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2 –3 –4 –5 –6 86. 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 87. 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 78. h(x) =# #1 __ x + 2 3 81. p(t) = −2 + 3t 84. r(x) = 4 88. 21 3 4 5 6 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 SECTION EXERCISES 307 NUMERIC For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data. x g(x) x k(x) 89. 92. 95. 5 13 x 2 f (x) −4 5 15 0 5 −10 −25 −40 10 10 28 4 16 20 58 6 36 25 73 8 56 90. 93. 96. x h(x) x g(x) x k(x) 0 5 5 30 10 105 15 230 2 0 6 6 −19 −44 −69 4 0 6 2 31 6 106 8 231 91. 94. 0 x f (x) −5 x h (x) 2 13 5 20 4 23 10 45 8 43 15 70 10 53 TECHNOLOGY For the following exercises, use a calculator or graphing technology to complete the task. 97. If f is a linear function, f (0.1) = 11.5, and f (0.4) = −5.9, find an equation for the function. 98. Graph the function f on a domain of [−10, 10] : f (x) = 0.02x − 0.01. Enter the function in a graphing utility. For the viewing window, set the minimum value of x to be −10 and the maximum value of x to be 10. 99. Graph the function f on a domain of [−10, 10] : f (x) = 2,500x + 4,
000 100. Table 3 shows the input, w, and output, k, for a linear function k. a. Fill in the missing values of the table. b. Write the linear function k, round to 3 decimal places. w −10 5.5 67.5 b k 30 −26 a −44 Table 3 101. Table 4 shows the input, p, and output, q, for a linear function q. a. Fill in the missing values of the table. b. Write the linear function k. p q 0.5 400 0.8 700 12 a b 1,000,000 Table 4 1 __ 102. Graph the linear function f on a domain of [−10, 10] for the function whose slope is and y-intercept is 8 31 __. 16 Label the points for the input values of −10 and 10. 103. Graph the linear function f on a domain of [−0.1, 0.1] for the function whose slope is 75 and y-intercept is −22.5. Label the points for the input values of −0.1 and 0.1. 104. Graph the linear function f where f (x) = ax + b on the same set of axes on a domain of [−4, 4] for the following values of a and b. a. a = 2; b = 3 b. a = 2; b = 4 c. a = 2; b = −4 d. a = 2; b = −5 EXTENSIONS 105. Find the value of x if a linear function goes through the following points and has the following slope: (x, 2), (−4, 6), m = 3 106. Find the value of y if a linear function goes through the following points and has the following slope: (10, y), (25, 100), m = −5 107. Find the equation of the line that passes through the following points: (a, b) and (a, b + 1) 108. Find the equation of the line that passes through the following points: (2a, b) and (a, b + 1) 109. Find the equation of the line that passes through the following points: (a, 0) and (c, d) 110. Find the equation of the line parallel to the line g(x) = −0.01x + 2.01 through the point (1, 2). Download the OpenStax text for free at http
://cnx.org/content/col11759/latest. 30 8 CHAPTER 4 LINEAR FUNCTIONS 111. Find the equation of the line perpendicular to the line g(x) = −0.01x + 2.01 through the point (1, 2). For the following exercises, use the functions f (x) = −0.1x + 200 and g(x) = 20x + 0.1. 112. Find the point of intersection of the lines f and g. 113. Where is f (x) greater than g (x)? Where is g (x) greater than f (x)? REAL-WORLD APPLICATIONS 114. At noon, a barista notices that she has $20 in her tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves n more customers during her shift? 115. A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session? 117. A phone company charges for service according to the formula: C(n) = 24 + 0.1n, where n is the number of minutes talked, and C(n) is the monthly charge, in dollars. Find and interpret the rate of change and initial value. 119. A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year. 121. Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: I(x) = 1,054x + 23,286, where x is the number of years after 1990. Which of the following interprets the slope in the context of the problem? a. As of 1990, average annual income was $23,286. b. In the ten-year period from 1990–1999, average annual income increased by a total of $1,054. c. Each year in the decade of the 1990s, average annual income increased by $1,054. d. Average annual income rose to a level of $23,286 by the end of 1999. 116. A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can charge per shirt
. In particular, historical data shows that 1,000 shirts can be sold at a price of $30, while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form p(n) = mn + b that gives the price p they can charge for n shirts. 118. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationship in the form y = mn + b that gives the yield when n stalks are planted. 120. A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, P(t), for the population t years after 2003. 122. When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C, the Celsius temperature, F (C). a. Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius. b. Find and interpret F (28). c. Find and interpret F (−40). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 MODELING WITH LINEAR FUNCTIONS 309 LEARNING OBJECTIVES In this section, you will: • Build linear models from verbal descriptions. • Model a set of data with a linear function. 4.2 MODELING WITH LINEAR FUNCTIONS Figure 1 (credit: EEK Photography/Flickr) Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models. Building Linear Models from Verbal Descriptions When building
linear models to solve problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them: 1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system. 2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value. 3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret. 4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem. 5. When needed, write a formula for the function. 6. Solve or evaluate the function using the formula. 7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically. 8. Clearly convey your result using appropriate units, and answer in full sentences when necessary. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 0 CHAPTER 4 LINEAR FUNCTIONS Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units. Output: M, money remaining, in dollars Input: t, time, in weeks So, the amount of money remaining depends on the number of weeks: M(t) We can also identify the initial value and the rate of change. Initial Value: She saved $3,500, so $3,500 is the initial value for M. Rate of Change: She anticipates spending $400 each week, so −$400 per week is the rate of change, or slope. Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week. The rate of change is constant, so we can start with the linear model M(t) = mt + b. Then we can substitute the intercept and slope provided. M(
t) = mt + b ↑ ↑ −400 3500 M(t) = − 400t + 3500 To find the x-intercept, we set the output to zero, and solve for the input. 0 = −400t + 3500 t = 3500 ____ 400 = 8.75 The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks. When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0 ≤ t ≤ 8.75. In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model. Using a Given Intercept to Build a Model Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is −$250 per month. We can then use the slopeintercept form and the given information to develop a linear model. Now we can set the function equal to 0, and solve for x to find the x-intercept. f (
x) = mx + b = −250x + 1000 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 MODELING WITH LINEAR FUNCTIONS 311 0 = −250x + 1000 1000 = 250x 4 = x x = 4 The x-intercept is the number of months it takes her to reach a balance of $0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan. Using a Given Input and Output to Build a Model Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output. How To… Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem. 1. Identify the input and output values. 2. Convert the data to two coordinate pairs. 3. Find the slope. 4. Write the linear model. 5. Use the model to make a prediction by evaluating the function at a given x-value. 6. Use the model to identify an x-value that results in a given y-value. 7. Answer the question posed. Example 1 Using a Linear Model to Investigate a Town’s Population A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues. a. Predict the population in 2013. b. Identify the year in which the population will reach 15,000. Solution The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2,000 years ago! To make computation a little nicer, we will define our input as the number of years since 2004: Input: t, years since 2004 Output: P(t), the town’s population To predict the population in 2013 (t = 9), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an
equation, we need the initial value and the rate of change, or slope. To determine the rate of change, we will use the change in output per change in input. m = change in output __ change in input The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t = 0, giving the point (0, 6200). Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to t = 5, giving the point (5, 8100). The two coordinate pairs are (0, 6200) and (5, 8100). Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 2 CHAPTER 4 LINEAR FUNCTIONS m = 8100 − 6200 __________ 5 − 0 = 1900 ____ 5 = 380 people per year We already know the y-intercept of the line, so we can immediately write the equation: P(t) = 380t + 6200 To predict the population in 2013, we evaluate our function at t = 9. P(9) = 380(9) + 6,200 = 9,620 If the trend continues, our model predicts a population of 9,620 in 2013. To fi nd when the population will reach 15,000, we can set P(t) = 15000 and solve for t. 15000 = 380t + 6200 8800 = 380t t ≈ 23.158 Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027. Try It #1 A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut. a. Write a linear model to repre sent the cost C of the company as a function of x, the number of doughnuts produced. b. Find and interpret the y-intercept. Try It #2 A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues. a. Predict
the population in 2014. b. Identify the year in which the population will reach 54,000. Using a Diagram to Model a Problem It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful. Example 2 Using a Diagram to Model Distance Walked Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact? Solution In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: “How long will it take them to be 2 miles apart?” Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 MODELING WITH LINEAR FUNCTIONS 313 In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables. Input: t, time in hours. Output: A(t), distance in miles, and E(t), distance in miles Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2. Anna walking east, 4 miles/hour Distance between them Emanuel walking south, 3 miles/hour Figure 2 Initial Value: They both start at the same intersection so when t = 0, the distance traveled by each person should also be 0. Thus the initial value for each is 0. Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A is 4 and the slope for E is 3. Using those values, we can
write formulas for the distance each person has walked. A(t) = 4t E(t) = 3t For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure. We can then define a third variable, D, to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure 3. Recall that we need to know how long it takes for D, the distance between them, to equal 2 miles. Notice that for any given input t, the outputs A(t), E(t), and D(t) represent distances. E A D Figure 3 Figure 3 shows us that we can use the Pythagorean Theorem because we have drawn a right angle. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 4 CHAPTER 4 LINEAR FUNCTIONS Using the Pythagorean Theorem, we get: D(t)2 = A(t)2 + E(t)2 = (4t)2 + (3t)2 = 16t2 + 9t2 = 25t2 D(t) = ± √ 25t2 — Solve for D(t) using the square root. In this scenario we are considering only positive values of t, so our distance D(t) will always be positive. We can simplify this answer to D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t) = 2 and solve for t. = ±5 \| t \| D(t) = 2 5t = 2 t =# #2 __ = 0.4 5 They will fall out of radio contact in 0
.4 hours, or 24 minutes. Q & A… Should I draw diagrams when given information based on a geometric shape? Yes. Sketch the figure and label the quantities and unknowns on the sketch. Example 3 Using a Diagram to Model Distance Between Cities There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough? Solution It might help here to draw a picture of the situation. See Figure 4. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates (30, 10), and Eastborough at (20, 0). Agritown (30, 10) (0, 0) Westborough (20, 0) Eastborough 20 miles Figure 4 Using this point along with the origin, we can find the slope of the line from Westborough to Agritown: m = 10 − 0 =# #1 __ ______ 30 − 0 3 Now we can write an equation to describe the road from Westborough to Agritown. W(x) =# #1 __ x 3 From this, we can determine the perpendicular road to Eastborough will have slope m = −3. Because the town of Eastborough is at the point (20, 0), we can find the equation: E(x) = −3x + b 0 = −3(20) + b b = 60 E(x) = −3x + 60 Substitute (20, 0) into the equation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 MODELING WITH LINEAR FUNCTIONS 315 We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal, 1 __ x = −3x + 60 3 10 __ x = 60 3 10x = 180 x = 18 y = W(18) = #1 __ (18) 3 = 6 Substitute this back into W(x). The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough
to the junction. distance#=# √ —— (x2 − x1)2 + (y2 − y1)2 — (18 − 0)2 + (6 − 0)2 = √ ≈ 18.974 miles Analysis One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points. Try It #3 There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson? Modeling a Set of Data with Linear Functions Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 5. Exactly one solution x g Infinitely many solutions y f (a) y y x g f x No solutions (c) (b) Figure 5 How To… Given a situation that represents a system of linear equations, write the system of equations and identify the solution. 1. Identify the input and output of each linear model. 2. Identify the slope and y-intercept of each linear model. 3. Find the solution by setting the two linear functions equal to one another and solving for x, or find the point of intersection on a graph. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 6 CHAPTER 4 LINEAR FUNCTIONS Example 4 Building a System of Linear Models to Choose a Truck Rental Company Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile[9]. When will Keep on Trucking, Inc. be the better choice for Jamal? Solution The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will
define two functions in Table 1. Input Outputs Initial Value Rate of Change d, distance driven in miles K(d): cost, in dollars, for renting from Keep on Trucking M(d) cost, in dollars, for renting from Move It Your Way Up-front fee: K(0) = 20 and M(0) = 16 K(d) = $0.59/mile and P(d) = $0.63/mile Table 1 A linear function is of the form f (x) = mx + b. Using the rates of change and initial charges, we can write the equations K(d) = 0.59d + 20 M(d) = 0.63d + 16 Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when K(d) < M(d). The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the K(d) function is smaller. These graphs are sketched in Figure 6, with K(d) in red. $ 120 110 100 90 80 70 60 50 40 30 20 10 0 K(d) = 0.59d + 20 (100, 80) M(d) = 0.63d + 16 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 d Figure 6 To find the intersection, we set the equations equal and solve: K(d) = M(d) 0.59d + 20 = 0.63d + 16 4 = 0.04d 100 = d d = 100 This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that K(d) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d > 100. Access this online resources for additional instruction and practice with linear function models. • Interpreting a Linear Function (http://openstaxcollege.org/l/interpretlinear) 9 Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ Download the OpenStax text for free at http://cn
x.org/content/col11759/latest. SECTION 4.2 SECTION EXERCISES 317 4.2 SECTION EXERCISES VERBAL 1. 2. 3. 4. ALGEBRAIC 5. yx=f x=+x f(x 6. x __ f(x=– x f(x 7. y 8. __ f(x=– x f(x xg x=f(x=x f x 9. 10. 11. 12. 13. 14. Ptt P P 16. P y 17. 18. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 8 CHAPTER 4 LINEAR FUNCTIONS : Th. Th s fi 19. 20. W, t W 21. W 22. W xy 23. 24. : Thffl ffl 25. C, t 26. C 27. C 28. C xy 29. 30. GRAPHICAL Figure 7fiy tt y Figure 7 t 31. yyt 32. y 33. x 34. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 SECTION EXERCISES 319 Figure 8fiy tt y t Figure 8 35. yyt 36. y 37. x 38. NUMERIC fl Table 2 Year Mississippi Table 2 Hawaii 39. 40. 41. ft r? (Th fl Table 3 Year Indiana Table 3 Alabama 42. 43. 44. ft r? (Th Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 0 CHAPTER 4 LINEAR FUNCTIONS REAL-WORLD APPLICATIONS 45. a. 46. a. b. b. c. d. e. P t c. d. e. Pt f. f. 47. ys a fl a. x 48. ys a fl a. x b. y c. b. y c. 49. a. 50. e 285. Th a. P P b. b. 51. Th a. Rt b. c. 52. a. Rt b. c. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 SECTION EXERCISES 321 53. o diff s. Th. Th
plus 54. o diff nies. Th Th 55. 56. 57. 58. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 2 CHAPTER 4 LINEAR FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Draw and interpret scatter plots. • Use a graphing utility to ﬁnd the line of best ﬁt. • Distinguish between linear and nonlinear relations. • Fit a regression line to a set of data and use the linear model to make predictions. 4.3 FITTING LINEAR MODELS TO DATA A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot. Drawing and Interpreting Scatter Plots A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure 1 shows a sample scatter plot. Final Exam Score vs. Age 100 90 80 70 60 50 40 30 20 10 10 20 30 40 50 Age Figure 1 A scatter plot of age and ﬁnal exam score variables. Notice this scatter plot does not indicate a linear relationship. The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam. Example 1 Using a Scatter Plot to Investigate Cricket Chirps Table 1 shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit[10]. Plot this data, and determine whether the data appears to be linearly related. Chirps Temperature 44 80.5 35 70.5 20.4 57 33 66 31 68 35 72 18.5 52 37 73.5 26 53 Table 1 Cricket Chirps vs Air Temperature Solution Plotting this data, as depicted in Figure 2 suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly
not perfectly so. 10 Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 FITTING LINEAR MODELS TO DATA 323 Cricket Chirps vs. Temperature ) 90 80 70 60 50 40 0 0 10 20 30 40 50 Cricket Chirps in 15 Seconds Figure 2 Finding the Line of Best Fit Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear function?” One way to approximate our linear function is to sketch the line that seems to best fit the data. Then we can extend the line until we can verify the y-intercept. We can approximate the slope of the line by extending it until we can estimate the rise _ run. Example 2 Finding a Line of Best Fit Find a linear function that fits the data in Table 1 by “eyeballing” a line that seems to fit. Solution On a graph, we could try sketching a line. Using the starting and ending points of our hand drawn line, points (0, 30) and (50, 90), this graph has a slope of m = = 1.2 60 __ 50 and a y-intercept at 30. This gives an equation of T(c) = 1.2c + 30 where c is the number of chirps in 15 seconds, and T(c) is the temperature in degrees Fahrenheit. The resulting equation is represented in Figure 3. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature 10 20 30 40 50 c, Number of Chirps Figure 3 Analysis This linear equation can then be used to approximate answers to various questions we might ask about the trend. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 4 CHAPTER 4 LINEAR FUNCTIONS Recognizing Interpolation or Extrapolation While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship will behave outside of the values for which we have data. We use a process known as interpolation when we predict a value inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain and range of
the data. Figure 4 compares the two processes for the cricket-chirp data addressed in Example 2. We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44. There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer applies after a certain point, it is sometimes called model breakdown. For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when x = 50, that is in 50 years, the model would not apply because we could not account for factors fifty years in the future. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature Extrapolation Interpolation 10 20 30 40 50 c, Number of Chirps Figure 4 Interpolation occurs within the domain and range of the provided data whereas extrapolation occurs outside. interpolation and extrapolation Different methods of making predictions are used to analyze data. The method of interpolation involves predicting a value inside the domain and/or range of the data. The method of extrapolation involves predicting a value outside the domain and/or range of the data. Model breakdown occurs at the point when the model no longer applies. Example 3 Understanding Interpolation and Extrapolation Use the cricket data from Table 1 to answer the following questions: a. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. b. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 FITTING LINEAR MODELS TO DATA 325 Solution a. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpol
ation. Using our model: T(30) = 30 + 1.2(30) = 66 degrees Based on the data we have, this value seems reasonable. b. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model: 40 = 30 + 1.2c 10 = 1.2c c ≈ 8.33 We can compare the regions of interpolation and extrapolation using Figure 5. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature Interpolation Extrapolation 10 20 30 40 50 c, Number of Chirps Figure 5 Analysis Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees. Try It #1 According to the data from Table 1, what temperature can we predict it is if we counted 20 chirps in 15 seconds? Finding the Line of Best Fit Using a Graphing Utility While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values[11]. One such technique is called least squares regression and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators[12]. Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression. 11 Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the data values. 12 For example, http://www.shodor.org/unchem/math/lls/leastsq.html Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 6 CHAPTER 4 LINEAR FUNCTIONS How To… Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression. 1. Enter the input in List 1 (L1). 2. Enter the output in List 2 (L2). 3. On a graphing utility, select Linear Regression (LinReg). Example 4 Finding a Least Squares Regression Line Find the least squares regression line using the cricket-
chirp data in Table 2. Solution 1. Enter the input (chirps) in List 1 (L1). 2. Enter the output (temperature) in List 2 (L2). See Table 2. L1 L2 44 80.5 35 70.5 20.4 57 33 66 31 68 35 72 18.5 52 37 73.5 26 53 Table 2 3. On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation: T(c) = 30.281 + 1.143c Analysis Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to: T(30) = 30.281 + 1.143(30) = 64.571 ≈ 64.6 degrees The graph of the scatter plot with the least squares regression line is shown in Figure 90 80 70 60 50 40 30 Number of Cricket Chirps vs. Temperature 0 10 20 30 40 50 c, Number of Chirps Figure 6 Q & A… Will there ever be a case where two different lines will serve as the best fit for the data? No. There is only one best fit line. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 FITTING LINEAR MODELS TO DATA 327 Distinguishing Between Linear and Non-Linear Models As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the correlation coefficient, which is a measure of how closely the line fits the data. Many graphing calculators require the user to turn a “diagnostic on” selection to find the correlation coefficient, which mathematicians label as r. The correlation coefficient provides an easy way to get an idea of how close to a line the data falls. We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless. To get a sense for the relationship between the value of r
and the graph of the data, Figure 7 shows some large data sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis shows the output. 1.0 0.8 0.4 0.0 −0.4 −0.8 −1.0 1.0 1.0 1.0 −1.0 −1.0 −1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 Figure 7 Plotted data and related correlation coefﬁcients. (credit: “DenisBoigelot,” Wikimedia Commons) correlation coefficient The correlation coefficient is a value, r, between −1 and 1. • r > 0 suggests a positive (increasing) relationship • r < 0 suggests a negative (decreasing) relationship • The closer the value is to 0, the more scattered the data. • The closer the value is to 1 or −1, the less scattered the data is. Example 5 Finding a Correlation Coefﬁcient Calculate the correlation coefficient for cricket-chirp data in Table 1. Solution Because the data appear to follow a linear pattern, we can use technology to calculate r. Enter the inputs and corresponding outputs and select the Linear Regression. The calculator will also provide you with the correlation coefficient, r = 0.9509. This value is very close to 1, which suggests a strong increasing linear relationship. Note: For some calculators, the Diagnostics must be turned “on” in order to get the correlation coefficient when linear regression is performed: [2nd]> [0]> [alpha][x − 1], then scroll to DIAGNOSTICSON. Fitting a Regression Line to a Set of Data Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that only one such line is a best fit for the data. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 8 CHAPTER 4 LINEAR FUNCTIONS Example 6 Using a Regression Line to Make Predictions Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in Table 3[13]. Determine whether the
trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008. Year ‘94 ‘95 ‘96 ‘97 ‘98 ‘99 ‘00 ‘01 ‘02 ‘03 ‘04 Consumption (billions of gallons) 113 116 118 119 123 125 126 128 131 133 136 The scatter plot of the data, including the least squares regression line, is shown in Figure 8. Table ( 150 140 130 120 110 100 0 Gas Consumption vs. Year 0 1 2 3 4 6 5 7 Years After 1994 8 9 10 11 12 13 14 Figure 8 Solution We can introduce a new input variable, t, representing years since 1994. The least squares regression equation is: C(t) = 113.318 + 2.209t Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend. Using this to predict consumption in 2008 (t = 14), The model predicts 144.244 billion gallons of gasoline consumption in 2008. C(14) = 113.318 + 2.209(14) = 144.244 Try It #2 Use the model we created using technology in Example 6 to predict the gas consumption in 2011. Is this an interpolation or an extrapolation? Access these online resources for additional instruction and practice with fitting linear models to data. • Introduction to Regression Analysis (http://openstaxcollege.org/l/introregress) • Linear Regression (http://openstaxcollege.org/l/linearregress) 13 http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 SECTION EXERCISES 329 4.3 SECTION EXERCISES VERBAL 1. Describe what it means if there is a model breakdown when using a linear model. 2. What is interpolation when using a linear model? 3. What is extrapolation when using a linear model? 4. Explain the difference between a positive and a negative correlation coefficient. 5. Explain how to interpret the absolute value of a correlation coefficient. ALGEBRAIC 6. A regression was run to determine whether there is 7. A regression was run to determine whether there is a a relationship between hours of TV watched per day
(x) and number of sit-ups a person can do (y). The results of the regression are given below. Use this to predict the number of situps a person who watches 11 hours of TV can do. y = ax + b a = −1.341 b = 32.234 r = −0.896 relationship between the diameter of a tree (x, in inches) and the tree’s age (y, in years). The results of the regression are given below. Use this to predict the age of a tree with diameter 10 inches. y = ax + b a = 6.301 b = −1.044 r = −0.970 For the following exercises, draw a scatter plot for the data provided. Does the data appear to be linearly related? 8. 10. 0 2 8 4 −22 −19 −15 −11 −6 6 10 −2 100 12 250 12.6 300 13.1 450 14 600 14.5 750 15.2 9. 11. 1 46 1 1 2 50 3 9 3 59 5 28 4 75 7 65 5 100 6 136 9 125 11 216 12. For the following data, draw a scatter plot. If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? Eyeball the line, and estimate the answer. 13. For the following data, draw a scatter plot. If we wanted to know when the temperature would reach 28°F, would the answer involve interpolation or extrapolation? Eyeball the line and estimate the answer. Year 2010 Population 11,500 12,100 12,700 13,000 13,750 2000 2005 1990 1995 Temperature, °F Time, seconds 16 46 18 50 20 54 25 55 30 62 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 0 CHAPTER 4 LINEAR FUNCTIONS GRAPHICAL For the following exercises, match each scatterplot with one of the four specified correlations in Figure 9 and Figure 10. (a) (b) Figure 9 14. r = 0.95 16. r = −0.26 (c) (d) Figure 10 15. r = −0.89 17. r = −0.39 For the following exercises, draw a best-fit line for the plotted data. 18. 12 10 8 6 4 2 0 19. 10 10 0 2 4 6 8 10 10 10 Download the Open
Stax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 SECTION EXERCISES 331 20. 10 10 NUMERIC 21. 10 10 22. The U.S. Census tracks the percentage of persons 25 years or older who are college graduates. That data for several years is given in Table 4[14]. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the percentage exceed 35%? Year Percent Graduates 1990 21.3 1992 21.4 1994 22.2 1996 23.6 1998 24.4 2000 25.6 2002 26.7 2004 27.7 2006 28 2008 29.4 Table 4 23. The U.S. import of wine (in hectoliters) for several years is given in Table 5. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will imports exceed 12,000 hectoliters? Year Imports 1992 2665 1994 2688 1996 3565 1998 4129 2000 4584 2002 5655 2004 6549 2006 7950 2008 8487 2009 9462 Table 5 24. Table 6 shows the year and the number of people unemployed in a particular city for several years. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the number of unemployed reach 5 people? Year Number Unemployed 1990 750 1992 670 1994 650 1996 605 Table 6 1998 550 2000 510 2002 460 2004 420 2006 380 2008 320 TECHNOLOGY For the following exercises, use each set of data to calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to 3 decimal places of accuracy. 25. 26. x y x y 8 23 5 4 15 41 7 12 26 53 10 17 31 72 12 22 56 103 15 24 14 http://www.census.gov/hhes/socdemo/education/data/cps/historical/index.html. Accessed 5/1/2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 2 CHAPTER 4 LINEAR FUNCTIONS 3 21.9 10 18.54 4 22.22 11 15.76 5 22.74 12 13.68 6 22.26 13 14.1 7 20.78 14 14.02 8 17.6 15 11.94 9 16.52 16 12.76 4 44
.8 5 43.1 6 38.8 7 39 8 38 9 32.7 10 30.1 11 29.3 12 27 13 25.8 21 17 100 2000 900 70 25 11 80 1798 988 80 30 2 60 1589 1000 82 31 −1 55 1580 1010 84 40 −18 40 1390 1200 105 50 −40 20 1202 1205 108 27. 28. 29. 30. 31 EXTENSIONS 32. Graph f (x) = 0.5x + 10. Pick a set of 5 ordered 33. Graph f (x) = −2x − 10. Pick a set of 5 ordered pairs using inputs x = −2, 1, 5, 6, 9 and use linear regression to verify that the function is a good fit for the data. pairs using inputs x = −2, 1, 5, 6, 9 and use linear regression to verify the function. For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs shows dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years: (46, 600), (48, 550), (50, 505), (52, 540), (54, 495). 34. Use linear regression to determine a function P 35. Find to the nearest tenth and interpret the where the profit in thousands of dollars depends on the number of units sold in hundreds. x-intercept. 36. Find to the nearest tenth and interpret the y-intercept. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 SECTION EXERCISES 333 REAL-WORLD APPLICATIONS For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population and the year over the ten-year span, (population, year) for specific recorded years: (2500, 2000), (2650, 2001), (3000, 2003), (3500, 2006), (4200, 2010) 37. Use linear regression to determine a function y, 38. Predict when the population will hit 8,000. where the year depends on the population. Round to three decimal places of accuracy. For the following exercises, consider this scenario: The profit of a company increased steadily over a ten-
year span. The following ordered pairs show the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years: (46, 250), (48, 305), (50, 350), (52, 390), (54, 410). 39. Use linear regression to determine a function y, 40. Predict when the profit will exceed one million where the profit in thousands of dollars depends on the number of units sold in hundreds. dollars. For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs show dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span (number of units sold, profit) for specific recorded years: (46, 250), (48, 225), (50, 205), (52, 180), (54, 165). 41. Use linear regression to determine a function y, 42. Predict when the profit will dip below the $25,000 where the profit in thousands of dollars depends on the number of units sold in hundreds. threshold. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 4 CHAPTER 4 LINEAR FUNCTIONS CHAPTER 4 REVIEW Key Terms correlation coefficient a value, r, between −1 and 1 that indicates the degree of linear correlation of variables, or how closely a regression line fits a data set. decreasing linear function a function with a negative slope: If f (x) = mx + b, then m < 0. extrapolation predicting a value outside the domain and range of the data horizontal line a line defined by f (x) = b, where b is a real number. The slope of a horizontal line is 0. increasing linear function a function with a positive slope: If f (x) = mx + b, then m > 0. interpolation predicting a value inside the domain and range of the data least squares regression a statistical technique for fitting a line to data in a way that minimizes the differences between the line and data values linear function a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line model breakdown when a model no longer applies after a certain point parallel lines two or more lines with the same slope perpendicular lines two lines that intersect at right angles and have slopes that are negative reciprocals of
each other point-slope form the equation for a line that represents a linear function of the form y − y1 = m(x − x1) slope the ratio of the change in output values to the change in input values; a measure of the steepness of a line slope-intercept form the equation for a line that represents a linear function in the form f (x) = mx + b vertical line a line defined by x = a, where a is a real number. The slope of a vertical line is undefined. Key Concepts 4.1 Linear Functions • Linear functions can be represented in words, function notation, tabular form, and graphical form. See Example 1. • An increasing linear function results in a graph that slants upward from left to right and has a positive slope. A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. A constant linear function results in a graph that is a horizontal line. See Example 2. • Slope is a rate of change. The slope of a linear function can be calculated by dividing the difference between y-values by the difference in corresponding x-values of any two points on the line. See Example 3 and Example 4. • An equation for a linear function can be written from a graph. See Example 5. • The equation for a linear function can be written if the slope m and initial value b are known. See Example 6 and Example 7. • A linear function can be used to solve real-world problems given information in different forms. See Example 8, Example 9, and Example 10. • Linear functions can be graphed by plotting points or by using the y-intercept and slope. See Example 11 and Example 12. • Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See Example 13. • The equation for a linear function can be written by interpreting the graph. See Example 14. • The x-intercept is the point at which the graph of a linear function crosses the x-axis. See Example 15. • Horizontal lines are written in the form, f (x) = b. See Example 16. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 4 REVIEW 335 • Vertical lines are written in the form, x = b. See Example 17. • Parallel lines
have the same slope. Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See Example 18. • A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x- and y-values of the given point into the equation, f (x) = mx + b, and using the b that results. Similarly, the point-slope form of an equation can also be used. See Example 19. • A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope. See Example 20 and Example 21. 4.2 Modeling with Linear Functions • We can use the same problem strategies that we would use for any type of function. • When modeling and solving a problem, identify the variables and look for key values, including the slope and y-intercept. See Example 1. • Draw a diagram, where appropriate. See Example 2 and Example 3. • Check for reasonableness of the answer. • Linear models may be built by identifying or calculating the slope and using the y-intercept. ◦ The x-intercept may be found by setting y = 0, which is setting the expression mx + b equal to 0. ◦ The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See Example 4. ◦ A graph of the system may be used to identify the points where one line falls below (or above) the other line. 4.3 Fitting Linear Models to Data • Scatter plots show the relationship between two sets of data. See Example 1. • Scatter plots may represent linear or non-linear models. • The line of best fit may be estimated or calculated, using a calculator or statistical software. See Example 2. • Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data. See Example 3. • The correlation coefficient, r, indicates the degree of linear relationship between data. See Example 4. • A regression line best fits the data. See Example 5. • The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables. See Example 6. Download the OpenStax
text for free at http://cnx.org/content/col11759/latest. 33 6 CHAPTER 4 LINEAR FUNCTIONS CHAPTER 4 REVIEW EXERCISES LINEAR FUNCTIONS 1. Determine whether the algebraic equation is linear. 2. Determine whether the algebraic equation is linear. 2x + 3y = 7 6x 2 − y = 5 3. Determine whether the function is increasing or 4. Determine whether the function is increasing or decreasing. f (x) = 7x − 2 decreasing. g(x) = −x + 2 5. Given each set of information, find a linear equation that satisfies the given conditions, if possible. Passes through (7, 5) and (3, 17) 6. Given each set of information, find a linear equation that satisfies the given conditions, if possible. x-intercept at (6, 0) and y-intercept at (0, 10) 7. Find the slope of the line shown in the graph. 8. Find the slope of the line shown in the graph. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 6 x 9. Write an equation in slope-intercept form for the line shown. 10. Does the following table represent a linear function? If so, find the linear equation that models the data. x g(x) –4 18 0 –2 2 10 –12 –52 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x 11. Does the following table represent a linear function? If so, find the linear equation that models the data. x 6 8 12 26 g(x) –8 –12 –18 –46 12. On June 1st, a company has $4,000,000 profit. If the company then loses 150,000 dollars per day thereafter in the month of June, what is the company’s profit nth day after June 1st? For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: 13
. 2x − 6y = 12 −x + 3y = 1 14. 1 __ x − 2 y = 3 3x + y = − 9 For the following exercises, find the x- and y-intercepts of the given equation 15. 7x + 9y = −63 16. f (x) = 2x − 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 4 REVIEW 337 For the following exercises, use the descriptions of the pairs of lines to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 17. Line 1: Passes through (5, 11) and (10, 1) Line 2: Passes through (−1, 3) and (−5, 11) 18. Line 1: Passes through (8, −10) and (0, −26) Line 2: Passes through (2, 5) and (4, 4) 19. Write an equation for a line perpendicular to f (x) = 5x − 1 and passing through the point (5, 20). 20. Find the equation of a line with a y-intercept of (0, 2) and slope − #1 __. 2 21. Sketch a graph of the linear function f (t) = 2t − 5. 22. Find the point of intersection for the 2 linear x = y + 6 2x − y = 13 functions: 23. A car rental company offers two plans for renting a car. Plan A: 25 dollars per day and 10 cents per mile Plan B: 50 dollars per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? MODELING WITH LINEAR FUNCTIONS 24. Find the area of a triangle bounded by the y-axis, the line f (x) = 10 − 2x, and the line perpendicular to f that passes through the origin. 25. A town’s population increases at a constant rate. In 2010 the population was 55,000. By 2012 the population had increased to 76,000. If this trend continues, predict the population in 2016. 26. The number of people afflicted with the common cold in the winter months dropped steadily by 50 each year since 2004 until 2010. In 2004, 875 people were inflicted. Find the linear function that models the number of people afflicted with the common cold C as a
function of the year, t. When will no one be afflicted? For the following exercises, use the graph in Figure 1 showing the profit, y, in thousands of dollars, of a company in a given year, x, where x represents years since 1980. y 12,000 10,000 8,000 6,000 4,000 2,000 0 5 10 15 20 25 30 x Figure 1 27. Find the linear function y, where y depends on x, the number of years since 1980. 28. Find and interpret the y-intercept. For the following exercise, consider this scenario: In 2004, a school population was 1,700. By 2012 the population had grown to 2,500. 29. Assume the population is changing linearly. a. How much did the population grow between the year 2004 and 2012? b. What is the average population growth per year? c. Find an equation for the population, P, of the school t years after 2004. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 33 8 CHAPTER 4 LINEAR FUNCTIONS For the following exercises, consider this scenario: In 2000, the moose population in a park was measured to be 6,500. By 2010, the population was measured to be 12,500. Assume the population continues to change linearly. 30. Find a formula for the moose population, P. 31. What does your model predict the moose population to be in 2020? For the following exercises, consider this scenario: The median home values in subdivisions Pima Central and East Valley (adjusted for inflation) are shown in Table 1. Assume that the house values are changing linearly. Year 1970 2010 Pima Central 32,000 85,000 Table 1 East Valley 120,250 150,000 32. In which subdivision have home values increased at a higher rate? 33. If these trends were to continue, what would be the median home value in Pima Central in 2015? FITTING LINEAR MODELS TO DATA 34. Draw a scatter plot for the data in Table 2. Then determine whether the data appears to be linearly related. 0 –105 2 –50 4 1 6 55 8 105 10 160 Table 2 35. Draw a scatter plot for the data in Table 3. If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? Year Population 1990 5,
600 1995 5,950 2000 6,300 2005 6,600 2010 6,900 Table 3 36. Eight students were asked to estimate their score on a 10-point quiz. Their estimated and actual scores are given in Table 4. Plot the points, then sketch a line that fits the data. Predicted Actual 6 6 7 7 7 8 8 8 Table 4 7 9 9 10 10 10 10 9 37. Draw a best-fit line for the plotted data. y 120 100 80 60 40 20 0 0 2 4 6 8 10 12 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 4 REVIEW 339 For the following exercises, consider the data in Table 5, which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year. Year Percent Graduates 2000 6.5 2002 7.0 Table 5 2005 7.4 2007 8.2 2010 9.0 38. Determine whether the trend appears to be linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places. 39. In what year will the percentage exceed 12%? 40. Based on the set of data given in Table 6, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. 41. Based on the set of data given in Table 7, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. x y 17 15 20 25 23 31 26 37 29 40 x y 10 36 12 34 15 30 18 28 20 22 Table 6 Table 7 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the population and the year over the ten-year span (population, year) for specific recorded years: (3,600, 2000); (4,000, 2001); (4,700, 2003); (6,000, 2006) 42. Use linear regression to determine a function y, 43. Predict when the population will hit 12,000. where the year depends on the population, to three decimal places of accuracy. 44. What is the correlation coefficient for this model to 45. According to the model, what is the population three decimal places of accuracy? in 2014? Download the OpenStax text for free at http
://cnx.org/content/col11759/latest. 34 0 CHAPTER 4 LINEAR FUNCTIONS CHAPTER 4 PRACTICE TEST 1. Determine whether the following algebraic equation can be written as a linear function. 2x + 3y = 7 2. Determine whether the following function is increasing or decreasing. f (x) = −2x + 5 3. Determine whether the following function is increasing or decreasing. f (x) = 7x + 9 4. Given the following set of information, find a linear equation satisfying the conditions, if possible. Passes through (5, 1) and (3, −9) 5. Given the following set of information, find a linear equation satisfying the conditions, if possible. x-intercept at (−4, 0) and y-intercept at (0, −6) 6. Find the slope of the line in Figure 1. 7. Write an equation for line in Figure 2. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 6 x Figure 1 Figure 2 8. Does Table 1 represent a linear function? If so, find 9. Does Table 2 represent a linear function? If so, find a linear equation that models the data. a linear equation that models the data. x −6 14 g(x) 0 32 2 38 4 44 Table 1 x g(x) 1 4 3 9 7 19 11 12 Table 2 10. At 6 am, an online company has sold 120 items that day. If the company sells an average of 30 items per hour for the remainder of the day, write an expression to represent the number of items that were sold n after 6 am. For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: 11. 3 __ x − 9 y = 4 −4x − 3y = 8 13. Find the x- and y-intercepts of the equation 2x + 7y = −14. 12. −2x + y = 3 3 __ y = 5 3x + 2 14. Given below are descriptions of two lines. Find the slopes of Line 1 and Line 2. Is
the pair of lines parallel, perpendicular, or neither? Line 1: Passes through (−2, −6) and (3, 14) Line 2: Passes through (2, 6) and (4, 14) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 4 PRACTICE TEST 341 15. Write an equation for a line perpendicular to f (x) = 4x + 3 and passing through the point (8, 10). 16. Sketch a line with a y-intercept of (0, 5) and slope − #5 __. 2 17. Graph of the linear function f (x) = −x + 6. 18. For the two linear functions, find the point of intersection: x = y + 2 2x − 3y = −1 19. A car rental company offers two plans for renting 20. Find the area of a triangle bounded by the y-axis, a car. Plan A: $25 per day and $0.10 per mile Plan B: $40 per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? 21. A town’s population increases at a constant rate. In 2010 the population was 65,000. By 2012 the population had increased to 90,000. Assuming this trend continues, predict the population in 2018. the line f (x) = 12 − 4x, and the line perpendicular to f that passes through the origin. 22. The number of people afflicted with the common cold in the winter months dropped steadily by 25 each year since 2002 until 2012. In 2002, 8,040 people were inflicted. Find the linear function that models the number of people afflicted with the common cold C as a function of the year, t. When will less than 6,000 people be afflicted? For the following exercises, use the graph in Figure 3, showing the profit, y, in thousands of dollars, of a company in a given year, x, where x represents years since 1980. y 35,000 30,000 25,000 20,000 15,000 10,000 5,000 0 5 10 15 20 25 30 x Figure 3 23. Find the linear function y, where y depends on x, the number of years since 1980. 24. Find and interpret the y-intercept. 25. In 2004, a school population was 1250. By 2012 the population had dropped to 8
75. Assume the population is changing linearly. a. How much did the population drop between the year 2004 and 2012? b. What is the average population decline per year? c. Find an equation for the population, P, of the school t years after 2004. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 2 CHAPTER 4 LINEAR FUNCTIONS 26. Draw a scatter plot for the data provided in Table 3. Then determine whether the data appears to be linearly related. 0 2 −450 −200 4 10 6 265 8 500 10 755 Table 3 27. Draw a best-fit line for the plotted data. 35 30 25 20 15 10 5 0 y 0 2 4 6 8 10 12 x For the following exercises, use Table 4 which shows the percent of unemployed persons 25 years or older who are college graduates in a particular city, by year. Year Percent Graduates 2000 8.5 2002 8.0 Table 4 2005 7.2 2007 6.7 2010 6.4 28. Determine whether the trend appears linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places. 29. In what year will the percentage drop below 4%? 30. Based on the set of data given in Table 5, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient. Round to three decimal places of accuracy. x y 16 106 18 110 20 115 24 120 26 125 Table 5 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population (in hundreds) and the year over the ten-year span, (population, year) for specific recorded years: (4,500, 2000); (4,700, 2001); (5,200, 2003); (5,800, 2006) 31. Use linear regression to determine a function y, where the year depends on the population. Round to three decimal places of accuracy. 32. Predict when the population will hit 20,000. 33. What is the correlation coefficient for this model? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 5 Polynomial and Rational Functions Figure 1 35-mm ﬁlm, once the standard for capturing photographic images, has been made largely
obsolete by digital photography. (credit “ﬁlm”: modiﬁcation of work by Horia Varlan; credit “memory cards”: modiﬁcation of work by Paul Hudson) CHAPTER OUTLINE 5.1 Quadratic Functions 5.2 Power Functions and Polynomial Functions 5.3 Graphs of Polynomial Functions 5.4 Dividing Polynomials 5.5 Zeros of Polynomial Functions 5.6 Rational Functions 5.7 Inverses and Radical Functions 5.8 Modeling Using Variation Introduction Digital photography has dramatically changed the nature of photography. No longer is an image etched in the emulsion on a roll of film. Instead, nearly every aspect of recording and manipulating images is now governed by mathematics. An image becomes a series of numbers, representing the characteristics of light striking an image sensor. When we open an image file, software on a camera or computer interprets the numbers and converts them to a visual image. Photo editing software uses complex polynomials to transform images, allowing us to manipulate the image in order to crop details, change the color palette, and add special effects. Inverse functions make it possible to convert from one file format to another. In this chapter, we will learn about these concepts and discover how mathematics can be used in such applications. 343 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Gie the characteristics of parabolas. • Graph a quadratic function. • etermine a quadratic functions minimum or maimum alue. • ole problems inoling a quadratic function. 5.1 QUADRATIC FUNCTIONS Figure 1 An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr) Curved antennas, such as the ones shown in Figure 1 are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function. In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than
working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior. Recognizing Characteristics of Parabolas The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. These features are illustrated in Figure 2. y 6 4 2 Axis of symmetry x-intercepts –6 –4 –2 2 4 6 x –2 y–intercept 4 –6 Vertex Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 345 The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses the x-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of x at which y = 0. Example 1 Identifying the Characteristics of a Parabola Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown in Figure 3. y 10 8 6 4 2 –4 –2 2 4 6 8 x Figure 3 Solution The vertex is the turning point of the graph. We can see that the vertex is at (3, 1). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is x = 3. This parabola does not cross the x-axis, so it has no zeros. It crosses the y-axis at (0, 7) so this is the y-intercept. Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions The general form of a quadratic function presents the function in the form f(x) = ax 2 + bx + c where
a, b, and c are real numbers and a ≠ 0. If a > 0, the parabola opens upward. If a < 0, the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry. b2 − 4ac −b ± √ The axis of symmetry is defined by x = − # b _______________ __, to solve 2a 2a ax 2 + bx + c = 0 for the x-intercepts, or zeros, we find the value of x halfway between them is always x = − # b __ 2a equation for the axis of symmetry.. If we use the quadratic formula, x =, the — Figure 4 represents the graph of the quadratic function written in general form as y = x2 + 4x + 3. In this form, a = 1, b = 4, and c = 3. Because a > 0, the parabola opens upward. The axis of symmetry is x = − # 4 ____ = −2. This 2(1) also makes sense because we can see from the graph that the vertical line x = −2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, (−2, −1). The x-intercepts, those points where the parabola crosses the x-axis, occur at (−3, 0) and (−1, 0). y y = x2 + 4x + 3 8 6 4 2 –4 –6 Vertex Axis of symmetry –2 –4 Figure 4 x-intercepts x 2 4 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS The standard form of a quadratic function presents the function in the form f(x) = a(x − h)2 + k where (h, k) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function. As with the general form, if a > 0, the parabola opens upward and the vertex is a minimum. If a < 0, the parabola opens downward,
and the vertex is a maximum. Figure 5 represents the graph of the quadratic function written in standard form as y = −3(x + 2)2 + 4. Since x − h = x + 2 in this example, h = −2. In this form, a = −3, h = −2, and k = 4. Because a < 0, the parabola opens downward. The vertex is at (−2, 4). Vertex y 4 2 y = −3(x + 2)2 + 4 –6 –4 –2 2 4 6 x –2 –4 –6 –8 Figure 5 The standard form is useful for determining how the graph is transformed from the graph of y = x 2. Figure 6 is the graph of this basic function. y 10 8 6 4 2 y = x2 –6 –4 –2 2 4 6 x –2 Figure 6 If k > 0, the graph shifts upward, whereas if k < 0, the graph shifts downward. In Figure 5, k > 0, so the graph is shifted 4 units upward. If h > 0, the graph shifts toward the right and if h < 0, the graph shifts to the left. In Figure 5, h < 0, so the graph is shifted 2 units to the left. The magnitude of a indicates the stretch of the graph. If.a. > 1, the point associated with a particular x-value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if.a. < 1, the point associated with a particular x-value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure 5,.a. > 1, so the graph becomes narrower. The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form. a(x − h)2 + k = ax 2 + bx + c ax 2 − 2ahx + (ah2 + k) = ax 2 + bx + c For the linear terms to be equal, the coefficients must be equal. −2ah = b, so h = − # b __. 2a Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUN
CTIONS 347 This is the axis of symmetry we defined earlier. Setting the constant terms equal: ah2 + k = c k = c − ah2 = c − a ( b2 __ 4a = c − 2 b ) __ 2a In practice, though, it is usually easier to remember that k is the output value of the function when the input is h, so f(h) = k. forms of quadratic functions A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola. The general form of a quadratic function is f(x) = ax 2 + bx + c where a, b, and c are real numbers and a ≠ 0. The standard form of a quadratic function is f(x) = a(x − h)2 + k where a ≠ 0. The vertex (h, k) is located at h = − #b __ 2a, k = f(h) = f ( −b ). ___ 2a How To… Given a graph of a quadratic function, write the equation of the function in general form. 1. Identify the horizontal shift of the parabola; this value is h. Identify the vertical shift of the parabola; this value is k. 2. Substitute the values of the horizontal and vertical shift for h and k. in the function f(x) = a(x − h)2 + k. 3. Substitute the values of any point, other than the vertex, on the graph of the parabola for x and f (x). 4. Solve for the stretch factor,.a.. 5. If the parabola opens up, a > 0. If the parabola opens down, a < 0 since this means the graph was reflected about the x-axis. 6. Expand and simplify to write in general form. Example 2 Writing the Equation of a Quadratic Function from the Graph Write an equation for the quadratic function g in Figure 7 as a transformation of f(x) = x 2, and then expand the formula, and simplify terms to write the equation in general form. y 6 4 2 –6 –4 –2 2 4 x –2 –4 Figure 7 Solution We can see the graph of g is the graph of f(x) = x 2 shifted to the left 2 and down 3, giving a
formula in the form g(x) = a(x −(−2))2 − 3 = a(x + 2)2 − 3. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Substituting the coordinates of a point on the curve, such as (0, −1), we can solve for the stretch factor. −1 = a(0 + 2)2 − 3 2 = 4a a =# #1 __ 2 1 __ (x + 2)2 − 3. In standard form, the algebraic model for this graph is g (x) = 2 To write this in general polynomial form, we can expand the formula and simplify terms. 1 __ (x + 2)2 −#3 g(x) = 2 1 __ = (x + 2)(x + 2) −#3 2 1 __ = (x 2 + 4x + 4) −#3 2 1 __ = x 2 + 2x + 2 −#3 2 1 __ = x 2 + 2x − 1 2 Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions. 1 __ (x + 2) 2 − 3. Next, Analysis We can check our work using the table feature on a graphing utility. First enter Y1 = 2 select TBLSET, then use TblStart = − 6 and ΔTbl = 2, and select TABLE. See Table 1. x y −6 5 −4 −1 −2 −3 0 −1 2 5 Table 1 The ordered pairs in the table correspond to points on the graph. Try It #1 A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 8 Find an equation for the path of the ball. Does the shooter make the basket? Figure 8 (credit: modiﬁcation of work by Dan Meyer) How To… Given a quadratic function in general form, find the vertex of the parabola. 1. Identify a, b, and c. 2. Find h, the x-coordinate of the vertex, by substituting a and b into h = − b ). 3. Find k, the y-coordinate of the vertex, by evaluating
k = f (h) = f ( − __ 2a b __. 2a Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 349 Example 3 Finding the Vertex of a Quadratic Function Find the vertex of the quadratic function f (x) = 2x 2 − 6x + 7. Rewrite the quadratic in standard form (vertex form). Solution The horizontal coordinate of the vertex will be at h = − # b __ 2a −6 ____ 2(2) = − 6 __ = 4 3 __ = 2 k = f(h) The vertical coordinate of the vertex will be at 3 ) = f ( __ 2 3 ) = 2 ( __ 2 5 __ = 2 Rewriting into standard form, the stretch factor will be the same as the a in the original quadratic. First, find the horizontal coordinate of the vertex. Then find the vertical coordinate of the vertex. Substitute the values into standard form, using the “a” from the general form. 3 ) + 7 − 6 ( __ 2 2 f (x) = ax2 + bx + c f (x) = 2x2 − 6x + 7 The standard form of a quadratic function prior to writing the function then becomes the following: 3 ) f (x) = 2 ( x − __ 2 2 5 __ + 2 Analysis One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, k, and where it occurs, x. Try It #2 Given the equation g (x) = 13 + x 2 − 6x, write the equation in general form and then in standard form. Finding the Domain and Range of a Quadratic Function Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up or down. domain and range of
a quadratic function The domain of any quadratic function is all real numbers unless the context of the function presents some restrictions. The range of a quadratic function written in general form f(x) = ax2 + bx + c with a positive a value is ), ∞#). ), or [ f ( − #b f (x) ≥ f ( − #b __ __ 2a 2a ) ]. ), or ( −∞, f ( − #b The range of a quadratic function written in general form with a negative a value is f (x) ≤ f ( − #b __ __ 2a 2a The range of a quadratic function written in standard form f (x) = a(x − h)2 + k with a positive a value is f (x) ≥ k; the range of a quadratic function written in standard form with a negative a value is f (x) ≤ k. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS How To… Given a quadratic function, find the domain and range. 1. Identify the domain of any quadratic function as all real numbers. 2. Determine whether a is positive or negative. If a is positive, the parabola has a minimum. If a is negative, the parabola has a maximum. 3. Determine the maximum or minimum value of the parabola, k. 4. If the parabola has a minimum, the range is given by f(x) ≥ k, or [k, ∞). If the parabola has a maximum, the range is given by f (x) ≤ k, or (−∞, k]. Example 4 Finding the Domain and Range of a Quadratic Function Find the domain and range of f (x) = −5x2 + 9x − 1. Solution As with any quadratic function, the domain is all real numbers. Because a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x-value of the vertex. h = − # b __ 2a = − # 9 _____ 2(−5) = 9 __ 10 The maximum value is given by f (h). 9 f (
__ 10 2 9 ) ) = −5 ( __ 10 61 ___ 20 = 9 + 9 ( __ 10 ) − 1 The range is f (x) ≤ 61 ___ 20, or ( −∞, 61 ]. ___ 20 Try It #3 Find the domain and range of f (x) = 2 ( x −# #4 ) __ 7 2 + 8 __. 11 Determining the Maximum and Minimum Values of Quadratic Functions The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in Figure 9. y f (x) = (x − 2)2 + 1 6 4 2 (2, 1) (–3, 4) y 6 4 2 g(x) = −(x + 3)2 + 4 –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –2 –4 –6 (a) Minimum value of 1 occurs at x = 2 Figure 9 Minimum value of 4 occurs at x = −3 –2 –4 –6 (b) There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 351 Example 5 Finding the Maximum Value of a Quadratic Function A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side. a. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L. b. What dimensions should she make her garden to maximize the enclosed area? Solution Let’s use a diagram such as Figure 10 to record the given information. It is also helpful to introduce a temporary variable, W, to represent the width of the garden and the length of the fence section parallel to the backyard fence. Garden L W Backyard Figure 10 a. We know we have only 80 feet of fence available, and L + W + L = 80, or more simply, 2L + W = 80. This allows us to represent the width, W, in terms
of L. W = 80 − 2L Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so A = LW = L(80 − 2L) A(L) = 80L − 2L2 This formula represents the area of the fence in terms of the variable length L. The function, written in general form, is A(L) = −2L2 + 80L. b. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since a is the coefficient of the squared term, a = −2, b = 80, and c = 0. To find the vertex: h = − # b _ 2a h = − # 80 _ 2(−2) k = A(20) = 80(20) − 2(20)2 = 20 and = 800 The maximum value of the function is an area of 800 square feet, which occurs when L = 20 feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet. Analysis This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 11. (20, 800) A y 1000 900 800 700 600 500 400 300 200 100 ) A ( a e r A 0 10 30 20 Length (L) Figure 11 40 x 50 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS How To… Given an application involving revenue, use a quadratic equation to find the maximum. 1. Write a quadratic equation for a revenue function. 2. Find the vertex of the quadratic equation. 3. Determine the y-value of the vertex. Example 6 Finding Maximum Revenue The unit price of an item affects its supply and demand. That is, if the unit price goes up,
the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue? Solution Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, p for price per subscription and Q for quantity, giving us the equation Revenue = pQ. Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently p = 30 and Q = 84,000. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, p = 32 and Q = 79,000. From this we can find a linear equation relating the two quantities. The slope will be m =# 79,000 − 84,000 _____________ 32 − 30 = −5,000 _______ 2 = −2,500 This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept. Q = −2,500p + b Substitute in the point Q = 84,000 and p = 30 84,000 = −2,500(30) + b Solve for b b = 159,000 This gives us the linear equation Q = −2,500p + 159,000 relating cost and subscribers. We now return to our revenue equation. Revenue = pQ Revenue = p(−2,500p + 159,000) Revenue = −2,500p2 + 159,000p We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex. h = − 159,000 _________ 2(−2,500) = 31.8 The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function. maximum revenue = −2,500(31.8)2 + 159,000(31.8) = 2,528,
100 Analysis This could also be solved by graphing the quadratic as in Figure 12. We can see the maximum revenue on a graph of the quadratic function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 353 (31.80, 2,528.1) y 3,000 2,500 2,000 1,500 1,000 500 ) 10 20 30 40 50 60 70 80 x Price (p) Figure 12 Finding the x- and y-Intercepts of a Quadratic Function Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y-intercept of a quadratic by evaluating the function at an input of zero, and we find the x-intercepts at locations where the output is zero. Notice in Figure 13 that the number of x-intercepts can vary depending upon the location of the graph. y 5 4 3 2 1 –3 –2 –1 –1 –2 –3 y 5 4 3 2 1 –1 –1 –2 –3 21 3 4 5 x –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 21 3 4 5 x 21 3 4 5 x –3 –2 No x-intercept One x-intercept Two x-intercepts Figure 13 Number of x-intercepts of a parabola How To… Given a quadratic function f (x), find the y- and x-intercepts. 1. Evaluate f (0) to find the y-intercept. 2. Solve the quadratic equation f (x) = 0 to find the x-intercepts. Example 7 Finding the y- and x-Intercepts of a Parabola Find the y- and x-intercepts of the quadratic f(x) = 3x2 + 5x − 2. Solution We find the y-intercept by evaluating f (0). So the y-intercept is at (0, −2). For the x-intercepts, we find all solutions of f(x) = 0. 0 = 3x2 + 5x − 2 f(0) = 3(0)2 + 5(0) − 2 = −
2 In this case, the quadratic can be factored easily, providing the simplest method for solution. 1 __, 0 ) and (−2, 0). So the x-intercepts are at ( 3 0 = (3x − 1)(x + 2) 0 = 3x − 1 1 __ x = 3 0 = x + 2 or x = −2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analysis By graphing the function, we can confirm that the graph crosses the y-axis at (0, −2). We can also confirm that the graph crosses the x-axis 1 __, 0 ) and (−2, 0). See Figure 14. at ( 3 y (–2, 0) 1 f (x) = 3 x2 + 5 x − 2, 0) ) 1 3 –3 –2 –1 1 2 3 x (0, –2) –1 –2 –3 –4 –5 Figure 14 Rewriting Quadratics in Standard Form In Example 7, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form. How To… Given a quadratic function, find the x-intercepts by rewriting in standard form. 1. Substitute a and b into h = − b __. 2a 2. Substitute x = h into the general form of the quadratic function to find k. 3. Rewrite the quadratic in standard form using h and k. 4. Solve for when the output of the function will be zero to find the x-intercepts. Example 8 Finding the x-Intercepts of a Parabola Find the x-intercepts of the quadratic function f (x) = 2x2 + 4x − 4. Solution We begin by solving for when the output will be zero. 0 = 2x2 + 4x − 4 Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form. We know that a = 2. Then we solve for h and k. f (x) = a(x − h)2 + k So now we can rewrite in standard form.
h = − # b __ 2a = − # 4 ___ 2(2) = −1 k = f (−1) = 2(−1)2 + 4(−1) − 4 = −6 f(x) = 2(x + 1)2 − 6 We can now solve for when the output will be zero. 0 = 2(x + 1)2 − 6 6 = 2(x + 1)2 3 = (x + 1)2 x + 1 = ± √ — 3 The graph has x-intercepts at (−1 − √ 3, 0) and (−1 + √ — — 3 x = −1 ± √ 3, 0). — Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 QUADRATIC FUNCTIONS 355 We can check our work by graphing the given function on a graphing utility and observing the x-intercepts. See Figure 15. y 6 4 2 (0.732, 0) (−2.732, 0) –6 –4 –2 2 4 6 x –2 –4 –6 Figure 15 Analysis We could have achieved the same results using the quadratic formula. Identify a = 2, b = 4, and c = −4. x = x = — b2 − 4ac −b ± √ __ 2a — 42 − 4(2)(−4) −4 ± √ ___ 2(2) — x = −4 ± √ 48 __ 4 x = — −4 ± √ 3(16) __ 4 x = −1 ± √ — 3 So the x-intercepts occur at (−1 − √ — 3, 0) and (−1 + √ — 3, 0). Try It #4 In a separate Try It, we found the standard and general form for the function g(x) = 13 + x2 − 6x. Now find the y- and x-intercepts (if any). Example 9 Applying the Vertex and x-Intercepts of a Parabola A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H(t) = −16t 2 + 80t + 40. a. When does the ball reach the maximum height
? b. What is the maximum height of the ball? c. When does the ball hit the ground? Solution a. The ball reaches the maximum height at the vertex of the parabola. h = − # 80 ______ 2(−16) = 80 ___ 32 5 __ = 2 = 2.5 The ball reaches a maximum height after 2.5 seconds. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS b. To find the maximum height, find the y-coordinate of the vertex of the parabola. k = H ( − # b ) __ 2a = H(2.5) = −16(2.5)2 + 80(2.5) + 40 = 140 The ball reaches a maximum height of 140 feet. c. To find when the ball hits the ground, we need to determine when the height is zero, H(t) = 0. We use the quadratic formula. t = — −80 ± √ _______________________ 802 − 4(−16)(40) 2(−16) = — −80 ± √ 8960 _____________ −32 Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions. t = — 8960 −80 − √ _____________ −32 ≈ 5.458 or t = — 8960 −80 + √ _____________ −32 ≈ −0.458 The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 16. H 150 125 100 75 50 25 (2.5, 140) H(t) =− 16 t2 + 80t + 40 1 2 3 4 Figure 16. 5 6 t Notice that the graph does not represent the physical path of the ball upward and downward. Keep quantities on each axis in mind while interpreting the graph. Try It #5 A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation H(t) = −16t 2 + 96t + 112. a. When does the rock reach the maximum height? b. What is the maximum height of the rock? c. When does the
rock hit the ocean? Access these online resources for additional instruction and practice with quadratic equations. • Graphing Quadratic Functions in General Form (http://openstaxcollege.org/l/graphquadgen) • Graphing Quadratic Functions in Standard Form (http://openstaxcollege.org/l/graphquadstan) • Quadratic Function Review (http://openstaxcollege.org/l/quadfuncrev) • Characteristics of a Quadratic Function (http://openstaxcollege.org/l/characterquad) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 SECTION EXERCISES 357 5.1 SECTION EXERCISES VERBAL 1. Explain the advantage of writing a quadratic 2. How can the vertex of a parabola be used in solving function in standard form. real-world problems? 3. Explain why the condition of a ≠ 0 is imposed in the definition of the quadratic function. 4. What is another name for the standard form of a quadratic function? 5. What two algebraic methods can be used to find the horizontal intercepts of a quadratic function? ALGEBRAIC For the following exercises, rewrite the quadratic functions in standard form and give the vertex. 6. f (x) = x 2 − 12x + 32 7. g(x) = x 2 + 2x − 3 8. f (x) = x 2 − x 9. f (x) = x 2 + 5x − 2 10. h(x) = 2x 2 + 8x − 10 11. k(x) = 3x 2 − 6x − 9 12. f (x) = 2x 2 − 6x 13. f (x) = 3x 2 − 5x − 1 For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry. 14. y(x) = 2x 2 + 10x + 12 15. f(x) = 2x 2 − 10x + 4 17. f(x) = 4x 2 + x − 1 18. h(t) = −4t 2 + 6t − 1 16. f(x) = −x 2 + 4x + 3 1 __ x 2 + 3
x + 1 19. f(x) = 2 20. f(x) = − #1 __ x 2 − 2x + 3 3 For the following exercises, determine the domain and range of the quadratic function. 21. f(x) = (x − 3)2 + 2 22. f(x) = −2(x + 3)2 − 6 23. f(x) = x 2 + 6x + 4 24. f(x) = 2x 2 − 4x + 2 25. k(x) = 3x 2 − 6x − 9 For the following exercises, use the vertex (h, k) and a point on the graph (x, y) to find the general form of the equation of the quadratic function. 26. (h, k) = (2, 0), (x, y) = (4, 4) 27. (h, k) = (−2, −1), (x, y) = (−4, 3) 28. (h, k) = (0, 1), (x, y) = (2, 5) 29. (h, k) = (2, 3), (x, y) = (5, 12) 30. (h, k) = (−5, 3), (x, y) = (2, 9) 31. (h, k) = (3, 2), (x, y) = (10, 1) 32. (h, k) = (0, 1), (x, y) = (1, 0) 33. (h, k) = (1, 0), (x, y) = (0, 1) GRAPHICAL For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. 34. f(x) = x 2 − 2x 35. f(x) = x 2 − 6x − 1 36. f(x) = x 2 − 5x − 6 37. f(x) = x 2 − 7x + 3 38. f(x) = −2x 2 + 5x − 8 39. f(x) = 4x 2 − 12x − 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 35 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS For the
following exercises, write the equation for the graphed function. 42. y y 5 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4 –5 40. 436 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6 –7 NUMERIC 416 –5 –4 –3 –2 –1 –1 –2 1 2 3 4 x 44. y 45. 7 6 5 4 3 2 1 –3 –2 –1 –1 –2 –2 –1 –1 –7 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function. 46. x −2 −1 y 5 2 49. x −2 −1 y −8 − 47. x −2 −1 y 1 0 50. x −2 − TECHNOLOGY For the following exercises, use a calculator to find the answer. 48. x −2 −1 y −2 1 0 2 1 2 1 −2 51. Graph on the same set of axes the functions 1 __ f (x) = x2, f(x) = 2x 2, and f(x) = x 2. What appears 3 to be the effect of changing the coefficient? 53. Graph on the same set of axes f(x) = x 2, f(x) = (x − 2)2, f(x − 3)2, and f(x) = (x + 4)2. What appears to be the effect of adding or subtracting those numbers? 52. Graph on the same set of axes f(x) = x 2, f(x) = x 2 + 2 and f(x) = x 2, f(x) = x 2 + 5 and f(x) = x 2 − 3. What appears to be the effect of adding a constant? 54. The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given −32 ____ by the function h(x) = (80)2 x 2 + x where x is the horizontal distance traveled and h(x) is the height in feet. Use the [TRACE] feature of your calculator to
determine the height of the object when it has traveled 100 feet away horizontally. 55. A suspension bridge can be modeled by the quadratic function h(x) = 0.0001x 2 with −2000 ≤ x ≤ 2000 where ∣ x ∣ is the number of feet from the center and h(x) is height in feet. Use the [TRACE] feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.1 SECTION EXERCISES 359 EXTENSIONS For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function. 56. Vertex (1, −2), opens up. 58. Vertex (−5, 11), opens down. 57. Vertex (−1, 2) opens down. 59. Vertex (−100, 100), opens up. For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function. 60. Contains (1, 1) and has shape of f(x) = 2x 2. 61. Contains (−1, 4) and has the shape of f(x) = 2x 2. Vertex is on the y-axis. Vertex is on the y-axis. 62. Contains (2, 3) and has the shape of f(x) = 3x 2. 63. Contains (1, −3) and has the shape of f(x) = −x 2. Vertex is on the y-axis. Vertex is on the y-axis. 64. Contains (4, 3) and has the shape of f(x) = 5x 2. 65. Contains (1, −6) has the shape of f(x) = 3x 2. Vertex is on the y-axis. Vertex has x-coordinate of −1. REAL-WORLD APPLICATIONS 66. Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing. 67. Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing. 68. Find the dimensions of the rectangular corral producing the greatest enclosed area split into 3 pens of
the same size given 500 feet of fencing. 69. Among all of the pairs of numbers whose sum is 6, find the pair with the largest product. What is the product? 70. Among all of the pairs of numbers whose difference is 12, find the pair with the smallest product. What is the product? 71. Suppose that the price per unit in dollars of a cell phone production is modeled by p = $45 − 0.0125x, where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x ċ p. Find the production level that will maximize revenue. 72. A rocket is launched in the air. Its height, in meters above sea level, as a function of time, in seconds, is given by h(t) = −4.9t 2 + 229t + 234. Find the maximum height the rocket attains. 73. A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by h(t) = −4.9t2 + 24t + 8. How long does it take to reach maximum height? 74. A soccer stadium holds 62,000 spectators. With a 75. A farmer finds that if she plants 75 trees per acre, ticket price of $11, the average attendance has been 26,000. When the price dropped to $9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue? each tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS G C F. and. In this section, you will: • Identify the degree and leading coefficient of polynomial functions. • dentify the end behaior of polynomial functions. • Gien a graph utilie the relationship between turning points and degree of a polynomial to determine the least possible degree. • Use factoring to find eros of a polynomial. • dentify eros and their multiplicity. • Graph polynomial functions using eros multipl
icity and end behaior • rite a polynomial function that satisfies certain criteria. 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown in Table 1. Figure 1 (credit: Jason Bay, Flickr) Year Bird Population 2009 800 2010 897 Table 1 2011 992 2012 1,083 2013 1,169 The population can be estimated using the function P(t) = −0.3t3 + 97t + 800, where P(t) represents the bird population on the island t years after 2009. We can use this model to estimate the maximum bird population and when it will occur. We can also use this model to predict when the bird population will disappear from the island. In this section, we will examine functions that we can use to estimate and predict these types of changes. Identifying Power Functions Before we can understand the bird problem, it will be helpful to understand a different type of function. A power function is a function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number. (A number that multiplies a variable raised to an exponent is known as a coefficient.) As an example, consider functions for area or volume. The function for the area of a circle with radius r is A(r) = πr 2 and the function for the volume of a sphere with radius r is V(r) =# #4 __ πr 3 3 Both of these are examples of power functions because they consist of a coefficient, π or 4 _ 3 π, multiplied by a variable r raised to a power. power function A power function is a function that can be represented in the form f (x) = kx p where k and p are real numbers, and k is known as the coefficient. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 361 Q & A… Is f (x) = 2x a power function? No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to a variable power. This is called an exponential function, not a power function. Example 1 Identifying Power Functions Which of the following functions are power functions? f(x) = 1
f (x) = x f (x) = x 2 f (x) = x 3 f (x) =# #1 __ x 1 __ x2 — f (x) = √ x — x f (x) = f (x) = 3 √ Constant function Identify function Quadratic function Cubic function Reciprocal function Reciprocal squared function Square root function Cube root function Solution All of the listed functions are power functions. The constant and identity functions are power functions because they can be written as f (x) = x0 and f (x) = x1 respectively. The quadratic and cubic functions are power functions with whole number powers f (x) = x 2 and f (x) = x 3. The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written as f (x) = x −1 and f (x) = x −2. The square and cube root functions are power functions with fractional powers because they can be written as f (x) = x1/2 or f (x) = x 1/3. Try It #1 Which functions are power functions? f (x) = 2x 2 ċ 4x 3 g(x) = −x 5 + 5x 3 h(x) = 2x5 − 1_ 3x2 + 4 Identifying End Behavior of Power Functions Figure 2 shows the graphs of f (x) = x 2, g(x) = x 4 and h(x) = x 6, which are all power functions with even, whole-number powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin. f (x) = x6 y g(x) = x4 h(x) = x2 4 3 2 1 –2 –1 1 2 x Figure 2 Even-power functions Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS To describe the behavior as numbers become larger and larger, we use the idea of in fi nity. We use the symbol ∞ for positive infinity and −∞ for negative infinity. When we say that “x approaches infinity,” which can
be symbolically written as x → ∞, we are describing a behavior; we are saying that x is increasing without bound. With the positive even-power function, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that as x approaches positive or negative infinity, the f (x) values increase without bound. In symbolic form, we could write Figure 3 shows the graphs of f (x) = x 3, g(x) = x 5, and h(x) = x 7, which are all power functions with odd, whole-number powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the graphs flatten near the origin and become steeper away from the origin. as x → ±∞, f (x) → ∞ g(x) = x5 f (x) = x3 h(x) = x7 1 2 x y 4 2 –2 –4 –2 –1 Figure 3 Odd-power functions These examples illustrate that functions of the form f (x) = xn reveal symmetry of one kind or another. First, in Figure 2 we see that even functions of the form f (x) = xn, n even, are symmetric about the y-axis. In Figure 3 we see that odd functions of the form f (x) = xn, n odd, are symmetric about the origin. For these odd power functions, as x approaches negative infinity, f (x) decreases without bound. As x approaches positive infinity, f (x) increases without bound. In symbolic form we write as x → −∞, f (x) → −∞ as x → ∞, f (x) → ∞ The behavior of the graph of a function as the input values get very small (x → −∞) and get very large (x → ∞) is referred to as the end behavior of the function. We can use words or symbols to describe end behavior. Figure 4 shows the end behavior of power functions in the form f (x) = kxn where n is a non-negative integer depending on the power and the constant. Even power y Odd power y Positive constant k > 0 x x x → −∞, f (x) → ∞ and x → ∞, f (x) → ∞ y x →
−∞, f (x) → ∞ and x → ∞, f (x) → −∞ y Negative constant k < 0 x x x → −∞, f (x) → ∞ and x → ∞, f (x) → −∞ x → −∞, f (x) → −∞ and x → ∞, f (x) → −∞ Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 363 How To… Given a power function f (x) = kx n where n is a non-negative integer, identify the end behavior. 1. Determine whether the power is even or odd. 2. Determine whether the constant is positive or negative. 3. Use Figure 4 to identify the end behavior. Example 2 Identifying the End Behavior of a Power Function Describe the end behavior of the graph of f (x) = x8. Solution The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As x approaches infinity, the output (value of f (x)) increases without bound. We write as x → ∞, f (x) → ∞. As x approaches negative infinity, the output increases without bound. In symbolic form, as x → −∞, f (x) → ∞. We can graphically represent the function as shown in Figure 5. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 21 3 4 5 x Example 3 Identifying the End Behavior of a Power Function Describe the end behavior of the graph of f (x) = −x 9. Figure 5 Solution The exponent of the power function is 9 (an odd number). Because the coefficient is −1 (negative), the graph is the reflection about the x-axis of the graph of f (x) = x 9. Figure 6 shows that as x approaches infinity, the output decreases without bound. As x approaches negative infinity, the output increases without bound. In symbolic form, we would write as x → −∞, f (x) → ∞ as x → ∞, f (x) → −∞ 1 –1 –2 –3 –4 –5 –6 –7
–8 –5 –4 –3 –2 21 3 4 5 x Figure 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analysis We can check our work by using the table feature on a graphing utility. x −10 −5 0 5 10 f (x) 1,000,000,000 1,953,125 0 −1,953,125 −1,000,000,000 Table 2 We can see from Table 2 that, when we substitute very small values for x, the output is very large, and when we substitute very large values for x, the output is very small (meaning that it is a very large negative value). Try It #2 Describe in words and symbols the end behavior of f (x) = −5x 4. Identifying Polynomial Functions An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. We want to write a formula for the area covered by the oil slick by combining two functions. The radius r of the spill depends on the number of weeks w that have passed. This relationship is linear. We can combine this with the formula for the area A of a circle. A(r) = πr 2 Composing these functions gives a formula for the area in terms of weeks. r(w) = 24 + 8w Multiplying gives the formula. A(w) = A(r(w)) = A(24 + 8w) = π(24 + 8w)2 A(w) = 576π + 384πw + 64πw 2 This formula is an example of a polynomial function. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. polynomial functions Let n be a non-negative integer. A polynomial function is a function that can be written in the form f (x) = an x n +... + a2 x 2 + a1 x + a0 This is called the general form of a polynomial function. Each ai is a coefficient and can be any
real number other than zero. Each expression ai x i is a term of a polynomial function. Example 4 Identifying Polynomial Functions Which of the following are polynomial functions? f (x) = 2x 3 ċ 3x + 4 g(x) = −x(x2 − 4) h(x) = 5 √ — x + 2 Solution The first two functions are examples of polynomial functions because they can be written in the form f (x) = an xn +... + a2 x2 + a1 x + a0, where the powers are non-negative integers and the coefficients are real numbers. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 365 • f (x) can be written as f (x) = 6x4 + 4. • g(x) can be written as g(x) = −x3 + 4x. • h(x) cannot be written in this form and is therefore not a polynomial function. Identifying the Degree and Leading Coefﬁcient of a Polynomial Function Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order of power, or in general form. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading term is the term containing the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term. terminology of polynomial functions We often rearrange polynomials so that the powers are descending. Leading coefficient Degree f (x) = an x n + … + a2 x 2 + a1x + a0 Leading term When a polynomial is written in this way, we say that it is in general form. How To… Given a polynomial function, identify the degree and leading coefficient. 1. Find the highest power of x to determine the degree function. 2. Identify the term containing the highest power of x to find the leading term. 3
. Identify the coefficient of the leading term. Example 5 Identifying the Degree and Leading Coefﬁcient of a Polynomial Function Identify the degree, leading term, and leading coefficient of the following polynomial functions. f (x) = 3 + 2x 2 − 4x 3 g(t) = 5t 5 − 2t 3 + 7t h(p) = 6p − p 3 − 2 Solution For the function f (x), the highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, −4x3. The leading coefficient is the coefficient of that term, −4. For the function g(t), the highest power of t is 5, so the degree is 5. The leading term is the term containing that degree, 5t5. The leading coefficient is the coefficient of that term, 5. For the function h(p), the highest power of p is 3, so the degree is 3. The leading term is the term containing that degree, −p3; the leading coefficient is the coefficient of that term, −1. Try It #3 Identify the degree, leading term, and leading coefficient of the polynomial f (x) = 4x 2 − x 6 + 2x − 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Identifying End Behavior of Polynomial Functions Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the power function consisting of the leading term. See Table 3. Polynomial Function Leading Term Graph of Polynomial Function f (x) = 5x4 + 2x3 − x − 4 5x4 –5 –4 –3 –2 f (x) = −2x6 − x5 + 3x4 + x3 −2x6 –5 –4 –3 –2 f (x) = 3x5 − 4x4 +
2x2 + 1 3x5 –5 –4 –3 –2 f (x) = −6x3 + 7x2 + 3x + 1 −6x3 –5 –4 –3 –2 Table 3 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 x 21 3 4 5 x 21 3 4 5 x 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 367 Example 6 Identifying End Behavior and Degree of a Polynomial Function Describe the end behavior and determine a possible degree of the polynomial function in Figure 7. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 7 Solution As the input values x get very large, the output values f (x) increase without bound. As the input values x get very small, the output values f (x) decrease without bound. We can describe the end behavior symbolically by writing as x → −∞, as x → ∞, f (x) → −∞ f (x) → ∞ In words, we could say that as x values approach infinity, the function values approach infinity, and as x values approach negative infinity, the function values approach negative infinity. We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive. Try It #4 Describe the end behavior, and determine a possible degree of the polynomial function in Figure 8. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 8 Example 7 Identifying End Behavior and Degree of a Polynomial Function Given the function f (x) = −3x2(x −
1)(x + 4), express the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function. Solution Obtain the general form by expanding the given expression for f (x). f (x) = −3x2(x − 1)(x + 4) = −3x2(x2 + 3x − 4) = −3x4 − 9x3 + 12x2 The general form is f (x) = −3x4 − 9x3 + 12x2. The leading term is −3x4; therefore, the degree of the polynomial is 4. The degree is even (4) and the leading coefficient is negative (−3), so the end behavior is as x → −∞, as x → ∞, f (x) → −∞ f (x) → −∞ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Try It #5 Given the function f (x) = 0.2(x − 2)(x + 1)(x − 5), express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function. Identifying Local Behavior of Polynomial Functions In addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the function values change from increasing to decreasing or decreasing to increasing. We are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial is a function, only one output value corresponds to each input value so there can be only one y-intercept (0, a0). The x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one x-intercept. See Figure 9. y Turning point x-intercepts x Turning point ←y-intercept Figure 9 intercepts and turning points of polynomial functions A turning point of a graph is a point at which
the graph changes direction from increasing to decreasing or decreasing to increasing. The y-intercept is the point at which the function has an input value of zero. The x-intercepts are the points at which the output value is zero. How To… Given a polynomial function, determine the intercepts. 1. Determine the y-intercept by setting x = 0 and finding the corresponding output value. 2. Determine the x-intercepts by solving for the input values that yield an output value of zero. Example 8 Determining the Intercepts of a Polynomial Function Given the polynomial function f (x) = (x − 2)(x + 1)(x − 4), written in factored form for your convenience, determine the y- and x-intercepts. Solution The y-intercept occurs when the input is zero so substitute 0 for x. The y-intercept is (0, 8). f (0) = (0 − 2)(0 + 1)(0 − 4) = (−2)(1)(− 4) = 8 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 369 The x-intercepts occur when the output is zero. 0 = (x − 2)(x + 1)(x − 4) The x-intercepts are (2, 0), (–1, 0), and (4, 0). x = 2 or x − 2 = 0 or x + 1 = 0 x = −1 or or x − 4 = 0 x = 4 We can see these intercepts on the graph of the function shown in Figure 10. y-intercept (0, 85 –4 –3 –2 21 3 –1 –1 –2 –3 –4 Figure 10 x 4 5 x-intercepts (−1, 0), (2, 0), and (4, 0) Example 9 Determining the Intercepts of a Polynomial Function with Factoring Given the polynomial function f (x) = x4 − 4x2 − 45, determine the y- and x-intercepts. Solution The y-intercept occurs when the input is zero. The y-intercept is (0, −45). f (0) = (0)4 − 4(0)2 − 45 = −45 The x
-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial. f (x) = x4 − 4x2 − 45 = (x2 − 9)(x2 + 5) = (x − 3)(x + 3)(x2 + 5) 0 = (x − 3)(x + 3)(x2 + 5) x − 3 = 0 or x + 3 = 0 or x2 + 5 = 0 x = 3 or x = −3 or (no real solution) The x-intercepts are (3, 0) and (−3, 0). We can see these intercepts on the graph of the function shown in Figure 11. We can see that the function is even because f (x) = f (−x). y 120 100 80 60 40 20 –2 –1 –20 –40 –60 –80 –100 –120 –5 –4 –3 x-intercepts (−3, 0) and (3, 0) x 21 3 4 5 y-intercept (0, −45) Try It #6 Given the polynomial function f (x) = 2x3 − 6x2 − 20x, determine the y- and x-intercepts. Figure 11 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Comparing Smooth and Continuous Graphs The degree of a polynomial function helps us to determine the number of x-intercepts and the number of turning points. A polynomial function of nth degree is the product of n factors, so it will have at most n roots or zeros, or x-intercepts. The graph of the polynomial function of degree n must have at most n − 1 turning points. This means the graph has at most one fewer turning point than the degree of the polynomial or one fewer than the number of factors. A continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The graphs of polynomial functions are both continuous and smooth. intercepts and turning points of polynomials A polynomial of degree n will have, at most,
n x-intercepts and n − 1 turning points. Example 10 Determining the Number of Intercepts and Turning Points of a Polynomial Without graphing the function, determine the local behavior of the function by finding the maximum number of x-intercepts and turning points for f (x) = −3x 10 + 4x 7 − x 4 + 2x 3. Solution The polynomial has a degree of 10, so there are at most 10 x-intercepts and at most 9 turning points. Try It #7 Without graphing the function, determine the maximum number of x-intercepts and turning points for f (x) = 108 − 13x9 − 8x4 + 14x12 + 2x3 Example 11 Drawing Conclusions about a Polynomial Function from the Graph What can we conclude about the polynomial represented by the graph shown in Figure 12 based on its intercepts and turning points? y 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 12 Solution The end behavior of the graph tells us this is the graph of an even-degree polynomial. See Figure 13. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 x-intercepts 21 3 4 5 x Turning points Figure 13 The graph has 2 x-intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.2 POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS 371 Try It #8 What can we conclude about the polynomial represented by the graph shown in Figure 14 based on its intercepts and turning points? y –5 –4 –3 –2 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x Figure 14 Example 12 Drawing Conclusions about a Polynomial Function from the Factors Given the function f (x) = −4x(x + 3)(x − 4), determine the local behavior. Solution The y-intercept is found by evaluating f (0). f (0) = −4(0)(
0 + 3)(0 − 4) = 0 The y-intercept is (0, 0). The x-intercepts are found by determining the zeros of the function. 0 = −4x(x + 3)(x − 4) x = 0 or x + 3 = 0 or x − 4 = 0 x = 0 or x = −3 or x = 4 The x-intercepts are (0, 0), (−3, 0), and (4, 0). The degree is 3 so the graph has at most 2 turning points. Try It #9 Given the function f (x) = 0.2(x − 2)(x + 1)(x − 5), determine the local behavior. Access these online resources for additional instruction and practice with power and polynomial functions. • Find Key Information About a Given Polynomial Function (http://openstaxcollege.org/l/keyinfopoly) • End Behavior of a Polynomial Function (http://openstaxcollege.org/l/endbehavior) • Turning Points and x-intercepts of Polynomial Functions (http://openstaxcollege.org/l/turningpoints) • Least Possible Degree of a Polynomial Function (http://openstaxcollege.org/l/leastposdegree) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.2 SECTION EXERCISES VERBAL 1. Explain the difference between the coefficient 2. If a polynomial function is in factored form, what of a power function and its degree. 3. In general, explain the end behavior of a power function with odd degree if the leading coefficient is positive. 5. What can we conclude if, in general, the graph of a polynomial function exhibits the following end behavior? As x → −∞, f (x) → −∞ and as x → ∞, f (x) → −∞. ALGEBRAIC would be a good first step in order to determine the degree of the function? 4. What is the relationship between the degree of a polynomial function and the maximum number of turning points in its graph? For the following exercises, identify the function as a power function, a polynomial function, or neither. 6. f (x) =
x5 9. f (x) = x2 _____ x2 − 1 7. f (x) = (x2)3 8. f (x) = x − x4 10. f (x) = 2x(x + 2)(x − 1)2 11. f (x) = 3x + 1 For the following exercises, find the degree and leading coefficient for the given polynomial. 12. −3x 15. x(4 − x2)(2x + 1) 13. 7 − 2x2 16. x 2 (2x − 3)2 14. −2x2 − 3x5 + x − 6 For the following exercises, determine the end behavior of the functions. 17. f (x) = x4 20. f (x) = −x9 18. f (x) = x3 19. f (x) = −x4 21. f (x) = −2x4 − 3x2 + x − 1 22. f (x) = 3x2 + x − 2 23. f (x) = x2(2x3 − x + 1) 24. f (x) = (2 − x)7 For the following exercises, find the intercepts of the functions. 25. f (t) = 2(t − 1)(t + 2)(t − 3) 26. g(n) = −2(3n − 1)(2n + 1) 27. f (x) = x4 − 16 28. f (x) = x3 + 27 29. f (x) = x(x2 − 2x − 8) 30. f (x) = (x + 3)(4x2 − 1) GRAPHICAL For the following exercises, determine the least possible degree of the polynomial function shown. 31. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 32. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 33. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5
.2 SECTION EXERCISES 373 34. 37. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 –5 –4 –3 –2 35. 21 3 4 5 x –5 –4 –3 –2 38. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 36. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x 21 3 4 5 x For the following exercises, determine whether the graph of the function provided is a graph of a polynomial function. If so, determine the number of turning points and the least possible degree for the function. 42. y x 39. 43. y y y y 40. 44. x x 41. 45. x x y y x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS NUMERIC For the following exercises, make a table to confirm the end behavior of the function. 46. f (x) = −x3 49. f (x) = (x − 1)(x − 2)(3 − x) 47. f (x) = x4 − 5x2 50. f (x) = x5 __ 10 − x4 48. f (x) = x2(1 − x)2 TECHNOLOGY For the following exercises, graph the polynomial functions using a calculator. Based on the graph, determine the intercepts and the end behavior. 51. f (x) = x3(x − 2) 52. f (x) = x(x − 3)(x + 3) 53. f (x) = x(14 − 2x)(10 − 2x) 54. f (x) = x(14 − 2x)(10 − 2x)2 55. f (x) = x3 − 16x 56. f (x) = x3 − 27 58. f (x) = −x
3 + x2 + 2x 59. f (x) = x3 − 2x2 − 15x 57. f (x) = x4 − 81 60. f (x) = x3 − 0.01x EXTENSIONS For the following exercises, use the information about the graph of a polynomial function to determine the function. Assume the leading coefficient is 1 or −1. There may be more than one correct answer. 61. The y-intercept is (0, −4). The x-intercepts are (−2, 0), (2, 0). Degree is 2. End behavior: as x → −∞, f (x) → ∞, as x → ∞, f (x) → ∞. 62. The y-intercept is (0, 9). The x-intercepts are (−3, 0), (3, 0). Degree is 2. End behavior: as x → −∞, f (x) → −∞, as x → ∞, f (x) → −∞. 63. The y-intercept is (0, 0). The x-intercepts are (0, 0), (2, 0). Degree is 3. End behavior: as x → −∞, f (x) → −∞, as x → ∞, f (x) → ∞. 64. The y-intercept is (0, 1). The x-intercept is (1, 0). Degree is 3. End behavior: as x → −∞, f (x) → ∞, as x → ∞, f (x) → −∞. 65. The y-intercept is (0, 1). There is no x-intercept. Degree is 4. End behavior: as x → −∞, f (x) → ∞, as x → ∞, f (x) → ∞. REAL-WORLD APPLICATIONS For the following exercises, use the written statements to construct a polynomial function that represents the required information. 66. An oil slick is expanding as a circle. The radius of the circle is increasing at the rate of 20 meters per day. Express the area of the circle as a function of d, the number of days elapsed. 67. A cube has an edge of 3 feet. The edge is increasing at the rate of 2 feet per minute.
Express the volume of the cube as a function of m, the number of minutes elapsed. 68. A rectangle has a length of 10 inches and a width of 6 inches. If the length is increased by x inches and the width increased by twice that amount, express the area of the rectangle as a function of x. 69. An open box is to be constructed by cutting out square corners of x-inch sides from a piece of cardboard 8 inches by 8 inches and then folding up the sides. Express the volume of the box as a function of x. 70. A rectangle is twice as long as it is wide. Squares of side 2 feet are cut out from each corner. Then the sides are folded up to make an open box. Express the volume of the box as a function of the width (x). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 375 G C F.. In this section, you will: • dentify the degree and leading coefficient of polynomial functions. • dentify the end behaior of polynomial functions. • Gien a graph utilie the relationship between turning points and degree of a polynomial to determine the least possible degree. • Use factoring to find eros of a polynomial. • dentify eros and their multiplicity. • Graph polynomial functions using eros multiplicity and end behaior • rite a polynomial function that satisfies certain criteria. 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 1. Year 2006 2007 2008 2009 2010 2011 2012 2013 Revenues 52.4 52.8 51.2 49.5 48.6 48.6 48.7 47.1 The revenue can be modeled by the polynomial function Table 1 R(t) = −0.037t4 + 1.414t3 − 19.777t2 + 118.696t − 205.332 where R represents the revenue in millions of dollars and t represents the year, with t = 6 corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored
the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general. Recognizing Characteristics of Graphs of Polynomial Functions Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial. y y f x Figure 1 x f Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 1 Recognizing Polynomial Functions Which of the graphs in Figure 2 represents a polynomial function? y y f h y x x Figure 2 y g k x x Solution The graphs of f and h are graphs of polynomial functions. They are smooth and continuous. The graphs of g and k are graphs of functions that are not polynomials. The graph of function g has a sharp corner. The graph of function k is not continuous. Q & A… Do all polynomial functions have as their domain all real numbers? Yes. Any real number is a valid input for a polynomial function. Using Factoring to Find Zeros of Polynomial Functions Recall that if f is a polynomial function, the values of x for which f (x) = 0 are called zeros of f. If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros. We can use this method to find x-intercepts because at the x-intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases: 1. The polynomial can be factored using known methods: greatest common factor and trinomial factoring. 2. The polynomial is given in factored
form. 3. Technology is used to determine the intercepts. How To… Given a polynomial function f, find the x-intercepts by factoring. 1. Set f (x) = 0. 2. If the polynomial function is not given in factored form: a. Factor out any common monomial factors. b. Factor any factorable binomials or trinomials. 3. Set each factor equal to zero and solve to find the x-intercepts. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 377 Example 2 Finding the x-Intercepts of a Polynomial Function by Factoring Find the x-intercepts of f (x) = x 6 − 3x 4 + 2x2. Solution We can attempt to factor this polynomial to find solutions for f (x) = 0. x 6 − 3x 4 + 2x 2 = 0 Factor out the greatest common factor. x 2(x 4 − 3x2 + 2) = 0 Factor the trinomial. x 2(x 2 − 1)(x 2 − 2) = 0 Set each factor equal to zero. (x 2 − 1) = 0 (x 2 − 2) = 0 x 2 = 0 or x 2 = 1 or This gives us five x-intercepts: (0, 0), (1, 0), (−1, 0), ( √ even function because it is symmetric about the y-axis. — 2, 0 ), and ( − √ — 2, 0 ). See Figure 3. We can see that this is an (–, 0) √2 –2 f (x) = x6 – 3 x4 + 2x2 y 2 1 (, 0)√2 x 2 (–1, 0) –1 (0, 0) (1, 0) –2 Figure 3 Example 3 Finding the x-Intercepts of a Polynomial Function by Factoring Find the x-intercepts of f (x) = x 3 − 5x 2 − x + 5. Solution Find solutions for f (x) = 0 by factoring. x 3 − 5x 2 − x + 5 = 0 Factor by grouping. x 2(x − 5) − (x − 5) = 0 Factor
out the common factor. (x 2 − 1)(x − 5) = 0 Factor the difference of squares. (x + 1)(x − 1)(x − 5) = 0 Set each factor equal to zero. x + 1 = 0 or x − 1 = 0 or x − 5 = 0 x = −1 x = 1 x = 5 There are three x-intercepts: (−1, 0), (1, 0), and (5, 0). See Figure 4. y 18 12 6 f (x) = x3 − 5 x2 − x + 5 –6 –4 –2 2 4 6 x –6 –12 –18 Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 37 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 4 Finding the y- and x-Intercepts of a Polynomial in Factored Form Find the y- and x-intercepts of g(x) = (x − 2)2(2x + 3). Solution The y-intercept can be found by evaluating g(0). g(0) = (0 − 2)2(2(0) + 3) = 12 So the y-intercept is (0, 12). The x-intercepts can be found by solving g(x) = 0. (x − 2)2(2x + 3) = 0 (x − 2)2 = 0 (2x + 3) = 0 x − 2 = 0 or x = 2 x = − #3 __ 2 So the x-intercepts are (2, 0) and ( − #3, 0 ). __ 2 Analysis We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5. y 15 (0, 12) 12 9 6 3 (–1.5, 0) g(x) = (x − 2)2(2x + 3 ) (2, 0) –4 –3 –2 –1 1 2 3 4 x –3 Figure 5 Example 5 Finding the x-Intercepts of a Polynomial Function Using a Graph Find the x-intercepts of h(x) = x3 + 4x2 + x − 6. Solution This polynomial is not in factored form, has no common factors, and does not appear
to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities. Looking at the graph of this function, as shown in Figure 6, it appears that there are x-intercepts at x = −3, −2, and 1. h(x) = x3 + 4 x2 + x − 6 y 2 –4 –2 2 4 x –2 –4 –6 –8 Figure 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 379 We can check whether these are correct by substituting these values for x and verifying that Since h(x) = x 3 + 4x 2 + x − 6, we have: h(−3) = h(−2) = h(1) = 0. h(−3) = (−3)3 + 4(−3)2 + (−3) − 6 = −27 + 36 − 3 − 6 = 0 h(−2) = (−2)3 + 4(−2)2 + (−2) − 6 = −8 + 16 − 2 − 6 = 0 h(1) = (1)3 + 4(1)2 + (1 Each x-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form. h(x) = x3 + 4x2 + x − 6 = (x + 3)(x + 2)(x − 1) Try It #1 Find the y- and x-intercepts of the function f (x) = x 4 − 19x 2 + 30x. Identifying Zeros and Their Multiplicities Graphs behave differently at various x-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and "bounce" off. Suppose, for example, we graph the function shown. f (x) = (x + 3)(x − 2)2(x + 1)3. Notice in Figure 7 that the behavior of the function at each of the x-intercepts is different. f (x) = (x
+ 3)(x − 2)2(x + 1)3 2 4 x y 40 30 20 10 –10 –20 –30 –40 –4 –2 Figure 7 Identifying the behavior of the graph at an x-intercept by examining the multiplicity of the zero. The x-intercept x = −3 is the solution of equation (x + 3) = 0. The graph passes directly through the x-intercept at x = −3. The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function. The x-intercept x = 2 is the repeated solution of equation (x − 2)2 = 0. The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept. (x − 2)2 = (x − 2)(x − 2) The factor is repeated, that is, the factor (x − 2) appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor, x = 2, has multiplicity 2 because the factor (x − 2) occurs twice. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS The x-intercept x = −1 is the repeated solution of factor (x + 1)3 = 0. The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic— with the same S-shape near the intercept as the toolkit function f (x) = x 3. We call this a triple zero, or a zero with multiplicity 3. For zeros with even multiplicities, the graphs touch or are tangent to the x-axis. For zeros with odd multiplicities, the graphs cross or intersect the x-axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3. y y y p = Single
zero Zero with multiplicity 2 Zero with multiplicity 3 Figure 8 For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x-axis. For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x-axis. graphical behavior of polynomials at x-intercepts If a polynomial contains a factor of the form (x − h)p, the behavior near the x-intercept h is determined by the power p. We say that x = h is a zero of multiplicity p. The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities. The sum of the multiplicities is the degree of the polynomial function. How To… Given a graph of a polynomial function of degree n, identify the zeros and their multiplicities. 1. If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. 2. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. 3. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity. 4. The sum of the multiplicities is n. Example 6 Identifying Zeros and Their Multiplicities Use the graph of the function of degree 6 in Figure 9 to identify the zeros of the function and their possible multiplicities. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 381 y 240 160 80 –6 –4 –2 2 4 6 x –80 –160 –240 Figure 9 Solution The polynomial function is of degree 6. The sum of the multiplicities must be 6. Starting from the left, the first zero occurs at x = −3. The graph touches the x-axis, so the multiplicity of the zero must be even. The zero of −3 most likely has multiplicity 2. The next
zero occurs at x = −1. The graph looks almost linear at this point. This is a single zero of multiplicity 1. The last zero occurs at x = 4. The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6. Try It #2 Use the graph of the function of degree 5 in Figure 10 to identify the zeros of the function and their multiplicities. y 60 40 20 –6 –4 –2 2 4 6 x –20 –40 –60 Figure 10 Determining End Behavior As we have already learned, the behavior of a graph of a polynomial function of the form f (x) = an x n + an − 1 x n − 1 +... + a1 x + a0 will either ultimately rise or fall as x increases without bound and will either rise or fall as x decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say −100 or −1,000. Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, anxn, is an even power function, as x increases or decreases without bound, f (x) increases without bound. When the leading term is an odd power function, as x decreases without bound, f (x) also decreases without bound; as x increases without bound, f (x) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure 11 summarizes all four cases. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Even Degree Odd Degree Positive Leading Coefficient, an > 0 y Positive Leading Coefficient, an > 0 y x x End Behavior: x → ∞, f (x) → ∞ x → −∞, f (x) → ∞ End Behavior: x → ∞, f (x) → ∞ x → −∞, f (x) → ∞ Negative Leading Coefficient, an < 0 y Negative Leading Coefficient, an < 0 y x x End Behavior:
x → ∞, f (x) → −∞ x → −∞, f (x) → −∞ End Behavior: x → ∞, f (x) → −∞ x → −∞, f (x) → ∞ Figure 11 Understanding the Relationship Between Degree and Turning Points In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function f (x) = x4 − x3 − 4x2 + 4x in Figure 12. The graph has three turning points. y Increasing Decreasing Increasing Decreasing x Turning points Figure 12 This function f is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 383 interpreting turning points A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). A polynomial of degree n will have at most n − 1 turning points. Example 7 Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function Find the maximum number of turning points of each polynomial function. b. f (x) = −(x − 1)2(1 + 2x2) a. f (x) = −x 3 + 4x 5 − 3x 2 + 1 Solution a. f (x) = −x 3 + 4x 5 − 3x 2 + 1 First, rewrite the polynomial function in descending order: f (x) = 4x 5 − x 3 − 3x 2 + 1 Identify the degree of the polynomial function. This polynomial function is of degree 5. The maximum number of turning points is 5 − 1 = 4. b. f (x) = −(x − 1)2(1 + 2x2) First, identify the leading term of the polynomial function if the function were expanded. f (x) = −(x − 1)2(1
+ 2x2) an = −(x 2)(2x 2) − 2x 4 Then, identify the degree of the polynomial function. This polynomial function is of degree 4. The maximum number of turning points is 4 − 1 = 3. Graphing Polynomial Functions We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions. How To… Given a polynomial function, sketch the graph. 1. Find the intercepts. 2. Check for symmetry. If the function is an even function, its graph is symmetrical about the y-axis, that is, f (−x) = f (x). If a function is an odd function, its graph is symmetrical about the origin, that is, f (−x) = −f (x). 3. Use the multiplicities of the zeros to determine the behavior of the polynomial at the x-intercepts. 4. Determine the end behavior by examining the leading term. 5. Use the end behavior and the behavior at the intercepts to sketch a graph. 6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial. 7. Optionally, use technology to check the graph. Example 8 Sketching the Graph of a Polynomial Function Sketch a graph of f (x) = −2(x + 3)2(x − 5). Solution This graph has two x-intercepts. At x = −3, the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x-intercept. At x = 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept. The y-intercept is found by evaluating f (0). The y-intercept is (0, 90). f (0) = −2(0 + 3)2(0 − 5) = −2 ċ 9 ċ (−5) = 90 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Additionally, we can see the leading term, if this polynomial were multiplied out, would be −2x3, so the
end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure 13. y Figure 13 x To sketch this, we consider that: • As x → −∞ the function f (x) → ∞, so we know the graph starts in the second quadrant and is decreasing toward the x-axis. • Since f (−x) = −2(−x + 3)2 (−x − 5) is not equal to f (x), the graph does not display symmetry. • At (−3, 0), the graph bounces off of the x-axis, so the function must start increasing. At (0, 90), the graph crosses the y-axis at the y-intercept. See Figure 14. y (0, 90) (−3, 0) x Figure 14 Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at (5, 0). See Figure 15. y (0, 90) (−3, 0) (5, 0) x Figure 15 As x → ∞ the function f (x) → −∞, so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 385 Using technology, we can create the graph for the polynomial function, shown in Figure 16, and verify that the resulting graph looks like our sketch in Figure 15. f (x) = −2(x + 3 )2(x − 5 ) y 180 120 60 –6 –4 –2 2 4 6 x –60 –120 –180 Figure 16 The complete graph of the polynomial function f (x ) = −2(x + 3)2(x − 5) Try It #3 1 __ x(x − 1)4(x + 3)3. Sketch a graph of f (x) = 4 Using the Intermediate Value Theorem In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial function f whose graph is smooth and continuous. The
Intermediate Value Theorem states that for two numbers a and b in the domain of f, if a < b and f (a) ≠ f (b), then the function f takes on every value between f (a) and f (b). (While the theorem is intuitive, the proof is actually quite complicated and require higher mathematics.) We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function f at x = a lies above the x-axis and another point at x = b lies below the x-axis, there must exist a third point between x = a and x = b where the graph crosses the x-axis. Call this point (c, f (c)). This means that we are assured there is a solution c where f (c) = 0. In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis. Figure 17 shows that there is a zero between a and b. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 f (b) is positive f (c) = 0 x 5 21 3 4 f(a) is negative Figure 17 Using the Intermediate Value Theorem to show there exists a zero Intermediate Value Theorem Let f be a polynomial function. The Intermediate Value Theorem states that if f (a) and f (b) have opposite signs, then there exists at least one value c between a and b for which f (c) = 0. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 9 Using the Intermediate Value Theorem Show that the function f (x) = x3 − 5x2 + 3x + 6 has at least two real zeros between x = 1 and x = 4. Solution As a start, evaluate f (x) at the integer values x = 1, 2, 3, and 4. See Table 2. x f (x) 1 5 2 0 3 −3 4 2 Table 2 We see that one zero occurs at x = 2. Also, since f (3) is negative and f (4) is positive, by the Intermediate Value Theorem, there must be at
least one real zero between 3 and 4. We have shown that there are at least two real zeros between x = 1 and x = 4. Analysis We can also see on the graph of the function in Figure 18 that there are two real zeros between x = 1 and x = 4. –5 –4 –3 –2 y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 f (1) = 5 is positive f (4) = 2 is positive 21 3 4 5 f (3) = −3 is negative Figure 18 Try It #4 Show that the function f (x) = 7x5 − 9x4 − x2 has at least one real zero between x = 1 and x = 2. Writing Formulas for Polynomial Functions Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x-intercept where each factor is equal to zero, we can form a function that will pass through a set of x-intercepts by introducing a corresponding set of factors. factored form of polynomials If a polynomial of lowest degree p has horizontal intercepts at x = x1, x2, …, xn, then the polynomial can be written (x − x2)p2 … (x − xn)pn where the powers pi on each factor can be determined in the factored form: f (x) = a(x − x1)p1 by the behavior of the graph at the corresponding intercept, and the stretch factor a can be determined given a value of the function other than the x-intercept. How To… Given a graph of a polynomial function, write a formula for the function. 1. Identify the x-intercepts of the graph to find the factors of the polynomial. 2. Examine the behavior of the graph at the x-intercepts to determine the multiplicity of each factor. 3. Find the polynomial of least degree containing all the factors found in the previous step. 4. Use any other point on the graph (the y-intercept may be easiest) to determine the stretch factor. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 GRAPHS OF POLYNOMIAL
FUNCTIONS 387 Example 10 Writing a Formula for a Polynomial Function from the Graph Write a formula for the polynomial function shown in Figure 19. y 6 4 2 –6 –4 –2 2 4 6 x –2 –4 –6 Figure 19 Solution This graph has three x-intercepts: x = −3, 2, and 5. The y-intercept is located at (0, −2). At x = −3 and x = 5, the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At x = 2, the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us f (x) = a(x + 3)(x − 2)2(x − 5) To determine the stretch factor, we utilize another point on the graph. We will use the y-intercept (0, −2), to solve for a. f (0) = a(0 + 3)(0 − 2)2(0 − 5) −2 = a(0 + 3)(0 − 2)2(0 − 5) −2 = −60a The graphed polynomial appears to represent the function f (x) = (x + 3)(x − 2)2 (x − 5). a = 1 __ 30 1 __ 30 Try It #5 Given the graph shown in Figure 20, write a formula for the function shown. y 12 8 4 –6 –4 –2 2 4 6 x –4 –8 –12 Figure 20 Using Local and Global Extrema With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph. Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38
8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS local and global extrema A local maximum or local minimum at x = a (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around x = a. If a function has a local maximum at a, then f (a) ≥ f (x) for all x in an open interval around x = a. If a function has a local minimum at a, then f (a) ≤ f (x) for all x in an open interval around x = a. A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at a, then f (a) ≥ f (x) for all x. If a function has a global minimum at a, then f (a) ≤ f (x) for all x. We can see the difference between local and global extrema in Figure 21. –6 –5 –4 –3 –2 y Global maximum Local maximum 21 3 4 5 6 x Local minimum 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 Figure 21 Q & A… Do all polynomial functions have a global minimum or maximum? No. Only polynomial functions of even degree have a global minimum or maximum. For example, f (x) = x has neither a global maximum nor a global minimum. Example 11 Using Local Extrema to Solve Applications An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box. Solution We will start this problem by drawing a picture like that in Figure 22, labeling the width of the cut-out squares with a variable, w. w w w w Figure 22 w w w w Notice that after a square is cut out from each end, it leaves a (14 − 2w) cm by (20 − 2w) cm rectangle for the base of the box, and the box will be w cm tall. This gives the volume V(w) = (20 − 2w)(14 − 2w)w = 280w − 68w2 + 4w3 Download the OpenStax text for free at http://cnx.org/content/col11759/
latest. SECTION 5.3 GRAPHS OF POLYNOMIAL FUNCTIONS 389 Notice, since the factors are w, 20 − 2w and 14 − 2w, the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values w may take on are greater than zero or less than 7. This means we will restrict the domain of this function to 0 < w < 7. Using technology to sketch the graph of V(w) on this reasonable domain, we get a graph like that in Figure 23. We can use this graph to estimate the maximum value for the volume, restricted to values for w that are reasonable for this problem—values from 0 to 7. V(w) V(w) = 280w − 68w2 + 4w3 400 300 200 100 –2 2 –100 –200 4 6 8 10 12 w Figure 23 From this graph, we turn our focus to only the portion on the reasonable domain, [0, 7]. We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure 24. V(w) 340 339 338 337 336 335 334 333 332 331 330 2.4 2.6 2.8 3 Figure 24 w From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side. Try It #6 Use technology to find the maximum and minimum values on the interval [−1, 4] of the function f (x) = −0.2(x − 2)3(x + 1)2(x − 4). Access the following online resource for additional instruction and practice with graphing polynomial functions. • Intermediate Value Theorem (http://openstaxcollege.org/l/ivt) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.3 SECTION EXERCISES VERBAL 1. What is the difference between an x-intercept and
a zero of a polynomial function f? 2. If a polynomial function of degree n has n distinct zeros, what do you know about the graph of the function? 3. Explain how the Intermediate Value Theorem can 4. Explain how the factored form of the polynomial assist us in finding a zero of a function. helps us in graphing it. 5. If the graph of a polynomial just touches the x-axis and then changes direction, what can we conclude about the factored form of the polynomial? ALGEBRAIC For the following exercises, find the x- or t-intercepts of the polynomial functions. 6. C(t) = 2(t − 4)(t + 1)(t − 6) 7. C(t) = 3(t + 2)(t − 3)(t + 5) 8. C(t) = 4t(t − 2)2(t + 1) 9. C(t) = 2t(t − 3)(t + 1)2 10. C(t) = 2t4 − 8t3 + 6t2 11. C(t) = 4t4 + 12t3 − 40t2 12. f (x) = x4 − x2 13. f (x) = x3 + x2 − 20x 14. f (x) = x3 + 6x2 − 7x 15. f (x) = x3 + x2 − 4x − 4 16. f (x) = x3 + 2x2 − 9x − 18 17. f (x) = 2x3 − x2 − 8x + 4 18. f (x) = x6 − 7x3 − 8 19. f (x) = 2x4 + 6x2 − 8 20. f (x) = x3 − 3x2 − x + 3 21. f (x) = x6 − 2x4 − 3x2 22. f (x) = x6 − 3x4 − 4x2 23. f (x) = x5 − 5x3 + 4x For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval. 24. f (x) = x3 − 9x, between x = −4 and x = −2. 25. f (x)
= x3 − 9x, between x = 2 and x = 4. 26. f (x) = x5 − 2x, between x = 1 and x = 2. 27. f (x) = −x4 + 4, between x = 1 and x = 3. 28. f (x) = −2x3 − x, between x = −1 and x = 1. 29. f (x) = x3 − 100x + 2, between x = 0.01 and x = 0.1 For the following exercises, find the zeros and give the multiplicity of each. 30. f (x) = (x + 2)3(x − 3)2 32. f (x) = x3 (x − 1)3(x + 2) 31. f (x) = x2(2x + 3)5(x − 4)2 33. f (x) = x2(x2 + 4x + 4) 34. f (x) = (2x + 1)3(9x2 − 6x + 1) 35. f (x) = (3x + 2)5(x2 − 10x + 25) 36. f (x) = x(4x2 − 12x + 9)(x2 + 8x + 16) 37. f (x) = x6 − x5 − 2x4 38. f (x) = 3x4 + 6x3 + 3x2 40. f (x) = 2x4(x3 − 4x2 + 4x) 39. f (x) = 4x5 − 12x4 + 9x3 41. f (x) = 4x4(9x4 − 12x3 + 4x2) GRAPHICAL For the following exercises, graph the polynomial functions. Note x- and y-intercepts, multiplicity, and end behavior. 42. f (x) = (x + 3)2(x − 2) 43. g(x) = (x + 4)(x − 1)2 44. h(x) = (x − 1)3(x + 3)2 45. k(x) = (x − 3)3(x − 2)2 46. m(x) = −2x(x − 1)(x + 3) 47. n(x) = −3x(
x + 2)(x − 4) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.3 SECTION EXERCISES 391 For the following exercises, use the graphs to write the formula for a polynomial function of least degree. 48. f(x) 49. f(x) 50. f(x) 5 4 3 2 1 –1–1 –2 –3 –4 –5 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 51. f(x) 52. f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, use the graph to identify zeros and multiplicity. 54. 21 3 4 5 x –5 –4 –3 –2 55. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x 53. 56. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 –5 –4 –3 –2 For the following exercises, use the given information about the polynomial graph to write the equation. 57. Degree 3. Zeros at x = −2, x = 1, and x = 3. 58. Degree 3. Zeros at x = −5, x = −2, and x = 1. y-intercept at (0, −4). y-intercept at (0, 6) 59. Degree 5. Roots of multiplicity 2 at x = 3 and x = 1, and a root of multiplicity 1 at x = −3.
y-intercept at (0, 9) 60. Degree 4. Root of multiplicity 2 at x = 4, and roots of multiplicity 1 at x = 1 and x = −2. y-intercept at (0, −3). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 61. Degree 5. Double zero at x = 1, and triple zero at 62. Degree 3. Zeros at x = 4, x = 3, and x = 2. x = 3. Passes through the point (2, 15). y-intercept at (0, −24). 63. Degree 3. Zeros at x = −3, x = −2 and x = 1. y-intercept at (0, 12). 1 _ 65. Degree 4. Roots of multiplicity 2 at x = and roots 2 of multiplicity 1 at x = 6 and x = −2. y-intercept at (0,18). 64. Degree 5. Roots of multiplicity 2 at x = −3 and x = 2 and a root of multiplicity 1 at x = −2. y-intercept at (0, 4). 66. Double zero at x = −3 and triple zero at x = 0. Passes through the point (1, 32). TECHNOLOGY For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum. 67. f (x) = x3 − x − 1 68. f (x) = 2x3 − 3x − 1 69. f (x) = x4 + x 70. f (x) = −x4 + 3x − 2 71. f (x) = x4 − x3 + 1 EXTENSIONS For the following exercises, use the graphs to write a polynomial function of least degree. 72. 74. (0, 50,000,000) 100 200 300 400 500 600 700 x 6·107 5·107 4·107 3·107 2·107 1·107 –100 0 –1·107 –2·107 –3·107 –4·107 –5·107 –6·107 –7·107 f (x) 24 16 73. f (x), 0 2 3 (0, 8) 4 3, 0 –6 –4
–2 2 4 6 x –300 –200 –8 –16 –24 f(x) 2·105 1·105 (100, 0) (–300, 0) x 100 200 (0, –90,000) –400 –300 –200 –100 –1·105 –2·105 –3·105 –4·105 REAL-WORLD APPLICATIONS For the following exercises, write the polynomial function that models the given situation. 75. A rectangle has a length of 10 units and a width of 8 units. Squares of x by x units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a polynomial function in terms of x. 77. A square has sides of 12 units. Squares x + 1 by x + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms of x. 79. A right circular cone has a radius of 3x + 6 and a height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone is 1 _ V = πr 2h for radius r and height h. 3 76. Consider the same rectangle of the preceding problem. Squares of 2x by 2x units are cut out of each corner. Express the volume of the box as a polynomial in terms of x. 78. A cylinder has a radius of x + 2 units and a height of 3 units greater. Express the volume of the cylinder as a polynomial function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 DIVIDING POLYNOMIALS 393 LEARNING OBJECTIVES In this section, you will: • Use synthetic diision to diide polynomials. • Use synthetic diision to determine whether is a factor of a polynomial x + 5.4 DIVIDING POLYNOMIALS Figure 1 Lincoln Memorial, Washington, D.C. (credit: Ron Cogswell, Flickr) The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters ( m ), width 40 m, and height 30 m.[15] We can easily find the volume using elementary geometry. V = l
ċ w ċ h = 61.5 ċ 40 ċ 30 = 73,800 So the volume is 73,800 cubic meters (m3). Suppose we knew the volume, length, and width. We could divide to find the height. h = V ____ l ċ w = 73,800 _______ 61.5 ċ 40 = 30 As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing dimensions. We can also use the same method if any or all of the measurements contain variable expressions. For example, suppose the volume of a rectangular solid is given by the polynomial 3x4 − 3x3 − 33x2 + 54x. The length of the solid is given by 3x; the width is given by x − 2. To find the height of the solid, we can use polynomial division, which is the focus of this section. Using Long Division to Divide Polynomials We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division. Long Division 5 × 3 = 15 and 17 − 15 = 2 Step 1: Step 2: Bring down the 8 9 × 3 = 27 and 28 − 27 = 1 Step 3: 1 __ Answer: 59 R 1 or 59 3 59 3)178 −15 28 −27 1 15. National Park Service. “Lincoln Memorial Building Statistics.” http://www.nps.gov/linc/historyculture/lincoln-memorial-building-statistics.htm. Accessed 4/3/2014/ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic. dividend = (divisor ċ quotient) + remainder 178 = (3 ċ 59) + 1 = 177 + 1 = 178 We call this the Division Algorithm and will discuss it more formally after looking at an example. Division of polynomials that contain more than one term has similarities
to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide 2x3 − 3x2 + 4x + 5 by x + 2 using the long division algorithm, it would look like this: Set up the division problem. 2x3 divided by x is 2x2. Multiply x + 2 by 2x2. Subtract. Bring down the next term. −7x 2 divided by x is −7x. Multiply x + 2 by −7x. Subtract. Bring down the next term. x + 2)2x3 − 3x2 + 4x + 5 2x2 x + 2)2x3 − 3x2 + 4x + 5 2x2 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x 2x2 − 7x x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 + 14x) 18x + 5 2x2 − 7x + 18 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 + 14x) 18x + 5 −18x + 36 −31 18x divided by x is 18. Multiply x + 2 by 18. Subtract. We have found or 2x3 − 3x2 + 4x + 5 __ x + 2 = 2x 2 − 7x + 18 − 31 _ x + 2 2x3 − 3x2 + 4x + 5 __ x + 2 = (x + 2)(2x2 − 7x + 18) − 31 We can identify the dividend, the divisor, the quotient, and the remainder. 2x3 – 3x2 + 4 x + 5 = (x + 2) (2x2 – 7 x + 18) + (–31) Dividend Divisor Quotient Remain
der Writing the result in this manner illustrates the Division Algorithm. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 DIVIDING POLYNOMIALS 395 the Division Algorithm The Division Algorithm states that, given a polynomial dividend f (x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f (x), there exist unique polynomials q(x) and r(x) such that f (x) = d(x)q(x) + r(x) q(x) is the quotient and r(x) is the remainder. The remainder is either equal to zero or has degree strictly less than d(x). If r(x) = 0, then d(x) divides evenly into f (x). This means that, in this case, both d(x) and q(x) are factors of f (x). How To… Given a polynomial and a binomial, use long division to divide the polynomial by the binomial. 1. Set up the division problem. 2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor. 3. Multiply the answer by the divisor and write it below the like terms of the dividend. 4. Subtract the bottom binomial from the top binomial. 5. Bring down the next term of the dividend. 6. Repeat steps 2–5 until reaching the last term of the dividend. 7. If the remainder is non-zero, express as a fraction using the divisor as the denominator. Example 1 Using Long Division to Divide a Second-Degree Polynomial Divide 5x2 + 3x − 2 by x + 1. Solution x + 1)5x2 + 3x − 2 5x Set up division problem. x + 1)5x2 + 3x − 2 5x2 divided by x is 5x. 5x x + 1)5x2 + 3x − 2 −(5x2 + 5x) −2x − 2 5x − 2 x + 1)5x2 + 3x − 2 −(5x2 + 5x) −2x − 2 −(−2
x − 2) 0 Multiply x + 1 by 5x. Subtract. Bring down the next term. −2x divided by x is −2. Multiply x + 1 by −2. Subtract. The quotient is 5x − 2. The remainder is 0. We write the result as or 5x2 + 3x − 2 __________ x + 1 = 5x − 2 5x2 + 3x − 2 = (x + 1)(5x − 2) Analysis This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 2 Using Long Division to Divide a Third-Degree Polynomial Divide 6x3 + 11x2 − 31x + 15 by 3x − 2. Solution 2x2 + 5x − 7 3x − 2)6x3 + 11x2 − 31x + 1 −(6x3 − 4x2) 15x2 − 31x −(15x2 + 10x) −21x + 15 −(−21x + 14) 1 6x3 divided by 3x is 2x2. Multiply 3x − 2 by 2x2. Subtract. Bring down the next term. 15x2 divided by 3x is 5x. Multiply 3x − 2 by 5x. Subtract. Bring down the next term. −21x divided by 3x is −7. Multiply 3x − 2 by −7. Subtract. The remainder is 1. There is a remainder of 1. We can express the result as: 6x3 + 11x2 − 31x + 15 __________________ 3x − 2 = 2x 2 + 5x − 7 + 1 _____ 3x − 2 Analysis We can check our work by using the Division Algorithm to rewrite the solution. Then multiply. (3x − 2)(2x 2 + 5x − 7) + 1 = 6x 3 + 11x 2 − 31x + 15 Notice, as we write our result, • the dividend is 6x3 + 11x2 − 31x + 15 • the divisor is
3x − 2 • the quotient is 2x2 + 5x − 7 • the remainder is 1 Try It #1 Divide 16x 3 − 12x 2 + 20x − 3 by 4x + 5. Using Synthetic Division to Divide Polynomials As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1. To illustrate the process, recall the example at the beginning of the section. Divide 2x 3 − 3x 2 + 4x + 5 by x + 2 using the long division algorithm. The final form of the process looked like this: 2x2 + x + 18 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 − 14x) 18x + 5 −(18x + 36) −31 There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 DIVIDING POLYNOMIALS 397 2)2 −3 4 5 −2 −4 −7 14 18 −36 −31 Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to −2, multiply and add. The process starts by bringing down the leading coefficient. −2 2 2 −3 −4 −7 4 5 14 −36 18 −31 We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is 2x2 − 7x + 18 and the remainder is −31. The process will be made
more clear in Example 3. synthetic division Synthetic division is a shortcut that can be used when the divisor is a binomial in the form x − k where k is a real number. In synthetic division, only the coefficients are used in the division process. How To… Given two polynomials, use synthetic division to divide. 1. Write k for the divisor. 2. Write the coefficients of the dividend. 3. Bring the lead coefficient down. 4. Multiply the lead coefficient by k. Write the product in the next column. 5. Add the terms of the second column. 6. Multiply the result by k. Write the product in the next column. 7. Repeat steps 5 and 6 for the remaining columns. 8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on. Example 3 Using Synthetic Division to Divide a Second-Degree Polynomial Use synthetic division to divide 5x2 − 3x − 36 by x − 3. Solution Begin by setting up the synthetic division. Write k and the coefficients. 3 5 −3 −36 Bring down the lead coefficient. Multiply the lead coefficient by k. 3 5 −3 −36 15 5 Continue by adding the numbers in the second column. Multiply the resulting number by k. Write the result in the next column. Then add the numbers in the third column. The result is 5x + 12. The remainder is 0. So x − 3 is a factor of the original polynomial. 3 5 −3 −36 36 15 12 0 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 39 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analysis remainder. Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the (x − 3)(5x + 12) + 0 = 5x2 − 3x − 36 Example 4 Using Synthetic Division to Divide a Third-Degree Polynomial Use synthetic division to divide 4x 3 + 10x 2 − 6x − 20 by x + 2. Solution The binomial divisor is x + 2 so k = −2. Add each column, multiply the result by −
2, and repeat until the last column is reached. −2 4 4 −6 −20 10 −8 −4 20 2 −10 0 The result is 4x 2 + 2x − 10. The remainder is 0. Thus, x + 2 is a factor of 4x3 + 10x2 − 6x − 20. Analysis The graph of the polynomial function f (x) = 4x3 + 10x2 − 6x − 20 in Figure 2 shows a zero at x = k = −2. This confirms that x + 2 is a factor of 4x 3 + 10x2 − 6x − 20. −2 −1.8 –5 –4 –3 –2 y 14 12 10 8 6 4 2 –1 –2 –4 –6 –8 –10 –12 –14 –16 –18 –20 –22 321 4 5 x Figure 2 Example 5 Using Synthetic Division to Divide a Fourth-Degree Polynomial Use synthetic division to divide −9x4 + 10x3 + 7x2 − 6 by x − 1. Solution Notice there is no x-term. We will use a zero as the coefficient for that term. 1 −9 −9 10 −9 1 7 1 8 0 −6 8 8 2 8 The result is −9x3 + x2 + 8x + 8 + 2 _. x − 1 Try It #2 Use synthetic division to divide 3x4 + 18x3 − 3x + 40 by x + 7. Using Polynomial Division to Solve Application Problems Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 DIVIDING POLYNOMIALS 399 Example 6 Using Polynomial Division in an Application Problem The volume of a rectangular solid is given by the polynomial 3x4 − 3x3 − 33x2 + 54x. The length of the solid is given by 3x and the width is given by x − 2. Find the height, t, of the solid. Solution There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as
in Figure 3. Height Width x − 2 Length 3 x Figure 3 We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid. V = l ċ w ċ h 3x4 − 3x3 − 33x2 + 54x = 3x ċ (x − 2) ċ h To solve for h, first divide both sides by 3x. 3x ċ (x − 2) ċ h ____________ 3x = 3x4 − 3x3 − 33x2 + 54x ___________________ 3x Now solve for h using synthetic division. (x − 2)h = x3 − x2 − 11x + 18 h = x3 − x2 − 11x + 18 ________________ x − 2 2 1 −1 −11 2 1 18 2 −18 −9 0 1 The quotient is x2 + x − 9 and the remainder is 0. The height of the solid is x2 + x − 9. Try It #3 The area of a rectangle is given by 3x3 + 14x2 − 23x + 6. The width of the rectangle is given by x + 6. Find an expression for the length of the rectangle. Access these online resources for additional instruction and practice with polynomial division. • Dividing a Trinomial by a Binomial Using Long Division (http://openstaxcollege.org/l/dividetribild) • Dividing a Polynomial by a Binomial Using Long Division (http://openstaxcollege.org/l/dividepolybild) • Ex 2: Dividing a Polynomial by a Binomial Using Synthetic Division (http://openstaxcollege.org/l/dividepolybisd2) • Ex 4: Dividing a Polynomial by a Binomial Using Synthetic Division (http://openstaxcollege.org/l/dividepolybisd4) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.4 SECTION EXERCISES VERBAL 1. If division of a polynomial by a binomial results in a remainder of zero, what can be conclude? 2. If a polynomial of degree n is divided by a binomial of degree 1, what is the
degree of the quotient? ALGEBRAIC For the following exercises, use long division to divide. Specify the quotient and the remainder. 3. (x2 + 5x − 1) ÷ (x − 1) 6. (4x2 − 10x + 6) ÷ (4x + 2) 9. (2x2 − 3x + 2) ÷ (x + 2) 12. (x3 − 3x2 + 5x − 6) ÷ (x − 2) 4. (2x2 − 9x − 5) ÷ (x − 5) 7. (6x2 − 25x − 25) ÷ (6x + 5) 10. (x3 − 126) ÷ (x − 5) 13. (2x3 + 3x2 − 4x + 15) ÷ (x + 3) 5. (3x2 + 23x + 14) ÷ (x + 7) 8. (−x2 − 1) ÷ (x + 1) 11. (3x2 − 5x + 4) ÷ (3x + 1) For the following exercises, use synthetic division to find the quotient. 14. (3x3 − 2x2 + x − 4) ÷ (x + 3) 16. (6x3 − 10x2 − 7x − 15) ÷ (x + 1) 18. (9x3 − 9x2 + 18x + 5) ÷ (3x − 1) 20. (−6x3 + x2 − 4) ÷ (2x − 3) 22. (3x3 − 5x2 + 2x + 3) ÷ (x + 2) 24. (x3 − 3x + 2) ÷ (x + 2) 26. (x3 − 15x2 + 75x − 125) ÷ (x − 5) 28. (6x3 − x2 + 5x + 2) ÷ (3x + 1) 30. (x4 − 3x2 + 1) ÷ (x − 1) 32. (x4 − 10x3 + 37x2 − 60x + 36) ÷ (x − 2) 34. (x4 + 5x3 − 3x2 − 13x + 10) ÷ (x + 5) 36. (4x4 − 2x3

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