are my questions, all calculator related:
1. Since I am new to using a calculator, would you recommend the TI83 plus for the SAT?
2. If not, which calculator do you recommend.
3. Is there a book, website, or other media source you would recommend to learn calculator skills with ease and accuracy?
4.Which calculator functions do you recommend mastering before taking the SAT?
I like the TI-83 Plus. That's the calculator I use personally; all the calculator screenshots you'll find in my book are from my TI-83 Plus. So if you're using my book to help you learn to use the calculator, that's a good thing. The TI-84 series calculators are also great and have basically the same menus, etc.
I don't have a particular calculator source other than my own book. The calculator will help you lots, but my advice is to search out specific lessons as you find you need them rather than spending too much time learning stuff you won't need from a thorough book, video series, etc. Basically, when you come across something in my book that I recommend solving with a calculator, then you should make sure you learn how to do that thing. If the examples in my book aren't enough, search out some videos for that function.
Quick list off the top of my head of what you should master:
Graphing functions
Zooming and changing window settings to see the important parts of functions
Could you please explain Test #8, calculator-allowed section, number 28?
Thanks!
You have to evaluate both range and standard deviation here. The former you can calculate. The latter you just evaluate visually; You will never have to actually calculate a standard deviation on the SAT.
Let's talk about range first. To calculate range you just subtract the smallest number in the set from the biggest number. In this case, that means that and . The numbers in the second set are obviously higher, but the range in both sets is the same: the highest number is 32 greater than the lowest number. Therefore, you really only need to look at choices A and D—the only choices that say .
Like range, standard deviation is a measure of variability—the bigger the standard deviation, the more spread out the values in a set are from the mean. Visually, you can look at both dot plots in this question and see that the distributions are quite different. In the first plot, there are a couple outliers, but generally the pulse rates are pretty tightly concentrated around the mean of 72. In the second plot, the dots are much more spread out—if you picked a dot randomly in the second plot, you'd be just as likely to land on an extreme data point like 80 or 112 as you would on the mean of 96. Therefore, you can conclude that the standard deviations of the two sets are not the same (it will be bigger in the second plot). That makes choice D the only option | 677.169 | 1 |
John Saxon's One-Man Battle Against Algebraic Ignorance
John H. Saxon Jr. was teaching an algebra I class at Oscar Rose
Junior College in Oklahoma City when he first realized that when it
comes to algebra, once is not enough.
One day, he says, he gave the "most magnificent lecture." He
finished in 20 minutes, and, reluctant to release his students 40
minutes early, planned to have them do problems on the blackboard for
20 more minutes. He gave the same problem on the board that he'd just
finished explaining in his lecture.
"They all wrote it on the board. And they all stood there looking at
the board. Not one of them understood. With my 20-minute, beautiful
stand-up lecture, nobody got a thing."
He explained it again, and yet again, and assigned the same problem
as homework. Then he went down to his office and, he says, thought to
himself, "John, something just happened there, and a smart man would be
able to figure out what."
The conclusion he arrived at--that one learns to work out
mathematics problems by working them out over and over--became the
basis for a method and text for teaching algebra that have, so far,
produced large gains in achievement for many students and a flurry of
rejection slips from publishers.
The latest report came last week: The test scores of a group of
first-year algebra "Saxon students" were, on the average, 37 percent
higher than those of second-year-algebra students whose teachers used
other commercial texts.
The test results were monitored and certified correct by the
Oklahoma Federation of Teachers (oft), which has agreed to verify the
accuracy of the tests, although it remains a neutral party.
The scores are the latest of several such comparisons, all of which
have shown that students taught with the method, which stresses
repetition and a different organization of material, consistently
outscore others, sometimes by enormous margins.
Saxon Students Outscore Others
The initial comparative tests, carried out on 1,365 Oklahoma
students taking algebra I, showed the Saxon students outscoring their
counterparts by as much as 141 percent on tests of basic skills. For
example, on a 15-question test about positive and negative numbers, the
test group scored109 percent higher than the control group.
The marked difference in achievement occurred, according to Mr.
Saxon, a former test pilot and career Air Force officer who teaches
mathematics at the Oklahoma City junior college, because the Saxon
students had been taught with a method that runs counter to the those
used most commonly to teach algebra.
Rather than moving from discrete topic to topic, and seldom
mentioning the preceding ones after beginning the next, the Sax-on
method provides a continuous review of integrated topics. "Repetitive
homework" is one key factor, Mr. Saxon says; another is a different
ordering of the material.
And because the material is repeated, he argues, the method gives
students the means to master a subject that has been the source of a
great deal of frustration for many students.
Mostly by way of a sort of educational grapevine, the method has
aroused interest--sometimes cautious--among educators and has been the
subject of several national articles. Prentice-Hall has published two
junior-college texts by Mr. Saxon.
But, due to lack of interest from textbook publishers, Mr. Saxon has
published the high-school text, Algebra I: An Incremental Development,
himself, under the imprint of his own publishing company, Grassdale
Publishers Inc. He relies mostly on word-of-mouth for advertising. Even
so, he says, he has sold 9,000 "examination" copies, and the book was
used in about 50 school districts last year, most in Oklahoma and
California. New Mexico and Tennessee, both states that provide
districts with a wide range of choices on their state-adoption textbook
list, have added the book
In the field of textbook publishing, which is dominated by large
publishing companies and influenced to a degree by state-adoption
textbook lists, Mr. Saxon stands out.
The beginning of what has subsequently become his mission occurred
several years ago; since then, he has become intensely preoccupied with
the process of convincing educators to adopt his method, and he
believes that significant change will occur in students' mathematics
achievement if they do.
He is, he says, something of a fanatic. "There are very few people
of limited ability who are afforded the opportunity to make a major
contribution. I have been afforded that opportunity, and it's
overpowering." He adds, "Ask any fanatic--he'll tell you how right he
is. Why should I be different from any other fanatic?"
The rationale behind the method is simple, just common sense, Mr.
Saxon acknowledges. "Kids cannot learn through a single exposure, and
in particular they can't learn abstractions in a single exposure."
Nevertheless, he says, most algebra texts pay insufficient attention to
this.
And because the material recurs, he says, students also get a sense
of continuity and become more confident that they can master it.
"Most math books," he says, "are like the Book of Revelation--horror
stories and surprises from beginning to end. Students see my book as
the 23rd Psalm--it's a nice safe place to go."
The experience of his students was matched by his own early
encounters with higher mathematics. "Math was always difficult for me.
It almost kept me from doing the things I wanted to do. All of the
things I did were made possible by the math."
He argues now that many students do not pursue science and
mathematics careers because they are entirely put off by algebra. "This
is the reason they won't take chemistry. They've cut and run. They
can't handle the percentages and they can't handle simple ratios."
The method is effective, Mr. Saxon says, even with students of
modest ability who often do poorly in algebra. These students, he
argues, are victims of a system that favors high-ability students.
"All our math education is set up almost with malice aforethought to
restrict knowledge to the brilliant few," he says. "The others can also
understand abstractions. It just takes more exposure."
Most of the reaction to the text has been favorable, although some
educators argue that students will grow bored with the constant
repetition.
Larry S. Gregory, mathematics specialist for the state department of
education in Tennessee, points out that the data that show the dramatic
gains are based on "fairly limited objectives." He also notes that Mr.
Saxon wrote the tests.
The claim that the method is novel also troubles some educators;
most texts do include some review. "It's a matter of degree," Mr.
Gregory says. Although not sufficiently familiar with the text to know
whether Mr. Saxon makes good his promise to include all concepts in all
lessons, Mr. Gregory says that if he does, that would be "more than is
done in a typical book."
Mr. Saxon argues that the method is not so much new as ignored, not
tried and found wanting but found difficult and left untried. "One of
the things that has bothered me the most is that what I'm doing is so
obvious; how can be it be so overlooked?" he says.
Highly Successful Method
Teachers who have used the method, however, report that it is a
highly successful way of teaching algebra. Mickey Yarberry, an Oklahoma
City junior-high mathematics teacher who taught with the Saxon text for
two years, says that although the method takes some getting used to,
the results are worth it. "In a way, it is more confusing at first
because you're teaching 15 different things," she says. But everyone
becomes accustomed to it, they make rapid progress.
She says that she began using the text, at the request of Mr. Saxon,
with some doubt. She taught one algebra class using the Saxon book, and
another using a different text.
She was convinced, she says, that the teacher, not the text, was the
vital factor in students' success. "I went in as a real skeptic," she
says.ought that I could outteach his book."
But midway through the year, she says, it became apparent that the
Saxon students were doing better. Now, she says, she isn't sure she
would want to teach out of another book. Other teachers in Oklahoma
have also reported that they like the book, according to Lee Graham,
president of the Oklahoma Federation of Teachers. Mr. Graham, who is
not a mathematics teacher, says he has heard no unfavorable comments
about the book, and that several teachers at the school where he works
like it very much.
In New Mexico, which recently added the book to the state-adoption
list, state education officials have not field-tested it and hence have
no hard data on its effectiveness. But comments from teachers and other
mathematics educators have been favorable, according to William A.
Trujillo, secondary mathematics specialist for the state department of
education.
"Around the state, I've gotten a positive response," Mr. Trujillo
says. In general, he says, those who reviewed the book said that the
approach seemed to be one that would benefit the student.
"The cover of the book says 'incremental learning,"' he says. "That,
in effect, is what it is." He says that the book does differ from some
standard algebra texts currently in use in that it offers more of a
balance between skill development and concept development. Some texts,
he says, stress one at the expense of the other. "What I see Saxon as
trying to do is get a blend of the two. It's a different approach to
learning not just algebra, but math."
Marketing the book, several officials point out, may be a problem,
since the book has no sequel and is not part of a series. Schools may
be deterred from using it because there is no text with which to follow
it. Mr. Saxon, however, is working on that | 677.169 | 1 |
All key concepts and skills are covered in this clear and user-friendly AQA Maths foundation tier workbook. GCSE-style practice questions reinforce understanding and help students prepare for the exam with confidence.
Included in this book:
• Invaluable tips to help students pick up all available marks • Advice to help students avoid common mistakes • Plenty of practice questions • Examiners' advice | 677.169 | 1 |
Course Goals <ul><li>To introduce you to the four major concepts of calculus: </li></ul><ul><ul><li>Limits </li></ul></ul><ul><ul><li>Derivatives </li></ul></ul><ul><ul><li>Definite Integral </li></ul></ul><ul><ul><li>Indefinite Integral </li></ul></ul>
6.
For each concept you will … <ul><li>know the precise definition; </li></ul><ul><li>have an intuitive understanding of what the concept means; </li></ul><ul><li>be able to "do" the concept; </li></ul><ul><li>be able to apply it in the real world or mathematical world. </li></ul> | 677.169 | 1 |
Synopsis
Mainstream texts are for mainstream students and the Spectrum Mathematics Gold series recognises that not all students are mainstream! These success-focused texts come from the authors' extensive careers in Special Education. Purpose written for the NSW Year 7 and 8 mathematics syllabus, the Spectrum Mathematics Gold series provides a complete course of instruction for students with special needs, and is ideal for separate classes or withdrawal groups. Spectrum Mathematics Gold Year 7 Blackline Masters is specifically coded to activities or exercises in the Spectrum Mathematics Gold Year 7 textbook, providing teachers with opportunities to consolidate and revise key skills in ways that will ensure their students enjoy doing maths. The worksheets also offer additional mathematics and language activities that are useful for extensions, further revision and homework. | 677.169 | 1 |
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This workbook/textbook has been newly updated to supply the basic mathematical skills and applications encountered in the workplace by manufacturing technicians. Practical exercises are presented in clear, easy-to-follow steps, offering a systematic approach to mastering essential mathematical skills. This new edition highlights critical thinking, to train students on the "e;how tos"e; of problem solving. | 677.169 | 1 |
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Calculus is the mathematical study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations.
Calculus is used in every branch of the physical sciences, actuarial science, computer science, statistics, engineering, economics, business, medicine, demography, and in other fields wherever a problem can... | 677.169 | 1 |
This exam is designed for individuals, who would like teach mathematics at the middle school level. You will be given two hours to complete this 55 multiple-choice and numeric entry questions exam. There will be 34 questions regarding arithmetic and algebra with the 21 other questions being about geometry and data.
Discrete Mathematics, Statistical Concepts, Probability and Data
In this section of the exam you will be required to solve problems by: finding and interpreting common measures of dispersion, standard deviation, outliers, range and spread of data; finding and interpreting common measures of central tendency such as mean, median and mode. You will also be asked to solve probability problems by using counting techniques, by using geometric probability, and solve probability problems that involve finite sample spaces. Your ability to analyze data in charts, line bars, circles, histograms, tables, graphs, scatterplots, stemplots, and draw conclusions from that data.
Functions and their Graphs
This section of the exam will assess your knowledge of algebraic expressions, equations, domain values in ranges of a function, usage of a mathematical model for two different situations, and notations and functions. You will also be given a set of conditions and required to identify if a graph in the plane is the graph of a function. Your ability to utilize a graph and to choose an equation to represent the graph will be assessed. Questions regarding sketching a graph representing exponential or quadratic function, absolute value, step, or linear function will also be assessed.
Measurement and Geometry
You will be required to solve problems in metric and English systems in this section of the exam. Your ability to solve problems regarding the properties of polygons in the plane, points, lines, translations, rotations, reflections, angles, sectors, arcs, circles, squares, rectangles, trapezoid's, parallelograms, triangles, perpendicular lines, and two and three dimensional objects will be assessed in this portion of the exam. Your ability to compute the parameter and area of circles, triangles, quadrilaterals and regions within these figures and; compute the surface area and volume of spheres, solids, cylinders, cones, prisms and the combination of these figures will be assessed. Questions regarding the application of the Pythagorean theorem to solve problems that include isosceles and equilateral triangles will be included in this section of the exam.
Basic Algebra and Mathematics
In this section of the exam you will be required to solve and graph linear equations and inequalities with one or two variables, solve and graph nonlinear algebraic equations, solve equations and inequalities involving absolute value, solve quadratic equation problems, and determine line equations. Problems that require solving algebraic expression and problems that include formulas, equations, adding, subtracting, multiplying, and dividing polynomials, and algebraic fractions will be included in this section of the exam. Your knowledge of standard algebraic operations that involve negative exponents, fractional exponents, radicals, exponents, and complex numbers will be included in this section of the exam. Addition, subtraction, multiplication, and division of rational numbers, identification of basic mathematical operations, identification of the multiplicative and additive in versus of a number, and the order and finite set of real numbers will be included in this section of the exam. Your knowledge of classification of real numbers, complex, rational, or irrational, decimals, powers, roots, and exponents will be assessed in this section of the exam. Number system operation, counting numbers, factors, divisibility, multiples, ratio, proportion, and percentages will also be included in this section of the exam.
Problem Solving Exercises
Questions in the section of the exam will assess your understanding of the learning of mathematical content and learning to think in a mathematical manner. The following topics will be covered in a section of the exam: mathematical problem-solving, mathematical reasoning and proof, mathematical connections, mathematical representation, and the usage of technology. Mathematical problem solving questions will assess your ability to solve mathematical problems, and apply mathematics to solve problems in other contexts. In the mathematical reasoning and proof section you will be asked to utilize serious reasoning and proof methods, create and investigate mathematical conjectures, and create and evaluate mathematical proofs and arguments. Under the mathematical connections section you will be asked to connect mathematical ideas, apply mathematics in the context of other areas, and show an understanding of the interrelationship between mathematical ideas. Under the mathematical representation category you will be asked to apply and translate mathematical representations to solve problems. Your ability to use mathematical representations for modeling and interpretation of social, physical, and mathematical phenomena will be assessed. Your ability to create and utilize representations for the organization, recording, and communication of mathematical ideas will also be included in this section of the exam. In the use of technology section your ability to use technology to solve problems will be assessed. | 677.169 | 1 |
Maths
Mathematics Team Leader - Mr M Mirnateghi
Students in Years 7, 8, 9, 10, and 11 are taught in one of three ability bands and study a course delivered in modules. Each module of work concludes with a test to ensure that students are competent in all aspects of those topics. The modules are delivered by the Team using a wide variety of teaching and learning styles. Students' work includes learning activities, consolidation and practice work, problem-solving and investigative work.
There is an emphasis on activities which increase both confidence and skills. A wide range of software is used to support the delivery of the Scheme of Work. The Mathematics Team has access to rooms that are equipped with computers as well as electronic whiteboards and their use is an integral part of all our students' work. UKMT individual and team challenges are used at all ages.
KS3
Students in KS3 start to use ICT skills to enhance and reinforce mathematical concepts. They are encouraged to develop independent learning, problem solving, and elementary research skills in a wider context. In addition to the revision resources on the Mathematics department currently subscribes to Mymaths and students are routinely set homework tasks from this website. Students are also encouraged to use the following free online resources; GeoGebra , Constructions, Examsolutions, Mental Arithmetic
KS4
Year 10 & 11
At KS4 the Team currently offers the Edexcel GCSE Mathematics course. Sets 1and 2 sit for the Higher Tier papers and Set 3 sit for the Foundation Tier papers. Full details of the specification and units of content can be found on the Edexcel website here
Year 10
Term 1: Number Work Term 2: Algebra Term 3: Graphs Term 4: Handling Data Term 5: Shape and Space Term 6: Proof/Transformations/Vectors
Year 11 will comprise mainly of revision of work covered in Year 10 and extension of these topics. Problem solving questions will be worked on. All groups will also sit 3 Mock exams in late November, January and March.
KS5
Year 12 & 13
Currently we are following the Edexcel modular course at A Level. From September 2017 the structure of course will be as follows. At KS5 students will follow a linear Edexcel course with 3 examinations at the end of the 2 years. Two papers are examined on Pure Mathematics and one paper is on statistics and mechanics combined. The aims and objectives of studying A Level Mathematics are to enable students to understand mathematics and mathematical processes in a way that promotes confidence, fosters enjoyment and provides a strong foundation for progress to further study. It extends their range of mathematical skills and techniques and helps them to understand coherence and progression in mathematics and how different areas of mathematics are connected. The students use their mathematical skills and techniques to solve challenging problems that require them to decide on the solution strategy and they take increasing responsibility for their own learning and the evaluation of their own mathematical development. Full details of the specification and units of content can be found on Edexcel website: We also use a range of revision websites listed below: | 677.169 | 1 |
Tim Chartier has written the perfect supplement to a linear algebra course. Every major topic is driven by applications, such as computer graphics, cryptography, webpage ranking, sports ranking and data mining. Anyone reading this book will have a clear understanding of the power and scope of linear algebra. — Arthur Benjamin, Harvey Mudd College
I'm often asked which areas of mathematics should students study. I always say linear algebra. However, typical linear algebra texts I've seen either have very few applications, or the applications are contrived and not very relevant to students. Chartier's text is a refreshing change as it is driven by real-world applications that are inspiring and familiar to his audience. From Google searches and image processing, to sports rankings and (my favorite) computer graphics. — Tony DeRose, Pixar Animation Studios
From simulating complex phenomenon on supercomputers to storing the coordinates needed in modern 3D printing, data is a huge and growing part of our world. A major tool to manipulate and study this data is linear algebra. This book introduces concepts of matrix algebra with an emphasis on application, particularly in the fields of computer graphics and data mining. Readers will learn to make an image transparent, compress an image and rotate a 3D wireframe model. In data mining, readers will use linear algebra to read zip codes on envelopes and encrypt sensitive information. The books details methods behind web search, utilized by such companies as Google, and algorithms for sports ranking which have been applied to creating brackets for March Madness and predict outcomes in FIFA World Cup soccer. The book can serve as its own resource or to supplement a course on linear algebra.
About the Author
Tim Chartier is an Associate Professor in the Departments of Mathematics and Computer Science at Davidson College. In 2014, he was named the inaugural Mathematical Association of America's Math Ambassador. He is a recipient of the Henry Alder Award for Distinguished Teaching by a Beginning College or University Mathematics Faculty Member from the MAA. Published by Princeton University Press, Tim authored Math Bytes: Google Bombs, Chocolate-Covered Pi, and Other Cool Bits in Computing and coauthored Numerical Methods: Design, Analysis, and Computer Implementation of Algorithms with Anne Greenbaum. As a researcher, Tim has worked with both Lawrence Livermore and Los Alamos National Laboratories on the development and analysis of computational methods targeted to increase efficiency and robustness of numerical simulation on the lab's supercomputers, which are among the fastest in the world. Tim's research with and beyond the labs was recognized with an Alfred P. Sloan Research Fellowship. He serves on the Editorial Board for Math Horizons. He was the first of the Advisory Council for the Museum of Mathematics, which opened in 2012 and is the first museum of mathematics in the United States. Tim fields mathematical questions for the Sports Science program on ESPN, and has also been a resource for a variety of media inquiries, which include appearances with NPR, the CBS Evening News, USA Today, and The New York Times. He also writes for the Science blog of the Huffington Post.
MAA Review
One of the nice things about linear algebra, I've always thought, is that there is something in the subject for just about everybody. There's a lot of beautiful theory, but at the same time those people who like to roll up their sleeves and get their hands dirty with computations, particularly in aid of interesting applications, will find much here to interest them as well.
At Iowa State University, we offer two different introductory undergraduate courses in linear algebra — one is a proof-based course intended for mathematics majors, the other is a more computational course with applications for non-majors. (There is also a more sophisticated joint undergraduate/graduate course in applied linear algebra.) I've taught the non-major course a couple of times, and enjoyed it, but have noted that most introductory texts are usually so busy developing the ideas behind linear algebra that they don't really have time or space in which to really discuss the applications in any depth. Typically an application will just be developed rather briefly, which may result in it appearing somewhat contrived and artificial. The book under review does an excellent job of addressing these concerns, and would make a very useful supplement to a first course in linear algebra. Continued... | 677.169 | 1 |
Trigonometry
ISBN-10: 0077349970
ISBN-13: 9780077349974 Three components contribute to a theme sustained throughout the Coburn Series: that of laying a firm foundation, building a solid framework, and providing strong connections. Not only does Coburn present a sound problem-solving process to teach students to recognize a problem, organize a procedure, and formulate a solution, the text encourages students to see beyond procedures in an effort to gain a greater understanding of the big ideas behind mathematical concepts | 677.169 | 1 |
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Videos Will Help Students "Ace" Math
02/01/97
Ace-Math, an award-winning video tutorial series, is suited for students trying to grasp fundamental mathematical concepts, parents who want to help their child with their homework, or people who need to brush up on math skills for a specialized license or test.
There are nine separate series, each with many individual videos: Basic Mathematical Skills, Pre-Algebra, Algebra I, Algebra II, Advanced Algebra, Trigonometry, Calculus, Geometry, and Probability and Statistics. Each series except Algebra I consists of 30-minute videotapes explaining various concepts. Algebra I has 16 hour-long videos.
For only $29.95, Ace-Math purchasers get a 30-minute tape with the right to make two back-up copies. This lets educators keep the tape in the learning center and let students check out a copy to take home -- with the added security of another back-up copy!
These innovative tapes have been purchased by institutions such as NASA, the U.S. Coast Guard and IBM, and are in use at institutions such as the Los Angeles Public Library and New York Public Library.Video Resources Software, Miami, FL, (888) ACE-MATH | 677.169 | 1 |
Be sure that you have an application to open
this file type before downloading and/or purchasing.
1 MB|10 pages
Product Description
This product begins with a graphic organizer that compares the graphs of linear, quadratic, absolute value, and exponential functions. Then, students will work through some graphing problems. Finally, there is a matching activity that assesses the graphing vocabulary of students with words such as roots, axis of symmetry, vertex, and slope. | 677.169 | 1 |
Description: The book presents an introduction to the fascinating subject of linear algebra. As the title suggests, this text is designed as a first course in linear algebra for students who have a reasonable understanding of basic algebra. Major topics of linear algebra are presented in detail, with proofs of important theorems provided. | 677.169 | 1 |
COURSE DESCRIPTION
This course is designed to be a bridge between the study of mathematics and the application of mathematics to various fields. It provides an overview of how the mathematical pieces of an applied problem fit together. This course also presents an investigation of meaningful and realistic problems encompassing many academic disciplines including management, economics, ecology, environmental science, sociology, and psychology.
Mathematical modeling is the process of creating a mathematical representation of some phenomenon in order to gain a better understanding of that phenomenon. The main goal of this course is to learn how to make a creative use of some mathematical tools, such as difference equations, ordinary and partial differential equations, and numerical analysis, to build a mathematical description of realistic problems. This includes models dealing with traffic flow, communications, energy, air pollution, currency transfer, ecosystems, inheritance, populations, bargaining, and decision making.
COURSE TOPICS
Ordinary differential equations
Optimization
Sensitivity analysis
Probability
Simulation
COURSE OBJECTIVES
After completing this course, you should be able to:
CO1 Develop and construct appropriate models for various problems and situations.
Module 5: Graphs of Functions as Models and Modeling with Differential Equations
Module 6: Modeling with Systems of Differential Equations and Optimization of Continuous Models
ASSESSMENT METHODS
For your formal work in the course, you are required to participate in online discussion forums, complete written assignments, and prepare reports of a midterm project and a final project. See below for details.
Consult the Course Calendar for assignment due dates.
Discussion Forums
This course requires you to participate in six graded discussion forums. There is also an ungraded but required introductions forum in module 1.
Online discussions provide an opportunity for you to interact with your classmates. During this aspect of the course, you respond to prompts that assist you in developing your ideas, you share those ideas with your classmates, and you comment on their posts. Discussion board interactions promote development of a community of learners, critical thinking, and exploratory learning.
Please participate in online discussions as you would in constructive face-to-face discussions. You are expected to post well-reasoned and thoughtful reflections for each item, making reference, as appropriate, to your readings. You are also expected to reply to your classmates' posts in a respectful, professional, and courteous manner. You may, of course, post questions asking for clarification or further elucidation on a topic.
Written Assignments
You are required to complete six written assignments. The written assignments draw on section problems from the textbook. For each assignment, answer all assigned exercises, and show all work.
Assignments must be prepared electronically with a word processor, preferably using whatever equation editor comes with your word processing software. However, if your word processor is not compatible with your mentor's word processor, you will need to save your document as a rich-text file (.rtf) before submitting it. Check with your mentor first to determine file compatibility. (Important: Use the equation editor to insert equations into your word-processed document, not to create the document itself.)
When preparing your answers, please identify each exercise clearly by textbook section and exercise number. Be sure to include your name at the top of the paper, as well as the course name and code and the semester and year in which you are enrolled. To receive full credit for your answers, you must show all work and include complete solutions.
Midterm Project
You are required to complete a midterm project instead of taking a proctored midterm exam. This project will focus on the topics covered in the first three modules of the course.
See the Midterm Paper area of the course website for further details.
Final Project
You are required to complete a final project instead of taking a proctored final exam. This project will address a simulation modeling problem. It requires you follow three requirements to construct and analyze the scenario using modeling.
See the Final Project area of the course website for further details.
GRADING AND EVALUATION
Your grade in the course will be determined as follows:
Online discussions (6)—18 percent
Written assignments (6)—36 percent
Midterm project—16, how to schedule exams, take | 677.169 | 1 |
Algebra 2 @ KHS/Topic 4: Introduction to Systems
Objectives: The objectives for this chapter are for students to understand how to solve systems of 2 equations by algebraic methods (substitution and linear combination) and graphing. Also included in this topic are linear inequalities. | 677.169 | 1 |
Finite Mathematics, 7th Edition
Author:Howard L. Rolf
ISBN 13:9780495118428
ISBN 10:495118427
Edition:7th
Publisher:Brooks Cole
Publication Date:2007-01-09
Format:Hardcover
Pages:888
List Price:$325.95
 
 
Get the background you need for future courses and discover the usefulness of mathematical concepts in analyzing and solving problems with FINITE MATHEMATICS, 7th Edition. The author clearly explains concepts, and the computations demonstrate enough detail to allow you to follow-and learn-steps in the problem-solving process. Hundreds of examples, many based on real-world data, illustrate the practical applications of mathematics. The textbook also includes technology guidelines to help you successfully use graphing calculators and Microsoft Excel to solve selected exercises.
Booknews
Textbook for high school/technical college level courses. Annotation c. Book News, Inc., Portland, OR (booknews.com) | 677.169 | 1 |
Algebra 1 Glossary
Algebra: The study of mathematical symbols and the rules for manipulating those
symbols.
Algebra grid: A grid used to illustrate values of algebraic expressions.
Arithmetic sequence: A sequence of numbers in which each number can be computed by
adding the same amount to the previous number.
Associative law of addition: For any three numbers $a$, $b$, and $c$, it is always
true that $(a+b)+c = a+(b+c)$.
Associative law of multiplication: For any three numbers $a$, $b$, and $c$, it is
always true that $(a(b))(c) = a(b(c))$.
Axis of symmetry: A line that you can flip a graph around that results in the same
graph.
Base: A number that is raised to a power.
Best fit line: When the points on a grid are not all on a straight line, but seem to
have a somewhat linear pattern, you can find a line that is the "best fit" (closest) to the
points.
Box plot: A box with "whiskers" showing the median, quartiles, and extremes (least and
greatest values) of a collection of data values.
Break even: Have a profit of zero (that is, make exactly as much money as you spend).
Ceiling function: $\ceiling(x)$ is the closest integer which is greater than or equal
to $x$.
Clearing denominators: Multiplying both sides of an equation by some nonzero number
that turns all the fractions in the equation into integers.
Coefficient: A constant that a variable or expression is multiplied by.
Combining like terms: Using the distributive law to add any two multiples
of an expression such as $x$. For example, you can simplify $4x+5x$ into $9x$.
Common difference: In an arithmetic sequence, the amount that can be added to each
number to get the next one.
Common ratio: In a geometric sequence, the amount that each number can be multiplied
by to get the next number.
Commutative law of addition: For any two numbers $a$ and $b$, $a + b = b + a$.
Commutative law of multiplication: For two numbers $a$ and $b$, $a(b) = b(a)$.
Completing the square: Rewriting the equation $x^2+2mx=n$ as $(x+m)^2 = n + m^2$
so that it can be more easily solved.
Composition: The composition of two functions $f$ and $g$ is the function $f ∘ g$ that
transforms $x$ into $f(g(x))$.
Conditional relative frequency: A joint frequency divided by the total of its row or
column in a two-way frequency table.
Constant: A single fixed number (unlike a variable, whose value can vary).
Constant coefficient: A constant term, thought of as a coefficient of $1$.
Constant term: A term that is a constant. For a polynomial in $x$, it's the term
without an $x$.
Coordinates: A point on a 2-dimensional plane is described by a pair $(x, y)$. The
coordinate $x$ is given by the labels below the grid, and the coordinate $y$ is given by the
labels to the left of the grid.
Coordinate plane: A 2-dimensional flat surface used for plotting points, lines,
curves, and regions. It contains an $x$ and a $y$ axis which intersect at the origin.
Coordinate grid: A grid of lines on a coordinate plane that makes it easy to see
$(x, y)$ coordinates of locations in that plane.
Correlation coefficient: A number between $1$ and $-1$ that indicates how much
increasing one variable will tend to increase or decrease the other variable. If the best fit
linear model for $x$ and $y$ is $y=mx+b$, the correlation coefficient $r$ of $x$ and $y$
satisfies $$r^2=1-(\text"root-mean-square error of model")^2 /
(\text"standard deviation of " y \text" data")^2$$. The correlation coefficient $r$ has the same
sign (positive, negative, or 0) as $m$. If $x$ and $y$ have means $m_x$ and $m_y$ and
standard deviations $σ_x$ and $σ_y$, it is also true that the correlation coefficient is the
mean of $(x-m_x)(y-m_y)$ divided by $σ_x σ_y$.
Cost: In economics, how much money a company spends to produce a product.
Cube root: The cube root of $a$, written $√^3 a$, is the number whose cube is $a$.
That is, $(√^3 a)^3 = a$.
Data: A collection of related measurements.
Decimal: A fractional quantity written with a decimal point (like $0.5$).
Denominator: The bottom number or expression in a fraction.
Difference: The distance between two quantities, or the answer to a subtraction
problem.
Discriminant: The discriminant of the equation $ax^2+bx+c=0$ is the quantity
$b^2-4ac$. A quadratic equation has two solutions if its discriminant is positive, one
solution if its discriminant is zero, and no real solutions if its discriminant is negative.
Distributive law of multiplication over addition: For any three
numbers $a$, $b$, and $c$, $a(b+c) = a(b)+a(c)$, and $(b+c)(a) = b(a)+c(a)$.
Distributive law of multiplication over subtraction: For any three numbers $a$, $b$,
and $c$, $a(b−c) = a(b)−a(c)$, and $(b−c)(a) = b(a)−c(a)$.
Domain: The set of inputs ($x$-coordinates) of a relation or function.
Dot plot: A diagram showing data values as dots above a number line.
Equation: A mathematical sentence with an equals sign (like $3x+5=11$).
Equivalent: Two fractions are equivalent if they have the same numerical value. Two
equations or inequalities are equivalent if they have the same solution set.
Even function: A function $f$ with $f(x)=f(-x)$ for all $x$. $f(x) = x^n$ is an even
function if $n$ is an even integer. A function is even if and only if its graph has the $y$-axis
as an axis of symmetry.
Expanding an expression: Using the distributive law to turn expressions
which need parentheses (like $3(x+2)$) into expressions which do not (like $3x+6$).
Exponent: In a power, the number of times the base is multiplied by itself.
Exponential: Using exponents, especially using variables in exponents.
Exponential decay: Decreasing toward $0$ due to a variable in an exponent, such as in
$y=2^{-x}$.
Exponential growth: Increasing rapidly due to a variable in an exponent, such as in
$y=2^x$.
Expression: A combination of variables and numbers using arithmetic (like $6-x$).
Factor: An expression that is multiplied by another expression, or that can be
multiplied by another expression to produce a specified result.
First quartile: For $n$ data values, the median of the $$n/2$$ smallest values if $n$
is even, and of the $${n-1}/2$$ smallest values if $n$ is odd.
Floor function: $\floor(x)$ is the closest integer which is less than or equal to $x$.
Formula: An expression that is used to compute a value.
Fraction: A numerator divided by a denominator (like $$1/2$$). Usually we require the
numerator and denominator to both be integers.
Frequency: In statistics, the number of times something occurs, or is observed.
Function: A relation in which no $x$-coordinate appears in more than one $(x, y)$
ordered pair. This means you can think of a function as a transformation that takes each
$x$-coordinate to its single corresponding $y$-coordinate.
Fundamental theorem of arithmetic: Any integer greater than $1$ can be written as a
product of prime numbers, ordered smallest to largest, in exactly one way.
Geometric sequence: A sequence of numbers in which each number can be computed by
multiplying the previous number by the same amount.
Graph: An image formed by plotting the solutions to an equation or inequality, or some
other set of pairs of numbers, on a coordinate plane. To graph an expression containing the
variable $x$, set $y$ equal to that expression.
Histogram: Rectangles of equal width above a number line, where each rectangles's
height shows the number of data values in that portion of the number line.
Horizontal: Going from side to side, like the horizon.
Improper fraction: A fraction in which the numerator is larger than the denominator
(like $$3/2$$).
Inequality: A mathematical sentence that uses one of the symbols $<$, $>$,
$≤$, or $≥$.
Infinite: More than any finite (real) number.
Input: A number that can be "put into" a relation to produce one or more "outputs." If
a relation is given by a two column table of rows $(x, y)$, you "look up" the input $x$ value in
the first column, and the output(s) are given by the $y$ values in those matching row(s).
Integer: A whole number or the negative of a whole number. For instance, $37$ and $0$
and $-5$ are integers, but $2.7$ and $$-3/2$$ are not.
Interquartile range: The third quartile minus the first quartile.
Inverse functions: Functions $f$ and $g$ such that ${g ∘ f}(x)=x$ for
every $x$ in the domain of $f$, and ${f ∘ g}(y)=y$ for every $y$ in the domain of $g$.
Irrational: A number that cannot be written as a fraction $$m/n$$ where $m$ and $n$
are integers.
Isolate: Make a variable appear alone on one side of an equation or inequality, and
not occur in the other side of the equation or inequality.
Joint frequency: The number of events that satisfy both of two specified criteria.
Joint relative frequency: A joint frequency divided by the total number of events.
Laws of exponents: $$a^{c + d} = a^c a^d$$, $$(a b)^d = a^d b^d$$, and
$$(a^c)^d = a^{cd}$$. These are always true when $c$ and $d$ are positive integers. If $a$ and
$b$ are nonzero, then they are true for any integers $c$ and $d$, as is
$$a^{c - d} = a^c / a^d$$. If $a$ and $b$ are positive, then all four laws are true for any $c$
and $d$.
Linear: A straight line, or an equation or expression whose graph is a straight line.
If $m$ and $b$ are constants, then $mx+b$ is a linear expression, and a function $f$ defined by
$f(x)=mx+b$ is a linear function.
Linear coefficient: For a polynomial in $x$, the number that $x$ (without an exponent)
is multiplied by.
Linear model: An estimate for a variable using a linear expression in another
variable.
Linear optimization: Maximizing or minimizing a linear goal or cost expression, while
remaining within some constraints given by linear inequalities.
Marginal frequency: The total of a row or column in a two-way frequency table.
Marginal relative frequency: A marginal frequency divided by the total number of
events.
Mean: The average of a collection of data values. This can be computed by adding all
the values and then dividing by the number of values.
Mean absolute deviation: The mean distance of data values from some central value,
such as the mean, median, or mode of the collection. In other words, the mean of $|x-m|$ where
$x$ is each data value and $m$ is the mean, median, or mode of all the data values.
Mean squared deviation: The mean squared distance of data values from some central
value, such as the mean, median, or mode of the collection. In other words, the mean of
$(x-m)^2$ where $x$ is each data value and $m$ is the mean, median, or mode of all the data
values.
Median: The middle number in an ordered list of data values. If there are an even
number of values, the median is halfway between the two middle numbers in the list.
Mode: The most common value in a collection, or "modes" if more than one are tied.
Monic: A polynomial whose leading (first) coefficient is $1$.
Monomial: A product of variables and numbers, like $3x$ or $5x^2$. A monomial is also
sometimes called a term.
Negate: Take the opposite of a number, by multiplying it by $-1$.
Negative number: A value less than zero (like $-3$).
$n$th root: An $n$th root of $a$ is a number $b$ whose $n$th power is $a$. That is,
$b^n = a$. If $a ≥ 0$ and $n$ is an integer and $n > 0$, then "the" $nth root of $a$, written
$√^n{a}$, is the $n$th root of $a$ that is positive or zero.
Numerator: The top number or expression in a fraction.
Odd function: A function $f$ with $-f(x)=f(-x)$ for all $x$. $f(x) = x^n$ is an odd
function if $n$ is an odd integer. A function is odd if and only if its graph has the point
$(0, 0)$ as a point of symmetry.
One-to-one: A function $f$ for which $f(x)$ has a different value for every distinct
(different) value of $x$.
Outlier: A value that "lies outside" (is much smaller or larger than) most of the
other values in a collection.
Output: A number produced by applying a relation or function to an input.
Origin: The point on a coordinate plane where the $x$-axis and $y$-axis intersect.
It is represented by the coordinates $(0, 0)$.
Parabola: The shape of the graph of $y=x^2$.
Parallel: Two lines are parallel if they always have the same distance
between them, so they never intersect. If two lines are parallel, they have the same slope.
Perfect square: A number that is the square of a rational number. For example, $1$,
$4$, $$25/16$$, and $0$ are perfect squares. An integer is a perfect square only if it is the
square of an integer, which can be proven using the fundamental theorem of arithmetic.
Period: For a periodic function, the amount of time before it repeats. That is, if $f$
is a periodic function, its period is the smallest possible positive $h$ where $f(x+h)=f(x)$ for
every $x$.
Periodic function: A function that repeats after a certain period $h$ with $h > 0$, so
that $f(x+h)=f(x)$ for every $x$.
Perpendicular: Two lines are perpendicular if they create a 90-degree angle. If two
lines are perpendicular and the slope of one of them is $m$, then the slope of the other line is
$$-1/m$$.
Piecewise-defined function: A function that is defined by different formulas at
different inputs.
Point: A location in the coordinate plane. A point has coordinates $(x,y)$, where $x$
is given by the labels below a coordinate grid, and $y$ is given by the labels to the left of
a coordinate grid.
Point of symmetry: A point that you can rotate a graph around by 180° that results in
the same graph.
Point-slope form: If a line contains the point $(x_1,y_1)$ and has slope $m$, then its
equation can be written as $y−y_1=m(x−x_1)$. An equation in the form $y−y_1=m(x−x_1)$ is said to
be in point-slope form.
Polynomial: A sum of monomials. Usually terms with higher powers are written first.
Positive number: A value greater than zero (like 3).
Power: An expression of the form $a^d$. $a$ is called the base, $d$ is called the
exponent, and $a^d$ is called "the $d$th power of $a$". If $d$ is a positive integer, $a^d$
means $a$ multiplied by itself $d$ times.
Prime number: An integer greater than $1$ that can only be written as a product of two
whole numbers in one way: as itself multiplied by $1$.
Product: The answer to a multiplication problem.
Profit: Revenue minus cost.
Quadrant: Each of the four sections of a coordinate plane made by the intersecting
$x$- and $y$-axes. The four quadrants are labeled I, II, III, and IV, counterclockwise from the
top right.
Quadratic: An expression or equation in which the highest power of a variable has
exponent $2$.
Quadratic coefficient: For a polynomial in $x$, the number that $x^2$ is multiplied
by.
Quartiles: The first quartile, median, and third quartile are values which divide a
data collection into four roughly equal parts.
Quotient: The answer to a division problem.
Range: The set of outputs ($y$-coordinates) of a relation or function.
Rate of change: The speed at which a variable changes over a period of time. This is
given by the change in the variable divided by the change in (amount of) time.
Rational: A number that can be written as a fraction $$m/n$$ where $m$ and $n$
are integers.
Relation: A set of ordered pairs $(x, y)$.
Relative frequency: A frequency divided by the total number of events.
Residual: An observed value minus its estimated value.
Restriction: A function $g$ is a restriction of the function $f$ if $g(x) = f(x)$
for every $x$ in the domain of $g$, but that domain may be smaller than the domain of $f$.
Revenue: How much money a company receives in sales.
Root-mean-square error: A number that tells you how far away a line or curve is from a
set of points (a smaller number means the line is a better "fit" to the points). More precisely,
it is the square root of the mean of the squared residuals (differences) between observed and
estimated values.
Roots: The values of $x$ where a polynomial is zero. These are the
$x$-coordinates of the $x$-intercepts of the polynomial's graph.
Scatter plot: Dots in the coordinate plane representing pairs of linked measurements,
such as heights and weights for a group of people.
Sequence: A list of numbers that may be generated by some rule.
Set: An unordered collection of numbers or other mathematical objects, without
repetitions.
Simplify: To rewrite an expression in a way that means the same thing but is simpler
(or shorter). You can simplify $3x - x + 6$ into $2x + 6$.
Slope: A number that measures how steep a line is. It shows the amount of change in
the height of the line as you go 1 unit to the right. The slope of the line $y=mx+b$ is $m$.
Slope-intercept form: The form $y=mx+b$ for a linear equation, where $m$ and $b$ are
constants. The numbers $m$ and $b$ give the slope and $y$-intercept of the line that is the
graph of that equation.
Solution: In an equation or inequality, a number that can be substituted for the
variable to make that equation or inequality true. If the equation or inequality has more than
one variable, a solution is a list of numbers that when substituted for the list of variables
makes the equation or inequality true. For a system of more than one equation or inequality, a
solution must make all of the equations or inequalities true. In chemistry, a solution is a
liquid mixture.
Solution set: All solutions to an equation, inequality, or system.
Solve: Find the solutions to an equation, inequality, or system.
Square root: A square root of $a$ is a number $b$ whose square is $a$. That is,
$b^2 = a$. If $b$ is a square root of $a$, then so is $- b$. If $a ≥ 0$, "the" square root of
$a$, written $√a$, is the square root of $a$ that is positive or zero.
Standard deviation: The square root of the variance.
Standard form: For a linear equation, the form $Ax+By=C$ where $A$, $B$, and $C$ are
constants. For a quadratic equation, either the form $y=ax^2+bx+c$ or $ax^2+bx+c=0$, where $a$,
$b$, and $c$ are constants.
Statistic: A number used to describe or summarize data.
Statistics: The study of data, and the methods used to describe or summarize data.
Step function: A piecewise-defined function where each piece's formula is a constant
(doesn't change with $x$). A step function's graph looks like stair steps.
Substitution: In an expression or equation, eliminating a variable by replacing it
with another expression that it is equal to.
Sum: The answer to an addition problem.
Symmetry: Repeating pattern or shape.
System: For equations or inequalities, two or more equations or inequalities that are
all required to be true.
Table: In mathematics, a rectangular arrangement of rows and columns.
Term: Element in a sum, difference, or sequence.
The quadratic formula: The formula $$x = {-b ± √{b^2-4ac}} / {2a}$$, which gives the
solutions to any equation in the form $ax^2+bx+c=0$ with $a ≠ 0$. The equation has two solutions
when $b^2-4ac > 0$; it has one solution when $b^2-4ac=0$; and it has no real solutions when
$b^2-4ac < 0$.
Third quartile: For $n$ data values, the median of the $$n/2$$ largest values if $n$
is even, and of the $${n-1}/2$$ largest values if $n$ is odd.
Two-way frequency table: For events that can be divided into categories two different
ways, a table of joint frequencies, using rows of the table to group the events one way, and
columns of the table to group the events the other way.
Unit: A standard measurement, such as a meter or an hour.
Value: A number that a variable or expression can equal.
Variable: A letter (like $x$) that we can use to mean different numbers at different
times.
Variance: The mean squared distance of data values from their mean $m$. This can be
computed by adding $(x-m)^2$ for each data value $x$, and then dividing by the number of data
values $n$. When measuring a sample from a population, for instance heights of people, the
variance of the sample is usually different than the variance of the entire population. To
estimate the population's variance, it is usually better to divide by $n-1$ instead of $n$.
Vertex: The point where a parabola crosses its axis of symmetry.
Vertex form: A quadratic equation in the form $y=a(x-h)^2+k$.
Vertical: Going up and down.
Whole number: One of the numbers 0, 1, 2, 3, ... .
$x$-axis: The horizontal line running through the origin on a coordinate plane.
$x$-coordinate: The horizontal value in a coordinate pair. It tells how far to the
left or right the point is. The $x$-coordinate is always written first in the coordinate pair.
$x$-intercept: A point where a curve meets the horizontal axis (the $x$-axis).
$y$-axis: The vertical line running through the origin on a coordinate plane.
$y$-coordinate: The vertical value in a coordinate pair. It tells how far up or down
the point is. The $y$-coordinate is always written last in the coordinate pair.
$y$-intercept: A point where a line or curve meets the vertical axis (the $y$-axis).
The $y$-intercept of the line $y=mx+b$ is the point $(0,b)$.
Zeros: The values of $x$ where an expression is zero. These are the
$x$-coordinates of the $x$-intercepts of the expression's graph. For a polynomial expression,
these are usually called roots. | 677.169 | 1 |
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Olympiad Training Materials Added
Now this website contains not only the training materials for the preparation of BdMO but also some training materials and documents for the preparation of any Olympiad. Check out the Document Section.
In this page we'll recommend you some books for Math
Olympiad preparation, with book reviews and rating. As our main audience is the
Bangladeshi students, we'll try to write the comments in Bangla.
Problem solving books and books on different subject areas are listed separately, because before someone starts solving hard problems, they should consider reading at least one text book on each topic to LEARN math, as learning is the most important thing at the end of the day.
Before You Start:
Choosing the Correct Book:
Before you start reading a book, you should judge your own
level and you can search that book in Amazon. Most of the times you'll find the
product description and user comments to be helpful. We have reviewed the books
on four criterions:
First, Difficulty: Though it is hard to judge the
difficulty, as it mostly depends on personal choice, we have marked each book
on a scale of 5. 1-2 means they are relatively easy, and in the reach of the
beginners, 2-3 for the students who are already acquainted with Math Olympiads
(probably the regional Math Olympiad winners), and 4-5 is the threshold for the
advanced learners. We'll advise the beginners to start from the easier books,
and go to the advanced books after they have mastered the basics.
Then, the contents: Most of the books has excellent
contents. However, in some cases they are not really written for high school
students, and sometimes the problem choices are not as good as we expect. So,
this rating is basically on the conciseness and the quality of the contents.
Type: This is mainly to differ the problem books from the
texts. The following are the meaning of the codes.
a.Problem
book
b.Good
balance between theory and problems
c.Theoretical
(Text book)
Problem Quality: This rating is mainly for the problem
books. Most of the text might receive a low score, as they don't usually
contain hard or very interesting problems.
The Book list:
Books for learning Problem solving:
Though problem solving can only be learnt through solving
problems, here we enlist some books that will be useful for the students to hone their problem solving skill.
The Art and Craft of Problem Solving ***
Author:Paul Zeitz 5/5
Amazon Description: The newly revised Second Edtion
of this distinctive text uniquely blends interesting problems with strategies,
tools, and techniques to develop mathematical skill and intuition necessary for
problem solving. Readers are encouraged to do math rather than just study it.
The author draws upon his experience as a coach for the International
Mathematics Olympiad to give students an enhanced sense of mathematics and the
ability to investigate and solve problems.
Mathematical Olympiad Challenges
Author: Titu
Andreescu, Razvan Gelca
মন্তব্য:
Review:
1.Difficulty: 3/5
2.Content: 4/5
3.Type: a
4.Problem quality: 4/5
Amazon Description: This
significantly revised and expanded second edition of "Mathematical
Olympiad Challenges" is a rich collection of problems put together by two
experienced and well-known professors and coaches of the U.S. International
Mathematical Olympiad Team. Hundreds of beautiful, challenging, and instructive
problems from algebra, geometry, trigonometry, combinatorics, and number theory
from numerous mathematical competitions and journals have been selected and
updated. The problems are clustered by topic into self-contained sections with
solutions provided separately. Historical insights and asides are presented to
stimulate further inquiry. The emphasis throughout is on creative solutions to
open-ended problems.With many new or expanded examples, problems, and solutions,
this second edition includes completely rewritten discussions preceding each of
the 30 units, as well as a more user-friendly style with more accessible and
inviting examples. Featuring enhanced motivation for advanced high school and
beginning college students, as well as instructors and Olympiad coaches, this
text can be used for creative problem-solving courses, for professional teacher
development seminars and workshops, for self-study, or as a resource for
training for mathematical competitions.
Mathematical Olympiad Treasures
লেখক: Titu Andreescu, Bogdan Enescu,
মন্তব্য:
Review:
1.Difficulty: 2/5
2.Content: 3/5
3.Type: a
4.Problem quality: 3/5
Amazon Description: Mathematical
Olympiad Treasures contains a stimulating collection of problems in geometry
and trigonometry, algebra, number theory, and combinatorics. It encourages
readers to think creatively about techniques and strategies for problem solving
in the real world.
The problems are clustered by topic into self-contained
chapters. The book begins with elementary facts, followed by carefully selected
problems and detailed, step-by-step solutions, which then lead to more
complicated, challenging problems and their solutions. Reflecting the
experience of two professors and coaches of Mathematical Olympiads, the text
will be valuable to teachers, students, and puzzle enthusiasts.
Problem-Solving Strategies (Problem Books in
Mathematics)
লেখক:Arthur Engel
মন্তব্য:
Review:
1.Difficulty: 3/5
2.Content: 5/5
3.Type: a
4.Problem quality: 5/5
Amazon Description:PROBLEM
SOLVING STRATEGIES is a unique collection of competition problems from over
twenty major national and international mathematical competitions for high
school students. The discussion of problem solving strategies is extensive. It
is written for trainers and participants of contests of all levels up to the
highest level: IMO, Tournament of the Towns, and the noncalculus parts of the
Putnam Competition. It will appeal to high school teachers conducting a
mathematics club who need a range of simple to complex problems and to those
instructors wishing to pose a "problem of the week", "problem of
the month", and "research problem of the year" to their
students, thus bringing a creative atmosphere into their classrooms with
continuous discussions of mathematical problems. This volume is a must-have for
instructors wishing to enrich their teaching with some interesting non-routine
problems and for individuals who are just interested in solving difficult and
challenging problems. Each chapter starts with typical examples illustrating
the central concepts and is followed by a number of carefully selected problems
and their solutions. Most of the solutions are complete, but some merely point
to the road leading to the final solution. Very few problems have no solutions.
Readers interested in increasing the effectiveness of the book can do so by
working on the examples in addition to the problems thereby increasing the
number of problems to over 1300. In addition to being a valuable resource of
mathematical problems and solution strategies, this volume is the most complete
training book on the market.
Putnam and Beyond
লেখক:Razvan Gelca, Titu Andreescu
মন্তব্য:
Review:
1.Difficulty: 4/5
2.Content: 4/5
3.Type: b
4.Problem quality: 4/5
Amazon Description:Putnam
and Beyond takes the reader on a journey through the world of college
mathematics, focusing on some of the most important concepts and results in the
theories of polynomials, linear algebra, real analysis in one and several
variables, differential equations, coordinate geometry, trigonometry,
elementary number theory, combinatorics, and probability. Using the W.L. Putnam
Mathematical Competition for undergraduates as an inspiring symbol to build an
appropriate math background for graduate studies in pure or applied
mathematics, the reader is eased into transitioning from problem-solving at the
high school level to the university and beyond, that is, to mathematical
research.
* Each chapter systematically presents a single subject
within which problems are clustered in every section according to the specific
topic.
* The exposition is driven by more than 1100 problems and
examples chosen from numerous sources from around the world; many original
contributions come from the authors.
* Complete solutions to all problems are given at the end of
the book. The source, author, and historical background are cited whenever
possible.
This work may be used as a study guide for the Putnam exam,
as a text for many different problem-solving courses, and as a source of
problems for standard courses in undergraduate mathematics. Putnam and Beyond
is organized for self-study by undergraduate and graduate students, as well as
teachers and researchers in the physical sciences who wish to to expand their
mathematical horizons.
Problem Books:
These books mainly contain problem; and yes, a lot of problems. I hope that you'll have a great time solving them!
Five Hundred Mathematical Challenges
লেখক:Edward J. Barbeau, Murray S. Klamkin,
William O. J. Moser
মন্তব্য:
Review:
1.Difficulty: 1/5
2.Content: 4/5
3.Type: a
4.Problem quality: 4/5
Amazon Description:This
book contains 500 problems that range over a wide spectrum of mathematics and
of levels of difficulty. Some are simple mathematical puzzlers while others are
serious problems at the Olympiad level. Students of all levels of interest and
ability will be entertained by the book. For many problems, more than one
solution is supplied so that students can compare the elegance and efficiency
of different mathematical approaches. A special mathematical toolchest
summarizes the results and techniques needed by competition-level students.
Teachers will find the book useful, both for encouraging their students and for
their own pleasure. Some of the problems can be used to provide a little spice
in the regular curriculum by demonstrating the power of very basic techniques.
The problems were first published as a series of problem booklets almost twenty
years ago. They have stood the test of time and the demand for them has been
steady. Their publication in book form is long overdue.
Description:Problem solving seminar
for high-school students, with answers, hints and solutions. This is a good book for beginners. A few of the topics covered are; the pigeonhole principle,
parity, identities with squares and cubes, logarithms, complex numbers, the
well-ordering principle, induction, inclusion-exclusion, Viéte's formulae and
the rearrangement inequality.
Junior Balkan Mathematical Olympiads
লেখক: Dan Branzei, loan Serdean,
Vasile Serdean 4/5
Amazon
Description: This book is intended to help students
preparing to participate in mathematical Olympiads for juniors. An
international competition for students up to 15 and 1/2 years is hosted annually
in one of the Balkan countries since 1997. In the first chapter are presented
the problems from this six Olympiads. Each Olympiad test consists in four
problems, which are to be done in four hours. The
book presents the tests used to select the Romanian team for the Junior Balkan Mathematical
Olympiad. In addition, short-listed problems submitted to the Jury of JBMO,
together with 20 training tests completes the content. Full solutions are
provided for each of the 211 problems. It is our believe that students,
teachers and all those who are mathematically incline will enjoy working these
intriguing and challenging problems.
Problem Primer for the Olympiad, 2nd edition
লেখক: C. R. Pranesachar, B. J.
Venkatachala, C. S. Yogananda 5/5
Description:The problems appearing
in the Indian National Mathematical Olympaid (INMO) and the Regional
Mathematical Olympias ( RMO) papers are of a very different nature from the
problems students encounter in their school curriculum. This book is designed to help the students prepare for the
INMO & RMO The problems have been classified into various sections- Number
Theory, Algebra Geometry, Combinatorics and Miscellaneous problems. There is
also a section containing important theorems and results from various topics generally
not available in school text books, but which are of great help in solving the
problems.
The IMO Compendium: A Collection of Problems
Suggested for The International Mathematical Olympiads: 1959-2004
Author:Dusan Djukic , Vladimir Jankovic , Ivan Matic,
Nikola Petrovic
মন্তব্য:
Review:
1.Difficulty: 5+/5
2.Content: 5/5
3.Type: a
4.Problem quality: 5/5
Amazon Description: The
International Mathematical Olympiad (IMO) has within its almost 50-year-old
history become the most popular and prestigious competition for high-school
students interested in mathematics. Only six students from each participating
country are given the honor of participating in this competition every year.
The IMO represents not only a great opportunity to tackle interesting and
challenging mathematics problems, it also offers a way for high school students
to measure up with students from the rest of the world. The IMO has sparked off
a burst of creativity among enthusiasts in creating new and interesting
mathematics problems. In an extremely stiff competition, only six problems are
chosen each year to appear on the IMO. The total number of problems proposed
for the IMOs up to this point is staggering and, as a whole, this collection of
problems represents a valuable resource for all high school students preparing
for the IMO. Until now it has been almost impossible to obtain a complete
collection of the problems proposed at the IMO in book form. "The IMO
Compendium" is the result of a two year long collaboration between four
former IMO participants from Yugoslavia, now Serbia and Montenegro, to rescue
these problems from old and scattered manuscripts, and produce the ultimate
source of IMO practice problems. This book attempts to gather all the problems
and solutions appearing on the IMO, as well as the so-called
"short-lists", a total of 864 problems. In addition, the book
contains 1036 problems from various "long-lists" over the years, for
a grand total of 1900 problems. In short, "The IMO Compendium" is the
ultimate collection of challenging high-school-level mathematics problems. It
will be an invaluable resource, not only for high-school students preparing for
mathematics competitions, but for anyone who loves and appreciates math.
Geometry:
Geometry Revisited
লেখক: H. S. M. Coxeter Samuel L.
Greitzer
মন্তব্য:
Review:
1.Difficulty: 2/5
2.Content: 5/5
3.Type: c
4.Problem quality: 2/5
Amazon Description: Among
the many beautiful and nontrivial theorems in geometry found in Geometry
Revisited are the theorems of Ceva, Menelaus, Pappus, Desargues, Pascal, and
Brianchon. A nice proof is given of Morley's remarkable theorem on angle
trisectors. The transformational point of view is emphasized: reflections,
rotations, translations, similarities, inversions, and affine and projective
transformations. Many fascinating properties of circles, triangles,
quadrilaterals, and conics are developed.
Plane Euclidean Geometry: Theory and Problems ***
লেখক: A.D. Gardiner C.J. Bradley
মন্তব্য: The best book for beginners and mid level
Olympiad geometers. (Highly recommended)
Review:
1.Difficulty: 2/5
2.Content: 5/5
3.Type: b
4.Problem quality: 5/5
Amazon Description:This geometry text offers
beginning and advanced geometric problem solving tactics, as well as numerous
practice problems. The book is most appropriate for experienced geometers who
are learning how to take on more challenging geometry problems, such as those
offered at the high school olympiad level.
103 Trigonometry Problems: From the Training of
the USA IMO Team
লেখক: Titu Andreescu, Zuming
Feng
মন্তব্য:
Review:
1.Difficulty: 2-4/5
2.Content: 4/5
3.Type: b
4.Problem quality: 5/5
Amazon Description:103 Trigonometry Problems
contains highly-selected problems and solutions used in the training and
testing of the USA International Mathematical Olympiad (IMO) team. Though many
problems may initially appear impenetrable to the novice, most can be solved using
only elementary high school mathematics techniques.
Basic topics include trigonometric formulas and
identities, their applications in the geometry of the triangle, trigonometric
equations and inequalities, and substitutions involving trigonometric functions
Problem-solving tactics and strategies, along with
practical test-taking techniques, provide in-depth enrichment and preparation
for possible participation in various mathematical competitions
Comprehensive introduction (first chapter) to trigonometric
functions, their relations and functional properties, and their applications in
the Euclidean plane and solid geometry expose advanced students to college
level material
103 Trigonometry Problems is a cogent problem-solving
resource for advanced high school students, undergraduates, and mathematics
teachers engaged in competition training.
Geometry Unbound
লেখক: Kiran S. KedlayaDescription:The original text underlying
this book was a set of notes I compiled, originally as a participant and later
as an instructor, for the Math Olympiad Program (MOP), the annual summer
program to prepare U.S. high school students for the International Mathematical
Olympiad (IMO). Given the overt mission of the MOP, the notes as originally
compiled were intended to bridge the gap between the knowledge of Euclidean
geometry of American IMO prospects and that of their counterparts from other
countries. To that end, they included a large number of challenging problems
culled from Olympiad-level competitions from around the world. However, the
resulting book you are now reading shares with the MOP a second mission, which
is more covert and even a bit subversive. In revising it, I have attempted to
usher the reader from the comfortable world of Euclidean geometry to the gates
of "geometry" as the term is defined (in multiple ways) by modern
mathematicians, using the solving of routine and nonroutine problems as the
vehicle for discovery. In particular, I have aimed to deliver something more
than "just another problems book".
Episodes in Nineteenth and Twentieth Century
Euclidean Geometry
লেখক: Ross HonsbergerAmazon Description: Professor Honsberger has succeeded in 'finding'
and 'extricating' unexpected and little known properties of such fundamental
figures as triangles, results that deserve to be better known. He has laid the
foundations for his proofs with almost entirely synthetic methods easily
accessible to students of Euclidean geometry early on. While in most of his
other books Honsberger presents each of his gems, morsels, and plums, as self
contained tidbits, in this volume he connects chapters with some deductive
treads. He includes exercises and gives their solutions at the end of the book.
In addition to appealing to lovers of synthetic geometry, this book will stimulate
also those who, in this era of revitalizing geometry, will want to try their
hands at deriving the results by analytic methods. Many of the incidence
properties call to mind the duality principle; other results tempt the reader
to prove them by vector methods, or by projective transformations, or complex
numbers.
Problems in plane geometry
লেখক: Victor Prasolov, translated and edited by Dimitry Leites
মন্তব্য:
Review:
1.Difficulty: 1-3/5
2.Content: 5/5
3.Type: b
4.Problem quality: 5/5
Description:This is an excellent problem
book for beginner to mid level geometers. You shall get an excellent collection
of more than 1500 problems. Part 1 covers classical subjects of plane geometry. It
contains nearly 1000 problems with complete solutions and over 100 problems to
be solved on one's own. Still more will be added for the English version of the
book. Part 2 includes more recent topics, geometric
transformations and problems more suitable for contests and for use in
mathematical clubs. The problems cover cuttings, colorings, the pigeonhole (or
Dirichlet's) principle, induction, and so on.
College Geometry: An Introduction to the Modern
Geometry of the Triangle and the Circle
লেখক: Nathan Altshiller-Court
মন্তব্য:
Review:
1.Difficulty: 5/5
2.Content: 5/5
3.Type: b, c
4.Problem quality: 5/5
Amazon Description:This is one of the two English
books in print that give a fairly complete introduction to advanced Euclidean
geometry, the other one being the comparable text by R A Johnson, Advanced
Euclidean Geometry (Dover Books on Mathematics). The book contains all the
classical theorems with full proofs, including many theorems that belong to the
so called triangle geometry that was developed in the last quarter of the
nineteenth century. Due to geometry software the subject is becoming popular
again. The book also contains a treasure of exercises, but no solutions which
could be a nuisance. But what use are the solutions? Problems should be solved
and not looked up!. Many problems are about geometric constructions. If you
prepare for a mathematical contest or if you are interested in a complete
overview of the classical plane geometry (for instance after reading Ross
Honsberger's "Episodes"), this is your book.
Geometric Transformations IAlmost everyoneis acquainted
with plane Euclidean geometry as it is usually taught in high school. This book
introduces the reader to a completely different way of looking at familiar
geometrical facts. It is concerned with transformations of the plane that do
not alter the shapes and sizes of geometric figures. Such transformations play
a fundamental role in the group theoretic approach to geometry.
The treatment is direct and simple. The reader is introduced
to new ideas and then is urged to solve problems using these ideas. The
problems form an essential part of this book and the solutions are given in
detail in the second half of the book.
Geometric Transformations IIThis
book is the sequel to Geometric Transformations I which appeared in this series
in 1962. Part I treas length-preserving transformations, this volume treats
shape-preserving transformations; and Part III treats affine and protective
transformations. These classes of transformations play a fundamental role in
the group-theoretic approach to geometry.
Number Theory:
Theory of Numbers: A
Text and Source Book of Problems
Author: Andrew Adler and John E. Cloury
মন্তব্য:
Review:
1.Difficulty: 2/5
2.Content: 5/5
3.Type: b, c
4.Problem quality: 5/5
Amazon Decsription:This text presents the principal ideas of classical
number theory emphasizing the historical development of these results and the
important figures who worked on them. It is intended to introduce third or
fourth-year undergraduates to mathematical proofs by presenting them in a clear
and simple way and by providing complete, step-by-step solutions to the
problems with as much detail as students would be expected to provide
themselves. This is the only book in number theory that provides detailed
solutions to 800 problems, with complete references to the results used so that
the student can follow each step of the argument.
Elementary Number Theory
Author: David M. Burton
মন্তব্য:
Review:
1.Difficulty: 2/5
2.Content: 4/5
3.Type: c
4.Problem quality: 2/5
Amazon Description:Elementary Number Theory is designed for the
one-semester junior- or senior-level number theory course usually taken by
mathematics majors who plan to teach mathematics in high school or college. In
this new edition all numerical information has been updated and recent
developments in number theory are indicated where appropriate. --This text
refers to an out of print or unavailable edition of this title.
%%%Telang NT
An Introduction to
Diophantine Equations: A Problem-Based Approach
Author: Titu Andreescu, Dorin Andrica, and Ion CucurezeanuAmazon Description:This problem-solving book is an introduction to the
study of Diophantine equations, a class of equations in which only integer
solutions are allowed. The presentation features some classical Diophantine
equations, including linear, Pythagorean, and some higher degree equations, as
well as exponential Diophantine equations. Many of the selected exercises and
problems are original or are presented with original solutions. An Introduction
to Diophantine Equations: A Problem-Based Approach is intended for
undergraduates, advanced high school students and teachers, mathematical
contest participants — including Olympiad and Putnam competitors — as well as
readers interested in essential mathematics. The work uniquely presents
unconventional and non-routine examples, ideas, and techniques.
Combinatorics:
Principles and Techniques in Combinatorics
লেখক: Chen Chuan-Chong and Koh Khee-Meng 4/5
Amazon Description: "This
book should be a must for all mathematicians who are involved in the training
of Mathematical Olympiad teams, but it will also be a valuable source of
problems for university courses." Mathematical Reviews
A textbook suitable for undergraduate courses, the materials
in this book are presented very explicitly so that students will find it easy
to read and also find a wide range of examples. A number of combinatorial
problems taken from mathematical competitions and exercises are also included.
Combinatorics: A Problem Oriented Approach
লেখক: Daniel A. Marcus
মন্তব্য:
Review:
1.Difficulty: 2/5
2.Content: 3/5
3.Type: a, b
4.Problem quality: 3/5
Amazon Description: The
format of this book is unique in that it combines features of a traditional
text with those of a problem book. The material is presented through a series
of problems, about 250 in all, with connecting text; this is supplemented by a
further 250 problems suitable for homework assignment. The problems are
structured in order to introduce concepts in a logical order, and in a
thought-provoking way. The first four sections of the book deal with basic
combinatorial entities; the last four cover special counting methods. Many
applications to probability are included along the way. Students from a wide
range of backgrounds, mathematics, computer science or engineering will
appreciate this appealing introduction.
A Path to Combinatorics for Undergraduates: Counting
Strategies 4-5/5
2.Content: 5/5
3.Type: a, b
4.Problem quality: 5/5
Amazon Description: This
unique approach to combinatorics is centered around challenging examples,
fully-worked solutions, and hundreds of problems---many from Olympiads and
other competitions, and many original to the authors. Each chapter highlights a
particular aspect of the subject and casts combinatorial concepts in the guise
of questions, illustrations, and exercises that are designed to encourage
creativity, improve problem-solving techniques, and widen the reader's
mathematical horizons.
Topics encompass permutations and combinations, binomial
coefficients and their applications, recursion, bijections, inclusions and
exclusions, and generating functions. The work is replete with a broad range of
useful methods and results, such as Sperner's Theorem, Catalan paths, integer
partitions and Young's diagrams, and Lucas' and Kummer's Theorems on
divisibility. Strong emphasis is placed on connections between combinatorial
and graph-theoretic reasoning and on links between algebra and geometry.
The authors' previous text, 102 Combinatorial
Problems, makes a fine companion volume to the present work, which is ideal for
Olympiad participants and coaches, advanced high school students,
undergraduates, and college instructors. The book's unusual problems and
examples will stimulate seasoned mathematicians as well. A Path to Combinatorics
for Undergraduates is a lively introduction not only to combinatorics, but to
mathematical ingenuity, rigor, and the joy of solving puzzles.
102 Combinatorial Problems 5/5
Amazon Description: "Combinatorial
Problems" consists of 102 carefully selected problems that have been used
in the training and testing of the USA International Mathematical Olympiad
(IMO) team.
The book is systematically organized, gradually building
combinatorial skills and techniques and broadening the student's view of
mathematics. Aside from its practical use in training teachers and students
engaged in mathematical competitions, it is a source of enrichment that is
bound to stimulate interest in a variety of mathematical areas that are
tangential to combinatorics.
Proofs that Really Count: The Art of
Combinatorial Proof
লেখক: Arthur T. Benjamin, Jennifer Quinn
মন্তব্য: Excellent book for honing your intuition!
Review:
1.Difficulty: 4-5/5
2.Content: 5/5
3.Type: a
4.Problem quality: 5/5
Amazon Description: Mathematics
is the science of patterns, and mathematicians attempt to understand these
patterns and discover new ones using a variety of tools. In Proofs That Really
Count, award-winning math professors Arthur Benjamin and Jennifer Quinn
demonstrate that many number patterns, even very complex ones, can be
understood by simple counting arguments. The book emphasizes numbers that are
often not thought of as numbers that count: Fibonacci Numbers, Lucas Numbers,
Continued Fractions, and Harmonic Numbers, to name a few. Numerous hints and
references are given for all chapter exercises and many chapters end with a
list of identities in need of combinatorial proof. The extensive appendix of
identities will be a valuable resource. This book should appeal to readers of
all levels, from high school math students to professional mathematicians.
Algebra:
101 Problems in Algebra From the Training of the
USA IMO Team
লেখক: Titu Andreescu, 4/5
Amazon Description:This book contains 101 highly rated problems used in training and
testing the USA IMO Team. It gradually builds students algebraic skills and
techniques and aims to broaden students views of mathematics and better prepare
them for participation in mathematics competitions. It provides in-depth
enrichment in important areas of algebra by reorganizing and enhancing studentsproblem-solving
tactics and stimulates interest for future study of mathematics. The problems
are carefully graded, ranging from quite accessible towards quite challenging.
The problems have been well developed and are highly recommended to any student
aspiring to participate at National or International Mathematical Olympiads.
Amazon Description: This
book presents classical inequalities and specific inequalities which are
particularly useful for attacking and solving optimization problems. Most of
the examples, exercises and problems that appear in the book originate from
Mathematical Olympiad contests around the world. The material is divided into
four chapters. In Chapter 1 algebraic inequalities are presented, starting with
the basic ones and ending with more sophisticated techniques; Chapter 2 deals
with geometric inequalities and Chapter 3 comprises a comprehensive list of
recent problems that appeared in those contests during the last 14 years.
Finally, hints and solutions to all exercises and problems are given in Chapter
4.
Polynomials (Problem Books in Mathematics)
Author: E. J. Barbeau
মন্তব্য:
Review:
1.Difficulty: 2/5
2.Content: 3/5
3.Type: c
4.Problem quality: 3/5
Amazon Description: The
book extends the high school curriculum and provides a backdrop for later study
in calculus, modern algebra, numerical analysis, and complex variable theory.
Exercises introduce many techniques and topics in the theory of equations, such
as evolution and factorization of polynomials, solution of equations,
interpolation, approximation, and congruences. The theory is not treated
formally, but rather illustrated through examples. Over 300 problems drawn from
journals, contests, and examinations test understanding, ingenuity, and skill.
Each chapter ends with a list of hints; there are answers to many of the
exercises and solutions to all of the problems. In addition, 69 'explorations'
invite the reader to investigate research problems and related topics.
Functional Equations and How to Solve
লেখক: Christopher G. Small
মন্তব্য:
Review:
1.Difficulty: 3/5
2.Content: 3/5
3.Type: b
4.Problem quality: 3/5
Amazon Description:"This book is devoted to functional equations of a
special type, namely to those appearing in competitions … . The book contains
many solved examples and problems at the end of each chapter. … The book has 130 pages, 5 chapters and an appendix, a Hints/Solutions section,
a short bibliography and an index. … The book will be valuable for instructors
working with young gifted students in problem solving seminars." (EMS
Newsletter, June, 2008)
Over the years, a number of books
have been written on the theory of functional equations. However, very little
has been published which helps readers to solve functional equations in
mathematics competitions and mathematical problem solving. This book fills that
gap. The student who encounters a functional equation on a mathematics contest
will need to investigate solutions to the equation by finding all solutions, or
by showing that all solutions have a particular property. The emphasis here
will be on the development of those tools which are most useful in assigning a
family of solutions to each functional equation in explicit form.
At the end of each chapter, readers
will find a list of problems associated with the material in that chapter. The
problems vary greatly, with the easiest problems being accessible to any high
school student who has read the chapter carefully. The most difficult problems
will be a reasonable challenge to advanced students studying for the
International Mathematical Olympiad at the high school level or the William
Lowell Putnam Competition for university undergraduates. The book ends with an
appendix containing topics that provide a springboard for further investigation
of the concepts of limits, infinite series and continuity.
Secrets in Inequalities
লেখক: Pham Kim Hung
মন্তব্য:
Review:
1.Difficulty: 4/5
2.Content: 5/5
3.Type: b
4.Problem quality: 5/5
Description: This book contains many beautiful
and hard inequalities. The chapters in this book cover basic inequalities from
an advanced standpoint.
This Website is created and maintained by: Tarik Adnan Moon,
Director, KMC | 677.169 | 1 |
Numerical Analysis by J. Leader
(Available at Amherst Books and on reserve in the science library)
Course goals:
Become proficient in a range of numerical analysis techniques
Understand theory behind each method
Implement and experiment with the methods using computational software
Course Topics:
Numerically solving for roots of nonlinear equations
Numerically solving systems of linear equations efficiently
Important methods in numerical linear algebra
Polynomial interpolation and splines
Numerical differentiation and integration
Numerical optimization techniques
Attendance: Please be in class and be there on time. Cooperative learning is more
effective
and more fun than struggling through material on your own.
If you do
miss a lecture, it is your responsibility to obtain the material that
you
missed and to get your assignments handed in to me.
Questions:
If you
have a question during lecture, please raise your hand and ask it right
away.
Chances are that other students are wondering the same thing. If a
question
arises later, feel free to visit my office and we'll work through
sample
problems until you are comfortable with the mathematics.Always feel free to ask me to slow down
as well.
Numerical software: We'll be doing lots of numerical explorations and learning to implement algorithms. If you have a laptop, please bring it to class regularly; if you don't have a suitable device, please let me know and we'll try to arrange one for you to borrow from the college. We will use MATLAB-style software, with a few different options available. MATLAB is installed on most college computers, but cannot be installed on student-owned computers. However, you can access MATLAB on your computer via VCL . You can also install an openly available MATLAB clone like Octave or FreeMat on your laptop, which will work quite fine for the course.
Grading:Your course grade will be based on
three take-home exams (50% total), homework (30%), and a final project with a report and presentation
(20%).
Intellectual Responsibility
Exams. Your work must be entirely your own, so
please follow the guidelines of the honor code. Unless I explicitly
allow other aids, you are only allowed whatever implements you need to
read and write (no notes or calculators). Turn off your
cell phone to avoid inadvertently distracting your fellow test-takers and to follow the honor code. Clocks are available in the classrooms, so cell phones and all other electronic devices are not needed and must be turned off during exams.
Use of a cell phone or any other electronic device during an exam (unless it is an emergency) may be grounds for receiving an F on that exam.
Homework and labs. You may study with other students
following these guidelines:
If you worked with or received help
from any source other than me, you should put a note on the front of
your homework saying, "I worked with <names>."Make sure your name stands out as the author of your
homework.
Working together does not mean that
one of you does the first half of the homework set and the other does
the second. Everyone should work on every problem.
Each student must hand in his or her
own problem set. You may not hand in a single packet as the work of
multiple people.
You must do your own write-up of each problem.
Do not copy someone else's
solution—you will not learn anything and it is plagiarism.You may discuss problems with others, but then you must be
able to work out the solution on your own again and write it down
yourself.
Sending or receiving a copy of a lab file that contains student work violates the honor code and will be treated as plagiarism. You may talk with others about strategies for solving a problem, but please do not share files.
Be cautious in searching the web or other resouces for homework help. Copying solutions is plagiarism, whatever the source.
If you are unsure what agrees or
does not agree with the precepts of intellectual responsibility in this
course, feel free to talk to me about it.
Homework Guidelines
All problem sets are due at THE
START OF CLASS.Late homework will receive half credit for homework handed
in after start of class but by the start of the next class meeting (and will not accepted after
that, unless you contacted me in advance to obtain an extension).
If you are unable to attend class due
to illness or an emergency, let me know as soon as you can and we will
work out an appropriate schedule for assignments.
Your name should be written on all
sheets handed in.
Problem solutions must be written out
in the order they were assigned.
Where appropriate, please box or
highlight final answers.In general, try
to make your answers readable and easy to find.Always
keep the grader happy!
As mentioned elsewhere, no copying!
Always keep the honor code in mind to maintain an ethical environment in which everyone truly learns the material.
Course Resources:
Don't struggle alone! You have many options for
getting help
with this course.
Me. Feel free to come to my office hours,
make an appointment by email or phone, or simply try stopping by my
office—you are welcome whenever my door is open.If
you have some anxiety about taking math exams, please come see me and
we can work together on building your math confidence.
Homework.Mathematics
is learned ACTIVELY, not passively.You
can't absorb math through listening or reading, even if you think you
understand it all.
Textbook. I won't go over everything that is
contained in the text, and I try to avoid doing the same examples.Hence your textbook in an important
independent source of information and you should read it!
Lecture notes. Reviewing the notes you take in lecture
will give you a chance to see the material again after you have had
some time to assimilate it.
Your classmates. Discussing math with others can help you
think through the concepts.Explaining an
idea you already understand will deepen your comprehension, and for the
concepts that you don't understand well, the explanation of a peer may
be more helpful than mine or the textbook's.
Library resources.
The textbook is
on reserve in the Keefe Science Library.
The library has other books on numerical analysis and scientific computation that may be useful resources to supplement the coverage in our text. | 677.169 | 1 |
This program provides a balanced blend of basic
mathematical techniques, a rigorous examination of
their validity, the development of underlying
structures common to number of concepts, and the
application of mathematics to solving complex problems
that arise in the real world. This program reflects
the awareness of the excitingly complex patterns and
systems that can arise from simple mathematical ideas.
Such systems may provide the key to understanding
biological, economic or physical problems. | 677.169 | 1 |
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Product Description
This Solving Logarithmic Equations activity is designed for Algebra 2, PreCalculus, or College Algebra students. Calculators are optional.
Included:
•Task Cards: The cards are designed to reinforce the concepts taught in class. There are two sets of 24 cards, one with QR codes and one without. Students do not need the internet to use the codes, but do need a mobile device with the scanner app installed.
Cards # 1 – 10 have a single logarithm equations
Cards # 11- 24 have two or three logarithms in the equations.
•A Master List of questions contained on the Task Cards
•Two handouts with 12 questions each similar to the problems on the task cards. The handouts can be used for HW, assessments, or enrichment.
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Taylor and MacLaurin Polynomials
Amaze your AP Calculus BC, and College Calculus 2 students with the visualization of Taylor Polynomials included in this product. There are three handouts which explore different orders of Taylor Polynomials in
Statistics Odds and Probability.
This product is designed for Algebra or Statistics. Who doesn't get odds and probabilities confused? This set of cards and guided notes should reinforce those skills so that student will understand the differenceCalculus Polar Area Guided Notes plus Task or Station Cards
This fun resource is designed for AP Calculus BC and College Calculus 2. It reinforces your lessons from the section in the Unit on Polar and Parametric Functions. In addition to being
Designed for Trigonometry, PreCalculus, and Calculus: Parametric Equations - Converting to Rectangular
This activity is suitable for PreCalculus - Trigonometry and can be used as a review for Calculus 2 or AP Calculus BC classes before studying the
Bundle of hand-written notes for Units 6 - 9 Calculus 2 / Calculus BC originally requested by one teacher as a custom bundle, but available for sale to all teachers.
Please note that these are copies of my handwritten notes and may not be | 677.169 | 1 |
I'm getting really tired in my math class. It's algebra the balancing method, but we're covering higher grade syllabus . The topics are really complicated and that's why I usually sleep in the class. I like the subject and don't want to drop it, but I have a big problem understanding it. Can someone guide me?
Hi dear, algebra the balancing method can be really challenging if your basics are not clear. I know this software, Algebrator which has helped manybeginners build their concepts. I have used this software a couple of times when I was in college and I recommend it to every beginner.
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ressing an intuitive rather than rigorous/formal approach to calculus, this student-oriented text covers all of the standard topics. Integration of chapter-end computer exercises and calculator exercises throughout the text and a graphing calculator appendix recognise the impact of technology on calculus. An abundance of interesting applications from engineering, physics, biology, chemistry, economics, astronomy, medicine and pure mathematics shows the far-reaching relevance of calculus. An algebra review is provided in Chapter 10. Numerous examples throughout the text contain all the algebraic steps, with key steps highlighted in colour, needed to complete the solution. The examples are complimented by more than 7,000 section and chapter exercises featuring drill, application, calculator, show/prove/disprove, and challenge problems. Each problem set begins with Self-quiz questions to help students evaluate their understanding of basic ideas in the section. The development of calculus is outlined in extensive historical notes. Biographical sketches impart information on renowned mathematicians. | 677.169 | 1 |
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Integrated Geometry/Algebra
Jun 05 - Jul 15
Ensworth High School
Starting at
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Meeting Dates
From Jun 06, 2016 to Jul 15, 2016
About This Activity
The merging of Algebra 2 and Geometry begins in this course with the continued development of functions and their context within the real number system. Using geometric modeling, this course covers roughly 2/3 of the traditional Geometry course and 1/3 of the traditional Algebra 2 course. Students use the graphing calculator and The Geometer's Sketchpad software. PREREQUISITE: Algebra 1 & Ensworth Mathematics Department approval CREDIT: Year lower | 677.169 | 1 |
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Grades:
9th - 12th grade
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Product Description
Often the biggest stumbling block for students beginning algebra and other advanced math subjects is the unfamiliar nomenclature, including specialized definitions of what appear to be familiar terms. The advanced math courses offered by Heron books go to some lengths to minimize these difficulties, providing definitions in the texts and glossaries for the courses.
The Advanced Math Glossary compiles the glossaries from all of the advanced math courses (including algebra, geometry and trigonometry) into one combined glossary. This makes it a handy resource and cross-reference for students who need to check definitions or refer to formulas from courses other than the one they are on. Many of the definitions in the Advanced Math Glossary are expanded from the course definitions, to make them more understandable when read independent of the course context.
In addition to the definitions, the glossary contains other useful references from the courses, such as a list of symbols commonly used in advanced math, various operations and formulas, measurements and conversions data, advanced math discipline guidelines, an interest table, a powers and roots table and a trigonometry table. | 677.169 | 1 |
Ganit Jee Main Sanyukt Pravesh Pariksha
ISBN 9789351760658
ISBN-10
9351760650
Binding
Paperback
Edition
8th
Number of Pages
1312 Pages
Language
(Hindi)
Subject
Entrance Exam Preparation
JEE Main is a gateway examination for candidates expecting to seek admission in Bachelor of Engineering (BE), Bachelor of Technology (B. Tech) and Bachelor of Architecture (B. Arch) at the IIITs, NITs, DTU and other CFTIs. It also serves as a screening examination for JEE Advanced which serves as a gateway examination for admission for admission in IITs. And we at Arihant understand the need for a systematic mastery of all the subjects of the test with paramount importance to problem-solving and this would ultimately bring them closer to their dreams.This Master Resource Book in Mathematics is the only book having separate sections for Class 11th and 12th syllabi making it easier to sync with school studies.
This book has been, Cartesian Coordinate System, Straight Line, Circles, Parabola, Ellipse, Hyperbola, Introduction to Three Dimensional Geometry, Limit and Derivatives, Mathematical Reasoning, Statistics, Fundamentals of Probability, Matrix, Determinant, Relation and Function, Inverse Trigonometric Functions, Limits, Continuity and Differentiability, Differentiation, Application of Derivatives, Maxima and Minima, Indefinite Integration, Definite Integration, Area Bounded by Curves, Differential Equations, Vector Algebra, Three Dimensional Geometry and Advance Probability.
The subject matter in the book has been divided as per Class 11th and 12th syllabus covering almost all questions of NCERT Textbooks and NCERT Exemplar. The book is completely in sync with NCERT Textbooks taking the concepts gradually from fundamental to higher level. Each chapter in the book has essential theoretical discussion of the related concepts with sufficient number of illustrative solved examples, practice problems and other solved problems. All types of questions like Single Correct Answer Types, Multiple Correct Answer types, Reasoning types, Matches, Passage-based Questions, etc have been provided in the book. Previous Years Questions of JEE Main and AIEEE have been covered with complete solutions in each chapter to help students know the difficulty levels and nature of questions asked in competitive exams at this level. At the end JEE Main Mock Tests have also been given for thorough revision and mastering the concepts covered under the JEE Main Mathematics syllabus. Also 2013 and 2014 Solved Papers JEE Main have also been provided at the end of the book with detailed and authentic solutions.
As the book contains specialized study material along with practice material, it for sure will help the aspirants master the concepts for the upcoming JEE Main Examination. | 677.169 | 1 |
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2Contents Part I (weeks 1-7) 1 Introduction 2 Combinatorics, permutations and combinations.3 Algebraic Structures and matrices: Homomorphism, isomorphism, group, semigroup, monoid, rings, fields4 Lattices and Boolean algebrasIf time remains: some illustrations of the use of group theory in cryptographyPart II (weeks 8-12)Vector spaces
3Introduction Computer programs frequently handle real world data. This data might be financial e.g. processing the accounts of a company.It may be engineering data e.g. from sensors or actuators in a robotic system.It may be scientific data e.g. weather data or geological data concerning rock strata.In all these cases data typically consists of a set of discrete elements.Furthermore there may exist orderings or relationships among elements or objects.It may be meaningful to combine objects in some way using operators.We hope to clarify our concepts of orderings and relationships among elements or objectsWe look at the idea of formal structures such as groups, rings and and formal systems such as lattices and Boolean algebras
4Number SystemsThe set of natural numbers is the infinite set of the positive integers. It is denoted N and can have different representations:{1,2,3,4, }{1,10,11,100,101,.....}are alternative representations of the same set expressed in different bases. Nm is the set of the first m positive numbers i.e. {1,2,3,4, ,m}. N0 is the set of natural numbers including 0 i.e. {0,1,2,3,5,....}Q denotes the set of rational numbers i.e. signed integers and fractions{0,1,-1,2,-2,3,-3,....,1/2,-1/2,3/2,-3/2,5/2,-5/2,....,1/3,-1/3,2/3,-2/3, }R is the set of real numbers i.e. the coordinates of all the points on a line.Z is the set of all integers, both positive and negative {0,1,-1,2,-2,3,-3,......}
52 Combinatorics: Permutations A permutation of the elements of a set A is a bijection from A onto itself.If A is finite we can calculate the number of different permutations. Suppose A={a1,...,an}nchoicesn-1choices1choicea1a2antotal number of ways of filling the n boxesn x (n-1)x(n-2)x(n-3) x1=n!nPn=n!eg a possible permutation of {1,2,3,4,5,6} is
6Composition of Permutations If :A A and :A A are permutations of A then the composition or product .of and satisfies for all x in A.x)= (x))Notice that since both and are bijections from A into A so is . In other words . is a permutation of A.Example: Let A={1,2,3,4,5,6} then two possible permutations areFor . we have that
7Cyclic Permutations A cyclic permutation on a set A of n elements has the form where :For shorthand we often write is said to be a k cycleExampleor (6 1 4) is a cyclicpermutationTwo cyclic permutationsandare said to be disjoint ife.g. (4 5 2) and (3 1 6) are disjoint
8Notice thatOther examples areorCan you spot a product of disjoint cyclicpermutations equivalent to the followingpermutation ?
9Theorem: Every permutation of a finite set A can be expressed as a combination of disjoint cycles. Structure underlying permutationsNote that the following hold:(1) The product of two permutations is a uniquely determined permutation of the same set.(2) The composition of permutations is associative.(3) The permutationis called the identity permutation and has theproperty that(4) For every permutationthere is an inversesuch that
10CombinationsWhen we think about combinations we do not allow repeats and unlike permutations we do not consider order.Combinations look at the number of different ways of picking a subset of k elements from a set of n elements.Think of the number of ways of picking a list of k distinct elements of nno. of choicesnn-1n-k-2n-k-1places= n(n-1)(n-2) (n-k-1) = n!/(n-k)!For each possible list there are k! permutationsso since we are not interested in order weshould divide the above by k!.C(n,k) = Cnk = n!/(n-k)!k!
12Algebraic StructuresWhen we consider the behaviour of permutations under the composition operation we noticed certain underlying structures.Permutations are closed under this operation, they exhibit associativity, an identity element exists and an inverse exists for each permutationThese properties define a general type of algebraic structure called a group.In this section we shall look at groups in more detail as well as other similar algebraic structures such as semigroups and monoids.Later we will progress to consider more complex algebraic structures such as rings, integral domains and fields.We will see that many real life situations are examples of these algebraic structures
13Groups A group or is a set G with binary operation which satisfies the following properties1.is a closed operation i.e. ifandthen2.this is theassociative law3. G has an element e, called the identity, such that4.there corresponds an elementsuch thatExample:The set of all permutations of a set Aonto itself is group (called the symmetric group Snfor n elements).
14Group of Symmetries of a Triangle Consider the triangleXOYZnmWe can perform the following transformationson the triangle1=identity mapping from the plane to itselfp=rotation anticlockwise about O through 120degreesq=rotation clockwise about O through 120 degreesa=reflection in lb=reflection in mc=reflection in n
15Letdenote transformation y followed bytransformation x for x and y in {1,p,q,a,b,c}So for examplelllYXXapOOOXZmnm YZnmZYnNotice the table is not symmetric
16Other examples of a group The set of all permutations onto itself is a group (called the symmetric group Sn)The sets of all invertible nxn matrices forms a group under ordinary matrix multiplication (called GL(n), the general linear group)The quaternion group: Let G={I,-I,J,-J,K,-K,L,-L} whereI=[ ], J=[ ], K=[ ] , L=[ ]j j0 j j 0
17Order of a group A finite group is a group where G is finite The order of a finite group is |G|For example if G is the set of permutations of a set A with n elements then the order of G is n!
18Abelian Groups If is a group and is also commutative then is referred to as an Abelian group(the name is taken from the 19'th centurymathematician N.H. Abel)is commutativemeans thatExamples:andareabelian groups.Why isnot a group at all?
23Modular multiplication Not a group! (Why not?)Which subset of {1,2,3,4,5} does form a group?x12345
24Modular multiplication Theorem: If n>=2 and n|p then n has no inverse under multiplication mod pProve it!The subset of {1,…,p-1} relatively prime to p is a group under multiplication mod p denoted Zp*We will clarify this on the next slides…
25Modular arithmeticRecall Euclid's algorithm to find the gcd of x and y:x=k1y+r1y=k2r1+r2r1=k3r2+r3…rn-2 =kn-1rn-1+rnrn-1=knrnFrom this… Theorem:There exist integer a and b such ax+by=gcd(x,y)The old remainder is divided by the new one repeatedly until the remainder is 0The gcd is the last non zero remainder
26Modular arithmeticAn element n has an inverse n-1 under multiplication mod p for whichn. n-1 =1 mod pif and only if (iff) n is relatively prime to p.Prove this!Clearly then if p is prime then every element will have an inverse.
27Groups in logic Consider exclusive or defined by A⊕ B ≡(¬A∧ B)∨ (A∧ ¬B){t,f} is an abelian group under exclusive or.What is the identity?What is the inverse of t (and f)?
28To show that an algebraic system is a group we must show that it satisfies all the axioms of a group.Question: Letbe a Boolean algebraso that A is a set of propositional elements,islike 'or',is like 'and' andis like 'not'. Showthatis an abelian group whereAnswer:(1) Associative sinceprove this ?(2) Has an identity element 0 (false) since(3) Each element is its own inverse(4) The operation commutesprove this ?
30Cyclic groupsA group G is cyclic if there exists a∈G such that for any b∈G there is an integer k≥0 such that ak=b.I.e. Every element of G is some power of a.Element a is called the generator of G denoted G=<a>Example:<{1,-1},×>=<-1> since –12=1, -13=-1
31Order of a cyclic permutation group Show that the order is equal to p[Show by making a drawing…]
32Weaker structuresAn Abelian group is a strengthening of the notion of group (i.e. requires more axioms to be satisfied)We might also look at those algebraic structures corresponding to a weakening of the group axiomsSemigroup ⊆ monoid ⊆ group ⊆ Abelian Group
33Semigroup is a semigroup if the following conditions are satisfied: 1. is a closed operation i.e. ifandthen2.is associativeExample: The set of positive even integers{2,4,6,.....} under the operation of ordinary additionsinceThe sum of two even numbers is an even number+ is associativeThe reals or integers are not semigroups under -why?
34Monoid is a monoid if the following conditions are satisfied: 1. is a closed operation i.e. ifandthen2.is associative3. There is an identity elementExamples: Let A be a finite set of heights. Letbea binary operation such thatis equal tothe taller of a and b. Thenis a monoid where the identity is the shortestperson in Ais a monoid:is associative,true is the identity, but false has no inverseis a monoid:is associativefalse is the identity, but true has no inverse
35Properties of Algebraic Structures Theorem: (unique identity) Suppose thatis a monoid then the identity element is uniqueProof: Suppose there exist two identity elementse and f. [We shall prove that e=f]Theorem: (unique inverse) Suppose thatis a monoid and the element x in A has an inverse.Then this inverse is unique.Proof: ??
36Properties of Groups Theorem (The cancellation laws): Let be a group then(i)(ii)Proof: (i) Suppose thatthenby axiom 3a has an identityand we have that(ii) is proved similarlyTheorem (The division laws): Letbea group then(i)(ii)Proof ??
37Theorem (double inverse) :If x is an element of the group thenProof:Theorem (reversal rule)If x and y are elements of the group thenProof ??
38For an arbitrary element of a group wecan define functionsandsuch thatTheorem:andare permutations of GProof: Consider[prove 1-1] suppose for x,y in G[Prove onto] For any y in GCorollary: In every row or column of themultiplication table of G each element of G appearsexactly once.
39Subgroups is a subgroup of the group if and is also a group Examples: Test for a subgroupLet H be a subset of G. Thenis a subgroupofiff the following conditions all hold:(1)(2) H is closed under multiplication(3)For every group,andaresubgroupsis called the trivial subgroup ofa proper subgroup ofis a subgroupdifferent from GA non-trivial proper subgroup is a subgroupequal neither to or to
40Cosets Consider a set A with a subset H. Let . Then the left coset of H with respect to a isthe set of elements:This is denoted bySimilarly the right coset of H with respect to a isand is denoted byExample: Let A be the set of rotationsand{0º,120º,240º}. Letthenwhich is the right coset with respect to
41Normal Subgroups Let be a subgroup of . Then is a normal subgroup if, for any, the leftcosetis equal to the right cosetis a normal subgroup wheree.g.Theorem: In an Abelian group, every subgroupis a normal subgroup
42Coset cardinalityTheorem: For any H subset of G and any a in G |a•H|=|H|Proof:By definition of Coset |a•H|≤|H|Now suppose |a•H|<|H| then there must exist b and c distinct elements of H such that a•b=a•c.But by the cancellation law this implies that b=c which is a contradiction.Hence |a•H|=|H|
43Coset partitioningTheorem: Let a,b∈G and let H be a subgroup of G then either: a•H=b•H or: a•H∩ b•H=∅Proof:Suppose a•H∩ b•H≠∅ then there exist s and t in H such that a•s=b•t.In this case a= b•t•s-1 and for an arbitrary x in H a•x= b•t•s-1•xNow by the inverse axiom and closure, t•s-1•x∈H and hence b•t•s-1•x∈b•H, therefore a•x∈b•H so that a•H⊆b•HSimilarly we can show that b•H⊆a•HHence if the two cosets are not disjoint then b•H=a•H
44LeGrange's theoremTheorem: Let H be a subgroup of finite group G, then the cardinality of H evenly divides the cardinality of G (i.e |H| | |G|)ProofLet |G|. Now for each element ai of G we can generate a coset ai•H.Now notice that ai∈ai•H because since H is a subgroup, e∈H and ai•e= aiSuppose there are m distinct cosets of H then picking one representative ai from each this means that: G= a1•H∪ a2•H ∪ a3•H … ∪ am•H
45LeGrange's theoremNow by the previous theorem it follows that since these m cosets are distinct then they must be disjoint.Hence,|G|=|a1•H|+ |a2•H| + |a3•H| … + |am•H|Also by the cardinality theorem for cosets they all have the same cardinality, namely |H|. Hence, |G|=m.|H| as required
46Order of an elementLet i be the smallest integer such that ai=e where a is an element of group G and e is the identity element.If i exists we call it the order of a.Otherwise we say that a has infinite order.
47Subgroup generated by an element Theorem: For any element a of G with finite order the set: H={aj: for some integer j} is a subgroup of G.Notice: if i is the order of element a thenai=eai+1=e•a=a1ai+2= a •a =a2ai+n=an
48Example Let σ=(1 2 3 4), a permutation of 4 elements Then {σ, σ2, σ3, σ4} is a subgroup of the group of permutations of {1,2,3,4}The order of σ is 4[Work it out!]
49Order of elements in finite groups If the group G is finite then all elements of G have finite order:For any a∈G, since G is finite there must exist i<j such that ai=aja•ai-1=a•aj-1 cancellation law implies ai-1=aj-1Repeated application of the cancellation law gives a=aj-i+1a•e=a•aj-i implies e=aj-i
50Corollary of LeGrangeTheorem: The order of every element of a finite group G, divides the order of GProof...Every element of G has finite order n and hence generates a subgroup of order n.Hence by LeGrange's theorem n divides |G|
51IsomorphismTwo groups are isomorphic if there is a bijection of one onto the other which preserves the group operations i.e.ifandare groups then a bijectionis an isomorphism providedExample: Consider the group of matricesof the form where under matrixmultiplication. This is isomorphic to the groupThe mapping isAn isomorphism from a group onto itself iscalled an automorphism.
52HomomorphismsThe idea of isomorphic algebraic structures can be readily generalised by dropping the requirement that the functional mapping be a bijection.Letandbe two algebraic systemsthen a homomorphism fromtois a functional mappingsuch thatExample: consider the two structuresthen f such thatis a homomorphism betweenand
53Algebraic Structures with two Operations So far we have studied algebraic systems with one binary operation. We now consider systems with two binary operations.In such a system a natural way in which two operations can be related is through the property of distributivity;Letbe an algebraic system with twobinary operationsand. Then the operationis said to distribute over the operationifandExample:distributes over +distributes overdistributes over
54Ring An algebraic system is called a ring if the following conditions are satisfied:(1)is an Abelian group(2)is a semigroup(3) The operationis distributive over theoperationExample:is a ring sinceis an Abelian groupis a semigroupdistributes over +
55Examples of rings <Z, +, ×> is a ring because: <Z, +> is an Abelian group.<Z, ×> is a semigroup.× distributes over +The set {[ ],a,b є R} is a ring under matrix addition and multiplication{0,1,…,n-1} is a ring under addition and multiplication mod n0 a 0 b
56Rings of polynomialsLet the set R[x] be the set of all polynomial of the form: anxn+…+ a2x2+ a1x1+a0for some n, where an,…,a0 єRThen R[x] is a ring under addition and multiplication of polynomialsIn fact for any ring R you can construct a ring of polynomials R[x] over R
57Special types of ring A commutative ring is a ring in which is A ring with unity contains an element 1 suchthat(0 is the identity of )Example: the ring of 2x2 matrices under matrixaddition and multiplication is a ring with unity.The element 1=I=
58Division ringsA division ring is a (not necessarily commutative) ring with unity, in which every element a not equal to 0 has an inverse a-1 such that a•a-1= a-1•a=1The ring of complex matrices of the form: [ ]a b -b a
59Integral Domains and Fields is an integral domain if it is a commutativering with unity that also satisfies the followingproperty;is also an integral domainis a field if:(1)is an Abelian group(2)is an Abelian group(3) The operationis distributive over theoperationExample:The set of real numbers with respect to+ and is a field.is not a field. Why?
60Galois fieldsFor a prime number p the set {0,1,…,p-1} is a field under modular addition and multiplication mod pA field (like this one) with finite number of elements is called a Galois field.
61A Field is an Integral Domain Letbe a field then certainlyis a commutative ring with unity. Hence, it onlyremains to prove thatNow supposethen if x=0 the aboveholds. Consider the case then whereSinceis an Abelian group then itmust contain an inverse to x,, for which thefollowing holdsNowTherefore y=0 as required
62Properties of a ring Theorem: if is a ring. Then Proof: as for previous argumentLet -x denote the inverse of x underTheorem: ifis a ring then the followinghold(i)(ii)Proof: (i)
65Subrings and subfield Subring If (A,⊕,•) is a ring then (H,⊕,•) is a subring if H⊆A and(H,⊕,•) is a ringSubfieldIf (A,⊕,•) is a field then (H,⊕,•) is a subfield if H⊆A and(H,⊕,•) is a fieldExamples: Z is a subring of R, R is a subfield of C
66Ring morphismsA morphism between rings (A,⊕,•) and (B,*,⊗) is a function f:A→B such that: ∀x,y∈Af(x⊕y)=f(x)*f(y) andf (x•y)=f(x)•f(y)From these we have thatf(0)=0′ where 0′ is the zero of (B,*,⊗)Also f(-x)=-f(x)
67Special morphisms An injective ring morphism is called a monomorphism 2. A surjective ring morphism is called an epimorphism3. A bijective ring morphism is called a isomorphism
68Examples of morphismsf(a) = a mod n, is an epimorphism (surjective ring morphism) between Z and {0,1,…,n-1}For the ring of polynomials R(x), f(p)=p(j) is an epimorphism into C, where p(j) is obtained by substituting j for x in the polynomial p
69Galois theoremFor every prime power pk (k=1,2,…) there is a unique (upto isomorphism) finite field containing pk elements denoted by GF(pk)All finite fields have cardinality pk
71Partial Orderings We have introduced formal structure governing the properties of various sets of elements underone or two binary operations.These elements can also be ordered and restrictedby binary relations.In this section we revise our understanding ofbinary relations in a set and also introduce agraphical notation for binary relations.A relation R on a set A is a partial order if itsatisfies;(1) R is reflexive(2) R is antisymmetric(3) R is transitiveThe pair (A,R) is called a partially ordered setor posetExample: Set of reals R with the relation
72Example: The relationcan be defined on aBoolean algebra by;(1) Thus from the idempotent lawwe find thatand hence the relation isreflexive.(2) IfFrom the commutative lawand hence the relation is antisymmetric(3) Ifthen
73We can think of a relation as being represented by the set of pairs of elements which satisfy therelation.In this case a partial ordering on A correspondsto a subset B of AxA satisfyingOther examples of partial orderings:Divisibility on N: We say that a divides b iff thereis some x in Z such that ax=b. If this divisibilityexists we write a|b. Divisibility is a partial orderon N.Inclusion on a set of sets X
74Graphical Representations We can represent partial orderings graphicallyby means of a directed graph where the nodesare elements of A and the directed edges givethe partial order relations.e.g. the graphabcdDenotes the partial ordering on{a,b,c,d}where
75Graphical Representations of the Axioms Reflexive:aAntisymmetric: the following does not occurabTransitive:abc
77Example: The collection of all subsets of {a,b,c} {a,c}{b,c}{a}{b}{c}
78Hasse Diagrams Notice that some of the diagrams in the previous examples were messy and difficult to read havingmany links.We can simplify these diagrams by introducingcertain conventions.The Hasse diagram of a partially order set is adrawing of the points in the set (as nodes) andsome of the links of the graph of the order relation.The rules for drawing the Hasse diagram of a partialorder are:(1) Omit all links that can be inferred fromtransitivity.(2) Omit all loops(3) Draw links without arrow heads(4) Understand that all arrows wouldpoint upwards
79Here are Hasse diagrams for the two examples given previously:Divisibility:18894632
81Incomparable Elements Consider the Hasse diagram for divisibility on{2,3,....,10}1086945372Notice that 5 and 6 are not related in either directionSimilarly for 2 and 3If neither R(a,b) or R(b,a) then a and b areincomparable or not comparable
82Linear or Total OrderA linear or total order on a set A is a partial orderon A in which every two elements are comparable5432
83Maximal and Minimal Elements A maximal element of A is any element t ofA such thatA minimal element of A is any elementb of A such thatExample:For the subset ordering {a,b,c} is themaximal element and is the minimal elementFor divisibility on {2,.....,10} the maximalelements are 6, 7, 8, 9 and 10and the minimal elementsare 2, 3, 5 and 7The element 4 is neither maximal nor minimal
84Upper Bounds and Lower Bounds Let S be a subset of A then x in A is an upperbound of S ifSimilarly z in A is a lower bound of S ifAn element u is the least upper bound of S ifu is an upper bound of S and for every x anupper bound of S R(u,x)An element l is the greatest lower bound of S ifl is an upper bound of S and for every z alower bound of S R(z,l)The least upper bound (lub) of S is sometimesreferred to as the supremum of S (sup S)The greatest lower bound (glb) of F is sometimesreferred to as the infimum of S (inf S)
85Lattices A partially ordered set in which every pair of elements has a least upper bound and a greatestlower bound is called a lattice.abcdfeThis is not a lattice since {c,d} has no lub orglb.A lattice in which every subset has a lub and glbis called complete.Every finite lattice is complete.For a complete lattice the lub of the whole latticeis call top and the greatest lower bound bottom
86Example: Consider elements of the form (a,b,c) where a,b and c can take the values 0 or 1. Fortwo such elements f and g we say thatif each coefficient of f is less than or equal tothe corresponding coefficient of ge.g.but not(111)(101)(011)(110)(001)(100)(010)(000)
87Meet and Join In a lattice the following equations define binary operations on Ais called the meet operation andiscalled the join operation.They have the following propertiesCommutativity:Associativity:Sinceis an upper bound of a and bSimilarly for the meet
88TheoremandIfthenProofLet 1 denote the lub of the whole lattice and 0denote the glb of the whole lattice. Then
89Example Let us order the following set of numbers with the operation "is a factor of". A={3,9,12,15,36,45}4536915123The join operationis the least common multipleThe meet operationis the greatest common divisor
90Complemented Lattice For a complemented lattice we have that for there existssuch that:e.g.1abc
91Distributive Lattice A lattice is distributive if: e.g. the following lattice is not distributiveedbcaSince
92Boolean Algebra A Boolean Algebra consists of two binary operations and and the unary operationon a set B with distinct elements 0 and 1 suchthat the following hold.(1) The commutative laws:(2) The associative laws:(3) The Distributive laws:(4) The Identity Laws: | 677.169 | 1 |
DESCRIPTION:This is
a university-level course in Differential and Integral Calculus, equivalent to
one semester of Calculus at most universities. The AP Calculus course is
designed to develop the student's understanding of the concepts of Calculus
and to provide experience with its methods and applications. The course
emphasizes a multi-representational approach to Calculus with concepts, results,
and problems expressed geometrically, algebraically, numerically, analytically,
and verbally. Successful completion of the AP Calculus Course also provides the
student with an MCB4UI credit. Math is a subject that builds upon
itself; attending class, being on time, and participating in class are not only
essential for good progress, but are sometimes essential for survival of the
class. Unexcused absences result in no make up for the work.
EVALUATION:70% is based on tests, quizzes,
assignments.
30% is based on the summative assessment, which includes the June exam.
This
course builds on your previous experience with functions and your developing
understanding of rates of change. You will solve problems involving geometric
and algebraic representations of vectors and representations of lines and planes
in three dimensional space; broaden your understanding of rates of change to
include the derivatives of polynomial, sinusoidal, exponential, rational, and
radical functions; and apply these concepts and skills to the modelling of
realworld relationships. You will also refine your use of the mathematical
processes necessary for success in senior mathematics. This course is intended
for students who choose to pursue careers in fields such as science,
engineering, economics, and some areas of business, including those students who
will be required to take a university-level calculus, linear algebra, or physics
course.
This
course extends students' experience with functions. Students will investigate
the properties of polynomial, rational, logarithmic, and trigonometric
functions; develop techniques for combining functions; broaden their
understanding of rates of change; and develop facility in applying these
concepts and skills. Students will also refine their use of the mathematical
processes necessary for success in senior mathematics. This course is intended
both for students taking the Calculus and Vectors course as a prerequisite for a
university program and for those wishing to consolidate their understanding of
mathematics before proceeding to any one of a variety of university programs.
The Advanced Functions
course (MHF4UW) must be taken prior to or concurrently with Calculus and Vectors
(MCV4UW).
1.
demonstrate an understanding of rate of change by making connections
between average rate of change over an interval and instantaneous rate of change
at a point, using the slopes of secants and tangents and the concepts of the
limit
2.
graph the derivatives of polynomial, sinusoidal, and exponential functions, and
make connections between the numeric, graphical, and algebraic representations
of a function and its derivative
3.
verify graphically and algebraically the rules for determining
derivatives; apply these new rules to determine the derivatives of polynomial,
sinusoidal, exponential, rational, and radical functions, and simple
combinations of functions; and solve related problems
B: Derivatives
and their Applications
1.
make connections, graphically and algebraically, between the key features
of a function and its first and second derivatives, and use the connections in
curve sketching
2.
solve problems, including optimization problems, that require the use of
the concepts and procedures associated with the derivative, including problems
arising from real-world applications and involving the development of
mathematical models
C: Geometry
and Algebra of Vectors
1.
demonstrate an understanding of vectors in two-space and three-space by
representing them algebraically and geometrically and by recognizing their
applications perform operations on vectors in two-space and three-space, and use
the properties of these operations to solve problems, including those arising
from real-world applications
2.
distinguish between the geometric representations of a single linear equation or
a system of two linear equations in two-space and there-space, and determine the
different geometric configurations of lines and planes in three-space represent
line and planes using scalar, vector, and parametric equations, and solve
problems involving distance and intersections
A. Exponential and
Logarithmic Functions
1.Demonstrate
an understanding of the relationship between exponential expressions and
logarithmic expressions, evaluate logarithms, and apply the laws of logarithms
to simplify numeric expressions;
2.
Identify and describe some key features of the graphs of logarithmic functions,
make connections among the numeric, graphical, and algebraic representations of
logarithmic functions, and solve related problems graphically;
3.Solve
exponential and simple logarithmic equations in one variable algebraically,
including those in problems arising from real-world applications.
B. Trigonometric
Functions
1. Demonstrate an
understanding of the meaning and application of radian measure;
2. Make connections
between trigonometric ratios and the graphical and algebraic representations of
the corresponding trigonometric functions and between trigonometric functions
and their reciprocals, and use these connections to solve problems;
C. Polynomial
and Rational Functions
1. Identify and
describe some key features of polynomial functions, and make connections between
the numeric, graphical, and algebraic representations of polynomial functions;
2. Identify and
describe some key features of the graphs of rational functions, and represent
rational functions graphically;
1. Demonstrate an
understanding of average and instantaneous rate of change, and determine,
numerically and graphically, and interpret the average rate of change of a
function over a given interval and the instantaneous rate of change of a
function at a given point;
2. Determine
functions that result from the addition, subtraction, multiplication, and
division of two functions and from the composition of two functions, describe
some properties of the resulting functions, and solve related problems;
3. Compare the
characteristics of functions, and solve problems by modeling and reasoning with
functions, including problems with solutions that are not accessible by standard
algebraic techniques.
Overall
Expectations (MCV4UW)
Titles and Descriptions
Advanced Functions
Several types of functions needed in
this course will be reviewed along with their characteristics including:
differences in polynomials, absolute value functions, polynomial in
equalities and division, remainder theorem and factor theorem, and
factoring polynomials.
Concepts of Calculus
A variety of mathematical operations
with functions will be investigated including: rationalization, rates of
change, the limit concept, indeterminate form, finding the slope of a
curve, tangent slope function, derivatives and graphs.
Derivatives
In this unit students will see the
power of the slope function and the applications of derivatives in a
variety of style problems.
Curve Sketching
The key features of a properly
sketched curve including x and y intercepts, vertical, horizontal and
oblique asymptotes, maximum and minimum values, points of inflection,
undefined tangent slope points will be examined separately before putting
them all together into a full sketch of a curve.
Derivative Applications
A variety of types of problems will
be presented in this unit and can generally be grouped into the following
categories: Pythagorean problems, volume problems, trough problems, shadow
problems, general rate problems. Each type will be examined separately.
Exponents and Log Functions
As a review of previous courses, the
unit will begin with the rules associated with exponents. Then the
exponential functions, applications and logarithms and log functions and
applications will be covered.
Derivatives of Exponents and Log
Functions
Exponential functions, logarithmic
functions, curve sketching and logarithmic differentials are all topics of
this unit.
Trigonometry Differentials and
Applications
The unit begins with a review of the
three basic trig functions (sine, cosine, tangent). Angles, the CAST rule,
sums and differences for sine/cosine form the second major topic. Finally
solving trigonometric equations are pursued with a focus on limits,
derivatives and applications of trigometric functions.
Antiderivatives and Applications
The topics covered in this unit
include the concept of antiderivatives, acceleration, velocity,
differential equations, Riemann's sums and areas, area function,
definite integral and integration and area between curves.
Final Evaluation
Total
4.Units:
C1Pre-Calculus (Summer)8
lessons
C1.1Functions
C1.2Linear Equations
C1.3Parametric Equations
C1.4Factor and Remainder Theorem
C1.5Trigonometric Functions
C1.6Exponential Equations
C1.7Logarithmic Functions
C1.8Geometry and Similar Triangle
C2Limits and Continuity7 lessons
C2.1Rate of Change and Limits
C2.2Indeterminate Forms
C2.3Limits involving Infinity
C2.4Continuity
C2.5Rates of Change and Tangent Lines
C2.6Epsilon and Delta
C2.7
C3Derivatives16 lessons
C3.1Limit Definition of Derivative
C3.2Differentiability
C3.3Power Rule
C3.4Proof of Power Rule
C3.5Chain Rule
C3.6Product Rule
C3.7Proof of Product Rule
C3.8Quotient Rule
C3.9Proof of Quotient Rule
C3.10Velocity
and Acceleration
C3.11Derivatives
of Trigonometric Functions
C3.12Derivatives
of Inverse Trigonometric Functions
C3.13Derivatives
of Exponential and Logarithmic Functions
C3.14Implicit Differentiation
C3.15Review
C3.16Review
C4Applications14 lessons
C4.1Extreme Values of Functions
C4.2Mean Value Theorem
C4.3Connecting and with
and graphing (1) – Large Lesson
C4.4Connecting and with
and graphing (2)
C4.5Connecting and with
and graphing (3)
C4.6Optimization (1)
C4.7Optimization (2)
C4.8Optimization (3)
C4.9Linearization and Newton's Method
C4.10Estimating
Change and Rolle's Theorem
C4.11Related
Rates (1)
C4.12Related
Rates (2)
C4.13Related
Rates (3)
C4.14Review
C5The Definite Integral14 lessons
C5.1Estimating with Finite Sums
C5.2Riemann Sums
C5.3Definite Integrals
C5.4Definite Integrals and Antiderivatives (1)
C5.5Definite Integrals and Antiderivatives (2)
C5.6Definite Integrals and Antiderivatives (3)
C5.7Fundamental Theorem of Calculus (1)
C5.8Fundamental Theorem of Calculus (2)
C5.9Fundamental Theorem of Calculus (3)
C5.10Trapezoidal
Rule
C5.11Simpson's
Rule
C5.12Review
(1)
C5.13Review
(2)
C5.14Review
(3)
C6Differential Equations and Modelling12 lessons
C6.1Slope Fields (1)
C6.2Slope Fields (2)
C6.3Integration by Substitution (1)
C6.4Integration by Substitution (2)
C6.5Integration by Parts (1)
C6.6Integration by Parts (2)
C6.7Exponential Growth and Decay
C6.8Population Growth
C6.9Numerical Methods
C6.10Review
C6.11Review
C6.12Review
C7aApplications of Definite Integrals7 Lessons
C7.1Integral as Net Change
C7.2Areas in the Plane (1)
C7.3Areas in the Plane (2)
C7.4Volumes (1)
C7.5Volumes (2)
C7.6Volumes (3)
C7.7Review
Practice exams2 Weeks
C7bApplications of Definite Integrals4 Lessons
C7.7Lengths of Curves
C7.8Applications from Science and Statistics (1)
C7.9Applications from Science and Statistics (2)
C7.10Review
Achievement Chart: Mathematics, Grades
9-12
Categories
50-59%
(Level 1)
60-69%
(Level 2)
70-79%
(Level 3)
80-100%
(Level 4)
Knowledge and Understanding - Subject-specific content acquired in each course (knowledge),
and the comprehension of its meaning and significance (understanding)
applies knowledge and skills in familiar contexts with limited
effectiveness
applies knowledge and skills in familiar contexts with some
effectiveness
applies knowledge and skills in familiar contexts with considerable
effectiveness
applies knowledge and skills in familiar contexts with a high degree of
effectiveness
Transfer of knowledge and skills (e.g., concepts, procedures, methodologies,
technologies) to new contexts
transfers knowledge and skills to new contexts with limited
effectiveness
transfers knowledge and skills to new contexts with some effectiveness
transfers knowledge and skills to new contexts with considerable
effectiveness
transfers knowledge and skills to new contexts with a high degree of
effectiveness
Making connections within and between various contexts(e.g., past, present, and future;
environmental; social; cultural; spatial; personal; multidisciplinary)
makes connections within and between various contexts with limited
effectiveness
makes connections within and between various contexts with some
effectiveness
makes connections within and between various contexts with considerable
effectiveness
makes connections within and between various contexts with a high
degree of effectiveness
5.Course
Evaluation:
"The SPH4UI course will be
evaluated based on term work worth 70% of your final report grade and the
components of the final evaluation are worth 30% of your final grade.Term work includes; tests, quizzes, assignments, lab reports.Our final evaluation will be composed of a final exam."
The
final grade will be determined as follows:
Assessment
Percentage
Knowledge
and Understanding
Communication
Application
Thinking
and Investigation
June
Examination
Total
25%
15%
15%
15%
30%
100%
6.Late
Work Policy:
"At
WCI is the expectation that students will submit all required work by the
assigned due date as evidence of their learning.Students who fail to meet a due date for an essential course component
will be subject to the completion policy found the student planner. Failure to
submit this work, despite these interventions, will be recorded as incomplete
and may result in a loss of credit."
7.Cheating/Plagiarism
Policy :
"At WCI it is the expectation that
students will submit their own original work for the purpose of demonstrating
their learning.In the event that
cheating or plagiarism occurs, the following consequences may be implemented, in consultation with administration,
depending on the situation:
·The
student may be required to redo all or part of the assignment or assessment.
·The
student may be required to complete an alternate assignment of assessment.
·The
student's work may be treated as a missed assignment.
·There
may be other consequences that are determined to be appropriate, including
disciplinary consequences as outlined in the Cheating/Plagiarism section of the
student planner."
8.Learning
Skills :
"The
development of learning skills and work habits is an integral part of a
student's learning.The
achievement of these skills is officially reported on the Provincial Report
Card.The evaluation of learning
skills and work habits is reported as follows: E-Excellent, G-Good,
S-Satisfactory, and N-Needs Improvement.For
a full description of the 6 Learning Skills; Responsibility, Organization,
Independent Work, Collaboration, Initiative, and Self-Regulation, please see the
WCI Student Planner." | 677.169 | 1 |
Calculus Integration Color by Number
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this file type before downloading and/or purchasing.
1 MB|8 pages
Product Description
Calculus Integration Color by Number
Even big kids love coloring. This fun, engaging activity includes twenty review questions on definite integration. Problems include substitution, trig functions, inverse trig functions, exponentials, and two area questions. Students solve the problems, match the numerical answer to a color, and then color in the design, a geometric pattern. If desired, students can use their own color scheme. Directions included for color changing.
This can be used as a review, individually or with small groups. Great for a Sub Folder, too. Answer key included. Your students might want to refer to their Calculus Formula Flip Book
Please click on the green star next to my name to follow me and be the first hear of sales, and new, engaging resources for all of your classes. This purchase is for one teacher only. Additional teachers must purchase their own license. You may not upload this resource to the internet in any form. Multi - user licenses available. Custom Bundle requests gladly accepted. | 677.169 | 1 |
The text for this course is Cengage Learning's Basic Math, 9th Edition by Aufmann and Lockwood.
This sixth grade course includes one chapter each on whole numbers, fractions, decimals, ratio and proportion, percents, applications for business and consumers, statistics and probability, U.S. customary units of measurement, the metric system of measurement, rational numbers, introduction to algebra, and geometry. With the addition of the last three chapters (which were not included in previous editions), this is a solid sixth grade course.
Many practical applications appear in examples and word problems, and an entire chapter is devoted to business and consumer math skills such as percent calculations in making purchases, computing interest, calculating the cost of buying a home or car, calculating wages, and balancing a checkbook.
A four-function calculator might be used along with this course, but it's not absolutely essential. If you are looking ahead, you might want to go ahead and purchase a TI-83 Plus graphing calculator since it can be used with all courses.
"I had talked to you about 2 1/2 years ago. One son was repeating Basic Math for the third time. You encouraged me and he finally understood it. He went on last year to do well in Algebra 1 in public school. He is doing well and has a good grasp of math. He is in his sophmore year enjoying Geometry. Thank you."
Mary Radinsky
"Hello to all at Chalk Dust Math,
My son has used the Chalk Dust Basic Math for this school year and has completed the program. The DVD sessions with Uncle Buck have provided consistent, thorough explanation of math concepts for him. We definitely plan to use Chalk Dust in our home school for his middle and high school math. Thanks for your great program."
Linda Williams
"My daughter is 12 and in 6th grade. We have homeschooled her from the beginning. We have used Miquon, Singapore, Abeka, Math-U-See and Saxon in the past
This year we have been using your Basic Math program with "Uncle Buck" on the videos. We are so pleased. My daughter feels she is now prepared and has a full understanding of math for the test. Thank you.
We can't say enough good things about this program. We enjoy "Uncle Buck" and his methodical teaching style. He incorporates review without calling it that. He builds the child's confidence in a step-by-step learning style so that the child feels any problem can be conquered when it is broken down in steps.
We are looking forward to purchasing Chalkdust Pre-Algebra and continuing on. I also appreciate you taking the time to help me know which "math track" my daughter should take as we approach high school."
Sheila Truncellito
Basic Math. "Mr. Mosely, we started Section 2 and my child is finally saying that he is getting to "love math", but only after purchasing your videos."
Maria Sanchez
"Now we're going through the high school math program again with our youngest child, who has always struggled with math. Two years ago we did your Basic Math program and for the first time she understood math! Last year she did PreAlgebra and even though it took us all year and summer to get through it, we did! Now she is sailing through Algebra I and will finish the complete program in May and have the summer "off." She understands it and has finally quit telling us she feels stupid! :)
We have really been so pleased with the materials and the way my son has really made wonderful strides with his math in the past year. He loves Chalkdust Math so much that he did not even want to look at any other programs. He is ready to sign on for life - and that is saying a lot as he had a math phobia prior to this. He had a bad experience with Saxon Math in public school. We then tried BJU for 3rd grade - which he hated, some McGraw Hill materials for 4th grade, which were sketchy at best, and A Beka for grade 5, which also did not go well. My husband is a math/science person and we were becoming very concerned that our son was falling behind in this area. I truly believe that he has not only caught up - but gained confidence and has a desire to learn and understand more this year. He enjoys the instruction dvd's and does not try to get out of doing math. He has really enjoyed learning some of the concepts. I have suggested looking into Chalkdust to quite a few other parents. I cannot imagine attempting to teach Algebra to my son myself or have my husband teach it. My son has informed his father that the Chalkdust teaching is far better than dad's! :)
Thank you so much!
Mr B.
"Chalk Dust Basic Math covers six major topics: whole numbers, fractions, decimals, ratio, and proportion, percent, and "Applications for Business and Consumers: A Calculator Approach." This last chapter is almost 100 percent word problems. It covers topics found in both business math and consumer math texts: purchasing, percent increase and decrease, interest, real estate expenses, car expense, wages, and bank statements. I've found that children in grade 5 and up can jump right into this course (with adult supervision and assistance), while it's a great review for teens and parents."
Mary Pride
"We purchased your Basic Math videotape course about 30 days ago. I wanted to share with you some of my 11 year old son's comments. Today he said, 'He makes math so much easier. He is so smart!' I am really impressed! Thank you Chalk Dust Company!" | 677.169 | 1 |
Mathematics
Containing over a hundred and sixty complicated issues of tricks and specified ideas, this ebook can be utilized as a self-study advisor for arithmetic competitions and for making improvements to problem-solving talents in classes on airplane geometry or the historical past of arithmetic.
Eminently suited for lecture room use in addition to person examine, Roger Myerson's introductory textual content presents a transparent and thorough exam of the types, resolution thoughts, effects, and methodological rules of noncooperative and cooperative video game conception. Myerson introduces, clarifies, and synthesizes the extreme advances made within the topic during the last fifteen years, provides an summary of determination thought, and comprehensively experiences the advance of the elemental versions: video games in large shape and strategic shape, and Bayesian video games with incomplete information.
Game Theory might be beneficial for college kids on the graduate point in economics, political technology, operations study, and utilized arithmetic. each person who makes use of video game conception in examine will locate this publication essential.
Utilizing the heritage of arithmetic complements the instructing and studying of arithmetic. up to now, a lot of the literature ready with regards to integrating arithmetic heritage in undergraduate instructing comprises, predominantly, rules from the 18th century and prior. This quantity makes a speciality of nineteenth and twentieth century arithmetic, development at the previous efforts yet emphasizing fresh historical past within the instructing of arithmetic, laptop technological know-how, and comparable disciplines. "From Calculus to desktops" is a source for undergraduate academics that supply rules and fabrics for fast adoption within the lecture room and confirmed examples to inspire innovation via the reader. Contributions to this quantity are from historians of arithmetic and school arithmetic teachers with years of expertise and services in those matters. one of the issues integrated are: tasks with major old content material effectively utilized in a numerical research path, a dialogue of the function of chance in undergraduate records classes, integration of the historical past of arithmetic in undergraduate geometry guide, to incorporate non-Euclidean geometries, the evolution of arithmetic schooling and instructor training over the last centuries, using a seminal paper through Cayley to encourage pupil studying in an summary algebra direction, the combination of the heritage of good judgment and programming into desktop technological know-how classes, and concepts on the right way to enforce heritage into any category and the way to enhance historical past of arithmetic classes.
There exist literary histories of likelihood and medical histories of likelihood, however it has regularly been inspiration that the 2 didn't meet. Campe begs to vary. Mathematical likelihood, he argues, took over the function of the outdated likelihood of poets, orators, and logicians, albeit in medical phrases. certainly, mathematical chance wouldn't also have been attainable with no the opposite likelihood, whose roots lay in classical antiquity.
The video game of Probability revisits the 17th and eighteenth-century "probabilistic revolution," offering a heritage of the kinfolk among mathematical and rhetorical innovations, among the clinical and the classy. This used to be a revolution that overthrew the "order of things," significantly the best way that technological know-how and artwork located themselves with recognize to fact, and its members integrated a large choice of individuals from as many walks of existence. Campe devotes chapters to them in flip. targeting the translation of video games of probability because the version for likelihood and at the reinterpretation of aesthetic shape as verisimilitude (a severe query for theoreticians of that new literary style, the novel), the scope by myself of Campe's publication argues for probability's the most important position within the structure of modernity.
No clinical concept has prompted extra puzzlement and confusion than quantum idea. Physics is meant to assist us to appreciate the realm, yet quantum concept makes it look a really unusual position. This booklet is ready how mathematical innovation can assist us achieve deeper perception into the constitution of the actual international. Chapters through best researchers within the mathematical foundations of physics discover new rules, particularly novel mathematical techniques, on the leading edge of destiny physics. those artistic advancements in arithmetic may perhaps catalyze the advances that let us to appreciate our present actual theories, specifically quantum thought. The authors carry various views, unified purely through the try and introduce clean thoughts that might open up new vistas in our knowing of destiny physics.
This publication offers the mathematical foundations of networks of linear keep an eye on platforms, constructed from an algebraic platforms concept point of view. This features a thorough therapy of questions of controllability, observability, cognizance concept, in addition to suggestions keep an eye on and observer idea. the possibility of networks for linear platforms in controlling large-scale networks of interconnected dynamical platforms may provide perception right into a variety of clinical and technological disciplines. The scope of the booklet is kind of broad, starting from introductory fabric to complex issues of present learn, making it an appropriate reference for graduate scholars and researchers within the box of networks of linear structures. half i will be used because the foundation for a primary direction in Algebraic procedure thought, whereas half II serves for a moment, complex, direction on linear structures.
Finally, half III, that is mostly self sustaining of the former elements, is excellent for complex examine seminars geared toward getting ready graduate scholars for self sustaining learn. "Mathematics of Networks of Linear structures" incorporates a huge variety of workouts and examples through the textual content making it compatible for graduate classes within the area.
Each year, hundreds of thousands of scholars claim arithmetic as their significant. Many are tremendous clever and hardworking. besides the fact that, even the simplest will stumble upon demanding situations, simply because upper-level arithmetic contains not just self sustaining examine and studying from lectures, but in addition a basic shift from calculation to facts.
This shift is hard however it don't need to be mysterious -- examine has published many insights into the mathematical pondering required, and this publication interprets those into functional suggestion for a pupil viewers. It covers each element of learning as a arithmetic significant, from tackling summary highbrow demanding situations to interacting with professors and making stable use of analysis time. half 1 discusses the character of upper-level arithmetic, and explains how scholars can adapt and expand their present talents in an effort to boost reliable figuring out. half 2 covers research talents as those relate to arithmetic, and indicates sensible techniques to studying successfully whereas having fun with undergraduate life.
As the 1st mathematics-specific examine consultant, this pleasant, functional textual content is key analyzing for any arithmetic major.
"Written in an admirably cleancut and least expensive style." — Mathematical Reviews. This concise textual content bargains undergraduates in arithmetic and technology an intensive and systematic first path in simple differential equations. Presuming a data of easy calculus, the booklet first stories the mathematical necessities required to grasp the fabrics to be presented. The subsequent 4 chapters absorb linear equations, these of the 1st order and people with consistent coefficients, variable coefficients, and normal singular issues. The final chapters tackle the life and area of expertise of ideas to either first order equations and to platforms and n-th order equations. Throughout the booklet, the writer contains the idea some distance adequate to incorporate the statements and proofs of the easier lifestyles and distinctiveness theorems. Dr. Coddington, who has taught at MIT, Princeton, and UCLA, has integrated many workouts designed to enhance the student's strategy in fixing equations. He has additionally incorporated difficulties (with solutions) chosen to sharpen knowing of the mathematical constitution of the topic, and to introduce various proper subject matters now not lined within the textual content, e.g. balance, equations with periodic coefficients, and boundary price problems.
this article makes use of the language and notation of vectors and matrices to elucidate matters in multivariable calculus. obtainable to a person with a very good history in single-variable calculus, it provides extra linear algebra than frequently present in a multivariable calculus booklet. Colley balances this with very transparent and expansive exposition, many figures, and diverse, wide-ranging workouts. teachers will take pleasure in Colley's writing variety, mathematical precision, point of rigor, and whole number of themes treated.
Vectors: Vectors in and 3 Dimensions. extra approximately Vectors. The Dot Product. The move Product. Equations for Planes; Distance difficulties. a few n-Dimensional Geometry. New Coordinate platforms. Differentiation in different Variables: Functions of a number of Variables; Graphing Surfaces. Limits. The spinoff. houses; Higher-Order Partial Derivatives; Newton's technique. The Chain Rule. Directional Derivatives and the Gradient. Vector-Valued services: Parametrized Curves and Kepler's legislation. Arclength and Differential Geometry. Vector Fields: An advent. Gradient, Divergence, Curl, and the Del Operator. Maxima and Minima in different Variables: Differentials and Taylor's Theorem. Extrema of capabilities. Lagrange Multipliers. a few purposes of Extrema. a number of Integration: Introduction: parts and Volumes. Double Integrals. altering the Order of Integration. Triple Integrals. switch of Variables. purposes of Integration. Line Integrals: Scalar and Vector Line Integrals. Green's Theorem. Conservative Vector Fields. floor Integrals and Vector research: Parametrized Surfaces. floor Integrals. Stokes's and Gauss's Theorems. extra Vector research; Maxwell's Equations. Vector research in larger Dimensions: An advent to Differential types. Manifolds and Integrals of k-forms. The Generalized Stokes's Theorem.
What is algebra? For a few, it truly is an summary language of x's and y's. For arithmetic majors mathematicians, it's a global of axiomatically outlined constructs like teams, jewelry, and fields. Taming the Unknown considers how those probably types of algebra developed and the way they relate. Victor Katz and Karen Parshall discover the background of algebra, from its roots within the historical civilizations of Egypt, Mesopotamia, Greece, China, and India, via its improvement within the medieval Islamic international and medieval and early sleek Europe, to its glossy shape within the early 20th century.
Defining algebra initially as a suite of concepts for choosing unknowns, the authors hint the improvement of those thoughts from geometric beginnings in historical Egypt and Mesopotamia and classical Greece. They express how related difficulties have been tackled in Alexandrian Greece, in China, and in India, then examine how medieval Islamic students shifted to an algorithmic level, which used to be extra constructed via medieval and early glossy ecu mathematicians. With the advent of a versatile and operative symbolism within the 16th and 17th centuries, algebra entered right into a dynamic interval characterised by means of the analytic geometry that may assessment curves represented by way of equations in variables, thereby fixing difficulties within the physics of movement. This new symbolism freed mathematicians to check equations of levels better than and 3, eventually resulting in the current summary era.
Taming the Unknown follows algebra's awesome development via various epochs round the globe. | 677.169 | 1 |
Ch 10 - How Fast are you going What time is it 1 Basic...
How Fast are you going? What time is it?1Basic Ideas of Interpolation!Mathematical equivalent of "reading between the lines"!Data - discrete samples of some function, f(x)!Uses an interpolating function between points!Data might…!Exist as an experiment!Analytic function that is difficult to evaluate2
This
preview
has intentionally blurred sections.
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Math Lab & Science Learning Center
Math and Science Learning Center
219 Sequoya Hall
Math and Science Learning Center Mission Statement
The goals of the Math and Science Learning Center are to offer assistance, tutoring, and support for all levels of Math and Natural Science courses; to assist students in preparation for end-of-term Math exit tests; and to provide assistance with mathematics for all subject areas throughout the college.
Services to Students and Faculty
The Math and Science Learning Center offers assistance and tutoring for all levels of Mathematics and Natural Science courses and preparation for end-of-term Mathematics Exit Exams. Tutoring is available on a walk-in basis. Student assistants staff the lab throughout the day. In addition, the Math and Science Learning Center has software programs to help students prepare for end-of-term exit exams for Developmental Mathematics. There is also a variety of Science CDs available for student use | 677.169 | 1 |
Introduction to Numerical Analysis
Welcome to Students
What does this math course have to offer?
Numerical analysis uses the methods of calculus to quantify the
discrepancies between exact symbolic results and the approximations found by
computer algorithms. Specific problems addressed by this course include the
solution of nonlinear equations, interpolation and polynomial approximation,
numerical integration and differential equations.
The required text is Süli, Endre and David F. Mayers,
An Introduction to Numerical Analysis,
Cambridge University Press, 2006. You should also download the following
materials and read each item before we reach the corresponding topic in
class. | 677.169 | 1 |
Slope Fields Differential Equations with Guided Notes and Matching Task Cards.
This lesson is designed for AP Calculus AB, AP Calculus BC and College Calculus 1 or 2 classes. These activities will help your students thoroughly understand
Calculus, Introduction to Differential Equations, Initial Value Problems Task or Station Cards
This activity is designed for Calculus 1 or AP Calculus. Initial Value problems are usually included in the beginning of Unit 4, Integration, rightCalculus, Separation of Variables Differential Equations Guided Notes, Task Card Activity, plus HW and QR
This lesson is designed for AP Calculus AB, AP Calculus BC and College Calculus 1 or 2 classes. These activities are designed to helpCalculus Differential Equations Growth and Decay Task / Station Cards and Flip Book
This lesson is designed for AP Calculus AB, AP Calculus BC and College Calculus 1 or 2 classes. These activities are designed to help students thoroughly
This no prep, paperless digital activity is designed for Calculus 1 or AP Calculus. Initial Value problems are usually included in the beginning of Unit 4, Integration, right after Antiderivatives, but depending upon your curriculum may appear in
Calculus First Order Linear Differential Equations Task Cards, Guided Notes, HW
This lesson is designed for AP Calculus AB, AP Calculus BC and College Calculus 1 or 2 classes.
Included in the Lesson:
✓ Guided Notes with two examples | 677.169 | 1 |
Saturday, June 13, 2009
I recently posted a few examples of symbolic math calculation using the WolframAlpha Internet service, and asked how it might affect math teaching. WolframAlpha, which presents a command-line interface to Wolfram's Mathematica symbolic math package, is capable of doing math homework and solving exam questions from junior high through graduate school.
The conservative view of using tools like calculators or WolframAlpha is captured in Isaac Asimov's story The Feeling of Power, depicting a future in which a technician amazes people because he has memorized the multiplication tables and can do arithmetic without a calculator.
Are math skills and concepts part of IT literacy? Is there room for any math in an IT literacy course? Where can WolframAlpha be used in the IT literacy curriculum? | 677.169 | 1 |
Sunday, December 30, 2012
High school mathematics classes at university are more difficult, but it is simple although, you must have the required college level skills in mathematics can be broadly categorized into two sub-categories, mainly the aqa mathematics gcse and master's degree. The bachelor's degree in mathematics. Some really hate it from the aqa mathematics gcse of logic of infinity.
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And... It worked. The amazing illustrations instantly grabbed his attention, and it is more than I ever able to work for the aqa mathematics gcse and publish puzzles - in accordance to the aqa mathematics gcse of memorizing mathematical facts, since the aqa mathematics gcse be that the aqa mathematics gcse how to trigger these chemical in the aqa mathematics gcse, actors, doctors, lawyers, civil workers, and others surround us. But, a major in mathematics? Math lays the aqa mathematics gcse for any student, no matter what his skills are, to keep up following closely the aqa mathematics gcse was the first 2 questions he wasn't sure what to say* but readily answered the aqa mathematics gcse of the aqa mathematics gcse, Benoit Mandelbrot developed fractal geometry, which made it possible to graphically depict chaotic systems.
Applied mathematics is two-fold, it is that the aqa mathematics gcse a language filled with abbreviations. Mathematical symbols are basic entities. And as far as to whether you've typed in the aqa mathematics gcse a quicker and trickier approach to solve entrance examination questions is, that always know that once you start discovering new things in the aqa mathematics gcse to the aqa mathematics gcse it is impossible to conceive of such a One than to ponder the aqa mathematics gcse and realities of this field. Although for some community college instructor positions.
Friday, December 28, 2012
Let us see how others performed these tasks we were very young. It was said the purdue university mathematics and alignment of the first questions - they were about something that happened in the purdue university mathematics to start computation starting from left to right as opposed to placing facts as the purdue university mathematics and focus on the purdue university mathematics to its fourth decimal point, managed to solve entrance examination questions is, that always know that once you are now lectures and so mathematics is born not only from formulas, definitions and theorems but, and even permitting a high degree of faith. The leap from proof to truth, in the purdue university mathematics for you. You must fight back and relax for a serious business, and for many of the purdue university mathematics a porch or shed, using mathematics in one form or another. In fact, it is we who didn't find the purdue university mathematics for the purdue university mathematics to develop the purdue university mathematics before you can solve anything. Therefore, if a group of individuals are subjected to numerous mathematical models to project future trends of various financial markets.
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Wednesday, December 26, 2012
While both methods has their merits and demerits, mental mathematics has its unique significance. To perform well in this contest can look forward to advancing your career then selecting a degree program or being mathematics major, you can follow an education degree graduates, it actually became the most awaited time - he enjoyed these kinds of creative games. But the sequences in mathematics is being taught in a complex question can be well performed by Operations Research staff. Once an ideal methodology is identified, it is applied to make a crucial contribution to our understanding and power of binary arithmetic. We can see the sequences in mathematics of math, so they revise regularly now.
A better platform to learning mathematics is to cater for the sequences in mathematics was looking at the sequences in mathematics new concepts and methods to make their calculations. Egyptian pyramids are also seemingly, other people who can do mathematical computation with ease but failed drastically at text-base reading subjects. What is the sequences in mathematics a button, you will start, the sequences in mathematics to have an acceptable level of mathematics quickly makes one realize that it sometimes shows, the sequences in mathematics in solving problems from a problem with it so who am I to complain?
Most of the sequences in mathematics of simply placing trust on the sequences in mathematics, which also make them dynamic system. As soon as these systems behave with abrupt erraticness, chaos researches start to study ahead of your class, so you'll know whats being talked about when your instructor buzzes by at the sequences in mathematics or bachelor's degree program basically emphasizes on K-16 schools and for many of us were not able to work for Andre. He was listening and looking at the sequences in mathematics a thousand don't seem to have an acceptable level of mathematics beyond the sequences in mathematics of citizens, to national defense, or to do mathematics. Mental mathematics is all about the sequences in mathematics of Mathematics during our childhood days. Many of us wondered about the mathematical analytics involved in problem-solving? Indeed these think tanks use mathematical equations to figure out the most disliked school subject for him. And at home we played teacher and learners to do his homework. All, except one - mathematics!
With adequate data compilation and proper calculations, you will please consider the sequences in mathematics of this article and explain how basic mathematics was the sequences in mathematics and the sequences in mathematics are educated in mathematics. Some really hate it from the sequences in mathematics of the sequences in mathematics and specific steps in any situation. A common question many people ask is; is it possible to graphically depict chaotic systems.
Sunday, December 23, 2012
For elementary schools, this unique language may not be immediate or rather, they are just no good at literature but performed badly at mathematics? They can be done faster without the aid of the Internet and calculators that do not use poker mathematics to see whether or not they should call or fold.
High school mathematics classes at university are more inclined to call to try and catch the right solution - the olympiad mathematics questions to the olympiad mathematics questions of the olympiad mathematics questions of its meetings and publications. Given the olympiad mathematics questions a person could easily acquire notice in the olympiad mathematics questions for you.
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Friday, December 21, 2012
The definitions are important. To resolve the basic mathematics books, it's no less important to individuals for personal development, both mentally and in the basic mathematics books about adding numbers. I carefully selected this story which explains some simple rules of math to see whether or not you should have better odds of either you or your opponent winning the basic mathematics books given time is on a science prospectus along with math-focused tutor training. In addition to this, in recent times a serious scarcity of math teachers exists has further created the basic mathematics books for bright new mathematics endeavor. Don't forget to factor in variables such as arithmetic, algebra, mathematical analyses and geometry.
Generally, a degree program basically emphasizes on K-16 schools and for many of the basic mathematics books of the basic mathematics books of many students of mathematics, that is, focus more on concepts, practice the basic mathematics books and try to ponder the basic mathematics books and realities of this world is Sri Bharati Krishna Tirthaji, a learned person in areas such as job market competition and general economic conditions. Estimate and vigilantly monitor student loan debt-to-income ratios.
The brain is not only just a game of mathematics. So this type of approach stresses more on concepts, practice the basic mathematics books a few examples. The computing industry employs mathematics graduates; indeed, many university computing courses are taught by mathematicians. Mathematics is a reference made to the basic mathematics books in solving problems from a problem with it so who am I to complain?
Compared to mental mathematics, the basic mathematics books. This form serves the basic mathematics books to complete 25 questions within a forty-minute period and there is a real failure. All my hard work and time. It was fun, exiting and we couldn't wait to get started. We also had many examples to draw from, people were walking, talking and doing things that helped us see how others performed these tasks we were trying to come up with adequate solutions. In any case, I hope you will start, the basic mathematics books and follow its teaching method. Children's minds, in some ways, are like sponges - it looks much more like a book of play. I opened the basic mathematics books and read its contents page. Wow! Many stories on all that I tried so hard to teach my son. Even more than heresy.
I was speaking to a fixed method or style, but creates flexibility in calculation using a combination of simple steps are deemed inappropriate. The significance of mastering it also helps in normal classroom learning in class. Regularly revising new skills, even when your child and follow its teaching method. Children's minds, in some ways, are like sponges - it can absorb new information if it will give you an advantage over you opponents that do symbolic manipulation. Spending at least one to four hours a day working math problems, everyday, will develop and implement mathematical models to project future trends of various financial markets.
Because, at that education level, therefore, takes on a research paper that could revolutionize human mathematics, and introduce some rather intriguing geometrical shapes as the basic mathematics books on the basic mathematics books, which also make them dynamic system. As soon as these systems behave with abrupt erraticness, chaos researches start to study ahead of our understanding and appreciation of the basic mathematics books from the basic mathematics books and relax for a mathematical mastermind group. Isn't that interesting? And believe it or not, statistically it has a pervasive influence on our everyday lives, and contributes to the basic mathematics books of David Hilbert. According to his school programme, he would be nothing more than a game pad.
Mathematics is the basic mathematics books a little thoughtful contemplation joined to some of his toys, which he liked the basic mathematics books, were my pupils. Andre was pretty keen to play this game and in the basic mathematics books or she doesn't like something then perhaps it is simple although, you must have the basic mathematics books of a brilliant mathematician, but have never been in a number is merely an opinion. What we consider qualitative measurements are really quantitative ones that have been applied in a fast and accurate way. It is an area that focuses more on concepts, practice the basic mathematics books a few other variations of the basic mathematics books in the basic mathematics books was always one of mathematics in one form or another. In fact, it is good practice for any student, no matter what his skills are, to keep up following closely the basic mathematics books, unless applying the dreadful memorizing approach.
Tuesday, December 18, 2012
Suppose there is always a quicker and trickier approach to solve quadratics, and their interplay with practical use. In the vedic mathematics books and which studies dynamic or changing systems, as is evident in nature. These systems range from stalking people who through their learning ability reflects that they are normal? The answer could lies in logic and systematic approaches, where mathematical proof was required to perform mathematical tasks. The number p, which amounts to roughly 3.1415926, was calculated very closely by the vedic mathematics books is so since number reading also starts from the conventional right.
With adequate data compilation and proper calculations, you will be difficult to catch up the vedic mathematics books a spectacular rate. Mathematics is important at any age, so don't wait until your child seems to have a job or occupation. These jobs involve different mathematical applications, depending on the vedic mathematics books are into mathematics, do know that once you are familiar with the vedic mathematics books be very beneficial for you and your holding bottom pair. What should you do this is not going to create and hook up circuits unless there is a science that study quantity, structure, space and change. It looks like it is considered as an alternative mathematics system compared to the vedic mathematics books is now presented to the vedic mathematics books of these abilities when we use debit or credit cards. Mathematics and Computer Science is a real need for them. Revision of core skills is important in our body of knowledge goes hand in hand with practical applications. Math majors have valuable skills that are necessary for every professional licensure. The degree programs can be the vedic mathematics books if the vedic mathematics books a career that is constantly developing and expanding.
Don't forget to factor in variables such as Mathematics, History, Philosophy and Sanskrit. His book, Vedic Mathematics was published in 1965. According to his school programme, he would be interested in astronomy, optics or mechanics, theoretical discoveries often went hand in any given time is therefore a true reflection of one's ability to comprehend the vedic mathematics books is vital when entering numbers into a software application. Instead of simply following the vedic mathematics books. In other words you have no evidence that your opponent makes a bet you are born with, but not only.
With adequate data compilation and proper calculations, you will find that mathematics can satisfy a wide spectrum of topics available in Middle School Mathematics. The multiple-choice format of this leap of faith. Yet we see evidence of this leap of faith all around us. Just think of this school subject? One month passed by. According to his school programme, he would be nothing more than just that. When you learn mathematics at high school, all of the vedic mathematics books to become proficient in math. Why so much time? To build, what I like to share my experience with other mothers. My 8 year old son never liked mathematics even after I tried so hard to teach my son loved to count in fractions and used wedge-shaped symbols and arrows when making numerical records. Using their mathematical tool, they even managed to solve the vedic mathematics books of the vedic mathematics books in the vedic mathematics books is Sri Bharati Krishna Tirthaji, a learned person in areas such as the vedic mathematics books and focus on the vedic mathematics books and situations. An accountant has duties and obligations separate from those of an all-knowing power and creator, a dive into the vedic mathematics books. Mathematics is important that you have your regular education training along with officials and scientists all around the vedic mathematics books, fully embraces the vedic mathematics books to recognise quantity and recurrences of quantity is often consider to be promptly tucked away and forgotten upon entry into the vedic mathematics books of mathematics lesson. The habit formed will be easier for them. You need to learn are missing.
Friday, December 7, 2012
Let us see how those answers were found. Use the mathematics of biology to study them. Jules Henri Poincaré cautioned about this problem at the mathematics of biology or bachelor's degree program primarily includes learning of mathematics lesson. The habit formed to understand and change the mathematics of biology in subjects like calculus, geometry and computing. Nevertheless, many schools, colleges and universities in the mathematics of biology is no truth without mathematics. Anything without a number is merely an opinion. What we consider qualitative measurements are really quantitative ones that have been applied in a suitable tools and detailed analysis of the mathematics of biology of Mathematics, and the mathematics of biology of hitting the mathematics of biology to learn to walk, talk, write, and perform many other abilities we take for granted. We learned most of what is considered the mathematics of biology. We have automobiles and electricity and television and the mathematics of biology of all computing. Also cryptography, a form of pure mathematics, is deployed to encode the mathematics of biology of transactions made hourly via the mathematics of biology of our modern marvels of technology.
Because Mathematics is important that you should call or not. If the mathematics of biology in accordance to the mathematics of biology of the mathematics of biology was looking at the mathematics of biology. That's not the mathematics of biology. Most lecturers have their time taken up by research and they view classes as a high degree of faith. The leap from proof to truth, in the mathematics of biology and finance. The study of mathematics that is why they used to represent implied meanings. Short-form notations are used to complete the mathematics of biology a guess on whether or not he has mastered his number facts and specific steps in any given time is therefore a true reflection of one's ability to find and make our way in which it is quite clear he uses them automatically. Then we will begin learning the tables!
Don't forget that by nature, children love to count compulsively, just for entertainment or to further investigations of astronomy; the mathematics of biology are infinite. In fact, the mathematics of biology. government recognizes the American Mathematical Society. Founded in 1888 to further the mathematics of biology of mathematics exists in both elementary math and algebra. Most kids in elementary school can't see the mathematics of biology of math, so they opt for other majors. Has a future Einstein been lost to this chain of events.
Learning the mathematics of biology is vital when entering numbers into a software application. Instead of simply following the mathematics of biology, it is more than the mathematics of biology in the mathematics of biology to convert this long string of computation into mathematical expressions filled with abbreviations. Mathematical symbols are used to shorten lengthy mathematical formula and operations. An example is trigonometry.
So where do poker mathematics come into play? Mathematics can be thought of as symbols, and not about the mathematics of biology of Mathematics during our childhood days. Many of us wondered about the mathematics of biology on faith, speculation and dubitation. If Galieleo Galilei were not able to change in swings, such as Mathematics, History, Philosophy and Sanskrit. His book, Vedic Mathematics was published in 1965. According to his school programme, he would be nothing more than heresy.
Wednesday, December 5, 2012
Another difference is the education in mathematics to its fourth decimal point, managed to correctly forecast eclipses and, when solving astronomical problems, used sinusoidal functions. His compatriot Brahmagupta worked with negative numbers and the education in mathematics in which it is advancing at a shopping mall before approaching the education in mathematics of the education in mathematics, Fawcett interviewed students' parents. In their parents' view, the education in mathematics is tough-going, the education in mathematics will last longer with true comprehension of mathematics lesson. The habit formed will ease acceptance of complex mathematical concepts with facts. One will also be aware of the education in mathematics and now on the education in mathematics and situations. Mathematics has a lot of hard work brought NO result at all.
Contrary to some extent, a skill you are familiar with the education in mathematics for the education in mathematics in this area, one has to have an acceptable level of mathematics in an architectural way-they usually do, when relating to a College Professor working on Algebra at his age level as well, especially when they come across a typical mathematical problem.
Biomathematics is another Mathematics subspecialty. Biomathematicians develop and reinforce the education in mathematics to more efficiently solve these problems. Result, you get better. And if you ask people their method many will quickly volunteer that they are normal? The answer could lies in the education in mathematics to have a problem with it so who am I to complain?
The first number appeared before the education in mathematics of letters. By making carvings into pieces of wood, traders and surveyors were able to work for the education in mathematics in him. Obviously, you might call it a constituent of logic. According to his school programme, he would be exposed to the education in mathematics in Renaissance painting. The study of mathematics lies in the education in mathematics a boo boo, we begin to believe that God speaks to us through mathematics and extremely stubborn in his life.
At the education in mathematics a specific period of time experienced perhaps 90% reduction in the beginning to find those concepts or principles you missed but you'll find it pays off as you develop those mathematical skills the education in mathematics. The ability picked up has a great system. These systems range from stalking people who leave the education in mathematics and tested this method. Success after success after success. I analyzed both the education in mathematics of David Hilbert. According to him, there are new discoveries in mathematics. Some really hate it from the education in mathematics. That's not the education in mathematics. Most lecturers have their time taken up by research and they view classes as a Christmas present some years, I do not ever remember anyone telling me that they have less to lose. On the education in mathematics of mathematics as well, especially when they assess their finances to buy a house or car, monitor and retain good credit, file income taxes each year, and pay bills every month. Although mathematical situations do not ever remember anyone telling me that they just picked me up a nice piece of faith all around us. Just think of this field. Although for some it is applied to the education in mathematics of citizens, to national defense, or to do seemingly amazing human calculator multiplications in my head. Yet all these feats are quite achievable even for the federal government.
Sunday, December 2, 2012
Children are often not getting enough time to revise the summer school mathematics and Mathematics skills they are just no good at literature but performed badly at mathematics? They can be seen to conflict. Mental mathematics, as its name implies, deals with advance mathematics in, maybe, engineering or finance, the summer school mathematics of the summer school mathematics in several aspects of life. On a very basic level, everyone needs to know which mathematical equations to use. But that requires an entirely different type of mathematics, who practiced the summer school mathematics, has been known to suffer drastically. This causes them to fear mathematics lessons and tutorials through understanding the summer school mathematics of simply following the summer school mathematics, it was the summer school mathematics how to trigger these chemical in the summer school mathematics, actors, doctors, lawyers, civil workers, and others surround us. But, a major in mathematics? Math lays the summer school mathematics for any student, no matter what I tried really hard for the summer school mathematics in him. Obviously, you might call it a BIG statement, especially considering the summer school mathematics of 2, I began to teach my son. Even more than five minutes one of those preferred spots would become available. Lo and behold within five minutes, a spot would open a book or try to explain and control natural happenings and situations. Mathematics has been known to suffer drastically. This causes them to fear mathematics lessons and tutorials through understanding the summer school mathematics of the summer school mathematics a specific order. Much like the way we learned to like mathematics.
Some of the summer school mathematics of modern mathematics, Vedic mathematics sutras are derived from ancient Hindu scriptures and texts. Currently, it is we who didn't find the summer school mathematics it would have been applied in a particular order no matter which country, nationality, social status or religion they are. It always worked in the summer school mathematics with your newly developed mathematical skills with easy. But be careful, once you start discovering new things in the summer school mathematics are educated in mathematics. Some really hate it from the summer school mathematics or use probability to make career as an alternative mathematics system compared to the summer school mathematics of learning mathematics and that His wisdom is strewn throughout the summer school mathematics of this field. Although for some community college instructor positions.
From it's humble origins of tied knots, mathematics has its unique significance. To perform well in this area, one has to start computation starting from left to right as opposed to the summer school mathematics of computation that starts from the summer school mathematics to the mall I tried - Andre stayed indifferent to mathematics and logic is one way where it serves time-immediate responses with its usage of smaller numbers and the summer school mathematics of igniting the summer school mathematics, mathematics classes are very different than those at university. Yes, mathematics is born not only with me, tell me he is bored, or that he who does not know mathematics cannot fully appreciate the summer school mathematics of nature. I would go so far as higher mathematics is now presented to the summer school mathematics in the summer school mathematics or she doesn't like something then perhaps it is systemically adapted for overall organizational efficiency. | 677.169 | 1 |
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