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Edexcel AS and A level Mathematics
Pure Mathematics
Year 1 /AS
Series Editor: Harry Smith
Authors: Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen/uni00A0Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Pledger, Harry Smith, Geoο¬ /uni00A0Staley, Robert Ward-Penny, Dave Wilkins11 β 19 PROGRESSION | [
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Edexcel AS and A level Mathematics
Pure Mathematics
Year 1 /AS
Series Editor: Harry Smith
Authors: Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen/uni00A0Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Pledger, Harry Smith, Geoο¬ /uni00A0Staley, Robert Ward-Penny, Dave Wilkins11 β 19 PROGRESSION | [
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iiContents
Overarching themes iv
Extra online c
ontent vi
1 Algebraic e
xpressions 1
1.1 Index law
s 2
1.2 Expanding brack
ets 4
1.3 Factorising 6
1.4 Negative and fractional indic
es 9
1.5 Surds 12
1.6 Rationalising denominators 13
Mixed ex
ercise 1 15
2 Quadratics 18
2.1 Solving quadratic equations 19
2.2 Completing the squar
e 22
2.3 Functions 25
2.4 Quadratic graphs 27
2.5 The discriminant 30
2.6 Modelling with quadratics 32
Mixed ex
ercise 2 35
3 Equations and inequalities 38
3.1 Linear simultaneous equations 39
3.2 Quadratic simultaneous equations 41
3.3 Simultaneous equations on graphs 42
3.4 Linear inequalities 46
3.5 Quadratic inequalities 48
3.6 Inequalities on graphs 51
3.7 Regions 53
Mixed ex
ercise 3 56
4 Graphs and trans
formations 59
4.1 Cubic graphs 60
4.2 Quartic graphs 64
4.3 Reciprocal gr
aphs 66
4.4 Points of int
ersection 68
4.5 Translating gr
aphs 71 Contents
4.6 Stretching graphs 75
4.7 Trans
forming functions 79
Mixed ex
ercise 4 82
Review ex
ercise 1 85
5 Straight line gr
aphs 89
5.1 y = mx
+ c 90
5.2 Equations of st
raight lines 93
5.3 Parall
el and perpendicular lines 97
5.4 Length and area 100
5.5 Modelling with straight lines 103
Mixed ex
ercise 5 108
6 Circles 113
6.1 Midpoints and perpendicular
bisectors 114
6.2 Equation of a cir
cle 117
6.3 Intersections of st
raight lines
and circles 121
6.4 Use tangent and chord pr
operties 123
6.5 Circles and t
riangles 128
Mixed ex
ercise 6 132
7 Algebraic methods 137
7.1 Algebraic fr
actions 138
7.2 Dividing polynomials 139
7.3 The factor theorem 143
7.4 Mathematical proof 146
7.5 Methods of proo
f 150
Mixed ex
ercise 7 154
8 The binomial expansion 158
8.1 Pascal
βs triangle 159
8.2 Factorial notation 161
8.3 The binomial expansion 163
8.4 Solving binomial problems 165 | [
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iiiContents
8.5 Binomial estimation 167
Mixed ex
ercise 8 169
9 Trigonometric r
atios 173
9.1 The cosine rul
e 174
9.2 The sine rule 179
9.3 Areas o
f triangles 185
9.4 Solving triangle pr
oblems 187
9.5 Graphs of sine, c
osine and tangent 192
9.6 Trans
forming trigonometric graphs 194
Mixed ex
ercise 9 198
10 Trigonometric identities and
equations 202
10.1 Angles in all four quadr
ants 203
10.2 Exact values o
f trigonometrical ratios 208
10.3 Trigonomet
ric identities 209
10.4 Simple trig
onometric equations 213
10.5 Harder trig
onometric equations 217
10.6 Equations and identities 219
Mixed ex
ercise 10 222
Review ex
ercise 2 226
11 Vectors 230
11.1 Vectors 231
11.2 Representing v
ectors 235
11.3 Magnitude and direction 239
11.4 Position v
ectors 242
11.5 Solving geometric pr
oblems 244
11.6 Modelling with vectors 248
Mixed ex
ercise 11 251
12 Differentiation 255
12.1 Gradients of cur
ves 256
12.2 Finding the derivative 259
12.3 Differentiating
xn 262
12.4 Differentiating quadr
atics 26412.5 Differentiating functions with t
wo
or more terms 266
12.6 Gradients, tang
ents and normal 268
12.7 Increasing and decr
easing functions 270
12.8 Second order deriv
atives 271
12.9 Stationary points 273
12.10 Sketching gr
adient functions 277
12.11 Modelling with differentiation 279
Mixed ex
ercise 12 282
13 Integration 287
13.1 Integr
ating xn 288
13.2 Indefinite integr
als 290
13.3 Finding functions 293
13.4 Definite integr
als 295
13.5 Areas under cur
ves 297
13.6 Areas under the
x-axis 300
13.7 Areas bet
ween curves and lines 302
Mixed ex
ercise 13 306
14 Exponentials and logarithms 311
14.1 Exponential functions 312
14.2 y = ex 314
14.3 Exponential modelling 317
14.4 Logarithms 319
14.5 Law
s of logarithms 321
14.6 Solving equations using logarithms 324
14.7 Working with natur
al logarithms 326
14.8 Logarithms and non-linear data 328
Mixed ex
ercise 14 334
Review ex
ercise 3 338
Practic
e exam paper 342
Answ
ers 345
Index 399 | [
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ivOverarching themes
The following three overarching themes have been fully integrated throughout the Pearson Edexcel
AS and A level Mathematics series, so they can be applied alongside your learning and practice.
1. Mathematical argument, language and proof
β’ Rigorous and consistent approach throughoutβ’ Notation boxes explain key mathematical language and symbolsβ’ Dedicated sections on mathematical proof explain key principles and strategiesβ’ Opportunities to critique arguments and justify methods
2. Mathematical problem solving
β’ Hundreds of problem-solving questions, fully integrated
into the main exercises
β’ Problem-solving boxes provide tips and strategiesβ’ Structured and unstructured questions to build confi denceβ’ Challenge boxes provide extra stretch
3. Mathematical modelling
β’ Dedicated modelling sections in relevant topics provide plenty of practice where you need it β’ Examples and exercises include qualitative questions that allow you to interpret answers in the
context of the model
β’ Dedicated chapter in Statistics & Mechanics Year 1/AS explains the principles of modelling in
mechanics Overarching themes
Each chapter starts with
a list of objectives
The Prior knowledge check
helps make sure you are ready to start the chapterThe real world applications of the maths you are about to learn are highlighted at the start of the chapter with links to relevant questions in the chapterFinding your way around the book
Access an online digital edition using the code at the front of the book.The Mathematical Problem-solving cycle
specify the problem
interpret resultscollect information
process and
represent information | [
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vOverarching themes
Every few chapters a Review exercise
helps you consolidate your learning with lots of exam-style questionsEach section begins
with explanation and key learning points
Step-by-step worked
examples focus on the key types of questions youβll need to tackleExercise questions are
carefully graded so they increase in diffi culty and gradually bring you up to exam standard
Problem-solving boxes provide hints, tips and strategies, and Watch out boxes highlight areas where students oft en lose marks in their examsExercises are packed with exam-style questions to ensure you are ready for the exams
A full AS level practice paper at the back of the book helps you prepare for the real thing
Exam-style questions are ο¬ agged with
Problem-solving
questions are ο¬ agged withE
PEach chapter ends with a Mixed exercise and a Summary of key pointsChallenge boxes give you a chance to tackle some more diffi cult questions | [
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viExtra online content
Whenever you see an Online box, it means that there is extra online content available to support you.
SolutionBank
SolutionBank provides a full worked solution for
every question in the book.
Download all the solutions
as a PDF or quickly fi nd the solution you need online Extra online content
Full worked solutions are
available in SolutionBank.Online | [
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viiExtra online content
Access all the extra online content for FREE at:
www.pearsonschools.co.uk/p1maths
You can also access the extra online content by scanning this QR Code:
GeoGebra interactives
Explore topics in more detail,
visualise problems and consolidate your understanding with GeoGebra-powered interactives.
Interact with the maths
you are learning using GeoGebra's easy-to-use tools
Explore the gradient of the
chord AP using GeoGebra.Online
Casio calculator support
Our helpful tutorials will guide
you through how to use your calculator in the exams. They cover both Casio's scientific and colour graphic calculators.
See exactly which
buttons to press and what should appear on your calculator's screen
Work out each coefficient
qui
ckly using the nCr and power
functions on your calculator.Online | [
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viiiPublished by Pearson Education Limited, 80 Strand, London WC2R 0RL.
www.pearsonschoolsandfecolleges.co.uk Copies of official specifications for all Pearson qualifications may be found on the website:
qualifications.pearson.com
Text Β© Pearson Education Limited 2017
Edited by Tech-Set Ltd, GatesheadTypeset by Tech-Set Ltd, GatesheadOriginal illustrations Β© Pearson Education Limited 2017 Cover illustration Marcus@kja-artists
The rights of Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen
Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Pledger, Harry Smith, Geoff Staley, RobertΒ Ward
-Penny, Dave Wilkins to be identified as authors of this work have been asserted
by them in accordance with the Copyright, Designs and Patents Act 1988.
First published 201720 19 18 17
10 9 8 7 6 5 4 3 2 1
British Library Cataloguing in Publication Data
A catalogue record for this book is available from the British Library
ISBN 978 1 292 20826 8 (Print)
Copyright notice
All rights reserved. No part of this publication may be reproduced in any form or by any means (including photocopying or storing it in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright owner, except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency, Barnards Inn 86 Fetter Lane, London EC4A 1EN (www.cla.co.uk). Applications for the copyright ownerβs written permission should be addressed to the publisher.
Printed in Slovakia by NeografiaPicture Credits
The publisher would like to thank the following for their kind permission to reproduce their photographs:
(Key: b-bottom; c-centre; l-left; r-right; t-top)123RF.com: David Acosta Allely 287, 338cr; Alamy Images: Utah Images 113, 226l, Xinhua 38,
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All other images Β© Pearson EducationISBN 978 1 292 20759 9 (PDF) | [
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1
Algebraic expressions
After completing this chapter you should be able to:
β Multiply and divide integer po
wers β pages 2β3
β Expand a single term over brackets and collect like
terms
β pages 3β4
β Expand the product of two or three expressions β pages 4β6
β Factorise linear, quadratic and simple cubic expressions β pages 6β9
β Know and use the laws of indices β pages 9β11
β Simplify and use the rules of surds β pages 12β13
β Rationalise denominators β pages 13β16Objectives
1 Simplify:
a 4m2n + 5mn2 β 2m2n + mn2 β 3mn2
b 3x2 β 5x + 2 + 3x2 β 7x β 12
β GCSE Mathematics
2 Write as a single power of 2:a
25 Γ 23 b 26 Γ· 22
c (23)2 β GCSE Mathematics
3 Expand:a
3(x
+ 4) b 5(2 β 3
x)
c 6(2x
β 5y) β GCSE Mathematics
4 Write down the highest common factor of:a
24 and 16 b 6x
and 8x2
c 4xy2 and 3xy β GCSE Mathematics
5 Simplify:
a 10x ____ 5 b 20x ____ 2 c 40x ____ 24
β GCSE MathematicsPrior knowledge check
Computer scientists use indices to describe
very large numbers. A quantum computer with 1000 qubits (quantum bits) can consider 2
1000
values simultaneously. This is greater than the number of particles in the observable universe.1 | [
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2
Chapter 1
1.1 Index laws
β You can use the laws of indices to simplify powers of the same base.
β’ am Γ an = am + n
β’ am Γ· an = am β n
β’ (am)n = amn
β’ (ab)n = anbn
Example 1
Example 2
Expand these expressions and simplify if possible:
a β3x
(7x β 4) b y2(3 β 2y3)
c 4x
(3x β 2x2 + 5x3) d 2x (5x + 3) β 5(2x + 3)Simplify these expressions:a
x2 Γ x5 b 2r2 Γ 3r3 c b7
__ b4 d 6x5 Γ· 3x3 e (a3)2 Γ 2a2 f (3x2)3 Γ· x4
x5Notation
This is the base .
This is the index, power or
exponent.
a x2Β ΓΒ x5 =Β x2 + 5 =Β x7
b 2r2Β ΓΒ 3 r3 =Β 2Β Γ Β 3Β ΓΒ r2Β ΓΒ r3
=Β 6Β ΓΒ r2 + 3 =Β 6 r5
c b7 __ b4 = b7 β 4 =Β b3
d 6x5Β Γ·Β 3 x3 =Β 6 __ 3 Β ΓΒ x5 ____ x3
=Β 2Β Γ
Β x2 = 2x2
e (a3)2Β ΓΒ 2 a2 =Β a6Β ΓΒ 2 a2
=Β 2Β ΓΒ a6Β ΓΒ a2 =Β 2 a8
f (3x2)3 _____ x4 =Β 33 Γ (x2)3 ____ x4
= 27
Γ x6 __ x4 =Β 27 x2Use the rule amΒ ΓΒ anΒ =Β am + n to simplify the index.
x5Β Γ·Β x3Β =Β x5 β 3Β =Β x2Rewrite the expression with the numbers
together and the r terms together.
2Β ΓΒ 3Β =Β 6
r2Β ΓΒ r3Β =Β r2 + 3
A min us sign outside
brackets changes the sign of every term inside the brackets.Watch outUse the rule am Γ· an = am β n to simplify the index.
Use the rule (am)nΒ =Β amn to simplify the index.
a6Β Γ a2 = a6 + 2 = a8
Use the rule (ab)nΒ =Β anbn to simplify the numerator.
(x2)3Β =Β x2 Γ 3Β =Β x6
x6 __ x4 Β =Β x6 β 4Β =Β x2 | [
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3Algebraic expressions
a β3x(7xΒ β 4 ) =Β β21x2Β +Β 12 x
b y2(3Β βΒ 2y3) =Β 3 y2Β βΒ 2y5
c 4x(3xΒ βΒ 2 x2Β +Β 5 x3)
=Β 12 x2Β βΒ 8 x3Β +Β 20 x4
d 2x(5xΒ +Β 3 )Β βΒ 5(2 xΒ +Β 3)
=Β 10 x2Β +Β 6 xΒ βΒ 10 xΒ βΒ 15
=Β 10 x2Β βΒ 4 xΒ βΒ 15
a x7 + x4 _______ x3 = x7 ___ x3 + x4 ___ x3
=
x7 β 3Β + x4 β 3 = x4Β + x
b 3x2 β 6x5 __________ 2x = 3 x 2 ____ 2x β 6 x 5 ____ 2x
= 3 __ 2 x 2 β 1 β 3x5 β 1 = 3x ___ 2 β 3 x 4
c 20x7 + 15x3 ____________ 5x2 = 20 x 7 _____ 5 x 2 + 15 x 3 _____ 5 x 2
= 4x7 β 2 + 3 x3 β 2 = 4 x 5 + 3 xExample 3β3xΒ ΓΒ 7xΒ ξ΅Β β21x1 +Β 1Β ξ΅Β β21x2
β3xΒ ΓΒ (β4)Β ξ΅Β ξ±12x
Remember a minus sign outside the brackets
changes the signs within the brackets.
Simplify 6xΒ βΒ 10x to give β4x.
Simplify these expressions:
a x 7 + x 4 ______ x 3 b 3 x 2 β 6 x 5 ________ 2x c 20 x 7 + 15 x 3 __________ 5 x 2 y2Β ΓΒ (β2y3)Β ξ΅Β β2y2 +Β 3Β ξ΅Β β2y5
Divide each term of the numerator by x 3 .
x1 is the same as x.
Divide each term of the numerator by 2x.
Simplify each fraction:
3 x 2 ____ 2x = 3 __ 2 Γ x 2 ___ x = 3 __ 2 Γ x2 β 1
β 6 x 5 ____ 2x = β 6 __ 2 Γ x 5 ___ x = β3 Γ x5 β 1
Divide each term of the numerator by 5x2.
1 Simplify these expressions:
a x3 Γ x4 b 2x3 Γ 3x2 c k3
__ k2
d 4p3
___ 2p e 3x3 ___ 3x2 f (y2)5
g 10x5 Γ· 2x3 h ( p3)2 Γ· p4 i (2a3)2 Γ· 2a3
j 8p4 Γ· 4p3 k 2a4 Γ 3a5 l 21a3b7 ______ 7ab4
m 9x2 Γ 3(x2)3 n 3x3 Γ 2x2 Γ 4x6 o 7a 4 Γ (3a 4)2
p (4y 3)3 Γ· 2y3 q 2a3 Γ· 3a2 Γ 6a5 r 3a4 Γ 2a5 Γ a3Exercise 1A | [
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4
Chapter 1
1.2 Expanding brackets
To find the product of two expressions you multiply each term in one expression by each term in the
other expression.
(x + 5)(4x β 2y + 3)x Γ
5 Γ= x(4x β 2y + 3) + 5(4x β 2y + 3)= 4x
2 β 2xy + 3x + 20x β 10y + 15
= 4x2 β 2xy + 23x β 10y + 15Multiplying each of the 2 terms in the first expression by each of the
3 terms in the second expression gives 2 Γ 3 = 6 terms.
Simplify your answer by collecting like terms.2 Expand and simplify if possible:
a 9(x β
2) b x(x
+ 9) c β3y
(4 β 3y)
d x(y
+ 5) e βx(3
x + 5) f β5x
(4x + 1)
g (4x
+ 5)x h β3y
(5 β 2y2) i β2x (5x β 4)
j (3x β
5)x2 k 3(x + 2) + (x β 7) l 5x β 6 β (3x β 2)
m 4(c +
3d 2) β 3(2c + d 2) n (r2 + 3t2 + 9) β (2r2 + 3t2 β 4)
o x(3x2 β 2x + 5) p 7y2(2 β 5y + 3y2) q β2y2(5 β 7y + 3y2)
r 7(x β
2) + 3(x + 4) β 6(x β 2) s 5x β
3(4 β 2x) + 6
t 3x2 β x(3 β 4x) + 7 u 4x( x + 3) β 2x(3x β 7) v 3x2(2x + 1) β 5x2(3x β 4)
3 Simplify these fractions:
a 6 x 4 + 10 x 6 _________ 2x b 3 x 5 β x 7 _______ x c 2 x 4 β 4 x 2 ________ 4x
d 8 x 3 + 5x ________ 2x e 7 x 7 + 5 x 2 ________ 5x f 9 x 5 β 5 x 3 ________ 3x
a (x + 5)(x + 2)
= x2 + 2x + 5 x + 10
=
x2 + 7x + 10
b (x
β 2y)(x2 + 1)
= x3 + x β 2x2y β 2 yExample 4
Expand these expressions and simplify if possible:
a (x
+ 5)(x + 2) b (x
β 2y)(x2 + 1) c (x β y)2 d (x + y)(3x β 2y β 4)
Multiply x by (x + 2) and then multiply 5 by (x + 2).
Simplify your answer by collecting like terms.
β2y Γ x2 = β2x2y
There are no like terms to collect. | [
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-0.0230973232537508,
-0.06870998442173004,
0.01709219068288803,
0.00... |
5Algebraic expressions
c (x β y)2
= (x β y)(x β y)
=
x2 β xy β xy + y2
= x2 β 2xy + y2
d (x + y)(3x β 2 y β 4)
= x(3x
β 2y β 4) + y (3x β 2 y β 4)
= 3x2 β 2xy β 4 x + 3 xy β 2 y2 β 4y
= 3x2 + xy β 4 x β 2 y2 β 4y
a x(2x + 3)(x β 7)
= (2x2 + 3 x)(x β 7)
= 2
x3 β 14 x2 + 3 x2 β 21x
= 2
x3 β 11 x2 β 21x
b x(5x β
3y)(2x β y + 4)
= (5x2 β 3 xy)(2x β y + 4)
= 5x2(2x β y + 4) β 3 xy(2x β y + 4)
= 10x3 β 5 x2y + 20 x2 β 6 x2y + 3 xy2
β 12 xy
= 10
x3 β 11 x2y + 20 x2 + 3 xy2 β 12 xy
c (x
β 4)( x + 3)( x + 1)
= (x2 β x β 12)( x + 1)
=
x2(x + 1) β x (x + 1) β 12( x + 1)
=
x3 + x2 β x2 β x β 12 x β 12
=
x3 β 13 x β 12Example 5
Expand these expressions and simplify if possible:
a x(2x
+ 3)(x β 7) b x(5x
β 3y)(2x β y + 4) c (x
β 4)(x + 3)(x + 1)
Be careful with minus signs. You need to change
every sign in the second pair of brackets when you multiply it out.
Choose one pair of brackets to expand first, for example:
(x β 4)(x + 3) = x
2 + 3x β 4x β 12
= x2 β x β 12
You multiplied together three linear terms, so the final answer contains an x
3 term.βxy β xy = β2xy
Multiply x by (3x β 2y β 4) and then multiply y by (3x β 2y β 4).
Start by expanding one pair of brackets:x(2x + 3) = 2x
2 + 3x
You could also have expanded the second pair of brackets first: (2x + 3)(x β 7) = 2x
2 β 11x β 21
Then multiply by x.(x β y)2 means (x β y) multiplied by itself.
1 Expand and simplify if possible:
a (x
+ 4)(x + 7) b (x
β 3)(x + 2) c (x
β 2)2
d (x β y)(2x + 3) e (x + 3y)(4x β y) f (2x β 4y)(3x + y)
g (2x
β 3)(x β 4) h (3x
+ 2y)2 i (2x + 8y)(2x + 3)
j (x
+ 5)(2x + 3y β 5) k (x
β 1)(3x β 4y β 5) l (x
β 4y)(2x + y + 5)
m (x
+ 2y β 1)(x + 3) n (2x
+ 2y + 3)(x + 6) o (4 β
y)(4y β x + 3)
p (4y
+ 5)(3x β y + 2) q (5y
β 2x + 3)(x β 4) r (4y
β x β 2)(5 β y)Exercise 1B | [
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6
Chapter 1
1.3 Factorising
You can write expressions as a product of their factors.
β Factorising is the opposite of expanding
brack
ets.4x(2x + y)
(x + 5)3
(x + 2y)(x β 5y)= 8x2 + 4xy
= x3 + 15x2 + 75x + 125
= x2 β 3xy β 10y2Expanding brackets
FactorisingExpand and simplify ( x + y )4. You can use the binomial expansion to expand
ex
pressions like ( x + y )4 quickly. β Section 8.3LinksChallenge2 Expand and simplify if possible:
a 5(x
+ 1)(x β 4) b 7(x
β 2)(2x + 5) c 3(x
β 3)(x β 3)
d x(x
β y)(x + y) e x(2x
+ y)(3x + 4) f y(x
β 5)(x + 1)
g y(3x
β 2y)(4x + 2) h y(7 β
x)(2x β 5) i x(2x
+ y)(5x β 2)
j x(x
+ 2)(x + 3y β 4) k y(2x
+ y β 1)(x + 5) l y(3x
+ 2y β 3)(2x + 1)
m x(2x
+ 3)(x + y β 5) n 2x
(3x β 1)(4x β y β 3) o 3x
(x β 2y)(2x + 3y + 5)
p (x
+ 3)(x + 2)(x + 1) q (x
+ 2)(x β 4)(x + 3) r (x
+ 3)(x β 1)(x β 5)
s (x
β 5)(x β 4)(x β 3) t (2x
+ 1)(x β 2)(x + 1) u (2x
+ 3)(3x β 1)(x + 2)
v (3x
β 2)(2x + 1)(3x β 2) w (x
+ y)(x β y)(x β 1) x (2x
β 3y)3
3 The diagram shows a rectangle with a square cut out.
The rectangle has length 3
x β y + 4 and width x + 7.
The square has length x β 2.Find an expanded and simplified expression for the shaded area.
x β 2x + 7
3x β y + 4
4 A cuboid has dimensions x + 2 cm, 2x β 1 cm and 2x + 3 cm.
Show tha
t the volume of the cuboid is 4x3 + 12x2 + 5x β 6 cm3.
5 Given tha
t (2x + 5y)(3x β y)(2x + y) = ax3 + bx2y + cxy2 + dy3, where a, b, c and d are
constants, find the values of a, b, c and d. (2 marks)P
Use the same strategy as you would use
if the lengths were given as numbers:
3cm6cm
10cmProblem-solving
P
E/P | [
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7Algebraic expressions
An ex pression in the form x2 β y2 is
called the difference of two squares.Notation= (x + 3)(2x β 1)β A quadratic expression has the form
ax2 + bx + c where a, b and c are real
numbers and a β 0.
To factorise a quadratic expression:
β’Find two fact
ors of ac that add up to b
β’Rewrite the
b term as a sum of these two
factors
β’Factorise each p
air of terms
β’Take out the c
ommon factor
β x2 β y2 = (x + y)(x β y)a 3x + 9 = 3( x + 3)
b x2 β 5 x = x(x β 5)
c 8x2 + 20 x = 4 x(2x + 5)
d 9x2y + 15 xy2 = 3 xy(3x + 5 y)
e 3x2 β 9 xy = 3 x(x β 3 y)Example 6
Factorise these expressions completely:
a 3x
+ 9 b x2 β 5 x c 8x2 + 20x d 9x2y + 15xy2 e 3x2 β 9xy
3 is a common factor of 3x and 9.
For the expression 2x2 + 5x β 3, ac = β6 = β1 Γ 6
and β1 + 6 = 5 = b.
2x2 β x + 6x β 3
= x(2x β 1) + 3(2x β 1)x is a common factor of x2 and β5x.
4 and x are common factors of 8x2 and 20x.
So take 4x outside the brackets.
3, x and y are common factors of 9x2y and 15xy2.
So take 3xy outside the brackets.
x and β3y have no common factors so this
expression is completely factorised.
Real n umbers are all the positive and
negative numbers, or zero, including fractions and surds.Notation
Example 7
Factorise:
a x2Β βΒ 5xΒ βΒ 6 b x2Β +Β 6xΒ +Β 8 c 6x2Β βΒ 11xΒ βΒ 10 d x2Β βΒ 25 e 4x2Β βΒ 9y2
a x2 β 5 x β 6
ac =
β6 and b = β 5
So x2 β 5 x β 6 ξ΅ x2 + x β 6 x β 6
= x(x
+ 1) β 6( x + 1)
= (x
+ 1)( x β 6)Here aΒ =Β 1, bΒ =Β β 5 and cΒ =Β β 6.
1 Work out the two factors of ac =Β β 6 which add
t
o give you bΒ =Β β5. β6 + 1Β =Β β5
2 Rewrite the b term using these two factors.
3 Factorise first two terms and last two terms.
4 xΒ + 1 is a factor of both terms, so take that
outside the brackets. This is now completely
factorised. | [
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8
Chapter 1
Example 8
Factorise completely:
a x3 β 2x2 b x3 β 25x c x3 + 3x2 β 10xb x2 + 6 x + 8
= x2 + 2 x + 4 x + 8
= x(x
+ 2) + 4( x + 2)
= (x
+ 2)( x + 4)
c 6x2 β 11 x β 10
= 6x2 β 15 x + 4 x β 10
= 3x(2x
β 5) + 2(2 x β 5)
= (2 x
β 5)(3 x + 2)
d x2 β 25
= x2 β 52
= (x + 5)( x β 5)
e 4x2 β 9 y2
= 22x2 β 32y2
= (2 x + 3y)(2x β 3 y)
a x3 β 2x2 = x2(x β 2)
b x3 β 25 x = x(x2 β 25)
= x(x2 β 52)
= x(x + 5)( x β 5)
c x3 + 3 x2 β 10x = x (x2 + 3 x β 10)
= x(x + 5)( x β 2)
1 Factorise these expressions completely:a
4x + 8 b 6x β 24 c 20x + 15
d 2x2 + 4 e 4x2 + 20 f 6x2 β 18x
g x2 β 7x h 2x2 + 4x i 3x2 β x
j 6x2 β 2x k 10y2 β 5y l 35x2 β 28x
m x2 + 2x n 3y2 + 2y o 4x2 + 12x
p 5y2 β 20y q 9xy2 + 12x2y r 6abΒ β 2ab2
s 5x2 β 25xy t 12x2y ξ± 8xy2 u 15y β 20yz2
v 12x2 β 30 w xy2 β x2y x 12y2 β 4yxExercise 1Cx2 β 25 is the difference of two squares.This is the difference of two squares as the two
terms are x2 and 52.
The two x terms, 5x and β 5x, cancel each other out.acΒ =Β β60 and 4Β βΒ 15Β =Β β11Β =Β b.
Factorise.
This is the same as (2x)2Β βΒ (3y)2.
You canβt factorise this any further.
x is a common factor of x3 and β25x.
So take x outside the brackets.
Write the expression as a product of x and a
quadratic factor.
Factorise the quadratic to get three linear factors.acΒ =Β 8 and 2Β +Β 4Β =Β 6Β =Β b.
Factorise. | [
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9Algebraic expressions
Write 4x4 β 13x2 + 9 as the product of four linear factors.Challenge2 Factorise:
a x2 + 4x b 2x2 + 6x c x2 + 11x + 24
d x2 + 8x + 12 e x2 + 3xΒ β 40 f x2 β 8x + 12
g x2 + 5x + 6 h x2 β 2xΒ β 24 i x2 β 3xΒ β 10
j x2 +Β xΒ β 20 k 2x2 + 5xΒ + 2 l 3x2 + 10x β 8
m 5x2 β 16xΒ + 3 n 6x2 β 8x β 8
o 2x2 + 7xΒ β 15 p 2x4 + 14x2 + 24
q x2 β 4 r x2 β 49
s 4x2 β 25 t 9x2 β 25y2 u 36x2 β 4
v 2x2 β 50 w 6x2 β 10xΒ + 4 x 15x2 + 42xΒ β 9
3 Factorise completely:a
x3 + 2x b x3 β x2 + x c x3 β 5x
d x3 β 9x e x3 β x2 β 12x f x3 + 11x2 + 30x
g x3 β 7x2 + 6x h x3 β 64x i 2x3 β 5x2 β 3x
j 2x3 + 13x2 + 15x k x3 β 4x l 3x3 + 27x2 + 60x
4 Factorise completel
y x4 β y4. (2 marks)
5 Factorise completel
y 6x3 + 7x2 β 5x. (2 marks) For part n , ta ke 2 out as a common
factor first. For part p , let yΒ =Β x2.Hint
Watch out for terms that can be written as a
function of a function: x4 = (x2)2Problem-solving P
E
1.4 Negative and fractional indic es
Indices can be negative numbers or fractions.
x 1 _ 2 Γ x 1 _ 2 = x 1 _ 2 + 1 _ 2 = x 1 = x,
similarly x 1 __ n Γ x 1 __ n Γ . . . Γ x 1 __ n = x 1 __ n + 1 __ n +...+ 1 __ n = x 1 = x
n terms
β You can use the laws of indices with any rational power.
β’ a 1 __ m = m ββ―__
a
β’ a n __ m = m ββ―___ a n
β’ a βm = 1 ___ a m
β’ a 0 = 1 β«
βͺβͺβ¬βͺβͺβ Ratio nal
numbers are those that
can be written as a __ b where
a and b are integers.Notation
a 1 _ 2 = ββ―__
a is the
positive square root of a .
For example 9 1 _ 2 = ββ―__
9 = 3
but
9 1 _ 2 β β3 .Notation | [
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10
Chapter 1
Example 9
Simplify:
a x 3 ___ x β3 b x1
2 Γ x32
c (x3)23
d 2x1.5Β Γ·Β 4xβ0.25 e 3 ββ―______ 125 x 6 f 2 x 2 β x _______ x 5
a x 3 ____ x β3 = x3 β (β3) = x6
b x1
2 Γ x3
2 = x1
2 ξ±Β 32 = x2
c (x3)23 =Β x3 ξ³Β 23 =Β x2
d 2x1.5Β ξ΄Β 4 xβ0.25 = 1 __ 2 x1.5Β β (β0 .25) = 1 __ 2 x1.75
e 3 β _____ 125 x 6 = ( 125 x6 ) 1 __ 3
= (12
5 ) 1 __ 3 (x6 ) 1 __ 3 = 3 ββ―_____ 125 ( x 6 Γ 1 __ 3 ) = 5 x2
f 2 x 2 β x ______ x 5 = 2 x 2 ____ x 5 β x ___ x 5
= 2 Γ
x2 β 5 β x1 β 5 = 2xβ3 β xβ4
= 2 ___ x 3 β 1 ___ x 4 Use the rule amΒ Γ·Β anΒ =Β am β n.
Evaluate:
a 9 1 _ 2 b 6 4 1 _ 3 c 4 9 3 _ 2 d 2 5 β 3 _ 2 Example 10
a 9 1 __ 2 = ββ―__
9 = 3
b 6 4 1 __ 3 = 3 ββ―___ 64 = 4
c 4 9 3 __ 2 = ( ββ―___ 49 ) 3
73 = 343
d 2 5 β 3 __ 2 = 1 ____
2 5 3 __ 2 = 1 ______
( ββ―___ 25 ) 3
= 1 ___ 53 = 1 _____ 125 Using a 1 __ m = m ββ―__
a . 9 1 _ 2 = ββ―__
9
Using a n __ m = m ββ―__ an .
This means the square root o
f 49, cubed.
Using aβm = 1 ___ am This could also be written as ββ―__
x .
Use the rule amΒ ΓΒ anΒ =Β am + n.
Use the rule (am)nΒ =Β amn.
Use the rule amΒ Γ·Β anΒ =Β am β n.
1.5Β βΒ (β0.25)Β =Β 1.75
Using a 1 __ m = m ββ―__
a .
Divide each term of the numerator by x5.
Using a βm = 1 ___ a m
This means the cube root of 64.
Use your calculator to enter
ne
gative and fractional powers.Online | [
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-0.0740... |
11Algebraic expressions
1 Simplify:
a x3 Γ· xβ2 b x5 Γ· x7 c x 3 _ 2 Γ x 5 _ 2
d (x2 ) 3 _ 2 e (x3 ) 5 _ 3 f 3x0.5 Γ 4xβ0.5
g 9 x 2 _ 3 Γ· 3 x 1 _ 6 h 5 x 7 _ 5 Γ· x 2 _ 5 i 3x4 Γ 2xβ5
j ββ―__
x Γ 3 ββ―__
x k ( ββ―__
x )3 Γ ( 3 ββ―__
x )4 l ( 3 ββ―__
x )2 _____ ββ―__
x
2 Eva
luate:
a 2 5 1 _ 2 b 8 1 3 _ 2 c 2 7 1 _ 3
d 4β2 e 9 β 1 _ 2 f (β 5)β3
g ( 3 _ 4 ) 0 h 129 6 3 _ 4 i ( 25 __ 16 ) 3
2
j ( 27 __ 8 ) 2
3 k ( 6 _ 5 ) β1 l ( 343 ___ 512 ) β2
3
3 Simplify:
a (64x10)1
2 b 5 x 3 β 2 x 2 ________ x 5 c (125x12)1
3 d x + 4 x 3 _______ x 3
e 2x + x 2 _______ x 4 f ( 4 __ 9 x4) 3
2 g 9 x 2 β 15 x 5 _________ 3 x 3 h 5x + 3 x 2 ________ 15 x 3
4 a Find the value of
8 1 1 _ 4 . (1 mark)
b Simplify x(2 x β 1 _ 3 )4. (2 marks)
5 Given tha
t y = 1 __ 8 x 3 express each of the following in the form k x n , where k and n are constants.
a y 1
3 (2 marks)
b 1 __ 2 y β2 (2 marks)E
EExercise 1Da y 1 _ 2 = ( 1 __ 16 x 2 ) 1 _ 2
= 1 ___ ββ―___ 16 x 2 Γ 1 _ 2 = x __ 4
b 4yβ1 = 4 ( 1 __ 16 x 2 ) β1
= 4 ( 1 __ 16 ) β1
x 2 Γ (β1) = 4 Γ 16 xβ2
= 64 xβ2Substitute y = 1 ___ 16 x 2 into y 1 _ 2 .
( 1 ___ 16 ) 1 _ 2
= 1 ____
β ___ 16 and ( x 2 ) 1 _ 2 = x 2 Γ 1 _ 2 Given that y = 1 __ 16 x2 express each of the following in the form k x n , where k and n are constants.
a y 1 __ 2 b 4 y β1 Example 11
Check that your answers are in the correct form.
If k and n are constants they could be positive or negative, and they could be integers, fractions or surds.Problem-solving ( 1 ___ 16 ) β1
= 16 and x2 Γ β1 = xβ2 | [
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12
Chapter 1
1.5 Surds
If n is an integer that is not a square number, then any multiple of ββ―__
n is called a surd.
Examples of surds are ββ―__
2 , ββ―___ 19 and 5 ββ―__
2 .
Surds are examples of irrational numbers.
The decimal expansion of a surd is never-ending and never repeats, for example
ββ―__
2 = 1.414213562...
You can use surds to write exact answers to calculations.
β You can manipulate surds using these rules:
β’ ββ―___ ab = ββ―__
a Γ ββ―__
b
β’ ββ―__
a __ b = ββ―__
a ___
ββ―__
b Irr ational numbers cannot be written
in the form a __ b where a and b are integers.
Surds are examples of irrational numbers .Notation
Simplify:
a ββ―___ 12 b ββ―___ 20 ____ 2 c 5 ββ―__
6 β 2 ββ―___ 24 + ββ―____ 294 Example 12
a ββ―___ 12 = ββ―_______ (4 Γ 3)
= ββ―__
4 Γ ββ―__
3 = 2 ββ―__
3
b ββ―___ 20 ____ 2 = ββ―__
4 Γ ββ―__
5 ________ 2
= 2 Γ ββ―__
5 _______ 2 = ββ―__
5
c 5 ββ―__
6 β 2 ββ―___ 24 + ββ―_____ 294
= 5 ββ―__
6 β 2 ββ―__
6 ββ―__
4 + ββ―__
6 Γ ββ―___ 49
= ββ―__
6 (5 β 2 ββ―__
4 + ββ―___ 49 )
= ββ―__
6 (5 β 2 Γ 2 + 7)
= ββ―__
6 (8)
= 8 ββ―__
6 ββ―__
6 is a common factor. ββ―__
4 = 2Look for a factor of 12 that is a square number.
Use the rule ββ―___ ab = ββ―__
a Γ ββ―__
b . ββ―__
4 = 2
ββ―___ 20 = ββ―__
4 Γ ββ―__
5
Cancel by 2.
Work out the square roots ββ―__
4 and ββ―___ 49 .
5 β 4 + 7 = 8 | [
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13Algebraic expressions
Expand and simplify if possible:
a ββ―__
2 (5 β ββ―__
3 ) b (2 β ββ―__
3 )(5 + ββ―__
3 ) Example 13
a ββ―__
2 (5 β ββ―__
3 )
= 5 ββ―__
2 β ββ―__
2 ββ―__
3
= 5 ββ―__
2 β ββ―__
6
b (2 β
ββ―__
3 )(5 + ββ―__
3 )
= 2(5
+ ββ―__
3 ) β ββ―__
3 (5 + ββ―__
3 )
= 10
+ 2 ββ―__
3 β 5 ββ―__
3 β ββ―__
9
= 7 β
3 ββ―__
3 β __
2 Γ 5 β β __
2 Γ β __
3
Using ββ―__
a Γ β __
b = β ____ ab
Collect like terms: 2 β __
3 β 5 β __
3 = β3 β __
3
Simplify any roots if possible: β __
9 = 3
1 Do not use your calcula tor for this exercise. Simplify:
a ββ―___ 28 b ββ―___ 72 c ββ―___ 50
d ββ―__ 32 e ββ―__ 90 f ββ―__ 12 ____ 2
g ββ―___ 27 ____ 3 h ββ―__ 20 + ββ―__ 80 i ββ―___ 200 + ββ―__ 18 β ββ―__ 72
j ββ―___ 175 + ββ―__ 63 + 2 ββ―__ 28 k ββ―__ 28 β 2 ββ―__ 63 + ββ―__
7 l ββ―__ 80 β 2 ββ―__ 20 + 3 ββ―__ 45
m 3 ββ―___ 80 β 2 ββ―___ 20 + 5 ββ―___ 45 n ββ―___ 44 ____ ββ―___ 11 o ββ―___ 12 + 3 ββ―___ 48 + ββ―___ 75
2 Expand and simplify if possible:
a β __
3 (2 + β __
3 ) b β __
5 (3 β β __
3 ) c β __
2 (4 β β __
5 )
d (2 β β __
2 )(3 + β __
5 ) e (2 β β __
3 )(3 β β __
7 ) f (4 + β __
5 )(2 + β __
5 )
g (5 β β __
3 )(1 β β __
3 ) h (4 + β __
3 )(2 β β __
3 ) i (7 β β ___ 11 )(2 + β ___ 11 )
3 Simplify β ___ 75 β β ___ 12 giving your answer in the form a β __
3 , where a is an integer. (2 marks) EExercise 1E
1.6 Rationalising denominators
If a fraction has a surd in the denominator, it is sometimes useful to rearrange it so that the
denominator is a rational number. This is called rationalising the denominator.
β The rules to rationalise denominators are:
β’For fractions in the f
orm 1 ___
ββ―__
a , multiply the numerator and denominat
or by ββ―__
a .
β’For fractions in the f
orm 1 ______
a + ββ―__
b , multiply the numerator and denominat
or by a β ββ―__
b .
β’For fractions in the f
orm 1 ______
a β ββ―__
b , multiply the numerator and denominat
or by a + ββ―__
b .Expand the brackets completely before you simplify. | [
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14
Chapter 1
Rationalise the denominator of:
a 1 ___ ββ―__
3 b 1 ______ 3 + ββ―__
2 c ββ―__
5 + ββ―__
2 _______ ββ―__
5 β ββ―__
2 d 1 ________ (1 β ββ―__
3 )2 Example 14
a 1 ___ ββ―__
3 = 1 Γ ββ―__
3 ________ ββ―__
3 Γ ββ―__
3
= ββ―__
3 ___ 3
b 1 _______ 3 + ββ―__
2 = 1 Γ (3
β ββ―__
2 ) ________________ (3 + ββ―__
2 )(3 β ββ―__
2 )
= 3 β ββ―__
2 ___________________ 9 β 3 ββ―__
2 + 3 ββ―__
2 β 2
= 3 β ββ―__
2 _______ 7
c ββ―__
5 + ββ―__
2 ________ ββ―__
5 β ββ―__
2 = ( ββ―__
5 + ββ―__
2 )( ββ―__
5 + ββ―__
2 ) __________________ ( ββ―__
5 β ββ―__
2 )( ββ―__
5 + ββ―__
2 )
= 5 + ββ―__
5 ββ―__
2 + ββ―__
2 ββ―__
5 + 2 _____________________ 5 β 2
= 7 + 2 ββ―___ 10 __________ 3
d 1 _________ (1 β ββ―__
3 ) 2 = 1 ________________ (1 β ββ―__
3 )(1 β ββ―__
3 )
= 1 __________________ 1 β ββ―__
3 β ββ―__
3 + ββ―__
9
= 1 ________ 4 β 2 ββ―__
3
= 1 Γ
(4 + 2 ββ―__
3 ) __________________ (4 β 2 ββ―__
3 )(4 + 2 ββ―__
3 )
= 4 +
2 ββ―__
3 ______________________ 16 + 8 ββ―__
3 β 8 ββ―__
3 β 12
= 4 +
2 ββ―__
3 ________ 4 = 2 +
ββ―__
3 _______ 2 Expand the brackets.
β __
3 Γ β __
3 = 3 ββ―__
3 Γ ββ―__
3 = ( ββ―__
3 )2 = 3Multiply the numerator and denominator by ββ―__
3 .
Multiply numerator and denominator by (3 β ββ―__
2 ) .
ββ―__
2 Γ ββ―__
2 = 2
9 β 2 = 7, β3 ββ―__
2 + 3 ββ―__
2 = 0
ββ―__
5 ββ―__
2 = ββ―___ 10 Multiply numerator and denominator by ββ―__
5 + ββ―__
2 .
β ββ―__
2 ββ―__
5 and ββ―__
5 ββ―__
2 cancel each other out.
Simplify and collect like terms. β __
9 = 3
Multiply the numerator and denominator by
4 + 2 β __
3 .
16 β 12 = 4, 8 β __
3 β 8 β __
3 = 0 | [
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-0.0334... |
15Algebraic expressions
1 Simplify:
a 1 ___ ββ―__
5 b 1 ____ ββ―___ 11 c 1 ___ ββ―__
2 d ββ―__
3 ____ ββ―___ 15
e ββ―__ 12 ____ ββ―__ 48 f ββ―__
5 ____ ββ―___ 80 g ββ―___ 12 _____ ββ―____ 156 h ββ―__
7 ____ ββ―___ 63
2 Rationa
lise the denominators and simplify:
a 1 ______ 1 + ββ―__
3 b 1 ______ 2 + ββ―__
5 c 1 ______ 3 β ββ―__
7 d 4 ______ 3 β ββ―__
5 e 1 _______ ββ―__
5 β ββ―__
3
f 3 β ββ―__
2 ______ 4 β ββ―__
5 g 5 ______ 2 + ββ―__
5 h 5 ββ―__
2 _______ ββ―__
8 β ββ―__
7 i 11 _______ 3 + ββ―___ 11 j ββ―__
3 β ββ―__
7 _______ ββ―__
3 + ββ―__
7
k ββ―___ 17 β ββ―___ 11 _________ ββ―___ 17 + ββ―___ 11 l ββ―___ 41 + ββ―___ 29 _________ ββ―___ 41 β ββ―___ 29 m ββ―__
2 β ββ―__
3 _______ ββ―__
3 β ββ―__
2
3 Rationa
lise the denominators and simplify:
a 1 ________ (3 β ββ―__
2 ) 2 b 1 ________ (2 + ββ―__
5 ) 2 c 4 ________ (3 β ββ―__
2 ) 2
d 3 ________ (5 + ββ―__
2 ) 2 e 1 ______________ (5 + β __
2 )(3 β β __
2 ) f 2 ______________ (5 β β __
3 )(2 + β __
3 )
4 Simplify 3 β 2 ββ―__
5 _______ ββ―__
5 β 1 giving your answer in the
form p + q ββ―__
5 , where p and q are rational
numbers. (4 marks)E/P
You can check that your answer is in the correct
form by writing down the values of p and q and checking that they are rational numbers.Problem-solving
1 Simplify:
a y3 Γ y5 b 3x2 Γ 2x5 c (4x2)3 Γ· 2x5 d 4b2 Γ 3b3 Γ b4
2 Expand and simplify if possible:a
(x
+ 3)(x β 5) b (2x
β 7)(3x + 1) c (2x
+ 5)(3x β y + 2)
3 Expand and simplify if possible:a
x(x
+ 4)(x β 1) b (x
+ 2)(x β 3)(x + 7) c (2x
+ 3)(x β 2)(3x β 1)
4 Expand the brackets:a
3(5y
+ 4) b 5x2(3 β 5x + 2x2) c 5x(2 x + 3) β 2x(1 β 3x) d 3x2(1 + 3x) β 2x(3x β2)Exercise 1F
Mixed exercise 1 | [
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16
Chapter 1
5 Factorise these expr
essions completely:
a 3x2 + 4x b 4y2 + 10y c x2 + xy + xy2 d 8xy2 + 10x2y
6 Factorise:
a x2 + 3x + 2 b 3x2 + 6x c x2 β 2x β 35 d 2x2 β x β 3
e 5x2 β 13x β 6 f 6 β 5 x β x2
7 Factorise:
a 2x3 + 6x b x3 β 36x c 2x3 + 7x2 β 15x
8 Simplify:
a 9x3 Γ· 3xβ3 b ( 4 3 _ 2 ) 1 _ 3 c 3xβ2 Γ 2x4 d 3 x 1 _ 3 Γ· 6 x 2 _ 3
9 Eva
luate:
a ( 8 ___ 27 ) 2 _ 3
b ( 225 ____ 289 ) 3 _ 2
10 Simplify:
a 3 ____ ββ―___ 63 b ββ―__ 20 + 2 ββ―__ 45 β ββ―__ 80
11 a Find the value of
35x2 + 2x β 48 when x = 25.
b By factorising the expression, sho
w that your answer to part a can be written as the product
of two prime factors.
12 Expand and simplify if possible:
a ββ―__
2 (3 + ββ―__
5 ) b (2 β ββ―__
5 )(5 + ββ―__
3 ) c (6 β ββ―__
2 )(4 β ββ―__
7 )
13 Rationa
lise the denominator and simplify:
a 1 ____ ββ―___
3 b 1 ______ ββ―__
2 β 1 c 3 ______ ββ―__
3 β 2 d ββ―___ 23 β ββ―___ 37 _________ ββ―___ 23 + ββ―___ 37 e 1 ________ (2 + ββ―__
3 )2 f 1 ________ (4 β ββ―__
7 )2
14 a Given tha
t x3 β x2 β 17x β 15 = (x + 3)(x2 + bx + c), where b and c are constants, work out
the values of b and c.
b Hence, fully factorise
x3 β x2 β 17x β 15.
15 Given tha
t y = 1 __ 64 x 3 express each of the following in the form k x n , where k and n are constants.
a y 1 _ 3 (1 mark)
b 4 y β1 (1 mark)
16 Show that 5 _________ ββ―___ 75 β ββ―___ 50 can be written in the form ββ―__
a + ββ―__
b , where a and b are integers. (5 marks)
17 Expand and simplify ( ββ―___ 11 β 5)(5 β ββ―___ 11 ) . (2 marks)
18 Factorise completel
y x β 64 x 3 . (3 marks)
19 Express 27 2x + 1 in the form 3 y , stating y in terms of x. (2 marks)E
E/P
E
E
E/P | [
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-0.04... |
17Algebraic expressions
20 Solve the equation 8 + x ββ―___ 12 = 8x ___ ββ―__
3
Give y
our answer in the form a ββ―__
b where a and b are integers. (4 marks)
21 A rectangle has a length of (1 + ββ―__
3 ) cm and area of ββ―___ 12 cm2.
Calculate the width of the rectangle in cm.
Express your answer in the form a + b ββ―__
3 , where a and b are integers to be found.
22 Show that (2 β ββ―__
x ) 2 ________ ββ―__
x can be written as 4 x β 1 _ 2 β 4 + x 1 _ 2 . (2 marks)
23 Given tha
t 243 ββ―__
3 = 3 a , find the value of a. (3 marks)
24 Given tha
t 4 x 3 + x 5 _ 2 ________ ββ―__
x can be written in the form 4 x a + x b , write down the value of a
and the value of b. (2 marks)E/P
P
E
E/P
E/P
1 You can use the laws of indices to simplify powers of the same base.
β am Γ an = am + n β am Γ· an = am β n
β (am)n = amn β (ab )n = anbn
2 Factorising is the opposite of expanding brackets.
3 A quadratic expr
ession has the form ax2 + bx + c where a, b and c are real numbers and a β 0.
4 x2 β y2 = (x + y)(x β y)
5 You can use the law
s of indices with any rational power.
β a 1 __ m = m ββ―__
a β a n __ m = m ββ―__
a n
β a βm = 1 ___ a m β a 0 = 1
6 You can manipulate sur
ds using these rules:
β ββ―____ ab = ββ―__
a Γ ββ―__
b β ββ―__
a __ b = ββ―__
a ___
ββ―__
b
7 The rules to r
ationalise denominators are:
β Fractions in the f
orm 1 ___
ββ―__
a , multiply the numerator and denominat
or by ββ―__
a .
β Fractions in the f
orm 1 ______
a + ββ―__
b , multiply the numerator and denominat
or by a β ββ―__
b .
β Fractions in the f
orm 1 ______
a β ββ―__
b , multiply the numerator and denominat
or by a + ββ―__
b .Summary of key pointsa Simplify ( ββ―__
a + ββ―__
b ) ( ββ―__
a β ββ―__
b ) .
b Hence show that 1 _______
ββ―__
1 + ββ―__
2 + 1 _______
ββ―__
2 + ββ―__
3 + 1 _______
ββ―__
3 + ββ―__
4 + ... +
1 _________
ββ―___ 24 + ββ―___ 25 = 4Challenge | [
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-0.03... |
18
Quadratics
After completing this chapter you should be able to:
β Solve quadratic equations using fact
orisation, the quadratic
formula and completing the square β pages 19 β 24
β Read and use f(x) notation when working with
functions β pages 25 β 27
β Sketch the graph and find the turning point of a quadratic function
β pages 27 β 30
β Find and interpret the discriminant o f a quadratic
expression β pages 30 β 32
β Use and apply models that involv e quadratic
functions β pages 32 β 35Objectives
1 Solve the following equations:
a 3x
+ 6 = x β 4
b 5(x
+ 3) = 6(2x β 1)
c 4x2 = 100
d (x β
8)2 = 64 β GCSE Mathematics
2 Factorise the following expressions:
a x2 + 8x + 15 b x2 + 3x β 10
c 3x2 β 14x β 5 d x2 β 400
β Section 1.3
3 Sketch the graphs of the following equations, labelling the points wher
e each
graph crosses the axes:
a y =
3x β 6 b y =
10 β 2x
c x +
2y = 18 d y =
x2
β GCSE Mathematics
4 Solve the following inequalities:
a x +
8 , 11 b 2x
β 5 > 13
c 4x
β 7 < 2 (x β 1) d 4 β
x , 11
β GCSE MathematicsPrior knowledge check
Quadratic functions are used to model projectile motion. Whenever an object is thrown or launched, its path will approximately follow the shape of a parabola.
β Mixed exercise Q112 | [
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19Quadratics
2.1 Solving quadratic equations
A quadratic equation can be written in the form ax2 + bx + c = 0, where a, b and c are real constants,
and a β 0. Quadratic equations can have one, two, or no real solutions.
β To solve a quadratic equation by factorising:
β’ Writ
e the equation in the form ax2 + bx + c = 0
β’ Factorise the left-hand side
β’ Set each factor equal to zero and solve to find the value(s) of x
Example 1 The s olutions to an
equation are sometimes called
the roots of the equation.Notation
The symbol β means βimplies thatβ .
This statement says βIf x + 3 = 0, then x = β3β.Notation
a x2 β 2x β 15 = 0
(x
+ 3)( x β 5) = 0
The
n either x + 3 = 0 β
x = β 3
or x β
5 = 0 β
x = 5
So x = β 3 and x = 5 are the two solutions
of the equation.
b x2 = 9x
x2 β 9 x = 0
x(x β
9) = 0
Th
en either x = 0
or x β
9 = 0 β
x = 9
The solutions are x = 0 and x = 9.
c 6x2 + 13 x β 5 = 0
(3x β
1)(2x + 5) = 0
The
n either 3 x β 1 = 0 β
x = 1 __ 3
or 2x
+ 5 = 0 β
x = β 5 __ 2
The s
olutions are x = 1 __ 3 and x = β 5 __ 2
d x2 β 5 x + 18 = 2 + 3 x
x2 β 8 x + 16 = 0
(x β
4)(x β 4) = 0
Th
en either x β 4 = 0 β
x = 4
or x β
4 = 0 β
x = 4
β x = 4Factorise. The s igns of the solutions are
opposite to the signs of the constant terms in
each factor.Watch outSolve the following equations:
a x2 β 2x β 15 = 0 b x2 = 9x
c 6x2 + 13x β 5 = 0 d x2 β 5x + 18 = 2 + 3xIf the product of the factors is zero, one of the
factors must be zero.Factorise the quadratic. β Section 1.3
A quadratic equation with two distinct factors has two distinct solutions.
Be careful not to divide both sides by x, since x may have the value 0. Instead, rearrange into the form ax
2 + bx + c = 0.
Factorise.
Solutions to quadratic equations do not have to be integers.
The quadratic equation (px + q)(rx + s) = 0 will have solutions x = β
q __ p and x = β s __ r .
When a quadratic equation has
e
xactly one root it is called a repeated root. You
can also say that the equation has two equal roots.NotationRearrange into the form ax2 + bx + c = 0.
Factorise. | [
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20
Chapter 2
In some cases it may be more straightforward to solve a quadratic equation without factorising.
Example 2
Solve the following equations
a (2x
β 3)2 = 25 b (x β 3)2 = 7
a (2x β 3)2 = 25
2x
β 3 = Β±5
2x = 3 Β±
5
The
n either 2x = 3 + 5 β
x = 4
or 2x = 3 β
5 β x = β 1
The solutions are x = 4 and x = β 1
b (x
β 3)2 = 7
x β
3 = Β± ββ―__
7
x = 3 Β±
ββ―__
7
The s
olutions are x = 3 + ββ―__
7 and
x = 3 β ββ―__
7 The symbol Β± lets you write two
statements in one line of working. You say
βplus or minusβ.Notation
Add 3 to both sides.
1 Solve the following equations using factorisation:
a x2 + 3x + 2 = 0 b x2 + 5x + 4 = 0 c x2 + 7x + 10 = 0 d x2 β x β 6 = 0
e x2 β 8x + 15 = 0 f x2 β 9x + 20 = 0 g x2 β 5x β 6 = 0 h x2 β 4x β 12 = 0
2 Solve the follo
wing equations using factorisation:
a x2 = 4x b x2 = 25x c 3x2 = 6x d 5x2 = 30x
e 2x2 + 7x + 3 = 0 f 6x2 β 7x β 3 = 0 g 6x2 β 5x β 6 = 0 h 4x2 β 16x + 15 = 0
3 Solve the follo
wing equations:
a 3x2 + 5x = 2 b (2x β 3)2 = 9 c (x β 7)2 = 36 d 2x2 = 8 e 3x2 = 5
f (x
β 3)2 = 13 g (3x β 1)2 = 11 h 5x2 β 10x2 = β7 + x + x2
i 6x2 β 7 = 11x j 4x2 + 17x = 6x β 2x2
4 This shape has an area of 44 m2.
Find the value of x.
5 Solve the equation 5
x + 3 = ββ―______ 3x + 7 .2x mx m
x m(x + 3) mP
PExercise 2A
Divide the shape into two sections:Problem-solvingTake the square root of both sides.
Remember 52 = (β5)2 = 25.
Take square roots of both sides.
You can leave your answer in surd form. | [
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21Quadratics
x = β (β7) Β± β ______________ (β7) 2 β 4 (3) (β1) _______________________ 2 Γ 3
x =
7 Β± β _______ 49 + 12 _______________ 6
x =
Β 7 Β± β ___ 61 ________ 6 Β
The
n x =
7 + β ___ 61 ________ 6 or x = 7 β β ___ 61 _______ 6
Or x
= 2.47 (3 s.f.) or x = β 0.135 (3 s.f.)Example 3
Solve 3x2 β 7x β1 = 0 by using the formula.
β4 Γ 3 Γ (β1) = +12a = 3, b = β7 and c = β1.
Put brackets around any negative values.Some equations cannot be easily factorised. You can also solve quadratic equations using the
quadratic formula.
β The solutions of the equation
ax2 + bx + c = 0 are given by the formula:
x = βb Β± ββ―________ b2 β 4ac _____________ 2a You n eed to rearrange the equation
into the form ax2 + bx + c = 0 before reading off
the coefficients.Watch out
1 Solve the follo
wing equations using the quadratic formula.
Give your answers exactly, leaving them in surd form where necessary.
a x2 + 3x + 1 = 0 b x2 β 3x β 2 = 0 c x2 + 6x + 6 = 0 d x2 β 5x β 2 = 0
e 3x2 + 10x β 2 = 0 f 4x2 β 4x β 1 = 0 g 4x2 β 7x = 2 h 11x2 + 2x β 7 = 0
2 Solve the follo
wing equations using the quadratic formula.
Give your answers to three significant figures.a
x2 + 4x + 2 = 0 b x2 β 8x + 1 = 0 c x2 + 11x β 9 = 0 d x2 β 7x β 17 = 0
e 5x2 + 9x β 1 = 0 f 2x2 β 3x β 18 = 0 g 3x2 + 8 = 16x h 2x2 + 11x = 5x2 β 18
3 For each of the equa
tions below, choose a suitable method and find all of the solutions.
Where necessary, give your answers to three significant figures.a
x2 + 8x + 12 = 0 b x2 + 9x β 11 = 0
c x2 β 9x β 1 = 0 d 2x2 + 5x + 2 = 0
e (2x +
8)2 = 100 f 6x2 + 6 = 12x
g 2x2 β 11 = 7x h x = Β ββ―_______ 8x β 15 You can use any method
yo
u are confident with to solve
these equations.HintExercise 2B | [
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22
Chapter 2
Given that x is positive, solve the equation
1 __ x + 1 _____ x + 2 = 28 ____ 195 Challenge Write the equation in the form
ax2 + bx + c = 0 before using the quadratic
formula or factorising.Hint
2.2 Completing the square
It is frequently useful to rewrite quadratic expressions by completing the square:
β x2 + bx = (x + b __ 2 ) 2
β ( b __ 2 ) 2
You can draw a diagram of this process when x and b
are positive:
The original rectangle has been rearranged into the
shape of a square with a smaller square missing. The two areas shaded blue are the same.b
2
x xx x =
b
2b
x2 + bx = (x + b __ 2 ) 2
β ( b __ 2 ) 2
a x2 + 8 x = ( x + 4)2 β 42
= (x + 4)2 β 16
b x2 β 3 x = (x β 3 _ 2 ) 2 β ( 3 _ 2 ) 2
= (x β 3 _ 2 ) 2 β 9 _ 4
c 2x2 β 12 x = 2( x2 β 6 x)
= 2(( x β 3)2 β 32)
= 2(( x β 3)2 β 9)
= 2(x β 3)2 β 18Example 4
Complete the square for the expressions:
a x2 + 8x b x2 β 3x c 2x2 + 12x A quadratic expression in the
form p(x + q)2 + r where p, q and r are real
constants is in completed square form.Notation
Begin by halving the coefficient of x. Using the
rule given above, b = 8 so b __ 2 = 4.
Expand the outer bracket by multiplying 2 by 9 to
get your answer in this form.4 This trapezium has an area of 50 m2.
Show that the height of the trapezium is equal to 5( ββ―__
5 β 1) m.
(x + 10) mx m
2x m
Height must be positive. You will have to discard
the negative solution of your quadratic equation.Problem-solvingP
Be careful if b __ 2 is a fraction. Here ( 3 __ 2 ) 2
= 3 2 __ 2 2 = 9 __ 4 .
Here the coefficient of x2 is 2, so take out a factor
of 2. The other factor is in the form (x2 + bx) so
you can use the rule to complete the square. | [
0.04041701555252075,
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-0.02522948756814003,
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0.02437373623251915,
-0.053487... |
23Quadratics
3x2 + 6 x + 1
= 3(x2 + 2x) + 1
= 3(( x + 1)2 β 12) + 1
= 3(x + 1)2 β 3 + 1
= 3(x + 1)2 β 2
So p = 3, q = 1 and r = β 2.Example 5
Write 3x2 + 6x + 1 in the form p(x + q)2 + r, where p, q and r are integers to be found.
1 Complete the square for the e
xpressions:
a x2 + 4x b x2 β 6x c x2 β 16x d x2 + x e x2 β 14
2 Complete the square for the e
xpressions:
a 2x2 + 16x b 3x2 β 24x c 5x2 + 20x d 2x2 β 5x e 8x β 2x2
3 Write each of these expressions in the form p(x + q)2 + r, where p, q and r are constants
to be found:
a 2x2 + 8x + 1 b 5x2 β 15x + 3 c 3x2 + 2x β 1 d 10 β 16 x β 4x2 e 2x β 8x2 + 10
4 Given tha
t x2 + 3x + 6 = (x + a)2 + b, find the values of the constants a and b. (2 marks)
5 Write 2
+ 0.8x β 0.04x2 in the form A β B(x + C)2, where A, B and C are constants to
beΒ determined. (3 marks)E
EExercise 2C
Solve the equation x2 + 8x + 10 = 0 by completing the square.
Give your answers in surd form.Example 6
x2 + 8 x + 10 = 0
x2 + 8 x = β10
(x + 4)2 β 42 = β10
(x + 4)2 = β10 + 16
(x + 4)2 = 6
(x + 4) = Β± ββ―__
6
x =
β4 Β± ββ―__
6
So th
e solutions are
x = β4 + ββ―__
6 and x = β 4 β ββ―__
6 .Check coefficient of x2 = 1.
Subtract 10 to get the LHS in the form x2 + bx.
Complete the square for x2 + 8x.
Add 42 to both sides.
Take square roots of both sides.
Subtract 4 from both sides.
Leave your answer in surd form.β a x 2 + bx + c = a (x + b ___ 2a ) 2
+ (c β b 2 ____ 4 a 2 )
You could also use the rule given above to
complete the square for this expression, but it is safer to learn the method shown here. This is an expression , so you canβt
divide every term by 3 without changing its value. Instead, you need to take a factor of 3 out of 3x
2 + 6x .Watch out
In question 3d ,
wr
ite the expression as
β4x2 β 16 x + 10 then
take a factor of β 4 out
of the first two terms to get β 4(x
2 + 4x) + 10.Hint | [
-0.01306331530213356,
0.07446865737438202,
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-0.047031763941049576,
0.03577741980552673,
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-0.05958917737007141,
0.0121842036023736,
0.0020413044840097427,
-0.005... |
24
Chapter 2
Solve the equation 2x2 β 8x + 7 = 0. Give your answers in surd form.Example 7
2x2 β 8 x + 7 = 0
x2 β 4 x + 7 __ 2 = 0
x2 β 4 x = β 7 __ 2
(x β
2)2 β 22 = β 7 __ 2
(x β
2)2 = β 7 __ 2 + 4
(x
β 2)2 = 1 __ 2
x β
2 = Β± ββ―__
1 __ 2
x =
2 Β± 1 ___ ββ―__
2
So th
e roots are
x = 2 + 1 ___ ββ―__
2 and x = 2 β 1 ___ ββ―__
2 Complete the square for x2 β 4x.
Add 22 to both sides.
Take square roots of both sides.
Add 2 to both sides.
1 Solve these quadratic equations by completing the square. Leave your answers in surd form.
a x2 + 6x + 1 = 0 b x2 + 12x + 3 = 0 c x2 + 4x β 2 = 0 d x2 β 10x = 5
2 Solve these quadra
tic equations by completing the square. Leave your answers in surd form.
a 2x2 + 6x β 3 = 0 b 5x2 + 8x β 2 = 0 c 4x2 β x β 8 = 0 d 15 β 6 x β 2x2 = 0
3 x2 β 14x + 1 = (x + p)2 + q, where p and q are constants.
a Find the values of
p and q. (2 marks)
b Using your answ
er to part a, or otherwise, show that the solutions to the equation
x2 β 14x + 1 = 0 can be written in the form r Β± s ββ―__
3 , where r and s are constants
to be found. (2 marks)
4 By completing the square, sho
w that the solutions to
the equation x2 + 2bx + c = 0 are given by the formula
x = β b Β± ββ―______ b 2 β c . (4 marks)E
E/P
Follow the same steps as you would
if the coefficients were numbers.Problem-solvingExercise 2D
Start by dividing the whole
equation by a .Hint
You can use this
methodΒ to prove the quadratic
formula. β Section 7.4Linksa Show that the solutions to the equation
ax2 + 2bx + c = 0 are given by x = β b __ a Β± ββ―______ b 2 β ac ______ a 2 .
b Hence, or otherwise, sho
w that the solutions to the
equation ax2 + bx + c = 0 can be written as
x = βb Β± ββ―_______ b 2 β 4ac ____________ 2a .Challenge Use your calculator to check
so
lutions to quadratic equations quickly.OnlineThis is an equation so you can divide every term
by the same constant. Divide by 2 to get x2 on its
own. The right-hand side is 0 so it is unchanged.Problem-solving | [
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-0.071272... |
25Quadratics
2.3 Functions
A function is a mathematical relationship that maps each value of a set of inputs to a single output.
The notation f(x) is used to represent a function of x.
β The set of possible inputs for a function is called the domain.
3DomainR ange
7
β7
2f(β7) = 49f(7) = 49f(3) = 9
f( 2) = 29
49
49
21
41
16 f( ) = 14 1
16β The set of possible outputs of a function is called the range.
This diagram sho
ws how the function f(x) = x2 maps five
values in its domain to values in its range.
β The roots of a function are the values of x for which f( x) = 0.
The functions f and g are given by f(x) = 2x β 10 If the i nput of a function,
x, can be any real number the
domain can be written as x β
β .
The s
ymbol β mean
s βis a member
ofβ and the symbol β rep
resents the
real numbers.Notation
and g(x) = x2 β 9, xΒ β β .
a Find the values of
f(5) and g(10).
b Find the value of
x for which f(x) = g(x).Example 8
a f(5) = 2(5) β 10 = 10 β 10 = 0
g
(10) = (10)2 β 9 = 100 β 9 = 91
b f(x) =
g(x)
2x β 10 = x2 β 9
x2 β 2 x + 1 = 0
(x β 1)2 = 0
x = 1To find f(5), substitute x = 5 into the function f(x).
Set f(x) equal to g(x) and solve for x.
The function f is defined as f(x) = x2 + 6x β 5, Β x β β .
a Write f(x
) in the form (x + p)2 + q.
b Hence, or otherwise, find the r
oots of f(x), leaving your answers in surd form.
c Write down the minim
um value of f(x), and state the value of x for which it occurs.Example 9
a f(x) = x2 + 6 x β 5
= (x + 3)2 β 9 β 5
= (x + 3)2 β 14
b f(x) =
0
(x + 3)2 β 14 = 0
(x + 3)2 = 14
x + 3 = Β± ββ―____ 14
x = β3 Β± ββ―____ 14
f(x) has two roots:
β3 + ββ―____ 14 and β 3 β ββ―____ 14 .Complete the square for x2 + 6x and then
simplify the expression.
You can solve this equation directly. Remember to
write Β± when you take square roots of both sides.To find the root(s) of a function, set it equal to zero. | [
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0.101796... |
26
Chapter 2
c (x + 3)2 > 0
So the minimum value of f( x) is β14.
This occurs when ( x + 3)2 = 0,
so when x = β 3A squared value must be greater than or equal to 0.
Find the roots of the function f(x) = x6 + 7x3 β 8, xΒ β β .Example 10
f(x) = 0
x6 + 7x3 β 8 = 0
(x3)2 + 7( x3) β 8 = 0
(x3 β 1)( x3 + 8) = 0
So x3 = 1 or x3 = β8
x3 = 1 β x = 1
x3 = β8 β x = β 2
The roots of f( x) are 1 and β 2.
Alternatively, let u = x3.
f(x) = x6 + 7x3 β 8
= (x3)2 + 7( x3) β 8
= u2 + 7u β 8
= (u β 1)( u + 8)
So when f( x) = 0, u = 1 or u = β 8.
If u = 1 β x3 = 1 β x = 1
If u = β 8 β x3 = β8 β x = β 2
The roots of f( x) are 1 and β 2.Treat x3 as a single variable and factorise.
Solve the quadratic equation to find two values
for x3, then find the corresponding values of x.
You can simplify this working with a substitution.
Replace x3 with u and solve the quadratic
equation in u.
The s olutions to the quadratic
equation will be values of u. Convert back to values of x using your substitution.Watch out
1 Using the functions f(x)
= 5x + 3, g(x) = x2 β 2 and h(x) = ββ―_____ x + 1 , find the values of:
a f(1) b g(3) c h(8) d f(1.5) e g ( ββ―__
2 )
f h (β1) g f(4) + g(2) h f(0) + g(0)
+ h(0) i g(4) ____ h(3)
2 The function f(x) is defined b
y f(x) = x2 β 2x, x β β .
Giv
en that f(a) = 8, find two possible values for a.
3 Find all of the r
oots of the following functions:
a f(x)
= 10 β 15x b g(x)
= (x + 9)(x β 2) c h(x)
= x2 + 6x β 40
d j(x)
= 144 β x2 e k(x) = x(x + 5)(x + 7) f m(x) = x3 + 5x2 β 24xSubstitute x = a into the function and
set the resulting expression equal to 8.Problem-solving PExercise 2Ef(x) can be written as a function of a function.
The only powers of x in f( x) are 6, 3 and 0 so you
can write it as a quadratic function of x3.Problem-solving(x + 3)2 > 0 so (x + 3)2 β 14 > β14 | [
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-0.06614738702774048,
-0.01590130478143692,
-0.00505... |
27Quadratics
4 The functions p and q are giv
en by p(x) = x2 β 3x and q(x) = 2x β 6, x β β .
Find the two v
alues of x for which p(x) = q(x).
5 The functions f and g are gi
ven by f(x) = 2x3 + 30x and g(x) = 17x2, Β x β β .
Find the three v
alues of x for which f(x) = g(x).
6 The function f is defined as f(x
) = x2 β 2x + 2, x β β .
a Write f(x
) in the form (x + p)2 + q, where p and q are constants to be found. (2 marks)
b Hence, or otherwise, e
xplain why f(x) > 0 for all values of x, and find the minimum
value of f(x). (1 mark)
7 Find all roots of
the following functions:
a f(x)
= x6 + 9x3 + 8 b g(x) = x4 β 12x2 + 32
c h(x)
= 27x6 + 26x3 β 1 d j(x) = 32x10 β 33x5 + 1
e k(x)
= x β 7 ββ―__
x + 10 f m(x) = 2 x 2 _ 3 + 2 x 1 _ 3 β 12
8 The function f is defined as f(x
) = 32x β 28(3x) + 27, x β β .
a Write f(x
) in the form (3x β a)(3x β b), where a and b are
real constants. (2 marks)
b Hence find the two roots of
f(x). (2 marks)E
The function in
par
t b has four roots.Hint
E/P
Consider f( x) as a
function of a function.Problem-solving
2.4 Quadratic graphs
When f(x) = ax2 + bx + c, the graph of y = f(x) has a curved shape called a parabola.
You can sketch a quadratic graph by identifying key features.
The coefficient of x2 determines the overall shape of the graph.
When a is positive the parabola will have this shape:
When a is negative the parabola will have this shape:
y
x Oy
x O1 The graph crosses the y-axis when
x
= 0. The y-coordinate is equal to c.
3 Quadratic graphs have one turning point. This can be a minimum or a maximum. Sinc
eΒ a parabola is symmetrical, the turning
point and line of symmetry are half-way between the two roots. 2
The graph crosses the x-axis when y = 0. The x-c
oordinates are roots of the function f(x). 11
2 2 2 2
33
β You can find the coordinates of the turning point
of a quadr
atic graph by completing the square.
If f(x) = a(x + p)2 + q, the graph of y = f(x) has a
turning point at (β p, q). The graph of y = a (x + p )2 + q
is a translation of the graph of
y = ax2 by ( βp q ) . β Section 4.5Links | [
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28
Chapter 2
As a = 1 is positive, the graph has a
shape and a minimum point.
When x = 0, y = 4, so the graph crosses
the y-axis at (0, 4).
When y = 0,
x2 β 5 x + 4 = 0
(x β 1)( x β 4) = 0
x = 1 or x = 4, so the graph crosses the x-axis at (1, 0) and (4, 0).
Completing the square:
x
2 β 5 x + 4 = (x β 5 _ 2 ) 2 β 25 __ 4 + 4
= (x β 5 _ 2 ) 2 β 9 _ 4
So th
e minimum point has coordinates
( 5 _ 2 , β 9 _ 4 ) .
Alternatively, the minimum occurs when
x is half-way between 1 and 4,
so x = 1 + 4 _______ 2 = 5 _ 2
y = ( 5 _ 2 ) 2 β 5 Γ ( 5 _ 2 ) + 4 = β 9 _ 4
so th
e minimum has coordinates ( 5 _ 2 , β 9 _ 4 ) .
The sketch of the graph is:
O xy
/four.ss01
1 /four.ss01
, β5
294()Use the coefficient of x2 to determine the general
shape of the graph.
This example factorises, but you may need to use
the quadratic formula or complete the square.
Complete the square to find the coordinates of the turning point.
You could use a graphic calculator or substitute some values to check your sketch.
When x = 5, y = 5
2 β 5 Γ 5 + 4 = 4.Sketch the graph of y = x2 β 5x + 4, and find the coordinates of its turning point.Example 11
If yo u use symmetry to find the
x-coordinate of the minimum point, you need to
substitute this value into the equation to find the
y-coordinate of the minimum point.Watch out
Explore how the graph of
y
= (x + p )2 + q changes as the values of p
and q change using GeoGebra.Online | [
0.03143851459026337,
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29Quadratics
As a = β 2 is negative, the graph has a
shape and a maximum point.
When x = 0, y = β 3, so the graph
crosses the y -axis at (0, β 3).
When y = 0,β2x
2 + 4x β 3 = 0
Using the quadratic formula,
x = β4 Β± β _____________ 4 2 β 4 (β2) (β3) ____________________ 2 Γ (β2)
x =
β4 Β± β ____ β8 __________ β4
The
re are no real solutions, so the graph
does not cross the x -axis.
Completing the square:
β2x2 + 4x β 3
= β2(x2 β 2x) β 3
= β2((x β 1)2 β 1) β 3
= β2(x β 1)2 + 2 β 3
= β2(x β 1)2 β 1
So the maximum point has coordinates
(1, β1).
The line of symmetry is vertical and goes through the maximum point. It has the equation x = 1.
O xy
(1, β1)
β3a = β2, b = 4 and c = β3Sketch the graph of y = 4x β 2x2 β 3. Find the coordinates of its turning point and write down the
equation of its line of symmetry.Example 12
A ske tch graph does not need to be
plotted exactly or drawn to scale. However you
should:β
dra
w a smooth curve by hand
β ide
ntify any relevant key points (such as
intercepts and turning points)
β lab
el your axes.Watch outItβs easier to see that a , 0 if you write the equation in the form y = β2x
2 + 4x β 3.
You would need to square root a negative number to evaluate this expression. Therefore this equation has no real solutions.
The coefficient of x2 is β2 so take out a factor of β 2 | [
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30
Chapter 2
1 Sketch the gra
phs of the following equations. For each graph, show the coordinates of the point(s)
where the graph crosses the coordinate axes, and write down the coordinate of the turning point
and the equation of the line of symmetry.
a y =
x2 β 6x + 8 b y = x2 + 2x β 15 c y = 25 β x2 d y = x2 + 3x + 2
e y =
βx2 + 6x + 7 f y = 2x2 + 4x + 10 g y = 2x2 + 7x β 15 h y = 6x2 β 19x + 10
i y =
4 β 7x β 2x2 j y = 0.5x2 + 0.2x + 0.02
2 These sketches are gr
aphs of quadratic functions of the form ax2 + bx + c.
Find the values of a, b and c for each function.
a y
x15
5 3y = f(x) b y
x10
β2
5y = g(x)
c y
x
β183 β3
y = h(x) d y
x
β14β1
y = j(x)
3 The graph of y = ax2 + bx + c has a minimum at (5, β3) and passes through (4, 0).
Find the values of a, b and c. (3 marks)P
E/PExercise 2F
2.5 The discriminant
If you square any real number, the result is greater than or equal to 0. This means that if y is negative,
ββ―__
y cannot be a real number. Look at the quadratic formula:
x = βb Β± ββ―_______ b 2 β 4ac ____________ 2a
β For the quadratic function f( x) = ax2 + bx + c, the expression b2 β 4ac is called the
discriminant. The value of the discriminant shows how many roots f( x) has:
β’If b2 β 4ac . 0 then f( x) has two distinct real roots.
β’If b2 β 4ac = 0 then f( x) has one repeated root.
β’If b2 β 4ac , 0 then f( x) has no real roots.Check your answers
by substituting values into the function. In part c the graph passes through (0, β 18), so h(0)
should be β 18.Problem-solving
If the value under the square root sign is negative, x cannot be a real number and there are no real solutions. If the value under the square root is equal to 0, both solutions will be the same. | [
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-0.013... |
31Quadratics
You can use the discriminant to check the shape of sketch graphs.
Below are some graphs of y = f(x) where f(x) = ax2 + bx + c.
a . 0
y
x O
y
x O y
x O
b2 β 4ac . 0 b2 β 4ac = 0 b2 β 4ac , 0
Two distinct real roots One repeated r oot No real roots
a
, 0
y
x O
y
x O
y
x O
Find the range of values of k for which x2 + 4x + k = 0 has two distinct real solutions.Example 14
x2 + 4x + k = 0
Here a = 1, b = 4 and c = k .
For two real solutions, b2 β 4 ac . 0
42 β 4 Γ 1 Γ k . 0
16 β 4 k . 0
16 . 4 k
4 . k
So k , 4This statement involves an inequality, so your
answer will also be an inequality.Find the values of k for which f(x) = x2 + kx + 9 has equal roots.Example 13
x2 + kx + 9 = 0
Here a = 1, b = k and c = 9
For equal roots, b2 β 4 ac = 0
k2 β 4 Γ 1 Γ 9 = 0
k2 β 36 = 0
k2 = 36
so k = Β± 6Use the condition given in the question to write a
statement about the discriminant.Problem-solving
Substitute for a, b and c to get an equation with one unknown.
Solve to find the values of k.
For any value of k less than 4, the equation will have 2 distinct real solutions.
Explore how the value of the
di
scriminant changes with k using GeoGebra.Online | [
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-0.120516836643219,
0.06300268322229385,
-0.11026746034622192,
0.007404150906950235,
-0.03229108825325966,
-0.08342619240283966,
-0.05215401574969292,
0.0336... |
32
Chapter 2
1 a Calcula
te the value of the discriminant for each of these five functions:
i f(x)
= x2 + 8x + 3 ii g(x) = 2x2 β 3x + 4 iii h(x) = βx2 + 7x β 3
iv j(x)
= x2 β 8x + 16 v k(x ) = 2x β 3x2 β 4
b Using your answ
ers to part a, match the same five functions to these sketch graphs.
i
x Oy ii
x Oy iii
Oy iv
x Oy v
x Oy
2 Find the values of k for which x2 + 6x + k = 0 has two real solutions. (2 marks)
3 Find the value of
t for which 2x2 β 3x + t = 0 has exactly one solution. (2 marks)
4 Given tha
t the function f(x ) = sx2 + 8x + s has equal roots, find the value of the positive
constant s. (2 marks)
5 Find the range of v
alues of k for which 3x2 β 4x + k = 0 has no real solutions. (2 marks)
6 The function g(x)
= x2 + 3px + (14p β 3), where p is an integer, has two equal roots.
a Find the value of
p. (2 marks)
b For this va
lue of p, solve the equation x2 + 3px + (14p β 3) = 0. (2 marks)
7 h(x)
= 2x2 + (k + 4)x + k, where k is a real constant.
a Find the discriminant of h(x ) in ter
ms of k . (3 marks)
b Hence or otherwise, pro
ve that h(x ) has two distinct
real roots for all values of k . (3 marks)E/P
E/P
E/P
E/P
E/P
E/PExercise 2G
2.6 Modelling with quadratics
A mathematical model is a mathematical description of a real-life situation. Mathematical
models use the language and tools of mathematics to represent and explore real-life patterns and relationships, and to predict what is going to happen next.
Models can be simple or complicated, and their results can be approximate or exact. Sometimes a model
is only valid under certain circumstances, or for a limited range of inputs. You will learn more about how models involve simplifications and assumptions in Statistics and Mechanics.
Quadratic functions can be used to model and explore a range of practical contexts, including
projectile motion.a Prove that, if the values of a and c are given and non-zero, it is always possible to choose a value of
b so that f(x) = ax2 + bx + c has distinct real roots.
b Is it alway
s possible to choose a value of b so that f(x) has equal roots? Explain your answer.ChallengeIf a question part says βhence or
otherwiseβ it is usually easier to use your answer to the previous question part.Problem-solving | [
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0.025897622108459473,
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-0.06933324784040451,
-0.059752628207206726,
0.02... |
33Quadratics
A spear is thrown over level ground from the top of a tower.
The height, in metres, of the spear above the ground after t seconds is modelled by the function:
h(t) = 12.25 + 14.7t β 4.9t2, t > 0
a Interpret the meaning of the constant ter
m 12.25 in the model.
b After how many seconds does the spear hit the gr
ound?
c Write h(t
) in the form A β B(t β C)2, where A, B and C are constants to be found.
d Using your answ
er to part c or otherwise, find the maximum height of the spear above the
ground, and the time at which this maximum height is reached.Example 15
a The tower is 12.25 m tal l, since
this is the height at time 0.
b Whe
n the spear hits the ground,
the height is equal to 0.
12.25 + 14.7 t β 4.9 t2 = 0
Using the formula, where a = β 4.9,
b = 14.7 and c = 12.25,
t = β14.7 Β± β ___________________ 14. 7 2 β 4 (β4.9 ) (12.25 ) _______________________________ (2 Γ β4.9)
t = β14.7 Β± β ______ 456.19 _________________ β9.8
t = Β β0.679 or t = 3.68 (to 3 s
.f.)
As t > 0, t = 3.68 seconds (to 3 s.f.).
c 12.
25 + 14.7 t β 4.9 t2
= β4.9(t2 β 3 t) + 12.25
= β4.9(( t β 1.5)2 β 2.25) + 12.25
= β4.9(( t β 1.5)2 + 11.025 + 12.25)
= 23.275 β 4.9( t β 1.5)2
So A = 23.275, B = 4.9 and C = 1.5.
d The m
aximum height of the spear is
23.275 metres, 1.5 seconds after
the spear is thrown.Give any non-exact numerical answers correct to
3 significant figures unless specified otherwise.
Always interpret your answers in the context of the model. t is the time after the spear was thrown so it must be positive.Read the question carefully to work out the meaning of the constant term in the context of the model. Here, t = 0 is the time the spear is thrown.Problem-solving
To solve a quadratic, factorise, use the quadratic formula, or complete the square.
4.9(t β 1.5)2 must be positive or 0, so h(t ) < 23.275
for all possible values of t .
The turning point of the graph of this function would be at (1.5, 23.275). You may find it helpful to draw a sketch of the function when working through modelling questions.
Explore the trajectory of the
sp
ear using GeoGebra.Online | [
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-0.... |
34
Chapter 2
1 The diagram sho
ws a section of a suspension bridge carrying a road over water.
The height of the cables above water level in metres can be modelled by the function
h(x)Β =Β 0.000 12x2 + 200, where x is the displacement in metres from the centre of the bridge.
a Interpret the meaning of the constant ter
m 200 in the model. (1 mark)
b Use the model to find the two v
alues of x at which the height is 346 m. (3 marks)
c Given tha
t the towers at each end are 346 m tall, use your ans
wer to part b to calculate the
length of the bridge to the nearest metre. (1 mark)
2 A car manufacturer uses a mode
l to predict the fuel consumption, y miles per gallon (mpg),
for a specific model of car travelling at a speed of x mph.
y = β0.01x2 + 0.975x + 16, x . 0
a Use the model to find two speeds a
t which the car has a fuel consumption of
32.5 mpg. (3 marks)
b Rewrite
y in the form A β B(x β C)2, where A, B and C are constants to be found. (3 marks)
c Using your answ
er to part b, find the speed at which the car has the greatest fuel
efficiency. (1 mark)
d Use the model to calcula
te the fuel consumption of a car travelling at 120 mph.
Comment on the va
lidity of using this model for very high speeds. (2 marks)
3 A fertiliser company uses a model to deter
mine how the amount of fertiliser used, fΒ kilograms
per hectare, affects the grain yield g, measured in tonnes per hectare.
g = 6 + 0.03f β 0.000 06fΒ 2
a According to the model, how much grain would each hectare yield without any fertiliser?
(1 mark)
b One farmer currentl
y uses 20 kilograms of fertiliser per hectare. How much more fertiliser
would he need to use to increase his grain yield by 1 tonne per hectare? (4 marks)
4 A football stadium has 25 000 seats. The f
ootball club know from past experience that they will
sell only 10 000 tickets if each tick
et costs Β£30. They also expect to sell 1000 more tickets every
time the price goes down by Β£1.
a The number of tick
ets sold t can be modelled by the linear equation t = M β 1000p,
where Β£p is the price of each ticket and M is a constant. Find the value of M. (1 mark)E/P
E/P
E/P
E/PExercise 2H
For part a , make sure your
answer is in the context of
the model. Problem-solving | [
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-0... |
35Quadratics
The total revenue, Β£r, can be calculated by multiplying the number of tickets sold by the price of
each ticket. This can be written as r = p(M β 1000p).
b Rearrange r into the f
orm A β B(p β C)2, where A, B and C are constants to be found. (3 marks)
c Using your answ
er to part b or otherwise, work out how much the football club should
charge for each ticket if they want to make the maximum amount of money. (2 marks)
1 Solve the follo
wing equations without a calculator. Leave your answers in surd form whereΒ necessary.
a y2 + 3y + 2 = 0 b 3x2 + 13x β 10 = 0 c 5x2 β 10x = 4x + 3 d (2x β 5)2 = 7
2 Sketch gra
phs of the following equations:
a y =
x2 + 5x + 4 b y = 2x2 + x β 3 c y = 6 β 10x β 4x2 d y = 15x β 2x2
3 f(x) = x2 + 3x β 5 and g(x) = 4x + k, where k is a constant.
a Given tha
t f(3) = g(3), find the value of k. (3 marks)
b Find the values of
x for which f(x) = g(x). (3 marks)
4 Solve the follo
wing equations, giving your answers correct to 3 significant figures:
a k2 + 11k β 1 = 0 b 2t2 β 5t + 1 = 0 c 10 β x β x2 = 7 d (3x β 1)2 = 3 β x2
5 Write each of these expressions in the form p (x + q)2 + r, where p , q and r are constants to beΒ found:
a x2 + 12x β 9 b 5x2 β 40x + 13 c 8x β 2x2 d 3x2 β (x + 1)2
6 Find the value k for which the equation 5x2 β 2x + k = 0 has exactly one solution. (2 marks)E
EMixed exercise 2Accident investigators are studying the stopping distance
of a particular car. When the car is travelling at 20
mph, its stopping distance
is 6 f
eet.
When the car is travelling at 30 mph, its stopping distance
is 14 f
eet.
When the car is travelling at 40 mph, its stopping distance
is 24 f
eet.
The investigators suggest that the stopping distance in feet, d, is a quadratic function of the speed in miles per hour, s.
a
Given that d(s
) = as2 + bs + c, find the values of the
constants a, b and c.
b At an accident sc
ene a car has left behind a skid that is
20 feet long.
Use your model to calculate the speed that this car was going at before the accident.Challenge Start by setting up three
si
multaneous equations. Combine
two different pairs of equations to
eliminate c . Use the results to find
the values of a and b first.Hint | [
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0.... |
36
Chapter 2
7 Given tha
t for all values of x:
3x2 + 12x + 5 = p(x + q)2 + r
a find the values of
p, q and r. (3 marks)
b Hence solve the equation 3
x2 + 12x + 5 = 0. (2 marks)
8 The function f is defined as f(x
) = 22x β 20(2x) + 64, x β β .
a Write f(x
) in the form (2x β a)(2x β b), where a and b are real constants. (2 marks)
b Hence find the two roots of
f(x). (2 marks)
9 Find, as surds, the r
oots of the equation:
2(x + 1)(x β 4) β (x β 2)2 = 0.
10 Use algebr
a to solve (x β 1)(x + 2) = 18.
11 A diver launches herse
lf off a springboard. The height of the diver, in metres, above the pool
tΒ seconds after launch can be modelled by the following function:
h(t) = 5t β 10t2 + 10, t > 0
a How high is the springboard a
bove the water? (1 mark)
b Use the model to find the time at w
hich the diver hits the water. (3 marks)
c Rearrange h(
t) into the form A β B(t β C)2 and give the values of the constants
A, B and C. (3 marks)
d Using your answ
er to part c or otherwise, find the maximum height of the diver, and
the time at which this maximum height is reached. (2 marks)
12 For this question, f(x
) = 4kx2 + (4k + 2)x + 1, where k is a real constant.
a Find the discriminant of f(x
) in terms of k. (3 marks)
b By simplifying your answ
er to part a or otherwise, prove that f(x) has two distinct
real roots for all non-zero values of k. (2 marks)
c Explain why f(
x) cannot have two distinct real roots when k = 0. (1 mark)
13 Find all of the r
oots of the function r(x) = x8 β 17x4 + 16. (5 marks)
14 Lynn is selling cushions as part of
an enterprise project. On her first attempt, she sold 80
cushions at the cost of Β£15 each. She hopes to sell more cushions next time. Her adviser
suggests that she can expect to sell 10 more cushions for every Β£1 that she lowers the price.
a The number of cushions sold
c can be modelled by the equation c = 230 β Hp, where
Β£p is the price of each cushion and H is a constant. Determine the value of H. (1 mark)
To model her tota
l revenue, Β£r, Lynn multiplies the number of cushions sold by the price of
each cushion. She writes this as r = p(230 β Hp).b
Rearrange
r into the form A β B( p β
C )2, where A, B and C are constants to be
found. (3 marks)
c Using your answ
er to part b or otherwise, show that Lynn can increase her revenue by Β£122.50
through lowering her prices, and state the optimum selling price of a cushion. (2 marks)E
E/P
E/P
E/P
E/P
E/P | [
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-0.0742... |
37Quadratics
1 To solve a quadratic equation by factorising:
β Write the equation in the f
orm ax2 + bx + c = 0
β Factorise the l
eft-hand side
β Set each factor equal to z
ero and solve to find the value(s) of x
2 The solutions of the equation ax2 + bx + c = 0 where a β 0 are given by the formula:
x = βb Β± ββ―_______ b 2 β 4ac _____________ 2a
3 x2 + bx = (x + b __ 2 ) 2
β ( b __ 2 ) 2
4 ax2 + bx + c = a (x + b ___ 2a ) 2
+ (c β b2 ___ 4a2 )
5 The set of possibl
e inputs for a function is called the domain.
The set of possible outputs of a function is called the range.
6 The roots
of a function are the values of x for which f(x) = 0.
7 You can find the coor
dinates of a turning point of a quadratic graph by completing the
square. If f(x) = a(x + p)2 + q, the graph of y = f(x) has a turning point at (βp, q).
8 For the quadratic function f(
x) = ax2 + bx + c = 0, the expression b2 β 4ac is called the
discriminant. The value of the discriminant shows how many roots f(x) has:
β If b2 β 4ac . 0 then a quadratic function has two distinct real roots.
β If b2 β 4ac = 0 then a quadratic function has one repeated real root.
β If b2 β 4ac , 0 then a quadratic function has no real roots
9 Quadratics can be used to model r
eal-life situations.Summary of key pointsa The ratio of the lengths a : b in this line is the same as the r atio
of the lengths b : c.
a
b c
Show that this ratio is 1 + ββ―__
5 ______ 2 : 1.
b Show also that the infinite squar
e root
ββ―_______________________ 1 + ββ―___________________ 1 + ββ―______________ 1 + ββ―__________ 1 + ββ―______ 1 + β¦ = 1 + ββ―__
5 ______ 2 Challenge | [
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0.011073657311499119,
0.0140... |
38
Equations and
inequalities
After completing this chapter you should be able to:
β Solve linear simultaneous equations using elimination or
substit
ution β pages 39 β 40
β Solve simultaneous equations: one linear and one quadratic
β pages 41 β 42
β Interpret algebraic solutions of equations graphically β pages 42 β 45
β Solve linear inequalities β pages 46 β 48
β Solve quadratic inequalities β pages 48 β 51
β Interpret inequalities graphically β pages 51 β 53
β Represent linear and quadratic inequalities graphically β pages 53 β 55Objectives
1 A = {factors of 12}
B = {factors of 20}
Write do
wn the
numbers in each of these sets:
a A β©
B b (A
βͺ B)9
β GCSE Mathematics
2 Simplify these expressions.
a ββ―___
75 b 2 ββ―___
45 + 3 ββ―___
32 ___________ 6
β Section 1.5
3 Match the equations to the correct graph. Label
the points of int
ersection with the axes and the
coordinates of the turning point.
a y =
9 β x2 b y = (x β 2)2 + 4
c y =
(x β 7)(2x + 5)
y
xii i
y
xiii
y
xO O
O
β Section 2.47AB
9
116
1231
25
10
20
j 134Prior knowledge check
Food scientists use regions on graphs to optimise athletesβ nutritional intake and ensure they satisfy the minimum dietary requirements for calories and vitamins.3 | [
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0.02276924... |
39
Equations and inequalities
3.1 Linear simultaneous equations
Linear simultaneous equations in two unknowns have one set of values that will make a pair of
equations true at the same time.
The solution to this pair of simultaneous equations is x = 5, y = 2
x + 3y = 11 (1)
4x
β 5y = 10 (2)
β Linear simultaneous equations can be solved using elimination or substit
ution.5 + 3(2) = 5 + 6 = 11 β
4(5) β 5(2) = 20 β 10 = 10 β
Example 1
Solve the simultaneous equations:
a 2x
+ 3y = 8 b 4x
β 5y = 4
3x
β y = 23 6x
+ 2y = 25
a 2x + 3 y = 8 (1)
3x
β y = 23 (2)
9x
β 3y = 69 (3)
11x
= 77
x = 7
14 + 3 y = 8
3y = 8 β 14
y = β2
The solution is x = 7, y = β 2.
b 4x β
5y = 4 (1)
6x
+ 2y = 25 (2)
12x
β 15 y = 12 (3)
12x
+ 4y = 50 (4)
β19y
= β38
y = 2
4x β 10 = 4
4x = 14
x = 3 1 __ 2
The s
olution is x = 3 1 __ 2 , y = 2.Remember to check your solution by substituting
into equation (2). 3(7) β (β2) = 21 + 2 = 23 β
Note that you could also multiply equation (1)
by 3 and equation (2) by 2 to get 6x in both equations. You could then subtract to eliminate x.
Multiply equation (1) by 3 and multiply equation (2) by 2 to get 12x in each equation.
Subtract, since the 12x terms have the same sign (both positive).
Substitute y = 2 into equation (1) to find x.First look for a way to eliminate x or y.
Multiply equation (2) by 3 to get 3y in each equation.
Number this new equation (3).
Then add equations (1) and (3), since the 3y terms
have different signs and y will be eliminated.
Substitute x = 7 into equation (1) to find y. | [
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40
Chapter 3
Example 2
1 Solve these simultaneous equations by elimination:
a 2x β
y = 6 b 7x +
3y = 16 c 5x
+ 2y = 6
4x
+ 3y = 22 2x +
9y = 29 3x β
10y = 26
d 2x β
y = 12 e 3x β
2y = β6 f 3x
+ 8y = 33
6x
+ 2y = 21 6x +
3y = 2 6x =
3 + 5y
2 Solve these simultaneous equa
tions by substitution:
a x +
3y = 11 b 4x β
3y = 40 c 3x
β y = 7 d 2y
= 2x β 3
4x β
7y = 6 2x +
y = 5 10x +
3y = β2 3y
= x β 1
3 Solve these simultaneous equa
tions:
a 3x β
2y + 5 = 0 b x β 2
y ______ 3 = 4 c 3y = 5(x β 2)
5(x
+ y) = 6(x + 1) 2x +
3y + 4 = 0 3(x β
1) + y + 4 = 0
4 3x
+ ky = 8
x β
2ky = 5
are simultaneous equa
tions where k is a constant.
a Show that
x = 3. (3 marks)
b Given tha
t y = 1 _ 2 determine the value of k. (1 mark)
5 2x
β py = 5
4x
+ 5y + q = 0
are simultaneous equa
tions where p and q are constants.
The solution to this pair of simultaneous equa
tions is x = q, y = β1.
Find the value of
p and the value of q. (5 marks) First rearrange
bo
th equations into
the same form
e.g. ax + by = c.Hint
E/P
k is a constant, so it has the same value in both equations.Problem-solving
E/PExercise 3ASolve the simultaneous equations:
2x β y = 1
4x + 2y = β30
2x β y = 1 (1)
4x
+ 2y = β30 (2)
y
= 2x β 1
4x + 2(2 x β 1) = β 30
4x + 4 x β 2 = β 30
8x = β28
x = β3 1 __ 2
y =
2(β3 1 __ 2 ) β 1 = β8
The solution is x = β 3 1 __ 2 , y = β8.Rearrange an equation, in this case equation (1),
to get either x = β¦ or y = β¦ (here y = β¦).
Solve for x.
Substitute x = β3 1 _ 2 into equation (1) to find the
value o
f y.Substitute this into the other equation (here into
equation (2) in place of y).
Remember to check your solution in equation (2). 4(β3.5) + 2(β8) = β14 β 16 = β30 β | [
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41
Equations and inequalities
3.2 Quadratic simultaneous equations
You need to be able to solve simultaneous equations where one equation is linear and one is quadratic.
To solve simultaneous equations involving one linear equation and one quadratic equation, you need
to use a substitution method from the linear equation into the quadratic equation.
β Simultaneous equations with one linear and one quadratic equation can hav
e up to two pairs
of solutions. You need to make sure the solutions are paired correctly.
The solutions to this pair of simultaneous equations are x = 4, y = β3 and x = 5.5, y = β1.5.
x β y = 7 (1)
y2 + xy + 2x = 5 (2)4 β (β3) = 7 β and 5.5 β (β1.5) = 7 β
(β3)2 + (4)(β3) + 2(4) = 9 β 12 + 8 = 5 β and
(β1.5)2 + (5.5)(β 1.5) + 2(5.5) = 2.25 β 8.25 + 11 = 5 β
Example 3
Solve the simultaneous equations:
x + 2y = 3
x2 + 3xy = 10
x + 2 y = 3 (1)
x2 + 3 xy = 10 (2)
x
= 3 β 2 y
(3 β 2 y)2 + 3y(3 β 2 y) = 10
9 β 12 y + 4 y2 + 9 y β 6 y2 = 10
β2y2 β 3y β 1 = 0
2y2 + 3y + 1 = 0
(2y + 1)( y + 1) = 0
y = β 1 __ 2 or y = β 1
So
x = 4 or x = 5
Solutions are x = 4, y = β 1 __ 2
and
x = 5, y = β 1.Rearrange linear equation (1) to get x = β¦ or
y = β¦ (here x = β¦).
Substitute this into quadratic equation (2) (here in place of x ).
Solve for y using factorisation.
Find the corresponding x-values by substituting the y-values into linear equation (1), x = 3 β 2y.
There are two solution pairs for x and y.The quadratic equation can contain terms involving y
2 and xy.
(3 β 2 y)2 means (3 β 2 y)(3 β 2 y) β Section 1.2
1 Solve the simultaneous equations:
a x +
y = 11 b 2x
+ y = 1 c y =
3x
xy =
30 x2 + y2 = 1 2y2 β xy = 15
d 3a
+ b = 8 e 2u
+ v = 7 f 3x
+ 2y = 7
3a2 + b2 = 28 uv = 6 x2 + y = 8
2 Solve the simultaneous equa
tions:
a 2x
+ 2y = 7 b x +
y = 9 c 5y
β 4x = 1
x2 β 4y2 = 8 x2 β 3xy + 2y2 = 0 x2 β y2 + 5x = 41Exercise 3B | [
0.07885614037513733,
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0.0587... |
42
Chapter 3
3 Solve the simultaneous equa
tions, giving your answers in their simplest surd form:
a x β
y = 6 b 2x
+ 3y = 13
xy =
4 x2 + y2 = 78
4 Solve the simultaneous equa
tions:
x +
y = 3
x2 β 3y = 1 (6 marks)
5 a By eliminating
y from the equations
y = 2 β 4x
3x2 + xy + 11 = 0
show that x2 β 2x β 11 = 0. (2 marks)
b Hence, or otherwise, solv
e the simultaneous equations
y = 2 β 4x3x
2 + xy + 11 = 0
giving your answers in the form a Β± b ββ―__
3 , where a and b are integers. (5 marks)
6 One pair of solutions for the sim
ultaneous equations
y = kx β 5
4x2 β xy = 6
is (1, p) where k and p are constants.
a Find the values of
k and p.
b Find the second pair of solutions for the sim
ultaneous equations. Use b rackets when you are
substituting an expression into an equation.Watch out
E/P
E/P
P
If (1, p ) is a solution, then x = 1, y = p
satisfies both equations.Problem-solving
3.3 Simultaneous equations on graphs
You can represent the solutions of simultaneous equations graphically. As every point on a line or
curve satisfies the equation of that line or curve, the points of intersection of two lines or curves satisfy both equations simultaneously.
β
The solutions to a pair o
f simultaneous equations represent the points of intersection of
their graphs.
Example 4
a On the same axes, dr aw the graphs of:
2x + 3y = 8
3x β y = 23
b Use your gra
ph to write down the solutions to the simultaneous equations.y β x = k
x2 + y2 = 4
Given that the simultaneous equations have exactly one pair of solutions, show that
k = Β± 2 ββ―__
2 Challenge | [
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-0... |
43
Equations and inequalities
a
β2 β/four.ss01 /four.ss01 8 2 6y
x123/four.ss01
β1
β2
β3
β/four.ss012x + 3 y = 8
3x β y = 23O
b The solution is (7, β2) or x = 7, y = β2.The point of intersection is the solution to the
simultaneous equations
2x + 3y = 8
3x β y = 23
This solution matches the algebraic solution to
the simultaneous equations.
Example 5
a On the same axes, dr aw the graphs of:
2x + y = 3
y = x2 β 3x + 1
b Use your gra
ph to write down the solutions to the simultaneous equations.
a y
x
β1β2β11 234 5 β31
O23456
(β1, 5)
(2, β1)
β2β3y = x2 β 3x + 1
2x + y = 3
b The solutions are ( β1, 5 ) or x = β 1,
y = 5 and (2, β 1) or x = 2, y = β 1.There are two solutions. Each solution will have
an x-value and a y-value.
Check your solutions by substituting into both
equations.
2(β1) + (5) = β2 + 5 = 3 β and
5 = (β1)2 β 3(β1) + 1 = 1 + 3 + 1 = 5 β
2(2) + (β1) = 4 β 1 = 3 β and β1 = (2)
2 β 3(2) + 1 = 4 β 6 + 1 = β1 β
The graph of a linear equation and the graph of a quadratic equation can either:
β’ intersect twice
β’ intersect once
β’ not intersect
Aft
er substituting, you can use the discriminant of the resulting quadratic equation to determine the
number of points of intersection. Find the point of intersection
gr
aphically using GeoGebra.Online
Plot the curve and the line using
Ge
oGebra to find the two points of intersection.Online | [
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0.03243115171790123,
0.... |
44
Chapter 3
β For a pair of simultaneous equations that produce a quadratic equation of the form
ax2 + bx + c = 0:
β’ b2 β 4ac > 0 β’ b2 β 4ac = 0 β’ b2 β 4ac < 0
two real solutions one real solution no real solutions
Example 6
The line with equation y = 2x + 1 meets the curve with equation kx2 + 2y + (k β 2) = 0 at exactly
one point. Given that k is a positive constant
a find the value of
k
b for this va
lue of k, find the coordinates of
the point of intersection.
a y = 2x + 1 (1)
kx2 + 2 y + ( k β 2) = 0 (2)
kx2 + 2(2 x + 1) + ( k β 2) = 0
kx2 + 4 x + 2 + k β 2 = 0
kx2 + 4 x + k = 0
42 β 4 Γ k Γ k = 0
16 β
4k2 = 0
k2 β 4 = 0
(k
β 2)( k + 2) = 0
k
= 2 or k = β 2
So k = 2
b 2x2 + 4 x + 2 = 0
x2 + 2x + 1 = 0
(x
+ 1)(x + 1) = 0
x = β1
y
= 2(β1) + 1 = β 1
Point of intersection is ( β1, β1).Substitute y = 2x + 1 into equation (2) and
simplify the quadratic equation. The resulting quadratic equation is in the form ax
2 + bx + c = 0
with a = k, b = 4 and c = k.
Factorise the quadratic to find the values of k.
The solution is k = +2 as k is a positive constant.
Substitute k = +2 into the quadratic equation kx
2 + 4x + k = 0. Simplify and factorise to find
the x-coordinate.
Check your answer by substituting into equation (2):
2x2 + 2y = 0
2(β1)2 + 2(β1) = 2 β 2 = 0 βYou are told that the line meets the curve at exactly one point, so use the discriminant of the resulting quadratic. There will be exactly one solution, so b
2 β 4ac = 0.Problem-solving
Substitute x = β1 into linear equation (1) to find the y-coordinate. Explore how the value of k affects
the l
ine and the curve using GeoGebra.Online | [
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-0.0059... |
45
Equations and inequalities
1 In each case:
i draw the gr
aphs for each pair of equations on the same axes
ii find the coordinates of
the point of intersection.
a y =
3x β 5 b y =
2x β 7 c y =
3x + 2
y =
3 β x y =
8 β 3x 3x
+ y + 1 = 0
2 a Use graph paper to draw accurately the graphs of 2y = 2x + 11 and y = 2x2 β 3x β 5 on the same axes.
b Use your graph to find the coordinates of the points of intersection.
c Verify your solutions b
y substitution.
3 a On the same axes sketch the curv
e with equation x2 + y = 9 and the line with equation 2x + y = 6.
b Find the coordinates of
the points of intersection.
c Verify your solutions b
y substitution.
4 a On the same axes sketch the curv
e with equation
y = (x β 2)2 and the line with equation y = 3x β 2.
b Find the coordinates of
the point of intersection.
5 Find the coordinates of
the points at which the line with equation y = x β 4 intersects the curve
with equation y2 = 2x2 β 17.
6 Find the coordinates of
the points at which the line with equation y = 3x β 1 intersects the curve
with equation y2 = xy + 15.
7 Determine the number of
points of intersection for these pairs of simultaneous equations.
a y =
6x2 + 3x β 7 b y = 4x2 β 18x + 40 c y = 3x2 β 2x + 4
y =
2x + 8 y =
10x β 9 7x
+ y + 3 = 0
8 Given the sim
ultaneous equations
2x β y = 1x
2 + 4ky + 5k = 0
where k is a non-zero constant
a show that
x2 + 8kx + k = 0. (2 marks)
Given tha
t x2 + 8kx + k = 0 has equal roots,
b find the value of
k (3 marks)
c for this va
lue of k, find the solution of the simultaneous equations. (3 marks)
9 A swimmer div
es into a pool. Her position, p m, underwater can be mode
lled
in relation to her horizontal distance, x m, from the point she entered the
water as a quadratic equation p = 1 _ 2 x2 β 3x.
The position of the bottom of the pool can be modelled by the linear
equation p = 0.3x β 6.
Determine whether this model predicts that the swimmer will touch the
bottom of the pool. (5 marks) You need to use algebra in
par
t b to find the coordinates.Hint
P
E/P
E/P p
xExercise 3C | [
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0.04552... |
46
Chapter 3
3.4 Linear inequalities
You can solve linear inequalities using similar methods to those for solving linear equations.
β The solution of an inequality is the set o
f all real numbers x that make the inequality true.
Example 7
Find the set of values of x for which:
a 5x
+ 9 > x + 20 b 12 β 3
x < 27
c 3(x
β 5) > 5 β 2(x β 8)
a 5x + 9 > x + 20
4x + 9 > 20
4x > 11
x > 2.75
b 12 β
3x < 27
β3x < 15
x > β5
c 3(x
β 5) > 5 β 2( x β 8)
3x β 15 > 5 β 2x + 16
5x > 5 + 16 + 15
5x > 36
x > 7.2Subtract 12 from both sides.
Divide both sides by β3. (You therefore need to
turn round the inequality sign.)
In set notation {x : x >
β5}.
Multiply out (note: β2 Γ β8 = +16).Rearrange to get x > β¦In set notation {x
: x >
7.2}.
β6 β4 β2 0 2 4 6
Here the solution sets are x < β1 or x > 3.
β6 β4 β2 0 2 4 6Here there is no overlap and the two inequalities have
to be written separately as x < β1 or x > 3.β is used for < and > and means the end value is not included.
β is used for < and > and means the end value is
included.
These are the only real values that satisfy both
equalities simultaneously so the solution is β2 < x < 4.You may sometimes need to find the set of
values for which two inequalities are true
together. Number lines can be useful to find your solution.
For example, in the number line below the
solution set is x > β2 and x < 4. In se t notation
x > β2 and x < 4 is written { x : β2 < x < 4}
or alternatively { x : x > β2} β { x : x < 4}
x < β1 or x > 3 is written { x : x < β1} β { x : x > 3}Notation You c an write the solution to this
inequality using set notation as { x : x >
2.75}.
This means the set of all values x for which x is
greater than or equal to 2.75.Notation
Rearrange to get x > β¦ | [
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0.009743... |
47
Equations and inequalities
Example 8
Find the set of values of x for which:
a 3x
β 5 < x + 8 and 5x > x β 8
b x β
5 > 1 β x or 15 β 3x > 5 + 2x.
c 4x
+ 7 > 3 and 17 < 11 + 2x.
a 3x β 5 < x + 8 5x > x β 8
2x β 5 < 8 4x >
β 8
2x < 13 x >
β2
x < 6.5
β/four.ss01 β2 0 /four.ss01 2 6 8
x < 6.5
x > β2
So the required set of values is β 2 < x < 6.5.
b x β
5 > 1 β x 15 β
3x > 5 + 2x
2x β 5 > 1 10 β
3x > 2x
2x > 6 10 >
5x
x > 3 2 >
x
x <
2
x < 3
x > 2β/four.ss01 β2 0 /four.ss01 2 6 8
The solution is x > 3 or x < 2.Draw a number line to illustrate the two
inequalities.
The two sets of values overlap (intersect) where
β2 < x < 6.5.
Notice here how this is written when x lies
between two values.
In set notation this can be written as
{x : β2 < x < 6.5}.
Draw a number line. Note that there is no overlap between the two sets of values.
In set notation this can be written as
{x : x < 2} β {x : x > 3}.
1 Find the set of va lues of x for which:
a 2x
β 3 < 5 b 5x
+ 4 > 39
c 6x
β 3 > 2x + 7 d 5x
+ 6 < β12 β x
e 15 β
x > 4 f 21 β 2
x > 8 + 3x
g 1 +
x < 25 + 3x h 7x
β 7 < 7 β 7x
i 5 β 0.5
x > 1 j 5x
+ 4 > 12 β 2xExercise 3D | [
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48
Chapter 3
3.5 Quadratic inequalities
β To solve a quadratic inequality:
β’ Rearr
ange so that the right-hand side of the inequality is 0
β’ Solve the corresponding quadratic equation to find the critical values
β’ Sketch the graph of the quadratic function
β’ Use your sketch to find the required set of values.
The sketch shows the graph of f(x) = x2 β 4x β 5
= (x + 1)(x β 5)
β1 5y
x OThe solutions to the quadratic inequality
x2 β 4x β 5 > 0 are the x-values when
the curve is above the x-axis (the darker
part of the curve). This is when x < β1 or x > 5. In set notation the solution is {x : x < β1} β {x : x > 5}.
The solutions to the quadratic inequality x
2 β 4x β 5 < 0 are the x-values when
the curve is below the x-axis (the lighter part of the curve). This is when x > β1 and x < 5 or β1 < x < 5. In set notation the solution is {x : β1 < x < 5}.The solutions to f(x) = 0 are x = β1 and x = 5. These are called the critical values.A = { x : 3x + 5 > 2} B = { x : x __ 2 + 1 < 3 } C = { x : 11 < 2x β 1}
Given that A β ( B β C ) = {x :
p < x < q } β {x : x > r }, find the values of p , q and r .Challenge2 Find the set of va lues of x for which:
a 2(x
β 3) > 0 b 8(1 β
x) > x β 1 c 3(x
+ 7) < 8 β x
d 2(x
β 3) β (x + 12) < 0 e 1 + 11(2
β x) < 10(x β 4) f 2(x
β 5) > 3(4 β x)
g 12x
β 3(x β 3) < 45 h x β
2(5 + 2x) < 11 i x(x
β 4) > x2 + 2
j x(5 β
x) > 3 + x β x2 k 3x + 2x(x β 3) < 2(5 + x2) l x(2x β 5) < 4x(
x + 3) ________ 2 β 9
3 Use set notation to describe the set of v
alues of x for which:
a 3(x
β 2) > x β 4 and 4x + 12 > 2x + 17
b 2x
β 5 < x β 1 and 7(x + 1) > 23 β x
c 2x
β 3 > 2 and 3(x + 2) < 12 + x
d 15 β
x < 2(11 β x) and 5(3x β 1) > 12x + 19
e 3x
+ 8 < 20 and 2(3x β 7) > x + 6
f 5x
+ 3 < 9 or 5(2x + 1) > 27
g 4(3x
+ 7) < 20 or 2(3x β 5) > 7 β 6
x ______ 2 | [
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49
Equations and inequalities
Example 9
Find the set of values of x for which:
3 β 5x β 2x2 < 0.
3 β 5 x β 2x2 = 0
2x2 + 5 x β 3 = 0
(2x β 1)( x + 3) = 0
x = 1 __ 2 or x = β 3
β3 1
2xy
O
So the required set of values is
x < β3 or x > 1 __ 2 .Multiply by β1 (so itβs easier to factorise).
1 _ 2 and β3 ar e the critical values.
Draw a sketch to show the shape of the graph
and the critical values.
Since the coefficient of x2 is negative, the graph
is βupside-down β-shapedβ. It crosses the x-axis at
β3 and 1 _ 2 . β Section 2.4
3 β 5x β 2x2 < 0 ( y < 0) for the outer parts of the
graph, belo
w the x-axis, as shown by the paler
parts of the curve.
In set notation this can be written as {x : x < β3} β {x : x >
1 _ 2 }.Quadratic equation.
Example 10
a Find the set of va lues of x for which 12 + 4x > x2.
b Hence find the set of va
lues for which 12 + 4x > x2 and 5x β 3 > 2.
a 12 + 4x > x2
0 > x2 β 4 x β 12
x2 β 4 x β 12 < 0
x2 β 4 x β 12 = 0
(x + 2)( x β 6) = 0
x = β2 or x = 6
Sketch of y = x2 β 4 x β 12
β26 xy
O
x2 β 4 x β 12 < 0
Solution: β 2 < x < 6You can use a table to check your solution.
β2 < x < 6
Use the critical values to split the real number
line into sets.
β26
x < β2 β2 < x < 6 x > 6
x + 2 β + +
x β 6 β β +
(x + 2)(x β 6) + β +
For each set, check whether the set of values makes the value of the bracket positive or negative. For example, if x < β2, (x + 2) is negative, (x β 6) is negative, and (x + 2)(x β 6) is (neg) Γ (neg) = positive.
In set notation the solution is {x : β2 < x < 6}. | [
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50
Chapter 3
Example 11
Find the set of values for which 6 __ x > 2 , x β 0b Solving 12 + 4 x > x2 gives β 2 < x < 6.
Solving 5 x β 3 > 2 gives x > 1.
β/four.ss01β202/four.ss0168
β2 < x < 6
x > 1
The two sets of values overlap where
1 < x < 6.
So the solution is 1 < x < 6.
6 __ x > x
6x
> 2x2
6x β 2x2 > 0
6x β 2x2 = 0
x(6 β 2 x) = 0
x = 0 or x = 3
3xy
O
The solution is 0 < x < 3.
1 Find the set of va lues of x for which:
a x2 β 11x + 24 < 0 b 12 β x β x2 > 0 c x2 β 3x β 10 > 0
d x2 + 7x + 12 > 0 e 7 + 13 x β 2x2 > 0 f 10 + x β 2x2 < 0
g 4x2 β 8x + 3 < 0 h β2 + 7x β 3x2 < 0 i x2 β 9 < 0
j 6x2 + 11x β 10 > 0 k x2 β 5x > 0 l 2x2 + 3x < 0
2 Find the set of va
lues of x for which:
a x2 < 10 β 3x b 11 < x2 + 10
c x(3 β
2x) > 1 d x(x
+ 11) < 3(1 β x2)Exercise 3EThis question is easier if you represent the
information in more than one way. Use a sketch graph to solve the quadratic inequality, and use a number line to combine it with the linear inequality.Problem-solving
In set notation this can be written as {x : 1 < x < 6}.
Solve the corresponding quadratic equation to find the critical values.
Sketch y = x (6 β 2 x). You are interested in the
values of x where the graph is above the x -axis.x = 0 can still be a critical value even though xΒ β Β 0. But it would not be part of the solution set, even if the inequality was > rather than > . x cou ld be either positive or negative,
so you canβt multiply both sides of this inequality by x . Instead, multiply both sides by x
2.
Because x2 is never negative, and x β 0 so x2 β 0,
the inequality sign stays the same.Watch out | [
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51
Equations and inequalities
3 Use set notation to describe the set of v alues of x for which:
a x2 β 7x + 10 < 0 and 3x + 5 < 17 b x2 β x β 6 > 0 and 10 β 2x < 5
c 4x2 β 3x β 1 < 0 and 4(x + 2) < 15 β (x + 7) d 2x2 β x β 1 < 0 and 14 < 3x β 2
e x2 β x β 12 > 0 and 3x + 17 > 2 f x2 β 2x β 3 < 0 and x2 β 3x + 2 > 0
4 Given tha
t x β 0, find the set of values of x for which:
a 2 __ x < 1 b 5 > 4 __ x c 1 __ x + 3 > 2
d 6 + 5 __ x > 8 __ x e 25 > 1 ___ x 2 f 6 ___ x 2 + 7 __ x < 3
5 a Find the range of
values of k for which the
equation x2 β kx + (k + 3) = 0 has no real roots.
b Find the range of v
alues of p for which the
roots of the equation px2 + px β 2 = 0 are real.
6 Find the set of va
lues of x for which x2 β 5x β 14 > 0. (4 marks)
7 Find the set of va
lues of x for which
a 2(3x
β 1) < 4 β 3x (2 marks)
b 2x2 β 5x β 3 < 0 (4 marks)
c both 2(3x
β 1) < 4 β 3x and 2x2 β 5x β 3 < 0. (2 marks)
8 Given tha
t x β 3, find the set of values for which 5 _____ x β 3 < 2 .
(6 marks)
9 The equation kx2 β 2kx + 3 = 0, where k is a constant, has no real roots.
Prove that k satisfies the inequality 0 < k < 3. (4 marks)P
The quadratic equation ax2 + bx + c = 0
has real roots if b2 β 4ac > 0. β Section 2.5Hint
E
E
Multiply both sides of the
inequality by ( x β 3)2.Problem-solvingE/P
E/P
3.6 Inequalities on graphs
You may be asked to interpret graphically the solutions to inequalities by considering the graphs of
functions that are related to them.
β The values of x for which the curve y = f(x) is below the curve y = g(x) satisfy the inequality
f(x) < g( x).
β The values of x for which the curve y = f(x) is above the curve y = g(x) satisfy the inequality
f(x) > g( x). | [
0.02889794111251831,
0.09589443355798721,
0.09131527692079544,
-0.06901653856039047,
-0.005281904712319374,
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0.04395867511630058,
-0.04577496275305748,
-0.07411419600248337,
0.06039924919605255,
0.0393194... |
52
Chapter 3
Example 12
L1 has equation y = 12 + 4x.
L2 has equation y = x2.
The diagram shows a sketch of L1 and L2 on the same axes.
a Find the coordinates of
P1 and P2, the points of intersection.
b Hence write down the solution to the inequality 12
+ 4x > x2.y
x OL1: y = 12 + 4x
L2: y = x2P1
P2
a x2 = 12 + 4 x
x2 β 4 x β 12 = 0
(x
β 6)( x + 2) = 0
x = 6 and x = β 2
substitute into y = x2
when x = 6, y = 36 P1 (6, 36)
when x = β 2, y = 4 P2 (β2, 4)
b 12 +
4x > x2 when the graph of L1 is
above the graph of L2
β2 < x < 6This is the range of values of x for which the
graph of y = 12 + 4x is above the graph of y = x2
i.e. between the two points of intersection.
In set notation this is {x : β2 < x < 6}.Equate to find the points of intersection, then
rearrange to solve the quadratic equation.
Factorise to find the x-coordinates at the points of intersection.
1 L1 has equation 2y + 3x = 6.
L2 has the equation x β y = 5.
The diagram shows a sketch of L1 and L2.
a Find the coordinates of
P, the point of intersection.
b Hence write down the solution to the inequality
2
y + 3x > x β y.y
x O
L1: 2y + 3x = 6L2: x β y = 5Exercise 3Fy
x 2 5 Oy = g(x)y = f(x)
The solutions to f(x) = g(x) are x = 2 and x = 5.f(x) is below g(x) when 2 < x < 5. These values of
x satisfy f(x) < g(x).f(x) is above g(x) when x < 2 and when x > 5.
These values of x satisfy f(x) > g(x). | [
0.07161044329404831,
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0.005367380101233721,
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0.044180333614349365,
0.03993464633822441,
-0.0017477524233981967,
0.0232989601790905,
0.0... |
53
Equations and inequalities
The sketch shows the graphs of
f(x) = x2 β 4x β 12
g(x) = 6 + 5 x β x2
a Find the coordinates of the points of intersection.
b Fin
d the set of values of x for which f( x) < g( x).
Give your answer in set notation.y
x O
y = g(x)y = f (x)Challenge
All the shaded points in this region satisfy the
inequality y > f(x).3.7 Regions
You can use shading on graphs to identify regions that satisfy linear and quadratic inequalities.
β y <
f(x) represents the points on the coordinate grid below the curve y = f(x).
β y >
f(x) represents the points on the coordinate grid above the curve y = f(x).
y = f (x)y
x OAll the unshaded points in this region satisfy the
inequality y < f(x).2 For each pair of functions:
i Sketch the gra
phs of y = f(x) and y = g(x) on the same axes.
ii Find the coordinates of
any points of intersection.
iii Write down the solutions to the inequa
lity f(x) < g(x).
a f(x
) = 3x β 7 b f(x
) = 8 β 5x c f(x
) = x2 + 5
g(x
) = 13 β 2x g(x
) = 14 β 3x g(x)
= 5 β 2x
d f(x
) = 3 β x2 e f(x ) = x2 β 5 f f(x ) = 7 β x2
g(x
) = 2x β 12 g(x)
= 7x + 13 g(x)
= 2x β 8
3 Find the set of va
lues of x for which the curve with equation y = f(x) is below the line with
equation y = g(x).a
f(x
) = 3x2 β 2x β 1 b f(x ) = 2x2 β 4x + 1 c f(x ) = 5x β 2x2 β 4
g(x
) = x + 5 g(x)
= 3x β 2 g(x)
= β2x β 1
d f(x
) = 2 __ x , x β 0 e f(x) = 3 __ x2 β 4 __ x , x β 0 f f(x) = 2 _____ x + 1 , x β β1
g(x
) = 1 g(x)
= β1 g(x)
= 8P | [
0.0902569517493248,
0.07046928256750107,
0.07022906839847565,
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0.04792482778429985,
-0.04281258583068848,
-0.026247045025229454,
-0.040408939123153687,
0.0278... |
54
Chapter 3
β If y
> f(x) or y < f(x) then the curve y = f(x) is not included in the region and is represented
by a dotted line.
β If y > f(x) or y < f(x) then the curve y = f(x) is included in the region and is represented by a
solid line.
Example 13
On graph paper, shade the region that satisfies the inequalities:
y > β2, x < 5, y < 3x + 2 and x > 0.
y
x
β5β2 β1 1234 567 β35
O10152025x = 0 x = 5
y = 3x + 2
y = β2Draw dotted lines for x = 0, x = 5.
Draw solid lines for y = β2, y = 3x + 2.Shade the required region.
Test a point in the region. Try (1, 2).For x = 1: 1 < 5 and 1 > 0 βFor y = 2: 2 > β2 and 2 < 3 + 2 β
Example 14
On graph paper, shade the region that satisfies the inequalities:
2y + x < 14
y > x2 β 3x β 4
y
x
β5β1 12 3/four.ss01 567 β2 β3 β/four.ss015
O1015202530
β10y = x2 β 3x β /four.ss01
2y + x = 1/four.ss01Draw a dotted line for 2y + x = 14 and a
solid line for y = x2 β 3x β 4.
Shade the required region.Test a point in the region. Try (0, 0). 0 + 0 < 14 and 0 > 0 β 0 β 4 β Explore which regions on
the g
raph satisfy which inequalities
using GeoGebra.Online | [
0.031383588910102844,
0.08265526592731476,
0.021256988868117332,
-0.05200735852122307,
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0.03958097845315933,
0.0057642157189548016,
0.03810950741171837,
0.0738588273525238,
0.08055... |
55
Equations and inequalities
1 On a coordinate grid, shade the r
egion that satisfies the inequalities:
y > x β 2, y < 4x and y < 5 β x.
2 On a coordinate grid, shade the r
egion that satisfies the inequalities:
x > β1, y + x < 4, 2x + y < 5 and y > β2.
3 On a coordinate grid, shade the r
egion that satisfies the inequalities:
y > (3 β x)(2 + x) and y + x > 3.
4 On a coordinate grid, shade the r
egion that satisfies the inequalities:
y > x2 β 2 and y < 9 β x2.
5 On a coordinate grid, shade the r
egion that satisfies the inequalities:
y > (x β 3)2, y + x > 5 and y < x β 1.
6 The sketch shows the gr
aphs of the straight lines
with equations:
y = x + 1, y = 7 β x and x = 1.
a Work out the coor
dinates of the points of
intersection of the functions.
b Write down the set of
inequalities that
represent the shaded region shown in the sketch.
7 The sketch shows the gr
aphs of the curves with
equations:
y = 2 β 5x β x2, 2x + y = 0 and x + y = 4.
Write down the set of inequalities that represent the
shaded region shown in the sketch.
8 a On a coordinate grid, shade the r
egion that satisfies
the inequalities
y < x + 4, y + 5x + 3 > 0, y > β1 and x < 2.
b Work out the coor
dinates of the vertices of the shaded region.
c Which of the v
ertices lie within the region identified by the
inequalities?
d Work out the ar
ea of the shaded region.Oy
x
β1β2β12 4 613 5712345678
y = 7 β xy = x + 1x = 1
Oy
xβ1
β2β3 1 β4β5β6 β2 2312345678
β1P
A vertex is only included if
both intersecting lines are included.Problem-solvingExercise 3G | [
0.013747014105319977,
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0.05208682641386986,
-0.09709397703409195,
-0.010302322916686535,
0.03112560696899891,
0.07... |
56
Chapter 3
1 2kx
β y = 4
4kx + 3y = β2
are two simultaneous equations, where k is a constant.a
Show that
y = β2. (3 marks)
b Find an expression f
or x in terms of the constant k. (1 mark)
2 Solve the simultaneous equa
tions:
x + 2y = 3
x2 β 4y2 = β33 (7 marks)
3 Given the sim
ultaneous equations
x β 2y = 13xy β y
2 = 8
a Show that 5
y2 + 3y β 8 = 0. (2 marks)
b Hence find the pairs (x,
y) for which the simultaneous equations are satisfied. (5 marks)
4 a By eliminating
y from the equations
x + y = 2x
2 + xy β y2 = β1
show that x2 β 6x + 3 = 0. (2 marks)
b Hence, or otherwise solve the sim
ultaneous equations
x + y = 2
x2 + xy β y2 = β1
giving x and y in the form a Β± b ββ―__
6 , where a and b are integers. (5 marks)
5 a Given tha
t 3x = 9y β 1, show that x = 2y β 2. (1 mark)
b Solve the simultaneous equa
tions:
x = 2y β 2
x2 = y2 + 7 (6 marks)
6 Solve the simultaneous equa
tions:
x + 2y = 3x
2 β 2y + 4y2 = 18 (7 marks)
7 The curve and the line giv
en by the equations
kx2 β xy + (k + 1)x = 1
β k __ 2 x + y = 1
where k is a non-zero constant, intersect at a single point.
a Find the value of
k. (5 marks)
b Give the coor
dinates of the point of intersection of the line and the curve. (3 marks)E
E
E
E
E
E
E/PMixed exercise 3 | [
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0.03844192996621132,
-0.0... |
57
Equations and inequalities
8 A person throws a ba
ll in a sports hall. The height of the ball, h m, h
x
can be modelled in relation to the horizontal distance from the
point it was thrown from by the quadratic equation:
h = β 3 __ 10 x2 + 5 _ 2 x + 3 _ 2
The hall has a sloping ceiling which can be mode
lled with equation
h = 15 __ 2 β 1 _ 5 x.
Determine w
hether the model predicts that the ball will hit the ceiling. (5 marks)
9 Give y
our answers in set notation.
a Solve the inequality 3
x β 8 > x + 13. (2 marks)
b Solve the inequality
x2 β 5x β 14 > 0. (4 marks)
10 Find the set of va
lues of x for which (x β 1)(x β 4) < 2(x β 4). (6 marks)
11 a Use algebr
a to solve (x β 1)(x + 2) = 18. (2 marks)
b Hence, or otherwise, find the set of
values of x for which (x β 1)(x + 2) > 18.
Give your answer in set notation. (2 marks)
12 Find the set of va
lues of x for which:
a 6x
β 7 < 2x + 3 (2 marks)
b 2x2 β 11x + 5 < 0 (4 marks)
c 5 < 20 ___ x (4 marks)
d both 6x
β 7 < 2x + 3 and 2x2 β 11x + 5 < 0. (2 marks)
13 Find the set of va
lues of x that satisfy 8 __ x2 + 1 < 9 __ x , x β 0 (5 marks)
14 Find the values of
k for which kx2 + 8x + 5 = 0 has real roots. (3 marks)
15 The equation 2x2 + 4kx β 5k = 0, where k is a constant, has no real roots.
Prove that k satisfies the inequality β 5 _ 2 < k < 0. (3 marks)
16 a Sketch the gra
phs of y = f(x) = x2 + 2x β 15 and g(x) = 6 β 2x on the same axes. (4 marks)
b Find the coordinates of
any points of intersection. (3 marks)
c Write down the set of
values of x for which f(x) > g(x). (1 mark)
17 Find the set of va
lues of x for which the curve with equation y = 2x2 + 3x β 15 is
below the line with equation y = 8 + 2x. (5 marks)
18 On a coordinate grid, shade the r
egion that satisfies the inequalities:
y > x2 + 4x β 12 and y < 4 β x2. (5 marks)
19 a On a coordinate grid, shade the r
egion that satisfies the inequalities
y + x < 6, y < 2x + 9, y > 3 and x > 0. (6 marks)
b Work out the ar
ea of the shaded region. (2 marks)E/P
E
E
E
E
E
E/P
E
E
E
E/P | [
0.12120036780834198,
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0.04... |
58
Chapter 3
1 Find the possible values of k for the quadratic equation 2 kx2 + 5kx + 5k β 3 = 0
to have real roots.
2 A strai
ght line has equation y = 2 x β k and a parabola has equation
y = 3 x2 + 2kx + 5 where k is a constant. Find the range of values of k for which
the line and the parabola do not intersect.Challenge
1 Linear simultaneous equations can be solved using elimination or substit ution.
2 Simultaneous equations with one linear and one quadratic equation can hav
e up to two pairs
of solutions. You need to make sure the solutions are paired correctly.
3 The solutions of a pair o
f simultaneous equations represent the points of intersection of their
graphs.
4 For a pair o
f simultaneous equations that produce a quadratic equation of the form
ax2 + bx + c = 0:
β’βb2 β 4ac > 0 two r eal solutions
β’βb2 β 4ac = 0 one real solution
β’βb2 β 4ac < 0 no real solutions
5 The solution of an inequality is the set o
f all real numbers x that make the inequality true.
6 To solve a quadr
atic inequality:
β’β Rearrange βsoβthatβtheβright-hand βsideβofβtheβinequality βisβ0
β’β Solv
eβtheβcorr
esponding βquadratic βequationβtoβfindβtheβcriticalβvalues
β’β
Sketchβtheβgraphβofβtheβquadratic βfunction
β’β Useβyourβsket
chβtoβfindβtheβrequir
edβsetβofβvalues
.
7 The values of
x for which the curve y = f(x) is below the curve y = g(x) satisfy the inequality
f(x) < g(x).
The values of x for which the curve y = f(x) is above the curve y = g(x) satisfy the inequality
f(x) > g(x).
8 y <
f(x) represents the points on the coordinate grid below the curve y = f(x).
y > f(x) represents the points on the coordinate grid above the curve y = f(x).
9 If y
> f(x) or y < f(x) then the curve y = f(x) is not included in the region and is represented by
a dotted line.
If y > f(x) or y < f(x) then the curve y = f(x) is included in the region and is represented by a solid line.Summary of key points | [
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0.... |
59
Graphs and
transformations
After completing this chapter you should be able to:
β Sketch cubic gr
aphs β pages 60 β 64
β Sketch quartic graphs β pages 64 β 66
β Sketch reciprocal graphs of the form y = a __ x and y = a __ x2 β pages 66 β 67
β Use intersection points of graphs to solve equations β pages 68 β 71
β Translate graphs β pages 71 β 75
β Stretch graphs β pages 75 β 78
β Transform graphs of unfamiliar functions β pages 79 β 81Objectives
1 Factorise these quadratic expressions:
a x2 + 6x + 5 b x2 β 4x + 3
β GCSE Mathematics
2 Sketch the graphs of the following functions:
a y =
(x + 2)(x β 3) b y =
x2 β 6x β 7
β Section 2.4
3 a Copy and complete the table of values for the
function y
= x3 + x β 2.
xβ2 β1.5 β1β0.5 0 0.5 1 1.5 2
yβ12β6.875 β2β1.375
b Use your table o
f values to draw the graph of
y = x3 + x β 2.
β GCSE Mathematics
4 Solve each pair of simultaneous equations:
a y =
2x b y =
x2
x + y = 7 y = 2x + 1
β Sections 3.1, 3.2Prior knowledge check
Many complicated functions can be understood by transforming simpler functions using stretches, reflections and translations. Particle physicists compare observed results with transformations of known functions to determine the nature of subatomic particles.4 | [
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0.04791460931301117,
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-0.... |
60
Chapter 4
4.1 Cubic graphs
A cubic function has the form f(x) = ax3 + bx2 + cx + d, where a, b, c and d are real numbers and a is
non-zero.
The graph of a cubic function can take several different forms, depending on the exact nature of the
function.
Oy
x Oy
x Oy
x Oy
x
β If p is a root of the function f( x), then the graph of y = f(x) touches or crosses the x-axis at
the point ( p, 0).
You can sketch the graph of a cubic function by finding the roots of the function.For these two functions a is positive. For these two functions a is negative.
Example 1
Sketch the curves with the following equations and show the points where they cross the coordinate axes.
a
y =
(x β 2)(1 β x)(1 + x) b y =
x(x + 1)(x + 2)
a y = (x β 2)(1 β x )(1 + x )
0 = (x
β 2)(1 β x )(1 + x )
So x = 2, x = 1 or x = β 1
So the curve crosses the x -axis at
(2, 0), (1, 0) and ( β1, 0).
When x = 0, y = β 2 Γ 1 Γ 1 = β 2
So the curve crosses the y -axis at (0, β 2).
x β β , y β β β
x β ββ, y β β
2 1 O
β2β1 xy
b y = x(x + 1)( x + 2)
0 = x(x + 1)( x + 2)
So x = 0, x = β 1 or x = β 2Find the value of y when x = 0.
Check what happens to y for large positive and
negative values of x.
The x3 term in the expanded function would be
x Γ (βx) Γ x = βx3 so the curve has a negative x3
coefficient.
Put y = 0 and solve for x. Explore the graph of
y
= (x β p )(x β q )(x β r ) where p , q and r are
constants using GeoGebra.Online
Put y = 0 and solve for x. | [
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0.... |
61Graphs and transformations
So the curve crosses the x -axis at
(0, 0), ( β1, 0) and ( β2, 0).
x β β , y β β
x β ββ, y β β β
β1 1 β2O xyYou know that the curve crosses the x-axis at
(0, 0) so you donβt need to calculate the y-intercept separately.
Check what happens to y for large positive and negative values of x.
Example 2
Sketch the following curves.
a y =
(x β 1)2(x + 1) b y = x3 β 2x2 β 3x c y = (x β 2)3
a y = (x β 1)2(x + 1)
0 = ( x β 1)2(x + 1)
So x = 1 or x = β 1
So the curve crosses the x -axis at ( β1, 0)
and touches the x -axis at (1, 0).
When x = 0, y = ( β1)2 Γ 1 = 1
So the curve crosses the y -axis at (0, 1).
x β β , y β β
x β ββ, y β β β
y
x O β1 11
b y = x3 β 2x3 β 3 x
= x(x2 β 2x β 3)
= x(x β 3)( x + 1)
0 = x(x β 3)( x + 1)
So x = 0, x = 3 or x = β 1
So the curve crosses the x -axis at (0, 0),
(3, 0) and ( β1, 0).
x β β , y β β
x β ββ, y β β β
y
x O β1 3Put y = 0 and solve for x.
(x β 1) is squared so x = 1 is a βdoubleβ repeated
root. This means that the curve just touches the x-axis at (1, 0).
Find the value of y when x = 0.
Check what happens to y for large positive and negative values of x.
This is a cubic curve with a positive coefficient of x
3 and three distinct roots.The x3 term in the expanded function would
be x Γ x Γ x = x3 so the curve has a positive x3
coefficient.
Check what happens to y for large positive and negative values of x.
x β β, y β β
x = 1 is a βdoubleβ repeated root.
x β ββ, y β ββ
First factorise. | [
-0.03504672273993492,
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0.03606843948364258,
0.0377005971968174,
-0.05392921715974808,
-0.001179443788714707,
0.... |
62
Chapter 4
Example 3
Sketch the curve with equation y = (x β 1)(x2 + x + 2).
y = ( x β 1)( x2 + x + 2)
0 = ( x β 1)( x2 + x + 2)
So x = 1 only and the curve crosses the
x-axis at (1, 0).
When x = 0, y = ( β1)(2) = β 2
So the curve crosses the y -axis at (0, β 2).
x β β , y β β
x β ββ, y β β β
y
x O
β21The quadratic factor x2 + x + 2 gives no solutions
since the discriminant b2 β 4ac = (1)2 β 4(1)(2) = β7.
β Section 2.5
A cubi c graph could intersect the
x-axis at 1, 2 or 3 points. Watch out
Check what happens to y for large positive and
negative values of x.
You havenβt got enough information y
x
to know the exact shape of the graph. It could also be shaped like this:
1 Sketch the following curves and indicate clearly the points of intersection with the axes:
a y =
(x β 3)(x β 2)(x + 1) b y =
(x β 1)(x + 2)(x + 3)
c y =
(x + 1)(x + 2)(x + 3) d y =
(x + 1)(1 β x)(x + 3)
e y =
(x β 2)(x β 3)(4 β x) f y =
x(x β 2)(x + 1)
g y =
x(x + 1)(x β 1) h y =
x(x + 1)(1 β x)
i y =
(x β 2)(2x β 1)(2x + 1) j y =
x(2x β 1)(x + 3)Exercise 4Ac y = (x β 2)3
0 = ( x β 2)3
So x = 2 and the curve crosses the x -axis
at (2, 0) only.
When x = 0, y = ( β2)3 = β8
So the curve crosses the y -axis at (0, β 8).
x β β , y β β
x β ββ, y β β β
y
x O
β82Check what happens to y for large positive and
negative values of x.
x = 2 is a βtripleβ repeated root. | [
-0.06215880811214447,
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0.03328702226281166,
-0.00296430173330009,
0.014659454114735126,
0.04... |
63Graphs and transformations
2 Sketch the curves with the f
ollowing equations:
a y =
(x + 1)2(x β 1) b y = (x + 2)(x β 1)2 c y = (2 β x)(x + 1)2
d y = (x β 2)(x + 1)2 e y = x2(x + 2) f y = (x β 1)2x
g y =
(1 β x)2(3 + x) h y = (x β 1)2(3 β x) i y = x2(2 β x)
j y =
x2(x β 2)
3 Factorise the follo
wing equations and then sketch the curves:
a y =
x3 + x2 β 2x b y = x3 + 5x2 + 4x c y = x3 + 2x2 + x
d y =
3x + 2x2 β x3 e y = x3 β x2 f y = x β x3
g y = 12x3 β 3x h y = x3 β x2 β 2x i y = x3 β 9x
j y =
x3 β 9x2
4 Sketch the following curves and indicate the coordinates of the points where the curves cross the
axes:
a y =
(x β 2)3 b y = (2 β x)3 c y = (x β 1)3 d y = (x + 2)3
e y =
β(x + 2)3 f y = (x + 3)3 g y = (x β 3)3 h y = (1 β x)3
i y = β (x β 2)3 j y = β (x β 1 _ 2 ) 3
5 The graph of y = x3 + bx2 + cx + d is shown opposite, where b, c and d y
1 β2β3 O x
β6
are real constants.
a Find the values of
b, c and d. (3 marks)
b Write down the coor
dinates of the point where the curve
crosses the y-axis. (1 mark)
6 The graph of
y = ax3 + bx2 + cx + d is shown opposite, where a, b, c and d y
x β12 3 O2
are real constants.
Find the values of a, b, c and d. (4 marks)
7 Given tha
t f(x) = (x β 10)(x2 β 2x) + 12x
a Express f(x
) in the form x(ax2 + bx + c) where a, b and c are real constants. (3 marks)
b Hence factorise f(x) complete
ly. (2 marks)
c Sketch the gra
ph of y = f(x) showing clearly the points where the graph intersects
the axes. (3 marks)P
Start by writing the equation in the form y = (x β p)(x β q)(x β r).Problem-solving
P
E | [
-0.01981847919523716,
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-0.032204121351242065,
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-0.03305774927139282,
0.008042678236961365,
0.0123... |
64
Chapter 4
Example 4
Sketch the following curves:
a y =
(x + 1)(x + 2)(x β 1)(x β 2) b y =
x(x + 2)2(3 β x) c y = (x β 1)2(x β 3)2
a y = (x + 1)( x + 2)( x β 1)( x β 2)
0 = ( x + 1)( x + 2)( x β 1)( x β 2)
So x = β 1, β2, 1 or 2
The curve cuts the x -axis at ( β2, 0), ( β1, 0),
(1, 0) and (2, 0).
When x = 0, y = 1 Γ 2 Γ ( β1) Γ (β2) = 4.
So the curve cuts the y -axis at (0, 4).
x β β , y β β
x β ββ, y β β
Oy
x21/four.ss01
β1β2We know the general shape of the quartic graph
so we can draw a smooth curve through the points.4.2 Quartic graphs
A quartic function has the form f(x) = ax4 + bx3 + cx2 + dx + e, where a, b, c, d and e are real
numbers and a is non-zero.
The graph of a quartic function can take several different forms, depending on the exact nature of the
function.
y
xy
xy
x
You can sketch the graph of a quartic function by finding the roots of the function.This is a
repeated root.
These roots are distinct.
For this function a is negative.For these two functions a is positive.
Check what happens to y for large positive and negative values of x.Substitute x = 0 into the function to find the coordinates of the y-intercept.Set y = 0 and solve to find the roots of the function. Explore the graph of
y
= (x β p )(x β q )(x β r )(x β s ) where p , q, r
and s are constants using GeoGebra.Online | [
-0.026368409395217896,
0.025389019399881363,
-0.037942059338092804,
-0.03609432652592659,
-0.0020098411478102207,
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0.004702458158135414,
... |
65Graphs and transformations
b y = x(x + 2)2(3 β x )
0 = x(x + 2)2(3 β x )
So x = 0, β 2 or 3
The curve cuts the x -axis at (0, 0), ( β2, 0)
and (3, 0)
x β β , y β β β
x β ββ, y β β β
Oy
x β2 3
c y = (x β 1)2(x β 3)2
0 = ( x β 1)2(x β 3)2
So x = 1 or 3
The curve touches the x -axis at (1, 0) and
(3, 0).
When x = 0, y = 9.So the curve cuts the y -axis at (0, 9).
x β β , y β β
x β ββ, y β β
Oy
x9
13The coefficient of x4 in the expanded function will
be negative so you know the general shape of the
curve.
These are both βdoubleβ repeated roots, so the curve will just touch the x-axis at these points.
The coefficient of x4 in the expanded function will
be positive.
There are two βdoubleβ repeated roots.
1 Sketch the following curves and indicate clearly the points of intersection with the axes:
a y =
(x + 1)(x + 2)(x + 3)(x + 4) b y =
x(x β 1)(x + 3)(x β 2)
c y =
x(x + 1)2(x + 2) d y = (2x β 1)(x + 2)(x β 1)(x β 2)
e y =
x2(4x + 1)(4x β 1) f y = β(x β 4)2(x β 2)2
g y = (x β 3)2(x + 1)2 h y = (x + 2)3(x β 3)
i y =
β(2x β 1)3(x + 5) j y = (x + 4)4
2 Sketch the following curves and indicate clearly the points of intersection with the axes:a
y =
(x + 2)(x β 1)(x2 β 3x + 2) b y = (x + 3)2(x2 β 5x + 6)
c y =
(x β 4)2(x2 β 11x + 30) d y = (x2 β 4x β 32)(x2 + 5x β 36) In part f the coefficient
of
x4 will be negative.Hint
Factorise the
qu
adratic factor first.HintExercise 4BThere is a βdoubleβ repeated root at x = β2 so the
graph just touches the x-axis at this point. | [
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0.004696281161159277,
-... |
66
Chapter 4
3 The graph of
y = x 4 + bx3 + cx2 + dx + e is shown opposite,
where b, c, d and e are real constants.
a Find the coordinates of
point P. (2 marks)
b Find the values of
b, c, d and e. (3 marks)
4 Sketch the gra
ph of y = (x + 5)(x β 4)(x2 + 5x + 14). (3 marks)E/P
OPy
x32 β2β1
E/P
Consider the discriminant of the quadratic factor.Problem-solving
4.3 Reciprocal graphs
You can sketch graphs of reciprocal functions such as y = 1 __ x , y = 1 __ x2 and y = β 2 __ x by considering their
as
ymptotes.
β The graphs of y = k __ x and y = k ___ x2 , where k is a real constant, have asymptotes at x = 0 and
y = 0.
An asy mptote is a line which the
graph approaches but never reaches.Notation
Oy
x Oy
x Oy
x Oy
x1
xy =β2
xy =2
x2y =β5
x2y =
y = k __ x with k > 0. y = k __ x with k < 0. y = k __ x2 with k > 0. y = k __ x2 with k < 0.The graph of y = ax 4 + bx3 + cx2 + dx + e
is shown, where a, b, c, d and e are real
constants.
Find the values of a, b, c, d and e.
Oy
x 3 β13Challenge | [
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0.04577682539820671,
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0.03172998130321503,
-0.06593326479196548,
-0.04839285835623741,
0.0421614... |
67Graphs and transformations
Example 5
Sketch on the same diagram:
a y =
4 __ x and y = 12 ___ x b y = β 1 __ x and y = β 3 __ x c y = 4 __ x2 and y = 10 ___ x2
a
O12
xy =
12
xy =/four.ss01
xy =
/four.ss01
xy =xy
b
c
10
x2 y =10
x2 y =
/four.ss01
x2 y =/four.ss01
x2 y =
x Oy1
xy = β1
xy = β
3
xy = β3
xy = β
x OyThis is a y = k __ x graph with k > 0
In this quadrant, x > 0
so for any values of x: 12 ___ x > 4 __ x
In this quadrant, x
< 0
so for any values of x: 12 ___ x < 4 __ x
This is a y = k __ x graph with k < 0
In this quadrant, x < 0
so for any values of x: β 3 __ x > β 1 __ x
In this quadrant, x
> 0
so for any values of x: β 3 __ x < β 1 __ x
This is a y = k __ x2 graph with k > 0.
x2 is always positive and k > 0 so the y-values are
all positive.
1 Use a separate dia gram to sketch each pair of graphs.
a y =
2 __ x and y = 4 __ x b y = 2 __ x and y = β 2 __ x c y = β 4 __ x and y = β 2 __ x
d y =
3 __ x and y = 8 __ x e y = β 3 __ x and y = β 8 __ x
2 Use a separate dia
gram to sketch each pair of graphs.
a y =
2 __ x2 and y = 5 __ x2 b y = 3 __ x2 and y = β 3 __ x2 c y = β 2 __ x2 and y = β 6 __ x2 Exercise 4C Explore the graph of y = a __ x for
different values of a in GeoGebra.Online | [
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-0.032653696835041046,
-0.04588412120938301,
0.03751184418797493,
-0.... |
68
Chapter 4
4.4 Points of intersection
You can sketch curves of functions to show points of intersection and solutions to equations.
β The x-coordinate(s) at the points of intersection of the curves with equations
y = f(x) and y = g(x) are the solution(s) to the equation f( x) = g( x).
a y
x
CBA
1 3Oy = x(x β 3)
y = x2(1 β x)
b From the graph there are three points
wh
ere the curves cross, labelled A, B
and C. The x -coordinates are given by the
solutions to the equation.
x(x β
3) = x2(1 β x)
x2 β 3 x = x2 β x3
x3 β 3 x = 0
x(x2 β 3) = 0
So
x = 0 or x2 = 3
So x = β ββ―__
3 , 0, ββ―__
3
Sub
stitute into y = x2 (1 β x)
T
he points of intersection are:
A(β ββ―__
3 , 3 + 3 ββ―__
3 )
B(0
, 0)
C( ββ―__
3 , 3 β 3 ββ―__
3 )A cubic curve will eventually get steeper than a
quadratic curve, so the graphs will intersect for some negative value of x.
There are three points of intersection so the equation x(x β 3) = x
2(1 β x) has three real roots.
Multiply out brackets.Collect terms on one side.Factorise.
The graphs intersect for these values of x, so you can substitute into either equation to find the y-coordinates.
Leave your answers in surd form.Example 6
a On the same diagram sk etch the curves with equations y = x(x β 3) and y = x2 (1 β x ).
b Find the coordinates of
the points of intersection.
Example 7
a On the same diagram sk etch the curves with equations y = x2(3x β a) and y = b __ x , where a and b
are positive constants.
b State, gi
ving a reason, the number of real solutions to the equation x2(3x β a) β b __ x = 0 | [
0.018378593027591705,
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-0.060521893203258514,
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0.02537... |
69Graphs and transformations
a y
x Oy = x2(3x β a)
ab
xy =b
x
1
3y =
b From the sketch there are only two points
of i
ntersection of the curves. This means
there are only two values of x where
x2 (3x β a) = b __ x
or x2 (3x β a) β b __ x = 0
So this equation has two real solutions.You can sketch curves involving unknown
constants. You should give any points of intersection with the coordinate axes in terms of the constants where appropriate.Problem-solving
Example 8
a Sketch the curves y = 4 ___ x 2 and y = x 2(x β 3) on the same axes.
b Using your sketch, sta
te, with a reason, the number of real solutions to the equation
x 4(x β 3) β 4 = 0.
a y
x
y = x2(x β 3)/four.ss01
x2y =
O 3
b There is a single point of intersection so the
equation x2(x β 3) = 4 ___ x2 has one real solution .
Rearranging:
x 4(x β 3) = 4
x 4(x β 3) β 4 = 0
So this equation has one real solution.Set the functions equal to each other to form an
equation with one real solution, then rearrange the equation into the form given in the question.Problem-solving
You would not be expected to solve this equation in your exam.
1 In each case:
i sketch the two curv
es on the same axes
ii state the number of
points of intersection
iii write down a suitab
le equation which would give the x-coordinates of these points.
(Y ou are not required to solve this equation.)Exercise 4D3x β a = 0 when x = 1 __ 3 a, so the graph of
y = x2(3x β a) touches the x-axis at (0, 0) and
intersects it at ( 1 __ 3 a, 0)
You only need to state the number of solutions.
You donβt need to find the solutions. | [
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-0.005618855822831392,
... |
70
Chapter 4
a y =
x2, y = x(x2 β 1) b y = x(x + 2), y = β 3 __ x c y = x2, y = (x + 1)(x β 1)2
d y = x2(1 β x), y = β 2 __ x e y = x(x β 4), y = 1 __ x f y = x(x β 4), y = β 1 __ x
g y =
x(x β 4), y = (x β 2)3 h y = βx3, y = β 2 __ x i y = βx3, y = x2
j y = β x3, y = βx(x + 2) k y = 4, y = x(x β 1)(x + 2)2 l y = x3, y = x2(x + 1)2
2 a On the same axes sketch the curv es given by y = x2(x β 3) and y = 2 __ x
b Explain how your sk
etch shows that there are only two real solutions to the equation
x 3(x β 3) = 2.
3 a On the same axes sketch the curv
es given by y = (x + 1)3 and y = 3x(x β 1).
b Explain how your sk
etch shows that there is only one real solution to the equation
x 3 + 6x + 1 = 0.
4 a On the same axes sketch the curv
es given by y = 1 __ x and y = βx(x β 1)2.
b Explain how your sk
etch shows that there are no real solutions to the equation
1 + x2(x β 1)2 = 0.
5 a On the same axes sketch the curv
es given by y = x2(x + a)
and y = b __ x where a and b are both positive
constants. (5 marks)
b Using your sketch, sta
te, giving a reason, the number of
real solutions to the equation x 4 + ax3 β b = 0. (1 mark)
6 a On the same set of axes sk
etch the graphs of
y =
4 __ x2 and y = 3x + 7. (3 marks)
b Write down the n
umber of real solutions to the equation 4 __ x2 = 3x + 7. (1 mark)
c Show that y
ou can rearrange the equation to give (x + 1)(x + 2)(3x β 2) = 0. (2 marks)
d Hence determine the exact coor
dinates of the points of intersection. (3 marks)
7 a On the same axes sketch the curv
e y = x3 β 3x2 β 4x and the line y = 6x.
b Find the coordinates of
the points of intersection.
8 a On the same axes sketch the curv
e y = (x2 β 1)(x β 2) and the line y = 14x + 2.
b Find the coordinates of
the points of intersection.
9 a On the same axes sketch the curv
es with equations y = (x β 2)(x + 2)2 and y = βx2 β 8.
b Find the coordinates of
the points of intersection.
10 a Sketch the gra
phs of y = x2 + 1 and 2y = x β 1. (3 marks)
b Explain why ther
e are no real solutions to the equation 2x2 β x + 3 = 0. (2 marks)
c Work out the r
ange of values of a such that the graphs of y = x2 + a and 2y = x β 1
have two points of intersection. (5 marks)E/P
Even though you donβt know
the values of a and b , you
know they are positive, so you know the shapes of the graphs. You can label the point a on the x -axis on your
sketch of
y = x2(x + a).Problem-solving
E
P
P
E/P | [
0.006487688515335321,
0.04751107469201088,
-0.014701266773045063,
-0.07448455691337585,
-0.022692671045660973,
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0.05339983105659485,
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0.024006370455026627,
0.05221424996852875,
-0.03407466039061546,
0.009938026778399944,
-0.0347... |
71Graphs and transformations
11 a Sketch the gra
phs of y = x2(x β 1)(x + 1) and y = 1 _ 3 x3 + 1. (5 marks)
b Find the number of r
eal solutions to the equation 3x2(x β 1)(x + 1) = x3 + 3. (1 mark)E/P
4.5 Translating graphs
You can transform the graph of a function by altering the function. Adding or subtracting a constant
βoutsideβ the function translates a graph vertically.
β The graph of y = f(x) + a is a translation of the graph y = f(x) by the vector ( 0 a ) .
Adding or subtracting a constant βinsideβ the function translates the graph horizontally.
β The graph of y = f(x + a) is a translation of the graph y = f(x) by the vector ( βa 0 ) .
y
x O2 4 1 3 β2 β4β5 β1 β3123456
β1y = f(x + 2) is a translation ( β2 0 ) , or 2 units in the
direction of the negative x-axis.y = f(x) + 1 is a translation ( 0 1 ) , or 1 unit in the
direction of the positive y-axis.
Example 9
Sketch the graphs of:
a y =
x2 b y = (x β 2)2 c y = x2 + 2
a y
x O
b y
y = (x β 2)2
x/four.ss01
2 OThis is a translation by vector ( 2 0 ) .
Remember to mark on the intersections with the
axes. | [
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0.02051602117717266,
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-0.02727763168513775,
-0.005836542695760727,
0.00... |
72
Chapter 4
c y = x2 + 2
y
x2
OThis is a translation by vector ( 0 2 ) .
Remember to mark on the y-axis intersection.
Example 10
f(x) = x3
g(x) = x(x β 2)
Sketch the following graphs, indicating any points where the curves cross the axes:a
y =
f(x + 1)
b y =
g(x + 1)
a The graph of f( x) is
y
x Oy = f( x) = x3
So the graph of y = f( x + 1) is
y
x Oy = f( x + 1) = ( x + 1)3
β11
b g(x) =
x(x β 2)
The curve is y = x (x β 2)
0 = x(x β 2)
So x = 0 or x = 2
y
xy = g( x) = x(x β 2)
2 OFirst sketch y = f(x).
This is a translation of the graph of y = f(x) by
vector ( β1 0 ) .
You could also write out the equation as
y = (x + 1)3 and sketch the graph directly.
First sketch g(x).Put y = 0 to find where the curve crosses the x-axis. Explore translations of the
gr
aph of y = x3 using GeoGebra.Online | [
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73Graphs and transformations
So the graph of y = g( x + 1) is
y
xy = g( x + 1)
= (x + 1)( x β 1)
1 β1β1O
β When you translate a function, any asymptotes are also translated.
Example 11
Given that h(x) = 1 __ x , sketch the curve with equation y = h(x) + 1 and state the equations of any
asymptotes and intersections with the axes
.
The graph of y = h( x) is
Oy
x1
xy =
So the graph of y = h( x) + 1 is
Oy
x1
The curve crosses the x -axis once.
y = h(x) +
1 = 1 __ x + 1
0 = 1 __ x + 1
β1 = 1 __ x
x = β1
So t
he curve intersects the x -axis at (β1, 0).
The horizontal asymptote is y = 1.
The vertical asymptote is x = 0.First sketch y = h(x ).
The curve is translated by vector ( 0 1 ) so the
asymptote is translated by the same vector.
Put y = 0 to find where the curve crosses the
x-axis.
Remember to write down the equation of the vertical asymptote as well. It is the y-axis so it has equation x = 0.This is a translation of the graph of y = g(x) by
vector ( β1 0 ) .
You could also write out the equation and sketch
the graph directly:
y = g(x
+ 1)
= (x + 1)(x + 1 β 2)
= (x + 1)(x β 1) | [
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... |
74
Chapter 4
1 Apply the f
ollowing transformations to the curves with equations y = f(x) where:
i f(x
) = x2 ii f(x) = x3 iii f(x) = 1 __ x
In each case state the coordina
tes of points where the curves cross the axes and in iii state the
equations of the asymptotes.
a f(x + 2) b f(x) + 2 c f(x
β 1)
d f(x)
β 1 e f(x
) β 3 f f(x
β 3)
2 a Sketch the curve
y = f(x) where f(x) = (x β 1)(x + 2).
b On separate dia
grams sketch the graphs of i y =
f(x + 2) ii y =
f(x) + 2.
c Find the equations of the curv
es y = f(x + 2) and y = f(x) + 2, in terms of x, and use these
equations to find the coordinates of the points where your graphs in part b cross the y-axis.
3 a Sketch the gra
ph of y = f(x) where f(x) = x2(1 β x).
b Sketch the curve with equa
tion y = f(x + 1).
c By finding the equation f(x
+ 1) in terms of x, find the coordinates of the point in part b
where the curve crosses the y-axis.
4 a Sketch the gra
ph of y = f(x) where f(x) = x(x β 2)2.
b Sketch the curves with equa
tions y = f(x) + 2 and y = f(x + 2).
c Find the coordinates of
the points where the graph of y = f(x + 2) crosses the axes.
5 a Sketch the gra
ph of y = f(x) where f(x) = x(x β 4).
b Sketch the curves with equa
tions y = f(x + 2) and y = f(x) + 4.
c Find the equations of the curv
es in part b in terms of x and hence find the coordinates of the
points where the curves cross the axes.
6 a Sketch the gra
ph of y = f(x) where f(x) = x2(x β 1)(x β 2).
b Sketch the curves with equa
tions y = f(x + 2) and y = f(x) β 1.
7 The point P(4,
β1) lies on the curve with equation y = f(x).
a State the coordina
tes that point P is transformed to on the curve with equation
y =
f(x β 2). (1 mark)
b State the coordina
tes that point P is transformed to on the curve with equation
y =
f(x) + 3. (1 mark)
8 The graph of
y = f(x) where f(x) = 1 __ x is translated so that the asymptotes are at x = 4 and
y = 0. Write down the equation for the transformed function in the form y = 1 _____ x + a (3 marks)
9 a Sketch the gra
ph of y = x3 β 5x2 + 6x, marking clearly the points of intersection with the axes.
b Hence sketch y
= (x β 2)3 β 5(x β 2)2 + 6(x β 2). E
E/P
PExercise 4E | [
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75Graphs and transformations
10 a Sketch the gra
ph of y = x2(x β 3)(x + 2), marking clearly the points of intersection with the axes.
b Hence sketch y
= (x + 2)2(x β 1)(x + 4).
11 a Sketch the gra
ph of y = x3 + 4x2 + 4x. (6 marks)
b The point with coordinates (
β1, 0) lies on the curve with
equation y = (x + a)3 + 4(x + a)2 + 4(x + a) where a is a
constant. Find the two possible values of a. (3 marks)
12 a Sketch the gra
ph of y = x(x + 1)(x + 3)2. (4 marks)
b Find the possible va
lues of b such that the point (2, 0) lies on the curve with equation
y = (x + b)(x + b + 1)(x + b + 3)2. (3 marks)P
E/P
E/P
1 Sketch the graph of y = (x β 3)3 + 2 and determine the coordinates of the point of inflection. β Section 12.9
2 The point Q (β5, β7) lies on the curve with equation y = f( x).
a Sta
te the coordinates that point Q is transformed to on the curve with equation y = f( x + 2) β 5.
b The c
oordinates of the point Q on a transformed curve are ( β3, β6). Write down the transformation in
the form y = f( x + a) β b.ChallengeLook at your sketch and
picture the curve sliding to the left or right.Problem-solving
4.6 Stretching graphs
Multiplying by a constant βoutsideβ the function stretches the graph vertically.
β The graph of y = af(x) is a stretch of the graph y = f(x) by a scale factor of a in the vertical
direction.
Oy
x β11y = f(x)1
2y = 2f(x)y = f(x)
Multiplying by a constant βinsideβ the function stretches the graph horizontally.
β The graph of y = f(ax) is a stretch of the graph y = f(x) by a scale factor of 1 __ a in the
horizontal direction.
Oy
x 2 64
y = f( x)1
3y = f(2x)
y = f(x)2f(x) is a stretch with scale factor 2 in the
y-direction. All y-coordinates are doubled.
1 _ 2 f(x) is a stretch with scale factor 1 _ 2 in the
y-direction.
All y-coordinates are halved.
y = f ( 1 _ 3 x) is a stretch with scale factor 3 in the
x-direction. All x-coordinates are tripled.y = f(2x) is a stretch with scale factor 1 _ 2 in the
x-direction.
All x-coordinates are halved. | [
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76
Chapter 4
Example 12
Given that f(x) = 9 β x2, sketch the curves with equations:
a y =
f(2x) b y =
2f(x)
a f(x) = 9 β x2
So f(x) = (3 β x)(3 + x )
The curve is y =
(3 β x )(3 + x )
0 = (3 β x )(3 + x )
So x = 3 or x = β 3
So the curve crosses the x -axis at (3, 0)
and (β3, 0).
When x = 0, y = 3 Γ 3 = 9
So the curve crosses the y -axis at (0, 9).
The curve y = f( x) is
3 β39
O xy
y = f(2 x) so the curve is
1.5 β1.59
O xy
b y = 2f(x) so the curve is
3 β3/one.ss018
x OyYou can factorise the expression.
Put y = 0 to find where the curve crosses the
x-axis.
Put x = 0 to find where the curve crosses the y-axis.
First sketch y = f(x).
y = f(ax) where a = 2 so it is a horizontal stretch with scale factor
1 _ 2 .
Check: The curve is
y = f(2x).
So y = (3 β 2x)(3 + 2x).
When y = 0, x = β1.5 or x = 1.5.So the curve crosses the x-axis at (β1.5, 0) and (1.5, 0).When x = 0, y = 9.So the curve crosses the y-axis at (0, 9).
y = af(x) where a = 2 so it is a vertical stretch with
scale factor 2.
Check: The curve is y = 2f(x).
So y = 2(3 β x)(3 + x).When y = 0, x = 3 or x = β3.So the curve crosses the x -axis at (β 3, 0) and (3, 0).
When x = 0, y = 2 Γ 9 = 18.So the curve crosses the y-axis at (0, 18). | [
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77Graphs and transformations
Example 13
a Sketch the curve with equation y = x(x β 2)(x + 1).
b On the same axes, sk
etch the curves y = 2x(2x β 2)(2x + 1) and y = βx(x β 2)(x + 1).
a
Oy
x 2y = x(x β 2)( x + 1)
β1
b
y = x(x β 2)( x + 1)y = 2 x(2x β 2)(2 x + 1)y = βx(x β 2)( x + 1)
Oy
x 2 β1y = βx(x β 2)(x + 1) is a stretch with scale factor
β1 in the y-direction. Notice that this stretch has the effect of reflecting the curve in the x-axis.
y = 2x(2x β 2)(2x + 1) is a stretch with scale factor
1 _ 2 in the x-dir ection.
You need to work out the relationship between
each new function and the original function. If
x(x β 2)(x + 1) = f( x) then
2x(2x β 2)(2 x + 1) = f(2 x), and
βx(x β 2)(x + 1) = β f(x).Problem-solving
β The graph of y = βf(x) is a reflection of the graph of y = f(x) in the x-axis.
β The graph of y = f(β x) is a reflection of the graph of y = f(x) in the y-axis.
Example 14
On the same axes sketch the graphs of y = f(x),
y = f(βx) and y = βf(x) where f(x) = x(x + 2).
f(x) = x(x + 2)
y
x O 2 β2
y = βf(x)y = f( x) y = f(βx)y = f(βx) is y = (βx)(βx + 2) which is y = x2 β 2x
or y = x(x β 2) and this is a reflection of the
original curve in the y-axis.
Alternatively multiply each x-coordinate by β1
and leave the y-coordinates unchanged.
This is the same as a stretch parallel to the x-axis
scale factor β1.
y = βf(x) is y = βx(x + 2) and this is a reflection of the original curve in the x-axis.
Alternatively multiply each y-coordinate by β1
and leave the x-coordinates unchanged.
This is the same as a stretch parallel to the y-axis
scale factor β1. Explore stretches of the graph
of
y = x (x β 2)( x + 1) using GeoGebra.Online | [
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... |
78
Chapter 4
1 Apply the f
ollowing transformations to the curves with equations y = f(x) where:
i f(x
) = x2 ii f(x) = x3 iii f(x) = 1 __ x
In each case show both f(x
) and the transformation on the same diagram.
a f(2x) b f(βx
) c f( 1 _ 2 x) d f(4x) e f( 1 _ 4 x)
f 2f(x
) g βf(x
) h 4f(x
) i 1 _ 2 f(x) j 1 _ 4 f(x)
2 a Sketch the curv
e with equation y = f(x) where f(x) = x2 β 4.
b Sketch the gra
phs of y = f(4x), 1 _ 3 y = f(x), y = f(βx) and y = βf(x).
3 a Sketch the curv
e with equation y = f(x) where f(x) = (x β 2)(x + 2)x.
b Sketch the gra
phs of y = f( 1 _ 2 x), y = f(2x) and y = βf(x).
4 a Sketch the curv
e with equation y = x2(x β 3).
b On the same axes, sk
etch the curves with equations:
i y =
(2x)2(2x β 3) ii y = βx2(x β 3)
5 a Sketch the curv
e y = x2 + 3x β 4.
b On the same axes, sk
etch the graph of 5y = x2 + 3x β 4.
6 a Sketch the gr
aph of y = x2(x β 2)2.
b On the same axes, sk
etch the graph of 3y = βx2(x β 2)2.
7 The point P(2,
β3) lies on the curve with equation y = f(x).
a State the coordina
tes that point P is transformed to on the curve with equation
y = f(2x). (1 mark)
b State the coordina
tes that point P is transformed to on the curve with equation
y = 4f(x). (1 mark)
8 The point Q(
β2, 8) lies on the curve with equation y = f(x).
State the coor dina
tes that point Q is transformed to on the curve with equation
y = f( 1 _ 2 x). (1 mark)
9 a Sketch the gr
aph of y = (x β 2)(x β 3)2. (4 marks)
b The graph of
y = (ax β 2)(ax β 3)2 passes through the point (1, 0).
Find two possible values for a . (3 marks) For part b, rearrange
the s
econd equation into
the form y = 3f( x).Hint
P
Let f( x) = x2(x β 3) and try to
write each of the equations
in part b in terms of f( x).Problem-solving
E
E
E/PExercise 4F
1 The point R(4, β 6) lies on the curve with equation y = f(x). State the coordinates
that point R is transformed to on the curve with equation y = 1 _ 3 f(2x).
2 The p
oint S(β4, 7) is transformed to a point S9(β8, 1.75). Write down the
transformation in the form y = af(bx).Challenge | [
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79Graphs and transformations
4.7 Transforming functions
You can apply transformations to unfamiliar functions by considering how specific points and features
are transformed.
Example 15
The following diagram shows a sketch of the curve f(x) which passes through the origin. The points A(1, 4) and B(3, 1) also lie on the curve.
Sketch the following:
a
y =
f(x + 1) b y =
f(x β 1) c y =
f(x) β 4
d 2y
= f(x) e y β
1 = f(x)
In each case you should show the positions of the images
of the points O, A and B.y
x Oy = f(x)
B (3, 1)A (1, 4)
a f(x + 1)
O(2, 1)/four.ss01
β1 xy
y = f( x + 1)
b f(x
β 1)
O(/four.ss01, 1)
1(2, /four.ss01)
xy
y = f( x β 1)
c f(x) β
4
y
x O
β/four.ss011
(3, β3)y = f( x) β 4Translate f(x) 1 unit in the direction of the
negative x-axis.
Translate f(x) 1 unit in the direction of the positive x-axis.
Translate f(x) 4 units in the direction of the negative y-axis. | [
-0.016790170222520828,
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... |
80
Chapter 4
d 2y = f(x) so y = 1 __ 2 f(x)
(3, )(1, 2)
1
2y = f(x)1
2y
x O
e y β 1 = f( x) so y = f( x) + 1
1y
x O(3, 2)(1, 5)
y = f( x) + 1Rearrange in the form y = β¦
Stretch f(x) by scale factor 1 _ 2 in the y-dir ection.
Rearrange in the form y = β¦
Translate f(x) 1 unit in the direction of the
positive y-axis.
1 The following diagram shows a sketch of the curve y
x O(4, 4)
C
B DA2
1 6
with equation y = f(x). The points A(0, 2), B(1, 0),
C(4, 4) and D(6, 0) lie on the curve.
Sketch the following graphs and give the coordinates
of the points, A, B, C and D after each transformation:
a f(x
+ 1) b f(x
) β 4 c f(x
+ 4)
d f(2x
) e 3f(x
) f f( 1 _ 2 x)
g 1 _ 2 f(x) h f(β x)
2 The curve y
= f(x) passes through the origin and y
x Ox = 1y = 2
has horizontal asymptote y = 2 and vertical
asymptote x = 1, as shown in the diagram.
Sketch the following graphs. Give the equations of
any asymptotes and give the coordinates of intersections with the axes after each transformation.
a
f(x
) + 2 b f(x
+ 1) c 2f(x
)
d f(x
) β 2 e f(2x
) f f( 1 _ 2 x)
g 1 _ 2 f(x) h βf(x )Exercise 4G | [
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81Graphs and transformations
3 The curve with equation
y = f(x) passes through the y
x OB
CD
A (β4, β6)β2 4
β3
points A(β4, β6), B(β2, 0), C(0, β3) and D(4, 0)
as shown in the diagram.
Sketch the following and give the coordinates of
the points A, B, C and D after each transformation.
a f(x
β 2) b f(x
) + 6 c f(2x
)
d f(x
+ 4) e f(x
) + 3 f 3f(x
)
g 1 _ 3 f(x) h f( 1 _ 4 x) i βf(x )
j f(β
x)
4 A sketch of the curv
e y = f(x) is shown in the y
x Ox = β21
diagram. The curve has a vertical asymptote
with equation x = β2 and a horizontal asymptote with equation y = 0. The curve crosses the y-axis at (0, 1).
a
Sketch, on separa
te diagrams, the
graphs of:
i 2f(x
) ii f(2x
) iii f(x
β 2)
iv f(x
) β 1 v f(β
x) vi βf(x
)
In each case state the equations of any
asymptotes and, if possible, points where the curve cuts the axes.
b
Suggest a possible equation f
or f(x).
5 The point P(2, 1) lies on the gr
aph with equation y = f(x).
a On the graph of
y = f(ax), the point P is mapped to the point Q(4, 1).
Determine the value of a. (1 mark)
b Write down the coor
dinates of the point to which P maps under each transformation
i f(x
β 4) ii 3f(x
) iii 1 _ 2 f(x) β 4 (3 marks)
6 The diagram sho
ws a sketch of a curve with equation y = f(x).
The points A(β1, 0), B(0, 2), C(1, 2) and D(2, 0) lie on the curve.
Sketch the following graphs and give the coordinates of the points
A, B, C and D after each transformation:
a y +
2 = f(x) b 1 _ 2 y = f(x)
c y β
3 = f(x) d 3y
= f(x)
e 2y
β 1 = f(x)E/P
OABC
Dy
xP
Rearrange each equation
into the form y = β¦Problem-solving | [
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82
Chapter 4
1 a On the same axes sketch the gr
aphs of y = x2(x β 2) and y = 2x β x2.
b By solving a suitable equa
tion find the points of intersection of the two graphs.
2 a On the same axes sketch the curv
es with equations y = 6 __ x and y = 1 + x.
b The curves intersect at the points
A and B . Find the coordinates of A and B .
c The curve C with equa
tion y = x2 + px + q, where p and q are integers, passes through A and B .
Find the values of p and q .
d Add C to y
our sketch.
3 The diagram sho
ws a sketch of the curve y = f(x). y
x O2
BA (3, 4)
y = 2
The point B (0, 0) lies on the curve and the point A (3, 4)
is a maximum point. The line y = 2 is an asymptote.
Sketch the following and in each case give the
coordinates of the new positions of A and B and
state the equation of the asymptote:
a f(2x) b 1 _ 2 f(x) c f(x) β 2
d f(x + 3) e f(x β
3) f f(x) + 1
4 The diagram sho
ws the curve with equation
y = 5 + 2x β x2 and the line with equation y = 2.
The curve and the line intersect at the points
A and B .
Find the x -coordinates of A and B . (4 marks)
5 f(x
) = x2(x β 1)(x β 3).
a Sketch the gra
ph of y = f(x). (2 marks)
b On the same axes, dr
aw the line y = 2 β x. (2 marks)
c State the number of
real solutions to the equation x2(x β 1)(x β 3) = 2 β x. (1 mark)
d Write down the coor
dinates of the point where the graph with equation
y = f(x) + 2 crosses the y-axis. (1 mark)
6 The figure shows a sk
etch of the curve with
equation y = f(x).
On separate axes sketch the curves with equations:a
y =
f(β x) (2 marks)
b y =
βf(x) (2 marks)
Mark on each sketch the
x-coordinate of any point,
or points, where the curve touches or crosses the x -axis.P
Ey
x Oy = 2 B A
y = 5 + 2x β x2
E/P
Ey
x O β2 2Mixed exercise 4 | [
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83Graphs and transformations
7 The diagram sho
ws the graph of the quadratic function f(x).
x O13
(2, β1)y
y = f(x)
The graph meets the x-axis at (1, 0) and (3, 0) and the
minimum point is (2, β1).
a Find the equation of the gr
aph in the form
y = ax2 + bx = c (2 marks)
b On separate ax
es, sketch the graphs of
i y =
f(x + 2) ii y =
(2x). (2 marks)
c On each graph la
bel the coordinates of the
points at which the graph meets the x-axis and
label the coordinates of the minimum point.
8 f(x
) = (x β 1)(x β 2)(x + 1).
a State the coordina
tes of the point at which the graph y = f(x ) intersects the y -axis. (1 mark)
b The graph of
y = af(x) intersects the y -axis at (0, β 4). Find the value of a . (1 mark)
c The graph of
y = f(x + b) passes through the origin. Find three possible values of b . (3 marks)
9 The point P(4, 3) lies on a curv
e y = f(x).
a State the coordina
tes of the point to which P is transformed on the curve with equation:
i y =
f(3x) ii 1 _ 2 y = f(x) iii y = f(x β 5) iv βy = f(x) v 2( y + 2) = f(x)
b P is transf
ormed to point (2, 3). Write down two possible transformations of f(x).
c P is transf
ormed to point (8, 6). Write down a possible transformation of f(x) if
i f(x
) is translated only
ii f(x
) is stretched only.
10 The curve C1 has equation y = β a __ x2 where a is a positive constant. The curve C2 has the
equation y = x2 (3x + b) where b is a positive constant.
a Sketch C1 and C2 on the same set of axes, showing clearly the coordinates of any
point where the curves touch or cross the axes. (4 marks)
b Using your sketch sta
te, giving reasons, the number of solutions to the equation
x 4 (3x + b) + a = 0. (2 marks)
11 a Factorise completel
y x3 β 6x2 + 9x. (2 marks)
b Sketch the curve of
y = x3 β 6x2 + 9x showing clearly the coordinates of the
points where the curve touches or crosses the axes. (4 marks)
c The point with coordinates (
β4, 0) lies on the curve with equation
y = (x β k)3 β 6(x β k)2 + 9(x β k) where k is a constant.
Find the two possible values of k. (3 marks)
12 f(x
) = x(x β 2)2
Sketch on separate axes the graphs of:
a y =
f(x) (2 marks)
b y =
f(x + 3) (2 marks)
Show on each sketch the coor
dinates of the points where each graph crosses or meets the axes.E/P
E/P
P
E/P
E/P
E | [
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84
Chapter 4
13 Given tha
t f(x) = 1 __ x , x β 0,
a Sketch the gra
ph of y = f(x) β 2 and state the equations of the asymptotes. (3 marks)
b Find the coordinates of
the point where the curve y = f(x) β 2 cuts a coordinate
axis. (2 marks)
c Sketch the gra
ph of y = f(x + 3). (2 marks)
d State the equations of
the asymptotes and the coordinates of the point where
the curve cuts a coordinate axis. (2 marks)E
The point R (6, β 4) l ies on the curve with equation y = f( x). State the coordinates
that point R is transformed to on the curve with equation y = f( x + c ) β d .Challenge
1 If p is a r oot of the function f( x), then the graph of y = f(x) touches or crosses the x-axis at
the point (p, 0).
2 The graphs of y = k __ x and y = k __ x2 , where k is a real constant, have asymptotes at x = 0 and y = 0.
3 The x-coordinate(s) at the points of intersection of the curves with equations y = f(x) and
y = g(x) are the solution(s) to the equation f( x) = g( x).
4 The graph of y = f(x) + a is a translation of the graph y = f(x) by the vector ( 0 a ) .
5 The graph of y = f(x + a) is a translation of the graph y = f(x) by the vector ( βa 0 ) .
6 When you translat
e a function, any asymptotes are also translated.
7 The graph of y = af(x) is a stretch of the graph y = f(x) by a scale factor of a in the vertical
direction.
8 The graph of y = f(ax) is a stretch of the graph y = f( x) by a scale factor of 1 __ a in the horizontal
direction.
9 The graph of y = βf( x) is a reflection of the graph of y = f(x) in the x-axis.
10 The graph of y = f(β x) is a reflection of the graph of y = f(x) in the y-axis.Summary of key points | [
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Review exercise
851
1 a Write down the value of 8 1 _ 3 . (1 mark)
b Find the value of
8 β 2 _ 3 . (2 marks)
β Section 1.4
2 a Find the value of 12 5 4 _ 3 . (2 marks)
b Simplify 24x2 Γ· 18 x 4 _ 3 . (2 marks)
β Sections 1.1, 1.4
3 a Express ββ―___ 80 in the form a ββ―__
5 ,
where a is an integer. (2 marks)
b Express (4 β
ββ―__
5 )2 in the form b + c ββ―__
5 ,
where b and c are integers. (2 marks)
β Section 1.5
4 a Expand and simplify
(4 +
ββ―__
3 )(4 β ββ―__
3 ). (2 marks)
b Express 26 ______ 4 + ββ―__
3 in the form a + b ββ―__
3 ,
where a and b are integers. (3 marks)
β Sections 1.5, 1.6
5 Here are three numbers:
1 β
β __
k , 2 + 5 β __
k and 2 β __
k
Given tha
t k is a positive integer, find:
a the mean of the three
n
umbers. (2 marks)
b the range of the thr
ee
numbers. (1 mark)
β Section 1.5
6 Given that y = 1 ___ 25 x 4 , express each of the
following in the form kxn, where k and n
are constants.
a yβ1 (1 mark)
b 5 y 1 _ 2 (1 mark)
β Section 1.4E
E
E
E/p
E7 Find the area of this tr apezium in cm2.
Give your answer in the form a + b β __
2 ,
where a and b are integers to be
found. (4 marks)
β Section 1.5
(5 + 3 2) cm3 + 2 cm
2 2 cm
8 Given that p = 3 β 2 β __
2 and q = 2 β β __
2 ,
find the value of p + q _____ p β q .
Give y
our answer in the form m + n β __
2 ,
where m and n are rational numbers to be
found. (4 marks)
β Sections 1.5, 1.6
9 a Factorise the expression x
2 β 10x +16. (1 mark)
b Hence, or otherwise, solv
e the equation
82y β 10(8y) + 16 = 0. (2 marks)
β Sections 1.3, 2.1
10 x2 β 8x β 29 (x + a)2 + b, where a and b
are constants.
a Find the value of
a and the value
of b. (2 marks)
b Hence, or otherwise, sho
w that the
roots of x2 β 8x β 29 = 0 are c Β± d ββ―__
5 ,
where c and d are integers. (3 marks)
β Sections 2.1, 2.2E/p
E
E/p
E | [
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86
Review exercise 1
11 The functions f and g are defined as
f(
x) = x(x β 2) and g(x) = x + 5, x β β .
Given tha
t f(a) = g(a) and a > 0,
find the value of a to three significant figures.
(3 marks)
β Sections 2.1, 2.3
12 An athlete launches a shot put from shoulder height. The height of the shot put, in metr
es, above the ground
tΒ seconds after launch, can be modelled by the following function:
h(t) = 1.7 + 10t β 5t
2 t > 0
a Give the ph
ysical meaning of the
constant term 1.7 in the context of the
model.
b Use the model to calcula
te how many
seconds after launch the shot put hits the ground.
c
Rearrange h(
t) into the form
AΒ βΒ B(tΒ βΒ C)2 and give the values of the
constants A, B and C.
d Using your answ
er to part c or
otherwise, find the maximum height of the shot put, and the time at which this maximum height is reached.
β Section 2.6
13 Given that f(x) = x2 β 6x + 18, x > 0,
a express f(
x) in the form (x β a)2 + b,
where a and b are integers. (2 marks)
The curve C
with equation y = f(x),
x > 0, meets the y-axis at P and has a minimum point at Q.
b
Sketch the gra
ph of C, showing the
coordinates of P and Q. (3 marks)
The line y
= 41 meets C at the point R.
c Find the x-coor
dinate of R, giving
your answer in the form p + q ββ―__
2 ,
where p and q are integers. (2 marks)
β Sections 2.2, 2.4
14 The function h(x) = x2 + 2 β __
2 x + k has
equal roots.
a Find the value of
k. (1 mark)
b Sketch the gra
ph of y = h(x), clearly
labelling any intersections with the
coordinate axes. (3 marks)
β Sections 1.5, 2.4, 2.5E/p
p
E/p
E15 The function g(x) is defined as
g(
x) = x9 β 7x6 β 8x3, x β β .
a Write g(x ) in the f
orm x3(x3 + a)(x3 + b),
where a and b are integers. (1 mark)
b Hence find the three roots
of
g(x). (1 mark)
β Section 2.3
16 Given that
x2 + 10x + 36 (x + a)2 + b,
where a and b are constants,
a find the value of
a and the value
of b. (2 marks)
b Hence show that the equa
tion
x2 + 10x + 36 = 0 has no
real roots. (2 marks)
The equation x2 + 10x + k = 0 has equal
roots.c
Find the value of
k. (2 marks)
d For this va
lue of k, sketch the graph
of y = x2 + 10x + k, showing the
coordinates of any points at which
theΒ graph meets the coordinate axes.
(3 marks)
β Sections 2.2, 2.4, 2.5
17 Given that x2 + 2x + 3 (x + a)2 + b,
a find the value of
the constants
a and b (2 marks)
b Sketch the gra
ph of y = x2 + 2x + 3,
indicating clearly the coordinates of any intersections with the coordinate axes.
(3 marks)
c Find the value of
the discriminant of
x2 + 2x + 3. Explain how the sign of
the discriminant relates to your sketch in part b.
(2 marks)
The equation x2 + kx + 3 = 0, where k is a
constant, has no real roots.
d Find the set of possible v
alues
of k, giving your answer in surd
form. (2 marks)
β Section 2.2, 2.4, 2.5E/p
E/p
E/p | [
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87
Review exercise 1
18 a By eliminating
y from the equations:
y = x β 4,
2x2 β xy = 8,
show that
x2 + 4x β 8 = 0. (2 marks)
b Hence, or otherwise, solv
e the
simultaneous equations:
y = x β 4,2x
2 β xy = 8,
giving your answers in the form a Β± b
ββ―__
3 , where a and b are
integers. (4 marks)
β Section 3.2
19 Find the set of va lues of x for which:
a 3(2x
+ 1) > 5 β 2x, (2 marks)
b 2x2 β 7x + 3 > 0, (3 marks)
c both 3(2
x + 1) > 5 β 2x and
2x2 β 7x + 3 > 0. (1 mark)
β Sections 3.4, 3.5
20 The functions p and q are defined as
p(x
) = β2(x + 1) and q(x) = x2 β 5x + 2,
x β β . Show alge
braically that there is no
value of x for which p(x ) = q(x). (3 marks)
β Sections 2.3, 2.5
21 a Solve the simultaneous equations:
y
+ 2x = 5
2x2 β 3x β y = 16. (5 marks)
b Hence, or otherwise, find the set of
values of x for which:
2x2 β 3x β 16 > 5 β 2x. (2 marks)
β Sections 3.2, 3.5
22 The equation x2 + kx + (k + 3) = 0, where
k is a constant, has different real roots.
a Show that
k2 β 4k β 12 > 0. (2 marks)
b Find the set of possible v
alues of k.
(2 marks)
β Sections 2.5, 3.5
23 Find the set of va lues for which
6 _____ x + 5 < 2, x β β5. (6 marks)
β Section 3.5E
E
E/p
E
E/p
E24 The functions f and g are defined as
f(
x) = 9 β x2 and g(x) = 14 β 6x, x β β .
a On the same set of axes
, sketch the
graphs of y = f(x) and y = g(x). Indicate
clearly the coordinates of anyΒ points where the graphs intersect with each other or the coordinate axes.
(5 marks)
b On your sketch, shade the r
egion that
satisfies the inequalities y > 0 and f(x)Β >Β g(x).
(1 mark)
β Sections 3.2, 3.3, 3.7
25 a Factorise completely x3 β 4x. (1 mark)
b Sketch the curve with equa
tion
y = x3 β 4x, showing the coordinates of
the points where the curve crosses the x-axis.
(2 marks)
c On a separate dia
gram, sketch the
curve with equation
y = (x β 1)3 β 4(x β 1)
showing the coordinates of the pointsΒ where the curve crosses the x-axis.
(2 marks)
β Sections 1.3, 4.1, 4.5
26
O
P(3, β2)y
x 2 4
The figure shows a sketch of the curve with equation y = f(x). The curve crosses the x-axis at the points (2, 0) and (4, 0). The minimum point on the curve is P(3, β2).
In separate diagrams, sketch the curves
with equation
a
y =
βf(x) (2 marks)
b y = f(2
x) (2 marks)
On each diagram, gi
ve the coordinates of
the points at which the curve crosses the
x-axis, and the coordinates of the image of P under the given transformation.
β Sections 4.6, 4.7E
E/p
E | [
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88
Review exercise 1
27
13
4
Oy
x
The figure shows a sketch of the curve
with equation y = f(x). The curve passes through the points (0, 3) and (4, 0) and touches the x-axis at the point (1, 0).
On separate diagrams, sketch the curves
with equations
a
y = f(
x + 1) (2 marks)
b y =
2f(x) (2 marks)
c y = f ( 1 _ 2 x) (2 marks)
On each diagram, sho
w clearly the
coordinates of all the points where the
curve meets the axes.
β Sections 4.5, 4.6, 4.7
28 Given that f(x) = 1 __ x , x β 0,
a sketch the gra
ph of y = f(x) + 3 and
state the equations of the
asymptotes (2 marks)
b find the coordinates of
the point
where y = f(x) + 3 crosses a coordinate axis.
(2 marks)
β Sections 4.3, 4.5
29 The quartic function t is defined as t(x
) = (x2 β 5x + 2)(x2 β 5x + 4), x β β .
a Find the four roots of
t(x), giving your
answers to 3 significant figures where necessary.
(3 marks)
b Sketch the gra
ph of y = t(x), showing
clearly the coordinates of all the pointsΒ where the curve meets the axes.
(2 marks)
β Sections 4.2, 2.1
30 The point (6, β8) lies on the gr aph of
yΒ =Β f(x). State the coordinates of the point to which P is transformed on the graph with equation:
a
y =
βf(x) (1 mark)
b y =
f(x β 3) (1 mark)
c 2y
= f(x) (1 mark)
β Section 4.7E
E
E
E31 The curve C1 has equation y = β a __ x , where
a is a positive constant.
The curve C2 has equation y = (x β b)2,
where b is a positive constant.a
Sketch C1 and C2 on the same set of
axes. Label any points where either
curve meets the coordinate axes, givingΒ your coordinates in terms of a and b.
(4 marks)
b Using your sketch, sta
te the number of
real solutions to the equation x(x β 5)
2 = 7. (1 mark)
β Sections 4.3, 4.4
32 a Sketch the graph of y = 1 __ x 2 β 4 ,
showing clearl
y the coordinates of
the points where the curve crosses
the coordinate axes and stating theΒ equations of the asymptotes.
(4 marks)
b The curve with y
= 1 _______ (x + k)2 β 4 passes
thr
ough the origin. Find the two
possible values of k. (2 marks)
β Sections 4.1, 4.5, 4.7E/p
E/p
1 a Solve the equation x2Β βΒ 10 xΒ +Β 9Β =Β 0
b Hen
ce, or otherwise, solve the equation
3x β 2(3x β 10) = β 1 β Sections 1.1, 1.3, 2.1
2 A rectangle has an area of 6 cm2 and a perimeter
of 8 ββ―__
2 cm. Find the dimensions of the
re
ctangle, giving your answers as surds in their
simplest form. β Sections 1.5, 2.2
3 Show algebraically that the graphs of
yΒ
= 3x3 + x2 β x and y = 2 x(x β 1)( x + 1) have
only one point of intersection, and find the coordinates of this point.
β Section 3.3
4 The quartic function f( x) = (x2 + x β 20)( x2 + x β 2)
has three roots in common with the function
g(x) = f(x β k), where k is a constant. Find the two
possible values of k . β Sections 4.2, 4.5, 4.7Challenge | [
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89
Straight line graphs
After completing this unit you should be able to:
β Calculat
e the gradient of a line joining a pair of points β pages 90 β 91
β Understand the link between the equation o f a line, and its gradient
and intercept β pages 91 β 93
β Find the equation of a line given (i) the gr adient and one point on
the line or (ii) two points on the line β pages 93 β 95
β Find the point of intersection f or a pair of straight lines
β pages 95 β 96
β Know and use the rules for parallel and perpendicular gradients
β pages 97 β 100
β Solve length and area problems on coordinate grids β pages 100 β 103
β Use straight line graphs to construct mathematical models
β pages 103 β 108Objectives
1 Find the point of intersection o f the
following pairs of lines.a
y =
4x + 7 and 3y = 2x β 1
b y =
5x β 1 and 3x + 7y = 11
c 2x
β 5y = β1 and 5x β 7y = 14
β GC SE Mathematics
2 Simplify each of the following:
a ββ―___ 80 b ββ―____ 200 c ββ―____ 125
β Sec tion 1.5
3 Make y the subject of each equation:
a 6x
+ 3y β 15 = 0 b 2x
β 5y β 9 = 0
c 3x
β 7y + 12 = 0 β GCSE MathematicsPrior knowledge check
Straight line graphs are used in mathematical
modelling. Economists use straight line graphs to model how the price and availability of a good affect the supply and demand.
βΒ ExerciseΒ 5HΒ Q95 | [
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90
Chapter 5
y
x O(x2, y2)
(x1, y1)x2 β x 1y2 β y 15.1 y = mx + c
You can find the gradient of a straight line joining two points
by considering the vertical distance and the horizontal distance between the points.
β The gradient m of a line joining the point with coordinates
( x 1 , y 1 ) to the point with coordinates ( x 2 , y 2 )
can be calculated using the f
ormula m = y2 β y1 ______ x2 β x1
a β (β5) ________ 4 β 2 = β1
So a +
5 ______ 2 = β1
a + 5 = β 2
a = β7Example 2
The line joining (2, β5) to (4, a) has gradient β1. Work out the value of a.
Use m = y2 β y1 ______ x2 β x1 . Here m = β1, (x1, y1) = (2, β5)
and (x2, y2) = (4, a).Example 1
Work out the gradient of the line joining (β2, 7) and (4, 5)
y
x O(β2, 7)
(4, 5)
m = 5 β 7 _______ 4 β (β2) = β 2 __ 6 = β 1 __ 3 Use m = y2 β y1 ______ x2 β x1 . Here (x1, y1) = (β2, 7) and
(x2, y2) = (4, 5)
1 Work out the gradients of the lines joining these pairs of points:
a (4, 2), (6, 3) b (β1, 3), (5, 4) c (β4, 5), (1, 2)
d (2, β3), (6, 5) e (β3, 4), (7,
β6) f (β
12, 3), (β2, 8)
g (β2,
β4), (10, 2) h ( 1 _ 2 , 2), ( 3 _ 4 , 4) i ( 1 _ 4 , 1 _ 2 ), ( 1 _ 2 , 2 _ 3 )
j (β2.4, 9.6), (0, 0) k (1.3, β2.2), (8.8,
β4.7) l (0, 5a), (10
a, 0)
m (3b
, β2b), (7b, 2b) n ( p,
p2), (q, q2)Exercise 5A Explore the gradient
fo
rmula using GeoGebra.Online | [
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0.028358004987239838,
0.00020366208627820015,
-0.0051992908120155334,
0.08801115304231644,
0.021469373255968094,
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0.03231723606586456,
0.0053368001244962215,
-0.010094579309225082,
0.08013003319501877,... |
91Straight line graphs
2 The line joining (3, β5) to (6,
a) has a gradient 4. Work out the value of a.
3 The line joining (5, b) to (8, 3) has gr
adient β3. Work out the value of b.
4 The line joining (c, 4) to (7, 6) has gr
adient 3 _ 4 . Work out the value of c.
5 The line joining (β1, 2
d ) to (1, 4) has gradient β 1 _ 4 . Work out the value of d.
6 The line joining (β3,
β2) to (2e, 5) has gradient 2. Work out the value of e.
7 The line joining (7, 2) to ( f, 3f ) has gradient 4. Wor
k out the value of f.
8 The line joining (3, β4) to (
βg, 2g) has gradient β3. Work out the value of g.
9 Show that the points
A(2, 3), B(4, 4) and
C(10,7) can be joined by a straight line.
10 Show that the points
A(β2a, 5a), B(0, 4a)
and points C(6a, a) are collinear. (3 marks)
β The equation of a straight line can be writt
en in the form
y = mx + c, where m is the gradient and c is the y-intercept.
β The equation of a straight line can also be writt
en in the
form ax + by + c = 0, where a, b and c are integers.P
E/P Poi nts are collinear if they all lie on
the same straight line.NotationFind the gradient of the line joining the points A
and B and the line joining the points A and C .Problem-solving
y
x Oy = mx + c
cm
1
a Gradient = β3 a nd y-intercept = (0, 2).
b y =
4 __ 3 x + 5 __ 3
Grad
ient = 4 __ 3 and y -intercept = (0, 5 __ 3 ).Example 3
Write down the gradient and y-intercept of these lines:
a y = β3
x + 2 b 4x β
3y + 5 = 0
Use f ractions rather than decimals
in coordinate geometry questions.Watch outRearrange the equation into the form y = mx + c.
From this m = 4 _ 3 and c = 5 _ 3 Compare y = β3x + 2 with y = mx + c.
From this, m = β3 and c = 2. | [
0.028725745156407356,
0.03366002440452576,
0.0209120512008667,
-0.06412093341350555,
-0.03791247680783272,
0.03455302119255066,
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0.023953501135110855,
0.01275207195430994,
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-0.013173... |
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