[PDF] VCE Maths Quest Math Methods - Solution Manual Year 12.pdf | Plain Text

First published 2010 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Offices also in Sydney and Melbourne Typeset in 10.5/12.5pt Times © John Wiley & Sons Australia, Ltd 2010 The moral rights of the author and contributors have been asserted. ISBN 978 1 74216 026 9 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher at the address above. Illustrated by Aptara and the Wiley Art Studio Typeset in India by Aptara Printed in Singapore by Markono Print Media Pte Ltd 10 9 8 7 6 5 4 3 2 1 The publishers would like to thank the following contributors: Ross Allen Joe Ardley Caroline Denney Elena Iampolsky Peter Matassa Carolyn Mews David Phillips Colin Shnier Sonja Stambulic Jenny Watson Ian Younger

Table of contents About eBook PLUS v Chapter 1 — Graphs and polynomials 1 Exercise 1A — The binomial theorem 1 Exercise 1B — Polynomials 2 Exercise 1C — Division of polynomials 4 Exercise 1D — Linear graphs 7 Exercise 1E — Quadratic graphs 8 Exercise 1F — Cubic graphs 11 Exercise 1G — Quartic graphs 13 Chapter review 15 Short answer 15 Multiple choice 16 Extended response 17 Chapter 2 — Functions and transformations 21 Exercise 2A — Transformations and the parabola 21 Exercise 2B — The cubic function in power form 22 Exercise 2C — The power function (the hyperbola) 25 Exercise 2D — The power function (the truncus) 29 Exercise 2E — The square root function in power form 32 Exercise 2F — The absolute value function 34 Exercise 2G — Transformations with matrices 38 Exercise 2H — Sum, difference and product functions 40 Exercise 2I — Composite functions and functional equations 41 Exercise 2J — Modelling 43 Chapter review 46 Short answer 46 Multiple choice 48 Extended response 49 Exam practice 1 52 Short answer 52 Multiple choice 52 Extended response 53 Chapter 3 — Exponential and logarithmic equations 54 Exercise 3A — The index laws 54 Exercise 3B — Logarithm laws 55 Exercise 3C — Exponential equations 56 Exercise 3D — Logarithmic equations using any base 58 Exercise 3E — Exponential equations (base e) 60 Exercise 3F — Equations with natural (base e) logarithms 62 Exercise 3G — Inverses 63 Exercise 3H — Literal equations 65 Exercise 3I — Exponential and logarithmic modelling 66 Chapter review 68 Short answer 68 Multiple choice 69 Extended response 70 Chapter 4 — Exponential and logarithmic graphs 72 Exercise 4A — Graphs of exponential functions with any base 72 Exercise 4B — Logarithmic graphs to any base 76 Exercise 4C — Graphs of exponential functions with base e 81 Exercise 4D — Logarithmic graphs to base e 86 Exercise 4E — Finding equations for graphs of exponential and logarithmic functions 90 Exercise 4F — Addition of ordinates 91 Exercise 4G — Exponential and logarithmic functions with absolute values 95 Exercise 4H — Exponential and logarithmic modelling using graphs 96 Chapter review 98 Short answer 98 Multiple choice 99 Extended response 100 Chapter 5 — Inverse functions 103 Exercise 5A — Relations and their inverses 103 Exercise 5B — Functions and their inverses 105 Exercise 5C — Inverse functions 109 Exercise 5D — Restricting functions 111 Chapter review 117 Short answer 117 Multiple choice 120 Extended response 121 Chapter 6 — Circular (trigonometric) functions 123 Exercise 6A — Revision of radians and the unit circle 123 Exercise 6B — Symmetry and exact values 124 Exercise 6C — Trigonometric equations 127 Exercise 6D — Trigonometric graphs 130 Exercise 6E — Graphs of the tangent function 135 Exercise 6F — Finding equations of trigonometric graphs 137 Exercise 6G — Trigonometric modelling 138 Exercise 6H — Further graphs 139 Exercise 6I — Trigonometric functions with an increasing trend 144 Chapter review 145 Short answer 145 Multiple choice 147 Extended response 148 Exam practice 2 148 Short answer 148 Multiple choice 150 Extended response 151 Chapter 7 — Differentiation 152 Exercise 7A — Review — gradient and rates of change 152 Exercise 7B — Limits and differentiation from first principles 154 Exercise 7C — The derivative of x n 158 Exercise 7D — The chain rule 160 Exercise 7E — The derivative of e x 164 Exercise 7F — The derivative of log e (x) 167 Exercise 7G — The derivatives of sin(x), cos(x) and tan(x) 172 Exercise 7H — The product rule 176 Exercise 7I — The quotient rule 179 Exercise 7J — Mixed problems on differentiation 183 Chapter review 190 Short answer 190 Multiple choice 192 Extended response 195 Chapter 8 — Applications of differentiation 196 Exercise 8A — Equations of tangents and normals 196 Exercise 8B — Sketching curves 199 Exercise 8C — Maximum and minimum problems when the function is known 209

Exercise 8D — Maximum and minimum problems when the function is unknown 210 Exercise 8E — Rates of change 213 Exercise 8F — Related rates 215 Exercise 8G — Linear approximation 217 Chapter review 217 Short answer 217 Multiple choice 219 Extended response 221 Chapter 9 — Integration 225 Exercise 9A — Antidifferentiation 225 Exercise 9B — Integration of e x, sin(x) and cos(x) 230 Exercise 9C — Integration by recognition 233 Exercise 9D — Approximating areas enclosed by functions 237 Exercise 9E — The fundamental theorem of integral calculus 238 Exercise 9F — Signed areas 244 Exercise 9G — Further areas 249 Exercise 9H — Areas between two curves 260 Exercise 9I — Average value of a function 270 Exercise 9J — Further applications of integration 271 Chapter review 276 Short answer 276 Multiple choice 278 Extended response 281 Exam practice 3 283 Short answer 283 Multiple choice 284 Extended response 285 Chapter 10 — Discrete random variables 286 Exercise 10A — Probability revision 286 Exercise 10B — Discrete random variables 289 Exercise 10C — Measures of centre of discrete random distributions 293 Exercise 10D — Measures of variability of discrete random distributions 297 Chapter review 302 Short answer 302 Multiple choice 304 Extended response 306 Chapter 11 — The binomial distribution 308 Exercise 11A — The binomial distribution 308 Exercise 11B — Problems involving the binomial distribution for multiple probabilities 311 Exercise 11C — Markov chains and transition matrices 315 Exercise 11D — Expected value, variance and standard deviation of the binomial distribution 318 Chapter review 321 Short answer 321 Multiple choice 323 Extended response 325 Chapter 12 — Continuous distributions 328 Exercise 12A — Continuous random variables 328 Exercise 12B — Using a probability density function to find probabilities of continuous random variables 330 Exercise 12C — Measures of central tendency and spread 336 Exercise 12D — Applications to problem solving 341 Exercise 12E — The normal distribution 343 Exercise 12F — The standard normal distribution 347 Exercise 12G — The inverse cumulative normal distribution 350 Chapter review 354 Short answer 354 Multiple choice 357 Extended response 358 Exam practice 4 360 Short answer 360 Multiple choice 361 Extended response 361 Solutions to investigations 363 Chapter 1 363 Investigation — Quartics and beyond 363 Chapter 2 363 Investigation — Goal accuracy 363 Chapter 9 363 Investigation — Definite integrals 363 Chapter 11 — 363 Investigation — Winning at racquetball! 363 Chapter 12 364 Investigation — Sunflower stems 364





Graphs and polynomials MM12-1 1 Exercise 1A — The binomial theorem 1 a (x + 3) 2 = 2 2 0 x    + 11 2 3 1 x    + 2 2 3 2    = x 2 + 6 x + 9 b ( x + 4) 5 = 5 5 0 x    + 4 5 4 1 x    + 32 5 4 2 x    + 23 5 4 3 x    + 4 5 4 4 x    + 5 5 4 5    = x 5 + 5 x 44 + 10 x 316 + 10 x 264 + 5 x256 + 1 × 1024 = x 5 + 20 x 4 + 160 x 3 + 640 x 2 + 1280 x + 1024 c ( x − 1) 8 = 8 8 0 x    + 7 8 (1) 1 x  −   + 62 8 (1) 2 x  −   + 53 8 (1) 3 x  −   + 44 8 (1) 4 x  −   + 35 8 (1) 5 x  −   + 26 8 (1) 6 x  −   + 7 8 (1) 7 x  −   + 8 8 (1) 8  −   = x 8 − 8x 7 + 28 x 6 − 56 x 5 + 70 x 4 − 56 x 3 + 28 x 2 − 8x + 1 d (2 x + 3) 4 = 4 4 (2 ) 0 x    + 3 4 (2 ) 3 1 x    + 22 4 (2 ) 3 2 x    + 3 4 (2 )3 3 x    + 4 4 3 4    = 16x 4 + 96 x 3 + 216 x 2 + 216 x + 81 e (7 − x) 4 = 4 4 7 0    + 322 44 7( ) 7( ) 12 x x   −+ −     + 34 44 7( ) ( ) 34 x x   −+ −     = 2401 − 1372 x + 294 x 2 − 28 x 3 + x 4 f (2 − 3x) 5 = 5 5 2 0    + 4 5 2( 3) 1 x  −   + 32 5 2( 3 ) 2 x  −   + 23 5 2( 3) 3 x  −   + 4 5 2( 3 ) 4 x  −   + 5 5 (3) 5 x  −   = 32 − 240 x + 720 x 2 − 1080 x 3 + 810 x 4 − 243 x 5 2 a 31 x x  +   = x 3 + 3 x 21 x + 3x 21 x    + 31 x    = x3 + 3 x + 3 x + 31 x b 72 3 x x  −   = (3 x) 7 − 7(3 x) 62 x + 21(3 x) 5 22 x    − 35(3 x) 4 32 x    + 35(3 x) 3 42 x    − 21(3 x) 2 52 x    + 7(3 x) 6722 x x   −     = 2187 x 7 − 10 206x 5 + 20 412x 3 − 22 680 x + 3 15 120 6048 x x − + 571344 128 x x − c 6 23 x x  +   = ( x2)6 + 6( x 2)53 x + 15( x 2)4 23 x    + 20(x 2)3 33 x    + 15(x 2)2 43 x    + 6(x 2) 53 x    + 63 x    = x12 + 18 x 9 + 135 x 6 + 540x 3 + 1215 + 31458 x + 6729 x d 5 2 3 2 x x  −   = 54 22 33 52 x xx   −     + 3 2 2 3 10 (2 ) x x    − 2 3 2 3 10 (2 ) x x    + 45 2 3 5(2) (2) x x x − = 10 7243 810 x x − + 41080 720 x x − + 240 x 2 − 32 x 5 3 ( r + 1) th term is n r   ( ax ) n − r br a (x − 7) 3 i x2 is the 2nd term ⇒ r = 1 Coefficient = 3 1   x2 (− 7) 1 = − 21 ii x3 is first term ⇒ r = 0 term = 3 0   x370 Coefficient = 1 iii x4 Coefficient = 0 b (2 x + 1) 5 i x2 is the 4 th term ⇒ r = 3 term = 5 3   (2 x) 213 Coefficient = 40 ii x3 is the third term ⇒ r = 2 term = 5 2   (2 x) 312 Coefficient = 80 iii x4 is the 2nd term ⇒ r = 1 term = 5 1   (2 x) 411 Coefficient = 80 c 5 2 3 x x  +   i Coefficient of x 2 = 0 Chapter 1 — Graphs and polynomials

MM12-1 2 Graphs and polynomials ii x3 is the 5 th term ⇒ r = 4 term = 1 5 2 4 x       (3 x) 4 Coefficient = 810 iii Coefficient of x 4 = 0 d 6 23 x x  −   i Coefficient of x 2 = 0 ii x3 is the 4 th term ⇒ r = 3 term = 6 3   ( x2)3 33 x −  Coefficient = − 540 iii Coefficient of x 4 = 0 e 6 2 3 7 x x  +   i Coefficient of x 2 = 0 ii x3 is the 2nd term ⇒ r = 1 term = 6 1   (7 x) 5 1 2 3 x    Coefficient = 302 526 iii Coefficient of x 4 is 0. 4 3 25 3 x x  −   x 3 is the 2nd term ⇒ r = 1 term = 3 1   (3 x 2)2 15 x −   = 3 × 9 x 4 × 15 x − = − 135 x 3 The answer is A. 5 When the expression for C is expanded it does not contain an x 5 term. The first three terms contain x 8, x6 and x 4 respectively. All the other expressions contain an x5 term. The answer is C. 6 5 3 2 2 x x  +   = ( x 3)5 + 5( x 3)4 22 x    + 10( x 3)3 2 2 2 x   + 10( x 3)2 3 2 2 x   + 5( x 3) 4 2 2 x   + 5 2 2 x   = 1x 15 + 10 x 10 + 40 x 5 + 80 + 380 x + 232 x ∴ = 1 + 10 + 40 + 80 + 80 + 32 = 243 The answer is D. 7 (2 x − 3) 4 = (2x)4 − 4(2 x) 33 + 6(2 x) 232 − 4(2 x)3 3 + 3 4 = 16x 4 − 96 x 3 + 216 x 2 − 216 x 3 + 81 The answer is D. 8 Fourth term = 6C3x3 × (3 y) 3 = 20 × x 3 × 27 y 3 = 540 x 3y3 9 Term 3 ⇒ r = 2 = 2 7 9 3 2 4 x   −    = 2 78 732 16 x = 2 19 683 4 x 10 x6, x3, x0 3rd term is independent of x. r = 2 = 2 4 2 6 2 (3 ) 2 x x      = 4860 11 Powers of x are ( x 2)5, (x 2)4 31 x    , ( x 2)3 2 3 1 x   x 10, x5, x0, The third term is independent of x. term = 5C2(x2)3 2 3 4 x −  = 10 × + 16 = 160 12 4 2 2 3 x x  +   Powers of x are ( x 2)4, (x 2)3 21 x    , ( x 2)2 2 2 1 x   x 8, x4, x0 The third term is independent of x. term = 4C2(x2)2 2 2 3 x   = 6 × 9 = 54 13 Expand ( p + 3) 5 = p 5 + 5 p 43 + 10 p 332 + 10 p 233 + 5 p3 4 + 3 5 = p 5 + 15 p 4 + 90 p 3 + 270 p 2 + 405 p + 243 ∴ (2p − 5)( p 5 + 15 p 4 + 90 p 3 + 270 p 2 + 405 p + 243) ⇒ 2 p 6 + 30 p 5 + 180 p 4 + 540 p 3 + 810 p 2 + 486 p − 5 p 5 − 75 p 4 − 450 p 3 − 1350 p 2 − 2025 p − 1215 Coefficient of 4 th term = 180 − 75 = 105 14 (2 a − 1) n 2nd term is nC1(2 a) n − 1(− 1) 1 coefficient: −n × 2 n − 1 = − 192 n × 2 n × 1 2 = 192 n × 2 n = 384 = 3 × 2 7 = 3 × 2 × 2 6 = 6 × 2 6 n = 6 Exercise 1B — Polynomials 1 Polynomial expressions cons ist of terms which have non– negative integer powers of x only. Not Polynomial: ii x4 + 3 x 2 − 2 x + x iii x7 + 3 x 6 − 2 xy + 5 x vi 2 x 5 + x 4 − x 3 + x 2 + 3 x − 2 x Polynomial: i x 3 − 2 x iv 3 x 8 − 2 x 5 + x 2 − 7 v 4 x 6 − x 3 + 2 x − 3 2 a P (x ) + Q (x ) = 8 − 3 x + 2 x 2 + x 4 + x 5 − 3 x 4 − 4 x 2 − 1 = x 5 − 2 x 4 − 2 x 2 − 3 x + 7 b Q (x ) − R (x ) = x 5 − 3 x 4 − 4 x 2 − 1 − (8 x 3 + 7 x 2 − 4 x) = x 5 − 3 x 4 − 4 x 2 − 1 − 8 x 3 − 7 x 2 + 4 x = x 5 − 3 x 4 − 8 x 3 − 11 x 2 + 4 x − 1 c 3 P(x ) − 2 R(x ). 3 P(x ) = 3(8 − 3 x + 2 x 2 + x 4) = 24 − 9 x + 6 x 2 + 3 x 4

Graphs and polynomials MM12-1 3 2 R(x ) = 2(8 x 3 + 7 x 2 − 4 x) = 16x 3 + 14 x 2 − 8 x ∴ 3P(x ) − 2 R(x ) = 24 − 9 x + 6 x 2 + 3 x 4 − (16 x 3 + 14 x 2 − 8 x) = 24 − 9 x + 6 x 2 + 3 x 4 − 16 x 3 − 14 x 2 + 8 x = 3x 4 − 16 x 3 − 8 x 2 − x + 24 d 2 P(x ) − Q (x ) + 3 R(x ) 2 P (x ) = 2(8 − 3 x + 2 x 2 + x 4) = 16 − 6 x + 4 x 2 + 2 x 4 3 R(x ) = 3(8 x 3 + 7 x 2 − 4 x) = 24x 3 + 21 x 2 − 12 x 2 P(x ) − Q (x ) + 3 R(x ) = 16 − 6 x + 4 x 2 + 2 x 4− ( x 5 − 3 x 4 − 4 x 2 − 1) + 24 x 3 + 21 x 2 − 12 x = 16 − 6 x + 4 x2 + 2 x 4 − x 5 + 3 x 4 + 4 x 2 + 1 + 24 x 3 + 21 x 2 − 12 x = 17 − 18 x + 29 x2 + 24 x 3 + 5 x 4 − x 5 3 a P (x ) = x 6 + 2 x 5 − x 3 + x 2 i degree = 6 ii P (0) = 0 6 + 2 × 0 5 − 0 3 + 0 2 = 0 iii P (2) = 2 6 + 2 × 2 5 − 2 3 + 2 2 = 124 iv P (− 1) = − 1 6 + 2 × − 1 5 − ( −1) 3 + ( −1) 2 = 1 b P (x ) = 3 x 7 − 2 x 6 + x 5 − 8 i degree = 7 ii P (0) = 3 × 0 7 − 2 × 0 6 + 0 5 − 8 = − 8 iii P (2) = 3 × 2 7 − 2 × 2 6 + 2 5 − 8 = 280 iv P (− 1) = 3 × (− 1) 7 − 2 × (− 1) 6 + ( −1) 5 − 8 = − 3 − 2 − 1 − 8 = − 14 c P(x ) = 5 x 6 + 3 x 4 − 2 x 3 − 6 x 2 + 3 i degree = 6 ii P (0) = 5 × 0 6 + 3 × 0 4 − 2 × 0 3 − 6 × 0 2 + 3 = 3 iii P (2) = 5 × 2 6 + 3 × 2 4 − 2 × 2 3 − 6 × 2 2 + 3 = 331 iv P (− 1) = 5 × ( −1) 6 + 3 × ( −1) 4 − 2 × ( −1) 3 − 6 × ( −1) 2 + 3 = 5 + 3 + 2 − 6 + 3 = 7 d P (x ) = − 7 + 2 x − 5 x 2 + 2 x 3 − 3 x 4 i degree = 4 ii P (0) = − 7 + 2 × 0 − 5 × 0 2 + 2 × 0 3 − 3 × 0 4 = − 7 iii P (2) = − 7 + 2 × 2 − 5 × 2 2 + 2 × 2 3 − 3 × 2 4 = − 55 iv P (− 1) = − 7 + 2 × (− 1) − 5 × ( −1) 2 + 2 × ( −1) 3 − 3( −1) 4 = − 7 − 2 − 5 − 2 − 3 = − 19 4 P(x ) = x 8 − 3 x 6 + 2 x 4 − x 2 + 3 P (− 2) = ( −2) 8 − 3 × ( −2) 6 + 2 × ( −2) 4 − ( −2) 2 + 3 = 95 The answer is B. 5 P (x ) = 2 x 7 + ax 5 + 3 x 3 + bx − 5 P (1) = 4 ∴ 4 = 2 × 1 7 + a × 1 5 + 3 × 1 3 + b × 1 − 5 4 = 2 + a + 3 + b − 5 4 = a + b [1] P (2) = 163 163 = 2 × 2 7 + a × 2 5 + 3 × 2 3 + b × 2 − 5 = 256 + 32 a + 24 + 2 b − 5 − 112 = 32 a + 2 b [2] [1] × 2 8 = 2 a + 2 b [3] [2] − [3] −120 = 30 a a = − 4 b = 8 6 f( x ) = ax 4 + bx 3 − 3 x 2 − 4 x + 7 f (1) = − 2 ∴ − 2 = a × (1) 4 + b × (1) 3 − 3 × 1 2 − 4 × 1 + 7 − 2 = a + b − 3 − 4 + 7 − 2 = a + b ∴ − 2 − b = a [1] f (2) = − 5 ∴ − 5 = a × 2 4 + b × 2 3 − 3 × 2 2 − 4 × 2 + 7 − 5 = 16 a + 8 b − 12 − 8 + 7 − 5 = 16 a + 8 b − 13 8 = 16 a + 8 b 8 = 8(2 a + b ) 1 = 2 a + b [2] Substitute [1] into [2] 1 = 2( −2 − b ) + b 1 = − 4 − 2 b + b 1 = − 4 − b b = − 5 If b = − 5, then [1] −2 − − 5 = a. 3 = a ∴ f(x ) = 3 x4 − 5x3 − 3 x2 − 4 x + 7 7 Q (x ) = x5 + 2 x 4 + ax 3 − 6 x + b Q (2) = 45 ∴ 45 = 25 + 2 × 2 4 + 2 3a − 6 × 2 + b 45 = 52 + 8 a + b − 7 = 8a + b − 7 − 8a = b [1] Q (0) = − 7 ∴ − 7 = 0 5 + 2 × 04 + a × 03 − 6 × 0 + b − 7 = b [2] Substitute [2] into [1]. − 7 − 8a = − 7 − 8a = 0 a = 0 ∴ Q(x ) = x5 + 2 x4 − 6 x − 7. 8 P (x ) = ax6 + bx4 + x3 − 6 If 3 P(1) = − 24 then 3 P(x ) = 3( ax 6 + bx4 + x3 − 6) − 24 = 3( a × 16 + b × 14 + 13 − 6) − 8 = a + b + 1 − 6 − 8 = a + b − 5 − 3 = a + b − 3 − a = b [1] If 3 P (−2) = 102 then 3 P(x ) = 3( ax 6 + bx4 + x3 − 6) 102 = 3[ a (−2) 6 + b(−2) 4 + (−2) 3 − 6) 34 = 64 a + 16b − 8 − 6 34 = 64 a + 16b − 14 48 = 64 a + 16b ( ÷ 16) 3 = 4 a + b [2] Substitute [1] into [2] 3 = 4 a + (− 3 − a) 3 = 4 a − 3 − a 6 = 3 a 2 = a If 2 = a then b = −3 − a b = − 3 − 2 b = − 5 ∴ P(x ) = 2 x6 − 5 x4 + x3 − 6 9 a P (x ) = ax4 − x3 + 3 x2 − 5 If P(1) = −1 then − 1 = a × (1) 4 − (1) 3 + 3 × (1) 2 − 5

MM12-1 4 Graphs and polynomials − 1 = a − 1 + 3 − 5 − 1 = a − 3 2 = a The answer is C. b f (x ) = xn − 2 x3 + x2 − 5x If f(2) = 10 then 10 = 2n − 2 × 23 + 22 − 5 × 2 10 = 2n − 16 + 4 − 10 10 = 2n − 22 32 = 2n 2 5 = 2n ∴ n = 5 The answer is D. Exercise 1C — Division of polynomials 1 a 2 32 32 22 213 252 4 4 25 28 13 2 13 52 50 xx xxx x xx xxxx x x ++  −+− −−  −    + −  −   −  −  −  Q (x ) = x 2 + 2 x + 13 R (x ) = 50 b 5432 54 43 43 32 32 2 2 03043 3 3 33 39 60 618 18 4 18 54 58 3 58 174 xxxxx x xx xx xx xx xx xx xxx x  +−+++ +−  +   −− −  −−   + −  +   −+ −  −−   +  −  +  43 2 3 6 18 58 171 xxx x −+ −+ − Q(x ) = x 4 − 3 x 3 + 6 x 2 − 18 x + 58 R (x ) = − 171 c 3 43 2 43 32 32 2617 6240 3 618 17 2 17 51 53 4 53 159 155 0 155 465 465 xx xx x x x xx xx xx xx xx x x +  −+ − + −−  −    + −  −   − −  −  +  −  −  2 53 155 x ++ Q (x ) = 6 x 3 + 17 x 2 + 53 x + 155 R (x ) = 465 d 32 432 43 32 32 7 7 1013927 360120 31 3 70 7 7 3 xx x xxx x x xx xx xx−++  −+ ++ +−  +   −+  − −−   2 2 7 12 3 77 39 1010 9 101 101 927 101 27 xx x x x x  +   −  +   +  −   +   − Q (x ) = x 3 − 7 3x2 + 7 9x + 101 27 R (x ) = 20 3 27 − 101 27 −  =   2 a i P (x ) = x 3 − 2 x 2 + 5 x − 2 P (4) = 4 3 − 2 × 4 2 + 5 × 4 − 2 = 50 ii P (x ) = x 4 + x 3 + 3 x 2 − 7 x P (1) = 1 + 1 + 3 − 7 = − 2 iii P(x ) = x 5 − 3 x 3 + 4 x + 3 P (− 3) = ( −3) 5 − 3 × (−3) 3 + 4 × (− 3) + 3 = − 171 iv P(x ) = 2 x 6 − x 4 + x 3 + 6 x 2 − 5 x P (− 2) = 2 × ( −2) 6 − ( −2) 4 + ( − 2) 3 + 6 × ( −2) 2 − 5 × (− 2) = 138 v P (x ) = 6 x 4 − x 3 + 2 x 2 − 4 x P (3) = 6 × 3 4 − 3 3 + 2 × 3 2 − 4 × 3 = 465 vi P (x ) = x 4 − 13 x 2 + 36 P (2) = 2 4 − 13 × 2 2 + 36 = 0 vii P (x ) = 3 x 4 − 6 x 3 + 12 x 1 3 P  −   = 3 × 41 3 −  − 6 × 31 3 −  + 12 × − 1 3 = − 3 20 27 viii P (x ) = x 5 + 3 x 3 − 4 x 2 + 6 x − 8 3 2 P    = 53 2   + 3 × 33 2   − 4 × 23 2   + 6 × 3 2 − 8 = 23 9 32 b The values obtained in 2 were the same as the remainder values obtained in 1. 3 a P(3) = 3 3 + 9 × 3 2 + 26 × 3 − 30 = 156 Since P(3) ≠ 0, x − 3 is not a factor b P (− 2) = ( −2) 4 − ( −2) 3 − 5 × ( −2) 2 − 2 × ( −2) − 8 = 0 Since P(− 2) = 0 then x + 2 is a factor.

Graphs and polynomials MM12-1 5 c P(+ 4) = 4 − 9 × + 4 + 6 × 4 2 − 13 × ( + 4) 3 − 12 × ( + 4) 4 + 3 × ( + 4) 5 = 4 − 36 + 94 − 832 − 3072 + 3072 = − 768 Since P(− 4) ≠ 0 then 4 − x is not a factor. d P 1 2  −   = 4 × 61 2 −  + 2 × 51 2 −  − 8 × 41 2 −  − 4 × 31 2 −  + 6 × 21 2 −  − 9 × − 1 2 − 6 = 0.0625 + − 0.0625 − 0.5 + 0.5 + 1.5 + 4.5 − 6 = 0 Since P 1 2  −   = 0 then 2 x + 1 is a factor. 4 a f( x ) = x 4 − 4 x 3 − x 2 + 16 x − 12 A x + 1 ⇒ f ( − 1) = (−1) 4 − 4 × ( −1) 3 − ( −1) 2 + 16 × (− 1) − 12 = 1 + 4 − 1 − 16 − 12 = − 24 B x ⇒ f(0) = − 12 C x + 2 ⇒ f ( − 2) = (−2) 4 − 4 × ( −2) 3 − ( −2) 2 + 16 × (− 2) − 12 = 16 + 32 − 4 − 32 − 12 = 0 Since f( − 2) = 0 then ( x + 2) is a factor. D x + 3 ⇒ f ( − 3) = (−3) 4 − 4 × ( −3) 3 − ( −3) 2 + 16 × (− 3) − 12 = 120 E x − 4 ⇒ f (4) = 4 4 − 4 × 4 3 − 4 2 + 16 × 4 − 12 = 36 The answer is C. b 32 432 43 32 32 2 2 6116 41612 2 2 6 612 11 16 11 22 612 612 0 xx x xxx x x xx xx xx xx xxx x −+−  −−+− +−  +   −− −  −−   + −  +   −−  −  −−  Test x = 1 into x 3 − 6 x 2 + 11 x − 6 = 1 3 − 6 × 1 + 11 − 6 = 0 ∴ x − 1 is a factor. 2 32 32 2 2 56 6116 1 511 55 66 66 0 xx xx x x xx xx xx x x −+  −+− −−  −   −+ −  −+   −  −  −  x 2 − 5 x + 6 = ( x − 3)( x − 2) f( x ) factorises to ( x + 2)( x − 1)( x − 3)( x − 2) The answer is B. 5 a P(x ) = x 3 + 4 x 2 − 3 x − 18 Test x = ± 1 P(x ) ≠ 0 x = 2, P(x ) = 0 ∴ (x − 2) is a factor 2 32 32 2 2 69 4318 2 2 63 612 918 918 0 xx xxx x xx xx xx x x ++  +−− −−  −    − −  −   −  −  −  ∴ (x − 2)( x 2 + 6 x + 9) = ( x − 2)( x + 3)2 b P (x ) = 3 x 3 − 13 x 2 − 32 x + 12 Test x = ± 1 P(x ) ≠ 0 x = ± 2 when x = − 2, P(x ) = 0 ∴ (x + 2) is a factor 2 32 32 2 23196 313 3212 2 36 19 32 19 38 612 612 0 xx xx x x xx xx xx x x +− +  −−+ +−  +   −− −  −−   +  −  +  ∴ (x + 2)(3 x 2 − 19 x + 6) = ( x + 2)(3 x − 1)( x − 6) c P(x ) = x 4 + 2 x 3 − 7 x 2 − 8 x + 12 Test x = − 2, P(x ) = 0 ∴ (x + 2) is a factor 32 432 43 2 2 076 27812 2 2 07 8 714 612 612 0 xxx xxxx x xx xx xx x x +−+  +−−+ +−  +   −− −  −−   +  −  +  ∴ (x + 2)( x 3 + 0 x 2 − 7 x + 6) Test x = 2, P(x ) = 0 ∴ (x − 2) is a factor 2 32 32 2 2 23 076 2 2 27 24 36 36 0 xx xxx x xx xx xx x x +−  +−+ −−  −    − −  −   −+  −  −+  ∴ (x + 2)( x − 2)( x 2 + 2 x − 3) ( x + 2)( x − 2)( x + 3)( x − 1)

MM12-1 6 Graphs and polynomials d P (x ) = 4 x 4 + 12 x 3 − 24 x 2 − 32 x Test x = − 1, P(x ) = 0 ∴ x + 1 is a factor 32 43 2 43 32 32 2 24832 41224 320 1 44 824 88 32 32 32 32 0x xx xxx x x xx xx xx xx xx+−  +− −+ +−  +   − −  +   −− −  −−   ∴ (x + 1)(4 x 3 + 8 x 2 − 32 x) Take out factor of 4 x. 4 x(x + 1)( x 2 + 2 x − 8) ∴ 4x(x + 1)( x − 2)( x + 4) 6 a 3 x 3 + 3 x 2 − 18 x = 0 Test x = 2, f( x ) = 0 ∴ (x − 2) is a factor 2 32 32 2 239 33180 2 36 918 918 0x x xx x x xx xx xx +  +−+ −−  −   − −  −   ∴ (x − 2)(3 x 2 + 9 x) = 0 3 x(x − 2)( x + 3) = 0 ∴ x = 0, 2, or −3 b 2 x 4 + 10 x 3 − 4 x 2 − 48 x = 0 Test x = 2, f( x ) = 0 ∴ (x − 2) is a factor 32 432 43 32 32 2 2214 24 210 4 48 2 24 14 4 14 28 24 48 24 48 0x xx x xx x x xx xx xx x x x x ++  +−− −−  −    − −  −   − −  −   ∴ (x − 2)(2 x 3 + 14 x 2 + 24 x) Take out common factor of 2 x: 2 x(x − 2)( x 2 + 7 x + 12) = 2 x(x − 2)( x + 3)( x + 4) = 0 ∴ x = 2, −3, 0, and −4 c 2 x 4 + x 3 − 14 x 2 − 4 x + 24 = 0 Test x = 2, f( x ) = 0 ∴ (x − 2) is a factor 32 43 2 43 32 32 2 225412 214424 2 24 514 510 44 48 xx x xx x x x xx xx xx xx xx +−−  +− − + −−  −   − −  −   −− −  −+   12 24 12 24 0 x x −+  −  −+  ∴ (x − 2)(2 x 3 + 5 x 2 − 4 x − 12) Test x = − 2, f( − 2) = 0 ∴ (x + 2) is a factor 2 32 32 2 226 25412 2 24 4 2 612 612 0 xx xx x x xx xx xx x x +−  +−− +−  +   − −  −   −−  −  −−  ∴ (x − 2)( x + 2)(2 x 2 + x − 6) ( x − 2)( x + 2)(2 x − 3)( x + 2) x = 2, −2, or 3 2 d x4 − 2 x 2 + 1 = 0 Test x = + 1, f( x ) = 0 ∴ (x − 1) is a factor 32 432 43 32 32 2 2 1 0201 1 2 0 1 1 0 x xx x xxx x xx xx xx xx xx x x +−−  +−++ −−  −   − −  −   −+ −  −+   −+  −  −+  ∴ (x − 1)( x 3 + x 2 − x − 1) Test x = − 1, f( x ) = 0 ∴ (x + 1) is a factor 2 32 32 1 1 1 0 1 1 0 x x xx x xx x x−  +−− +−  +   −−  −  −−  ∴ (x − 1)( x + 1)( x 2 − 1) = 0 ( x − 1)( x + 1)( x − 1)( x + 1) = 0 x = ± 1 7 If ( x − 2) is a factor then when x = 2, f( x ) = 0 0 = x 3 + ax 2 − 6 x − 4 f (2) = 0 = 2 3 + a 2 2 − 6 × 2 − 4 0 = 8 + 4 a − 12 − 4 0 = 4 a − 8 8 = 4 a 2 = a 8 Let P(x ) = x3 + x2 − ax + 3 P(1) = 1 + 1 − a + 3 = 0 (x − 1) is a factor a = 5

Graphs and polynomials MM12-1 7 9 If ( x + 3) is a factor then when x = − 3, f( x ) = 0 f (− 3) = 0 = 2( −3) 4 + a (− 3) 3 − 3 × (− 3) + 18 0 = 162 − 27 a + 9 + 18 0 = 189 − 27 a 27 a = 189 a = 7 10 If ( x + 1) is a factor then when x = − 1, f( x ) = 0 f (− 1) = 0 = − a − 4 − b − 12 0 = − a − b − 16 a = − b − 16 [1] If ( x − 2) is a factor then when x = 2, f( x ) = 0 f (2) = 0 = 8 a − 16 + 2 b − 12 0 = 8 a + 2 b − 28 28 = 8 a + 2 b 14 = 4 a + b [2] Sub [1] into [2] 14 = 4( −b − 16) + b 14 = − 4b − 64 + b 14 = − 3b − 64 78 = − 3b − 26 = b ∴ a = + 26 − 16 a = 10 11 (2 x − 3) and ( x + 2) are factors of 2 x 3 + ax2 + bx + 30 P(x ) = 2 x3 + ax2 + bx + 30 P (−2) = 2(−2) 3 + a(−2) 2 + b(−2) + 30 = 0 −16 + 4a − 2 b + 30 = 0 4 a − 2b = − 14 [1] P 3 2    = 2 33 2    + a 23 2    + b 3 2    + 30 = 0 2 × 27 84 + 9 4a + 3 2b + 30 = 0 27 + 9 a + 6b + 120 = 0 9 a + 6b = − 147 3 a + 2b = − 49 [2] [1] + [2] 7 a = −63 a = −9 Substitute into [1] 4 × − 9 − 2b = − 14 − 2b = 22 b = −11 a = −9, b = −11. Exercise 1D — Linear graphs 1 a 2 x + 3 y = 12 x-intercept when y = 0 2 x = 12 x = 6 y-intercept when x = 0 3 y = 12 y = 4 b 2 y − 5 x − 10 = 0 x-intercept when y = 0 − 5x = 10 x = − 2 y-intercept when x = 0 2 y = 10 y = 5 c 2 x − y = 1 x-intercept when y = 0 2 x = 1 x = 1 2 y-intercept when x = 0 − y = 1 y = − 1 2 a y = mx + c y = 3 x + c find c in (2, 1) 1 = 3 × 2 + c − 5 = c ∴ y = 3 x − 5 − 3x + y + 5 = 0 b y = mx + c y = − 2x + c find c, sub in ( −4, 3) 3 = − 2 × − 4 + c 3 = 8 + c − 5 = c ∴ y = − 2x − 5 2 x + y + 5 = 0 3 a ( −3, −4), ( −1, −10) m = 10 4 13 −+ −+ = 6 2 − = − 3 y = − 3x + c sub in ( −3, −4) − 4 = − 3 × − 3 + c − 13 = c y = − 3x − 13 3 x + y + 13 = 0 b (7, 5), (2, 0) m = 50 72 − − = 5 5 = 1 y = x + c sub in (2, 0) 0 = 2 + c − 2 = c y = x − 2 − x + y + 2 = 0 4 2 y − 3 x − 6 = 0 A 2 × 6 − 3 × 2 − 6 = 0 12 − 6 − 6 = 0 B 2 × 0 − 3 × − 2 − 6 = 0 0 + 6 − 6 = 0 C 2 × 3 − 3 × 0 − 6 = 0 6 − 0 − 6 = 0 D 2 × 2 − 3 × 1 − 6 = 0 4 − 3 − 6 ≠ 0 E 2 × 9 − 3 × 4 − 6 = 0 18 − 12 − 6 = 0 The answer is D. 5 a i −2 = 5 12 b − + − 2 = 5 3 b − − 6 = b − 5 − 1 = b ii y − x = 7 ∴ y = x + 7 m = 1 1 = 5 12 b − + 3 = b − 5 8 = b b parallel to y = 3 x − 4 ∴ m = 3 y = 3 x + c sub in (4, 5) 5 = 3 × 4 + c − 7 = c ∴ y = 3 x − 7 0 = 3 x − y − 7 c 2 y − x + 1 = 0 2 y = x − 1 y = 1 2x − 12 m = − 2 gradient of perpendicular line y − y1 = m (x − x1) Sub in (− 2, 4) y − 4 = −2( x + 2) y − 4 = −2x − 4 2 x + y = 0 6 i x + 2 y + 4 = 0 • x-intercept when y = 0 x = − 4 • y-intercept when x = 0 2y = − 4 y = − 2 Graph e ii y = 3 − Graph f iii y − 2 x − 2 = 0 • x-intercept when y = 0 − 2 = 2 x − 1 = x • y-intercept when x = 0 y = 2 Graph a iv 3 y + 2 x = 6 • x-intercept when y = 0 2 x = 6 x = 3 • y-intercept when x = 0 3 y = 6 y = 2 Graph c v y − 2 x = 0 • x- and y-intercepts occur at the origin. Graph b vi x = − 2 − Graph d.

MM12-1 8 Graphs and polynomials 7 a y ≥ − 2 or [ −2, ∞) b y ≥ − 5 or ( −5, ∞) c −2 ≤ y < 3 or [ −2, 3) d −2 ≤ y ≤ 3 or [ −2, 3] e R f −∞ < y < 6 or ( −∞, 6) 8 a 4y + 3 x = 24 x ∈ [− 12, 12] x-intercept 3 x = 24 x = 8 y-intercept 4 y = 24 y = 6 when x = − 12 y = 15 when x = 12 y = − 3 i domain [ −12, 12] ii range [ −3, 15] b 2 x − 5 y = 10 x < 5 x-intercept 2 x = 10 x = 5 y-intercept − 5y = 10 y = − 2 when x = 5, y = 0 i domain ( −∞, 5) ii range ( −∞, 0) c 4 x − 3 y − 6 = 0 x ∈ [2, 5) x-intercept 4 x = 6 x = 3 2 y-intercept − 3y = 6 y = − 2 when x = 2 y = 2 3 when x = 5 y = 14 3 i Domain [2, 5) ii Range 214 , 33     9 a Parallel ∴ m = − 2 y = − 2x + c sub in (2, 5) 5 = − 4 + c 9 = c ∴ y = − 2x + 9 2 x + y − 9 = 0 b Perpendicular ∴ m = − 1 3 y = − 1 3 x + c sub in (2, 5) ( y − 5) = − 1 3 ( x − 2) y − 5 = 3 x− + 2 3 3 y − 15 = − x + 2 3 y + x − 17 = 0 x + 3 y − 17 = 0 10 a Parallel to 4 x − 13 = 2 y 2 x − 13 2 = y m = 2 ∴ y = 2 x + c sub in ( −3, 1) 1 = 2 × − 3 + c 7 = c y = 2 x + 7 − 2x + y − 7 = 0 b 4 x − 2 y = 13 4 x − 13 = 2 y 2 x − 13 2 = y ∴ m = 2 Perpendicular m = − 1 2 sub in ( −3, 1) y − 1 = − 1 2( x + 3) y − 1 = − 1 2x − 3 2 y = − 1 2x − 1 2 2 y = − x − 1 x + 2 y + 1 = 0 11 3 x − y = − 2 3 x + 2 = y m = 3 ax + 2 y = 3 2 y = − ax + 3 y = 2a −x + 3 2 ∴ 3 = 2a − 6 = − a − 6 = a The answer is E. 12 5 x + y − 3 = 0 bx − y − 2 = 0 y = − 5x + 3 y = bx − 2 gradient −5 gradient b = 1 5 The answer is B. Exercise 1E — Quadratic graphs 1 b2 − 4 ac = ∆ a f( x ) = x 2 − 3 x + 4 a = 1, b = − 3, c = 4 ∆ = 9 − 16 ∆ = − 7 b 2 − 4 ac < 0. No x-intercepts b f( x ) = x 2 + 5 x − 8 a = 1 b = 5 c = − 8 ∆ = 25 + 32 = 57 b 2 − 4 ac > 0. Two x-intercepts c f( x ) = 3 x 2 − 5 x + 9 a = 3 b = − 5 c = 9 ∆ = 25 − 108 ∆ = − 83 b 2 − 4 ac < 0. No x-intercepts d f( x ) = 2 x 2 + 7 x − 11 a = 2 b = 7 c = − 11 ∆ = 49 + 88 = 137 b 2 − 4 ac > 0. Two x-intercepts e f( x ) = 1 − 6 x − x 2 a = − 1 b = − 6 c = 1 ∆ = 36 + 4 = 40 b 2 − 4 ac > 0. Two x-intercepts f f( x ) = 3 + 6 x + 3 x 2 a = 3 b = 6 c = 3 ∆ = 36 − 36 = 0 b 2 − 4 ac = 0. One x-intercept 2 a f( x ) = x 2 − 6 x + 8 y-intercept x = 0 y = 8 x-intercept(s) 0 = ( x − 4)( x − 2) ∴ x = 4 or 2 b f( x ) = x 2 − 5 x + 4 y-intercept x = 0 y = 4 x-intercept y = 0 0 = ( x − 4)( x − 1) x = 4 or 1 c f( x ) = 10 + 3 x − x 2 y-intercept x = 0 y = 10 x-intercept y = 0 0 = (5 − x )(2 + x ) x = 5 or −2 d f( x ) = 6 x 2 − x − 12 y-intercept x = 0 y = − 12

Graphs and polynomials MM12-1 9 x-intercept(s) y = 0 0 = (3 x + 4) (2 x − 3) x = − 4 3 or 3 2 3 a f( x ) = x 2 − 6 x + 8 = x2 − 6 x + 3 2 − 3 2 + 8 = (x − 3) 2 − 9 + 8 = (x − 3)2 − 1 TP is (3, −1) b f( x ) = x 2 − 5 x + 4 = x 2 − 5 x + 2255 22   −     + 4 = 25 2 x  −   −25 4 + 4 = 25 2 x  −   −25 4 + 16 4 = 25 2 x  −   − 9 4 TP is 59 , 24  −   c f( x ) = 10 + 3 x − x 2 = − (x 2 − 3 x − 10) = − 22 233 310 22 xx    −+ − −        = − 239 10 24 x   −−−      = − 23940 244 x   −−−      = − 23 2 x  −   + 49 4 TP is 349 , 24  +   = 11 1,12 24    d f( x ) = 6 x 2 − x − 12 = 2 62 6x x  −−   = 22 2 11 62 612 12x x    −+ − −        = 2 1 1 288 6 12 144 144 x   −−−      = 2 1289 6 12 144 x   −−      = 2 1 6 12 x  −   −289 24 TP is 11 ,12 12 24    4 a i y = 2 − x 2 a = − 1, h = 0, k = 2 TP = (0, 2) ii Domain = R iii Range = ( −∞ , 2] b i y = ( x − 6) 2 a = 1, h = 6, k = 0 TP = (6, 0) ii Domain = R iii Range = [0, ∞) c i y = − (x + 2) 2 a = − 1, h = − 2, k = 0 TP = ( −2, 0) ii Domain = R iii Range = ( −∞ , 0] d i y = 2( x + 3) 2 − 6 a = 2, h = − 3, k = − 6 TP = ( −3, −6) ii Domain = R iii Range = [ −6, ∞) 5 Using y = A ( x + B) 2 + C a i TP = (1, − 2) ∴ B = − 1 and C = − 2 Assume A = 1 ⇒ y = 1( x − 1) 2 − 2 y = x 2 − 2 x + 1 − 2 y = x2 − 2 x − 1 ii Domain = R iii Range [ −2, ∞) b i TP = (2, −3) ∴ B = − 2 C = − 3 Assume A = 1 ⇒ y = 1 ( x − 2) 2 − 3 = x 2 − 4 x + 4 − 3 = x2 − 4 x + 1 ii Domain = [ −1, ∞ ) iii Range = [ −3, ∞) c i TP = (1, 9) ∴ B = − 1 and C = 9 Assume A = − 1 ⇒ y = − 1( x − 1) 2 + 9 y = − 1( x 2 − 2 x + 1) + 9 y = − x 2 + 2 x − 1 + 9 y = − x2 + 2 x + 8 ii Domain = [ −4, 4) iii Range = [ −16, 9] 6 a y = 2 x 2 + 3 TP = (0, 3) y-intercept when x = 0 y = 3 x-intercepts when y = 0 0 = 2 x 2 + 3 There are no x-intercepts. b y = 1 − 4(2 − x ) 2 TP = (2, 1) y-intercept when x = 0 y = 1 − 4 × 4 = −15 x-intercepts when y = 0 0 = 1 − 4(2 − x ) 2 0 = 1 − 4(4 − 4 x + x 2) = 1 − 16 + 16 x − 4 x 2 = − 4x 2 + 16 x − 15 x = 3 2 and x = 5 2 c y = (2 x − 3) 2 − 8 TP = 3 ,8 2  −   y-intercept when x = 0 y = ( −3) 2 − 8 = 1 x-intercepts when y = 0 0 = (2 x − 3) 2 − 8 = 4 x 2 − 12 x + 9 − 8 = 4 x2 − 12 x + 1 From the graphics calculator, x = 2.91 and x = 0.09 7 a y = x 2 − 2 x − 3 x-intercepts y = 0 0 = ( x − 3)( x + 1) x = 3 or −1 (3, 0)( −1, 0) The answer is B. b y = x 2 − 2 x − 3 y = x2 − 2 x + 1 2 − 1 2 − 3 y = ( x − 1) 2 − 4 TP = (1, −4) The answer is C. 8 f (x ) = − (x + 3) 2 + 4 TP = ( −3, 4) ∴ range ( −2, 4] The answer is D. 9 y = ( x − 4) 2 x ∈ [0, 6] TP = (4, 0) When x = 0 y = ( −4) 2 = 16 ∴ range [0, 16] When x = 6 y = (6 − 4) 2 = 2 2 = 4 But x = 4 y = 0 The answer is A. 10 a f( x ) = ( x − 2) 2 − 4 TP = (2, −4) y -int x = 0 y = (0 − 2) 2 − 4 y = ( −2) 2 − 4 y = 0

MM12-1 10 Graphs and polynomials x -int y = 0 0 = ( x − 2 − 2)( x − 2 + 2) 0 = ( x − 4)( x) ∴ x = 4 or 0 b f( x ) = − (x + 4) 2 + 9 TP = ( −4, 9) y int x = 0 y = − (0 + 4) 2 + 9 y = − 16 + 9 y = − 7 x -int y = 0 0 = 9 − ( x + 4) 2 0 = (3 − ( x + 4))(3 + ( x + 4)) 0 = (3 − x − 4) (3 + x + 4) 0 = ( −x − 1)(7 + x ) x = − 1 or −7 c y = x 2 + 4 x + 3 y = x2 + 4 x + 4 − 4 + 3 y = ( x + 2) 2 − 1 TP = ( −2, −1) y-int x = 0 y = 3 x-int y = 0 0 = ( x + 2 − 1)( x + 2 + 1) 0 = ( x + 1)( x + 3) ∴ x = − 1 or −3 d y = 2 x 2 − 4 x − 6 y = 2[ x2 − 2 x − 3] = 2[x2 − 2 x + 1 − 1 − 3] = 2[(x − 1) 2 − 4] y = 2( x − 1)2 − 8 TP = (1, −8) y -int x = 0 y = − 6 x -int y = 0 0 = 2[( x − 1 − 2)( x − 1 + 2)] 0 = 2( x − 3)( x + 1) ∴ x = 3 or −1 11 a y = x 2 − 2 x + 2 x ∈ [−2, 2] y = x 2 − 2 x + 1 − 1 + 2 y = ( x − 1) 2 + 1 ∴ TP = (1, 1) i Domain = [ −2, 2] ii Range: When x = − 2 y = 10 When x = 2 y = 2 but TP = (1, 1) ∴ [1, 10] b y = − x 2 + x − 1 x ∈ R + y = − (x 2 − x + 1) y = − 22 211 1 22 xx     −+ − +        y = − 213 24 x    −+         y = − 21 2 x  −   −3 4 ∴ TP = 13 , 24  −   i Domain = R + ii Range = ( −∞ , − 3 4] c f( x ) = x 2 − 3 x − 2 x ∈ [−10, 6] f (x ) = x − 3 x + 23 2    − 23 2    − 2 = 23 2 x  −   − 9 4 − 8 4 = 23 2 x  −   − 17 4 ∴ TP = 317 , 24  −   i Domain = [ −10, 6] ii Range: When x = − 10 y = 128 ∴ 17 ,128 4  −    d f( x ) = − 3x 2 + 6 x + 5 x ∈ [−5, 3) = 2 5 32 3 xx   −−−     = 2 5 3211 3 xx  −−+−−   = 2 8 3( 1) 3 x  −−−   = − 3( x − 1) 2 + 8 TP = (1, 8) i Domain = [ −5, 3) ii Range: When x = − 5, y = − 100 ∴ [ −100, 8] 12 V(t) = 2 t 2 − 16 t + 40 t ∈ [0, 10] V (t) = 2( t 2 − 8 t + 20) = 2[t2 − 8 t + 16 − 16 + 20] = 2[(t − 4) 2 + 4] = 2(t − 4)2 + 8 ∴ TP = (4, 8) When t = 0 V(t) = 40 When t = 10 V (t) = 2 × 6 2 + 8 = 80 a minimum V = 8 m 3 b maximum V = 80 m 3 13 h (t) = − 3t 2 + 12 t + 36 h (t) = − 3[ t2 − 4 t − 12] = − 3[ t2 − 4 t + 4 − 4 − 12] = − 3[( t − 2) 2 − 16] = − 3( t − 2)2 + 48 ∴ TP = (2, 48) a maximum height = 48 m b When h(t) = 0 0 = − 3[( t − 2 − 4)( t − 2 + 4)] 0 = − 3( t − 6)( t + 2) ∴ t = 6 or −2 Since time ⇒ 6 seconds c Domain [0, 6] Range [0, 48] 14 a h(t) = t 2 − 12 t + 48, t ∈ [0, 11] The lowest point is the y-coordinate of the turning point h(t) = t 2 − 12 t + 36 − 36 + 48 = ( t − 6) 2 + 12

Graphs and polynomials MM12-1 11 TP = (6, 12) Lowest point is 12 m above the ground. b Time taken is the x-coordinate of the turning point. t = 6 seconds c Check the end points of the domain h(0) = 48 h(11) = 11 2 − 12 × 11 + 48 = 37 The highest point above the ground is 48 m. d Domain = [0, 11] Range = [12, 48] e Exercise 1F — Cubic graphs 1 a Positive cubic so a = 1. Goes through origin so x is a factor. y = x (x − a )( x − b ) = x (x + 6)( x − 5) b Positive cubic in form y = a (x − m )(x − n ) 2 ∴ a = 1, m = 1, n = − 2 ∴ y = 1( x − 1)( x + 2) 2 2 a Positive cubic in form y = ( x − l )( x − m )(x − n ) l = − 3, m = 1, n = 4 ∴ y = ( x + 3)( x − 1)( x − 4) ∴ ( v) b Negative cubic in form y = a (x − m )(x − n ) 2 ∴ a = − 1, m = 5, n = − 2 ∴ y = − 1( x − 5)( x + 2) 2 y = (5 − x )( x + 2) 2 ( iv) c Negative cubic in form y = a (x − l )( x − m )(x − n ) a = − 1, l = − 3, m = 1, n = 4 ∴ y = − 1( x + 3)( x − 1)( x − 4) y = ( x + 3)(1 − x )( x − 4) (ii) d Positive cubic in form y = a (x − t ) 3 a = 1, t = 3 ∴ y = ( x − 3) 3 (i) e Positive cubic in form y = a (x − l )( x − m )(x − n ) a = 1, l = − 4, m = − 2, n = 1 ∴ y = ( x + 4)( x + 2)( x − 1) ( vi) f Positive cubic in form y = a (x − m )(x − n ) 2 a = 1, m = 5, n = − 2 y = ( x − 5)( x + 2) 2 ( viii ) g Negative cubic in form y = a (x − t ) 3 a = − 1, t = 3 ∴ y = − 1( x − 3) 3 y = (3 − x ) 3 ( vii) h Negative cubic in form y = a (x − l )( x − m )(x − n ) a = − 1, l = − 4, m = − 2, n = 1 ∴ y = − 1( x + 4)( x + 2)( x − 1) y = ( x + 4)( x + 2)(1 − x ) ( iii) 3 a y = x 3 + x 2 − 4 x − 4 y-intercept x = 0 y = − 4 Factorise to find x-intercepts Test x = − 1, y = 0 ∴ x + 1 is a factor 2 32 32 4 44 1 44 44 0 x xx x x xx x x−  +−− +−  +   −−  −  −−  ∴ y = ( x + 1)( x 2 − 4) y = ( x + 1)( x − 2)( x + 2) If y = 0, x = − 1, 2, or −2 b y = 2 x 3 − 8 x 2 + 2 x + 12 y-int x = 0 y = 12 Factorise to find x-intercepts Test x = − 1 so y = 0 ∴ (x + 1) is a factor 2 32 32 2 221012 28 212 1 22 10 2 10 10 12 12 12 12 0 xx xx x x xx xx xx x x −+  −++ +−  +   −+ −  −+   +  −  +  ∴ y = ( x + 1)(2 x 2 − 10 x + 12) y = 2( x + 1)( x − 2)( x − 3) If y = 0, then x = − 1, 2 or 3 c y = − 2x 3 + 26 x + 24 y-int x = 0 y = 24 Factorise to find x-intercepts Test x = − 1 so y = 0 ∴ (x + 1) is a factor. 2 32 32 2 22224 2 0 26 24 1 22 226 22 24 24 24 24 0 xx xx x x xx xx xx x x −++  −+ + + +−  −−   + −  +   +  −  +  ∴ y = ( x + 1)( −2x 2 + 2 x + 24) y = 2( x + 1)( −x − 3)( x − 4) If y = 0, then x = − 1, −3 or 4. d y = − x 3 + 8 x 2 − 21 x + 18 y-int x = 0 y = 18 x-intercept Factorise: Test x = 3 so y = 0 ∴ (x − 3) is a factor. 2 32 32 2 2 56 82118 3 3 521 515 618 618 0 xx xx x x xx xx xx x x −+ −  −+ − + −−  −+   − −  −   −+  −  −+  ∴ y = ( x − 3)( −x 2 + 5 x − 6) = − (x − 3)( x2 − 5 x + 6) y = − 1( x − 3)( x − 3)( x − 2) ∴ x = 3 or 2 4 a x3 + 6 x 2 + 12 x + 8 = y Test x = − 2 so y = 0 ∴ (x + 2) is a factor 2 32 32 2 44 6128 2 2 412 48 48 48 0 xx xx x x xx xx xx x x ++  +++ +−  +    + −  +  +  −  +  ∴ y = ( x + 2)( x 2 + 4 x + 4) = ( x + 2)( x + 2) 2 y = ( x + 2) 3 The answer is B.

MM12-1 12 Graphs and polynomials b In form y = a (x − t ) 3 a = 1, t is intercept The answer is E. 5 Function graph is a negative cubic so a = − 1 Point of inflection (2, 2) The answer is C. y = (2 − x ) 3 + 2 6 f (x ) = 5( x + 1) 3 − 3 Point of inflection ( −1, −3) Graph is a positive cubic \bhe answer is A. 7 Positive cubic Turning point at (1, 0) because of repeated factor x-intercept at ( −3, 0) y-intercept at (0, 6) The answer is B. 8 Negative cubic A or D Point of inflection is ( a, b) a < 0 so, y = − (x + a ) 3 + b The answer is D. 9 y = − h(x − a ) 2 ( x − c ) x = 0, y = − ha2(− c) b = ha2c h = 2b ac y = 2b ac − ( x − a ) 2(x − c ) The answer is E. 10 a f( x ) = x 3 + x 2 − 10 x + 8 x ∈ [2, ∞) a > 1 positive b > 1 so 2 turning points. y- intercept x = 0 y = 8 When x = 2 y = 2 3 + 2 2 − 20 + 8 = 0 Closed end point = (2, 0) i Domain [2, ∞) ii range [0, ∞) b f( x ) = 3 x 3 − 5 x 2 − 4 x + 4 for x ∈ [−2, −1] a > 1 so positive b ≠ 0 ∴ 2 turning points. y-intercept x = 0 y = 4 When x = − 2 y = 3 × ( −2) 3 − 5 × ( −2) 2 − 4 × − 2 + 4 = − 32 When x = − 1 y = 3 × ( −1) 3 − 5 × ( −1) 2 − 4 × − 1 + 4 = 0 Closed end point ( −2, −32) Closed end point ( −1, 0) i Domain [ −2, −1] ii Range [ −32, 0] c f( x ) = − 3x 3 + 4 x 2 + 27 x − 36 x ∈ (0, 1] a < 1 ∴ negative b ≠ 0 ∴ 2 turning points y-intercept x = 0 y = − 36 When x = 0 y = − 36 Open end point (0, −36) When x = 1 y = − 3 × 1 3 + 4 × 1 2 + 27 × 1 − 36 = − 8 Closed end point (1, −8) i Domain (0, 1] ii Range ( −36, −8] d f( x ) = − x 3 − 3 x for x ∈ [ − 1, 2) a < 1 negative b = 0 c < 1 y- int x = 0 y = 0 When x = − 1, y = 4 closed end point x = 2, y = − 14 open end point i Domain − [ −1, 2) ii Range − ( −14, 4] e f( x ) = x 3 + 2 x for x ∈ [−2, −1) ∪ (0, 3] a > 1 positive b = 0 c > 1 y- int x = 0 y = 0 When x = − 2, y = − 12 closed end point x = − 1, y = − 3 open end point x = 0 y = 0 open end point x = 3 y = 33 closed end point i Domain [ −2, −1) ∪ (0, 3] ii Range [ −12, −3) ∪ (0, 33] f f( x ) = − 2x 3 − x for x ∈ (−1, 1) ∪ [2, 3) a < 1 negative b = 0 c < 1 y- int x = 0, y = 0 When x = − 1, y = 3 open end point x = 1, y = − 3 open end point x = 2, y = − 18 closed end point x = 3, y = − 57 open end point i Domain ( −1, 1) ∪ [2, 3) ii Range ( −3, 3) ∪ (−57, −18] 11 f (x ) = x 3 + ax 2 + bx − 64 0 = (− 2) 3+ (− 2) 2a + (− 2)b − 64 = − 8 + 4 a − 2 b − 64 72 = 4 a − 2 b 36 = 2 a − b b = 2 a − 36 [1] f( x ) = x 3 + ax 2 + bx − 64 0 = 4 3 + 16 a + 4 b − 64 0 = 16 a + 4 b 0 = 4 a + b [2] Sub [1] into [2] 0 = 4 a + 2 a − 36 0 = 6 a − 36 36 = 6 a 6 = a ∴ b = 12 − 36 = − 24 12 0 = − 1 − 2 − a + 10 0 = 7 − a [1] ∴ a = 7 ∴ y = 6 + (7 + b )x − 4 x 2 − x 3 0 = 6 + (7 + b ) × (− 1) − 4 × 1 − (− 1) 0 = 6 − 7 − b − 4 + 1 0 = − 4 − b b = − 4 13 a f( x ) = a (x + b ) 3 + c point of inflection (3, 3) ⇒ b = − 3 and c = 3 f (x ) = a (x − 3) 3 + 3 When x = 2, f( x ) = 0 0 = a (2 − 3) 3 + 3 0 = a × (− 1)3 + 3 0 = − a + 3 ∴ a = 3 ∴ f ( x ) = 3( x − 3) 3 + 3 b Point of inflection due to reflection = (−3, 3) ∴ g (x ) = − 3( x + 3) 3 + 3 domain [ −4, −2] range [0, 6] c When f( x ) = 3.375 3.375 = 3( x − 3) 3 + 3 x = 3.5 ∴ width = 3.5 × 2 = 7 cm

Graphs and polynomials MM12-1 13 14 d (t) = at 2( b − t ) a (2, 3) and (5, 0) ∴ 3 = 4 a(b − 2) [1] 0 = 25 a(b − 5) [2] 3 = 4 ab − 8 a × 25 0 = 25 ab − 125 a × 4 75 = 100 ab − 200 a 0 = 100 ab − 500 a ∴ 75 = 300 a 1 4 = a Sub 1 4 = a into [1] 3 = 4 × 1 4 ( b − 2) 3 = 1( b − 2) 5 = b b ∴ Rule: d(t) = 2 4 t (5 − t ) for domain = [0, 5] c d d(t) = 235 44tt − d ′( t) = 2 10 3 44 tt − Let d ′( t) = 0 = 2 10 3 4 tt − 0 = 10 t − 3 t 2 0 = t (10 − 3 t) ∴ t = 0 or 10 − 3 t = 0 10 = 3 t ∴ time is 1 3 3 hours When time is 1 3 3, d (t) = 2 (3.3) 4  × 1.6  Maximum distance = 4.6 km Exercise 1G — Quartic graphs 1 a y = ( x − 2)( x + 3)( x − 4)( x + 1) y-intercept when x = 0 y = − 2 × + 3 × − 4 × + 1 y = 24 x-intercepts when y = 0 x = − 3, −1, 2 and 4 b y = 2 x 4 + 6 x 3 − 16 x 2 − 24 x + 32 y-intercept when x = 0 y = 32 x-intercept when y = 0 Test x = 1, y = 0 ∴ (x − 1) is a factor Test x = + 2 so y = 0 ∴ (x − 2) is a factor ( x − 1)( x − 2) = x 2 − 3 x + 2 2 43 2 2 432 32 32 2 221216 2 6 16 24 32 32 264 12 20 24 12 36 24 16 48 32 16 48 32 0 xx xx x x xx xxx xxx xx x xx xx++  +− − + −+−  −+    −− −  −−   −+ −  −+   ∴ (x − 1)( x − 2)(2 x 2 + 12 x + 16) = y ( x − 1)( x − 2)2( x 2 + 6 x + 8) = y 2( x − 1)( x − 2)( x + 4)( x + 2) = y When using N.F.L, x = 1, 2, −4, −2 then y = 0. c y = x 4 − 4 x 2 + 4 y-intercept when x = 0 y = 4 x-intercept when y = 0 Let a = x 2 y = a 2 − 4 a + 4 y = ( a − 2) 2 Sub a = x 2 back in y = ( x 2 − 2) 2 ∴ x = ± 2 d y = − 2x 4 + 15 x 3 − 37 x 2 + 30 x y-intercept, when x = 0 y = 0 x-intercepts when y = 0 Test x = 2 ∴ y = 0 Test x = 3 ∴ y = 0 (x − 2)( x − 3) are factors (x − 2)( x − 3) = x 2 − 5 x + 6 2 43 2 2 432 32 3225 21537 30 56 21012 52530 52530 0x x x xxx xx xxx x xx x xx −+  −+ − + −+−  −+ −   −+ −  −+   ∴ (x − 2)( x − 3)( −2x 2 + 5 x) = y x (x − 2)( x − 3)( −2x + 5) = y Using N.F.L, x = 0, 2, 3 or 5 2 e y = 6 x 4 + 11 x 3 − 37 x 2 − 36 x + 36 y-intercept when x = 0 y = 36 x-intercepts when y = 0 Test x = − 3 ∴ y = 0 Test x = 2 (x + 3)( x − 2) are factors (x + 3)( x − 2) = x 2 + x − 6 2 43 2 2 43 2 32 32 2 2656 61137 3636 6 6636 536 5530 6636 6636 0 xx xx x x xx xx x xx x xx x xx xx+−  +− −+ +− −  +−   −− −  +−   −−+ −  −−+   ∴ (x + 3)( x − 2)(3 x − 2)(2 x + 3) = y Using N.F.L x = − 3, 2, 2 3, − 3 2 2 a y = x 2( x − 2)( x − 3) y = 0, x2(x − 2)( x − 3) = 0 Turning point (0, 0) Intercepts at x = 2 and x = 3 y-intercept when x = 0 y = 0( −2)( −3) = 0 Positive quartic Maximum turning point at (1.16, 2.08) Minimum turning points at (0, 0) and (2.59, −1.62) b y = − (x + 1) 2(x − 1) 2 x = 0, y = 1 y-intercept is 1 y = 0, ( x + 1)2(x − 1) 2 = 0 x = − 1, x = 1 Minimum turning points at ( −1, 0) and (1, 0). Maximum turning point (0, 1)

MM12-1 14 Graphs and polynomials c y = ( x − 1) 2(x + 1)( x + 3) x = 0, y = 3 y-intercept is 3 y = 0, (x − 1) 2(x + 1)( x + 3) = 0 x = 1, − 1, −3 Positive quartic Minimum turning points (−2.28, −9.91) and (1, 0) Maximum turning point (−0.22, 3.23) d y = ( x + 2) 3(1 − x ) x = 0, y = 8 y-intercept is 8 y = 0, (x + 2) 3(1 − x ) = 0 x = − 2, 1 Point of inflection (2, 0) x-intercept is 1 Negative quartic (1 − x ) = − (x − 1 ) Maximum turning point (0.25, 8.54) 3 a f(x ) = x 4 − 8 x 2 + 16 Let x 2 = a f (x) = a2 − 8 a + 16 = (a − 4) 2 Substitute x2 = a back in: f (x) = ( x 2 − 4) 2 ∴ (x − 2) 2(x + 2) 2 The answer is E. b f(x ) = x 4 − 8 x 2 + 16 y -int when x = 0 y = 16 (0, 16) x -int when y = 0 2, −2 ∴ The answer is B. c range = [0, 16] The answer is A. d 25 = x 4 − 8 x 2 + 16 0 = x 4 − 8 x 2 − 9 Let x 2 = a 0 = a2 − 8 a − 9 0 = ( a − 9)( a + 1) Substitute x 2 = a back in 0 = ( x 2 − 9)( x 2 + 1) ∴ x = ± 3 ∴ (−3, +3) is the restricted domain. The answer is D. 4 a x = − 2, −1, 1, 3 y = a(x + 2)( x + 1)( x − 1)( x − 3) (0, 6) 6 = a(0 + 2)(0 + 1)(0 − 1)(0 − 3) 6 = a × 6 a = 1 y = ( x + 2)( x + 1)( x − 1)( x − 3) b x = 4, 2, − 1 Repeated factor at x = − 2. y = a(x − 4)( x − 2) 2(x + 1) (0, 8) 8 = a(0 − 4)(0 − 2) 2(0 + 1) 8 = − 16a a = − 1 2 y = 1 2 −(x − 4)( x − 2) 2(x + 1) 5 a y = (2 − x )( x 2 − 4)( x + 3) x ∈ [2, 3] y-int when x = 0 y = 2 × − 4 × 3 = − 24 x-int when y = 0 x = 2, − 2, −3. when x = 2, y = 0 closed end point x = 3, y = − 15 closed end point i Domain [2, 3] ii Range [ −15, 0] b y = 9 x 4 − 30 x 3 + 13 x 2 + 20 x + 4x ∈ (−2, −1] y-int when x = 0 y = 4 x-int when y = 0 Test x = 2 y = 0 ∴ (x − 2) factor 32 432 43 32 32 2 2912 93013204 2 918 12 13 12 24 11 20 11 22 24 24 0 xx xxxx x xx xx xx xx xxx x −−  −+++ −−  −   −+ −  −+   −+ −  −+   −+  −  −+  11 2 x− y = ( x − 2)(9 x 3 − 12 x 2 − 11 x − 2) Try for 2nd factor of x − 2 Test x = 2, 9x 3 − 12 x 2 − 11 x − 2 = 0 x − 2 is a factor. So turning point at (2, 0) 2 32 32 2 2961 2 9 12 11 2 918 6 11 6 12 2 xx xxxx xxxx xxx ++ −−−− −− −− y = ( x − 2) 2(9 x 2 + 6 x + 1) y = ( x − 2) 2(9 x 2 + 6 x + 1) = (x − 2) 2(3 x + 1) 2 Turning points at (2, 0) and 1,0 3  −   when x = − 2, y = 400 open end point when x = − 1, y = 36 closed end point i Domain (− 2, −1] ii Range [36, 400) c y = − (x − 2) 2(x + 1) 2 x ∈ (−∞ , −2] y-int when x = 0 y = − 1 × 4 × 1 y = − 4 x-int when y = 0 0 = − 1(x − 2) 2(x + 1) 2 ∴ x = − 1 or 2 When x = − 2, y = − 16. y = − (− 2 − 2) 2(− 2 + 1) 2 y = − (− 4) 2(−1) 2 y = − 16 × 1 y = − 16 i Domain (−∞ ,− 2] ii Range ( −∞,−16] d y = − x 4 + 4 x 2 x ∈ [−3, −2] y-int when x = 0 y = 0 x-int when y = 0 0 = x 2(− x 2 + 4) 0 = x 2(4 − x 2) 0 = x 2(2 − x )(2 + x ) ∴ x = 0 ± 2 When x = − 3, y = − 45 x = − 2, y = 0 Both of these are closed end points.

Graphs and polynomials MM12-1 15 i Domain [ −3, −2] ii Range [ −45, 0] 6 f (x ) = x 4 + ax 3 − 4 x 2 + bx + 6 (2, 0): 0 = 2 4 + 2 3a − 4 × 2 2 + 2 b + 6 16 + 8 a − 16 + 2 b + 6 = 0 6 + 8 a + 2 b = 0 8a + 2 b = − 6 Divide both sides by 2: 4a + b = − 3 − 4a − 3 = b [1] 4 a + 3 = − b ( −3, 0): 0 = (− 3) 4 + ( −3) 3a − 4 × (− 3) 2 − 3 b + 6 0 = 81 − 27 a − 36 − 3 b + 6 0 = 51 − 27 a − 3 b 0 = 17 − 9 a − b [2] Sub [1] into [2] 0 = 17 − 9 a + 4 a + 3, 0 = 20 − 5 a 5 a = 20 a = 4 If a = 4 then b = − 4 × 4 − 3 b = − 16 − 3 b = − 19 7 f (x ) = x 4 + ax 3 + bx 2 − x + 6 ( x − 1) is a factor P(1) = 1 + a + b − 1 + 6 a + b = − 6 [1] ( x + 3) is a factor P(− 3) = 81 − 27 a + 9 b + 3 + 6 = 0 27 a − 9 b = 90 3 a − b = 10 [2] [1] + [2] 4 a = 4 a = 1 b = − 7 8 y = ( a − 2 b)x 4 − 3 x − 2 Sub in (1, 0): 0 = ( a − 2 b)1 4 − 3 − 2 0 = a − 2 b − 5 5 = a − 2 b 5 + 2 b = a [1] Sub (1, 0) into equation. y = x 4 − x 3 + ( a + 5 b)x 2 − 5 x + 7 0 = 1 − 1 + ( a + 5 b)1 − 5 + 7 0 = a + 5 b + 2 − 2 = a + 5 b [1] Sub [1] into [2] − 2 = 5 + 2 b + 5 b − 7 = 7 b − 1 = b If −1 = b then 5 − 2 = a ∴ 3 = a Chapter review Short answer 1 a (2 y − 3 x) 5 = (2y) 5 + 5(2 y) 4(− 3x) + 10(2 y) 3(− 3x) 2 + 10(2 y) 2(− 3x) 3 + 5(2 y)( −3x) 4 + ( −3x) 5 = 32y 5 + 5 × 16 × ( −3) y 4x + 10 × 8 × 9 y 3x2 + 10 × 4 × (− 27) y 2x3 + 5 × 2 × 81 yx 4 − 243 x 5 = 32y 5 − 240 y 4x + 720 y 3x2 − 1080 y 2x3 + 810 yx 4 − 243 x 5 b 82 2 x x  −   = 8 2 x   + 8 7 2 2 x x  −   + 28 62 2 2 x x   −     + 56 53 2 2 x x   −     + 70 44 2 2 x x   −     + 56 35 2 2 x x   −     + 28 26 2 2 x x   −     + 8 52 2 x x   −     + 82 x  −   = 86 256 8 x x − + 47 4x − 14 x 2 + 70 − 3224 x + 4448 x − 6512 x + 8256 x 2 ( x 2 − 1) = ( x − 1)( x + 1) ∴ solutions are x = − 1 or 1 ∴ If x = − 1 then 0 = − 7 − a + 5 − 15 + b 0 = − 17 − a + b 17 + a = b [1] If x = 1, then 0 = − 7 + a + 5 + 15 + b 0 = 13 + a + b [2] Sub [1] into [2] 0 = 13 + a + 17 + a 0 = 30 + 2 a − 30 = 2 a − 15 = a ∴ b = 17 − 15 b = 2 3 a x3 − 12 x 2 + 17 x + 90 = y Test x = − 2. y = 0 ∴ (x + 2) is a factor 2 32 32 2 2 14 45 12 17 90 2 2 14 17 14 28 45 90 45 90 0 xx xx x x xx xx xx x x −+  −++ +−  +   −+ −  −−   +  −  +  ∴ (x + 2)( x 2 − 14 x + 45) ( x + 2)( x − 9)( x − 5) b 2 x 4 + 7 x 3 − 31 x 2 + 0 x + 36 = y Test x = − 1 ∴ y = 0 Test x = 2 ∴ y = 0 ∴ (x + 1)( x − 2) are factors ∴ (x 2 − x − 2) is a factor 2 43 2 2 432 32 322918 2731036 2 224 927 0 9918 xx xx xx xx xxx xxx xx x +−  +− ++ −− −  −−   −+ −  −−   2 2 18 18 36 18 18 36 0 xx xx  −++ −  −++   ∴ (x + 1)( x − 2) (2 x 2 + 9 x − 18) ( x + 1)( x − 2) (2 x − 3)( x + 6) 4 a (−5, 6), (1, −1) m = 61 51 + −− = 7 6 − ∴ y = 7 6 −x + c Sub in (1, −1) to find c: − 1 = 7 6 − × 1 + c − 1 + 7 6 = c 1 6 = c ∴ y = 7 6 −x + 1 6 6 y = − 7 x + 1 7 x + 6 y − 1 = 0 b 2x − y + 10 = 0 2 x + 10 = y Perpendicular ⇒ m = 1 2 − ∴ y = 1 2 −x + c Sub in point (3, 3) 3 = 3 2 − + c 3 + 3 2 = c so c = 9 2. ∴ y = 1 2 −x + 9 2 2 y = − x + 9 x + 2 y − 9 = 0 5 y = − x2 − 2 x + 8 y-int when x = 0 ∴ y = 8 x-int when y = 0 0 = − 1( x2 + 2 x − 8) 0 = − 1( x + 4)( x − 2) ∴ x = − 4 or 2 TP ⇒ y = − 1( x2 + 2 x − 8) = − 1( x2 + 2 x + 1 − 1 − 8) = − 1[( x + 1) 2 − 9] = − 1( x + 1) 2 + 9 TP = ( −1, 9) Domain = R Range = ( −∞ , 9] 6 y = 3 x2 + 8 x − 3 x ∈ [−3, 0) y-int when x = 0 y = − 3 x-intercepts when y = 0 0 = 3 x2 + 8 x − 3 0 = (3 x − 1)( x + 3)

MM12-1 16 Graphs and polynomials ∴ x = 1 3 or −3 TP ⇒ y = 2 8 31 3 xx +−   Now x2 + 8 3x − 1 = 2 2 8 8 64 36 3 6 36 36 xx  ++ −−     ∴ y = 28100 3 636 x  +−     = 28 100 3 62 x  +−   TP = 8 100 , 612 −−    = 425 , 33 −−    Domain [ −3, 0) Range: When x = 3, y = 0 closed end point When x = 0, y = − 3 open end point 25 ,0 3 −   7 a f(x) = − x3 + bx2 + ax − 18 0 = − (− 3) 3 + b(− 3) 2 + a × ( −3) − 18 0 = 27 + 9 b − 3 a − 18 0 = 9 + 9 b − 3 a 0 = 3 + 3 b − a a = 3 + 3 b [1] g (x) = ax2 + bx − 75 0 = a(− 3) 2 + b × ( −3) − 75 0 = 9 a − 3 b − 75 25 = 3 a − b [2] Sub [1] into [2] 25 = 3(3 + 3 b) − b 25 = 9 + 9 b − b 16 = 8 b 2 = b ∴ a = 3 + 6 = 9 b f(x) = − x3 + 2 x2 + 9 x − 18 y-int when x = 0 ∴ y = − 18 x-int when y = 0 ( x + 3) is a factor 2 32 32 2 2 56 2918 3 3 59 515xx xxx x xx xx xx −+ −  −+ + − +−  −−   + −  +   618 618 0x x−−−−− ∴ ( x + 3)( − x2 + 5 x − 6) = 0 − 1( x + 3)( x2 − 5 x + 6) = 0 − 1( x + 3)( x − 3)( x − 2) = 0 ∴ x = − 3, 3, or 2 8 f(x) = x4 − 7 x3 + 12 x2 + 4 x − 16 y-int when x = 0 y = − 16 x-int when y = 0 Test x = − 1 ∴ ( x + 1) is a factor x = 2 ∴ ( x − 2) is a factor 2 43 2 2 43 2 32 32 2 2 68 712 416 2 2 614 4 6612 8816 8816 0 xx xx xx xx xx x xxx xx x xx xx−+  −+ +− −− −  −−    −+ + −−+ +   −− −  −−   ∴ ( x + 1)( x − 2)( x2 − 6 x + 8) ( x + 1)( x − 2)( x − 4)( x − 2) ∴ x = − 1, 2, or 4. Multiple choice 1 (1 − 2 x)5 = 1 5 + 5 × 1 4 ( −2 x) + 10 × 1 3 ( −2 x)2 + 10 × 1 2 ( −2 x)3 + 5 × 1( −2 x)4 + (− 2 x)5 = 1 − 10 x + 40 x2 − 80 x3 + 80 x4 − 32 x5 The answer is E. 2 8 2 1 4 x x  −   = x8, 7 2x x = x5 ∴ x5 = 2nd term so r = 1 2nd term = 8 1    (4x)7 1 21 x −    = 8 × 16 384 x7 × 21 x − = − 131 072 x5 The answer is B. 3 101 3 x x  +   5 th = 10 4    (3x)6 41 x    = 210 × 729 x6 × 41 x = 153 090 x2 The answer is A. 4 D x4 + 5 x3 3 2 2x −+ 5 x − 3 Positive whole number indices only for polynomials. 5 P (− 3) = ( −3) 5 − 4 × ( −3) 3 − 3( −3) 2 + 10 × (− 3) + 1 = − 243 + 108 − 27 − 30 + 1 = − 191 The answer is C. 6 (5 − 6 x + x3 + 62 4 107)( 321x xx x −   + 2) The answer is C. 7 5432 54 43 43 32 32 2 24053 3 3 4 3 70 721 21 5 21 63 58 3 58 174 xxxxx x xx xx xx xx xx xx xxx x  +++−+ +−  +   −+ −  −−   + −  +   −− −  −−  +  −  −  43 2 72158 171 xx x x −+ − + − The answer is C. 8 The answer is B. If ( x − 2) was a factor, when x = 2 was substituted the expression would be zero. 2 4 − 2 × 2 3 − 6 × 2 2 − 8 × 2 + 2 16 − 16 − 24 − 16 + 2 = − 38 9 2x4 − 4 x3 − 10 x2 + 12 x To check if ( x − 3) is a factor, substitute in x = 3 2 × 3 4 − 4 × 3 3 − 10 × 3 2 + 12 × 3 = 162 − 108 − 90 + 36 = 0 The answer is E. 10 (4, 0), (0, 2) m = 20 04 − − = 2 4 − = 1 2 − y = 1 2 −x + c Sub in (4, 0): 0 = 4 2 − + c 0 = − 2 + c 2 = c ∴ y = 1 2 −x + 2 2 y = − x + 4 2 y + x − 4 = 0 The answer is D. 11 (2, b), ( −3, −5) m = 5 32 b −− −− = 2 5 5b −− − = 2 − 5 − b = − 10

Graphs and polynomials MM12-1 17 − 5 + 10 = b 5 = b The answer is A. 12 y = 2 x + c Sub in (2, 5) 5 = 4 + c 1 = c ∴ y-int (0, 1) The answer is D. 13 3x2 + 4 x − 5 = 0 b 2 − 4 ac = 4 2 − 4 × 3 × − 5 = 16 + 60 = 76 The answer is E. 14 y = 2 x2 + 8 x − 10 x ∈ (−6, 2) y-int when x = 0 y = − 10 x-int when y = 0 0 = 2( x2 + 4 x − 5) 0 = 2( x + 5)( x − 1) ∴ x = − 5, or 1 When x = − 6, y = 2 × 36 − 48 − 10 y = 14 open ended When x = 2, y = 8 + 16 − 10 = 14 open ended The answer is B. 15 y = 2( x2 + 4 x − 5) = 2( x2 + 4 x + 4 − 4 − 5) = 2(( x + 2) 2 − 9) = 2( x + 2) 2 − 18 TP = ( −2, −18) Range = [ −18, 14) The answer is C. 16 y = − 3 x3 a < 1 ∴ negative b = 0 c = 0 \b he answer is B. 17 f(x) = − 2 x3 + 3 x2 + 11 x − 6 Test x = − 2, f(x) = 0 x + 2 is a factor 2 32 32 2 2273 (2) 3 11 6 2 24 711 714 36 36 0xx xx x x xx xx xx x x−+− −++− +−−−   + −  +   −−−−− f(x) = ( x + 2)( −2 x2 + 7 x − 3) = (x + 2)( −2 x + 1)( x − 3) ∴ x = − 2, 1 , 2 3 are x-intercepts and (0, −6) is the y-intercept. The answer is A. 18 TP at ( x + 1) 2 Other intercept ( x − 3) \b he answer is E. f (x) − ( x − 3)( x + 1) 2 19 Two intercepts x = 2 or 0 ∴ x(x − 2) 3 The answer is D. 20 ( −3, 0) is a TP so ( x + 3) 2 is a factor. (1, 0) infers ( x − 1) is a factor and (3, 0) means ( x − 3) is a factor. a is positive. \b he answer is A. 21 y = x2 Translate 3 units down ⇒ y = x2 − 3 Translate 2 units to the right ⇒ y = ( x − 2) 2 − 3 ∴ C Extended response 1 a y = a(x − b)2 + c Turning point ( −2.5, 4) y = a(x + 2.5) 2 + 4 (0, 18) 18 = 2.5 2a + 4 6.25 a = 14 a = 2.24 y = 2.24( x + 2.5) 2 + 4 b Domain [ −5, 0] Range [4, 18] c y = 5 5 = 2.24( x + 2.5) 2 + 4 ( x + 2.5) 2 = 0.446 x + 2.5 = 0.668 ( x < 0) x = − 1.83 Coordinates are ( −1.83, 5) d x + 2.5 = − 0.668 x = − 3.17 f:[ − 3.17, −1.83] → R, f(x) = 5 (Check answers using a graphics calculator.) 2 a Initial distance is the day of launch. t = 0, d = 0 km b t = 0, d = 0 km t = 1, d = 4000 km t = 2, d = 2000 km t = 3, d = 0 km c d(t) = at3 + bt2 + ct + d t = 0, d = 0 d(t) = at3 + bt2 + ct When t = 1, d = 4 4 = a + b + c [1] When t = 2, d = 2 2 = 8 a + 4 b + 2 c [2] When t = 3, d = 0 0 = 27 a + 9 b + 3 c [3] Solving [1], [2] and [3] simultaneously, then a = 1, b = − 6 and c = 9 so, d(t) = t3 − 6 t2 + 9 t d When t = 8, d(8) = 8 3 − 6 × 8 2 + 9 × 8 = 200 Distance is 200 000 km from Earth. The moon is 240 000 km from Earth. So, the satellite is closer by 40 000 km. e 490 = t3 − 6 t2 + 9 t t3 − 6 t2 + 9 t − 490 = 0 When t = 10, d(10) = 490 So, the satellite will self-destruct 490 000 km from Earth, 10 days after launching. Thus, the ‘life span’ of the satellite is 10 days. f Domain is [0, 10] Range is [0, 490] 3 a Turning Point form: y = ax2 + 3 (2, 0) ⇒ 0 = 4 a + 3 −3 = 4 a a = − 0.75 ∴ y = − 0.75 x2 + 3 b Assuming symmetry, x = 1.5 y = − 0.75(1.5) 2 + 3 y = 1.3125 m and therefore cannot fit c 1.7 – 1.3125 = 0.3875 m ∴ Remove 0.4 m (to 1 dp) 4 Liney • Starts 2.4 m in front of start line moving forwards at a constant speed of 0.75 m/min. • Passed by Cubric after 0.3 mins and 2.6 m from the start line, meets Cubric coming back towards Limey at 3.8 min and 5.24 m from the start line. Quadder • Starts at start line, travelling 1.3 m the wrong way for 2.55 mins, stopping momentarily then moving forward with increasing speed • Meets Cubric at the start line after 5.1 mins. Cubric • Starts at start line moving very fast towards the finish passing Liney at (0.3, 2.6), then slowing, stopping momentarily at (2.1, 8.8), then moving back towards the start. • Meets Liney at (3.8, 5.24) and Quadder at the start line at 5.1 min. • Slows down, stopping at (7.37, −5.8) and then speeding for the finish.

MM12-1 18 Graphs and polynomials The Finish: Quadder overtakes Limey at (10.04, 9.9) Cubric overtakes Limey at (10.0899, 9.97) Quadder finishes race in 10.07 min, Cubric in 10.092 min, Liney in 10.13 min. Quadder wins by 1.2 sec, Cubric second by 2.4 sec to Liney. 5 a C = (3, 0) and D = (2.25, −8.54) b y = ax3(x − 3) − 8.54 = 2.25 3(2.25 − 3) a a = 38.54 2.25 0.75 − ×− ≈ 0.9997 a ≈ 1 y = x3(x − 3) so, y = x4 − 3 x3 c Find y when x = 1 to obtain B. y = x4 − 3 x3 y = (1) 4 − 3(1) 3 = 1 − 3 = − 2 ∴ B is (1, −2) Distance from road at A to river at B is 2 km. d Distance OD = 22 2.25 8.54 + = 5.0625 72.9316+ = 77.99 ≈ 8.83 Distance CD = 22 0.75 8.54 + = 0.5625 72.9316+ = 73.49 ≈ 8.57 Total distance is 17.4 km Yes because the straight route from O to D to C is approximately 17.4 km and the river course is longer than this. 6 a x-intercepts −3, −1, 1, 3 y = a(x + 3)( x + 1)( x − 1)( x − 3) 1 0, 2 4    9 4 = a × 9 a = 1 4 y = 1 4(x + 3)( x + 1)( x − 1)( x − 3) b x = 4, y = 1 4(4 + 3)(4 + 1)(4 − 3)(4 − 1) = 1 4(7 × 5 × 1 × 3) = 1 26 4 Coordinates are 11 4, 26 4, 26 44   −     c Domain is [ −4, 4] d (2.236, −4)( −2.236, −4) Range 1 4, 26 4   −    e y = 1 4(x + 3)( x + 1)( x − 3)( x − 1) + 4 = 1 4(x2 + 4 x + 3)( x2 − 4 x + 3) + 4 = 1 4(x4 − 4 x3 + 3 x2 + 4 x3 − 16 x2 + 12 x + 3 x2 − 12 x + 9) + 4 = 1 4(x4 − 10 x2 + 25) or y = 1 4(x − 2.236) 2(x + 2.236) 2 f Domain [ −4, 4] Range 1 0, 30 4       7 Since smooth landing, the grap h could have a turning point at (0, 0). So, a factor of 2 (0)x− is possible. A further x intercept (to the right of 50 km) of b, indicates a factor of ( x – b). A dilation factor of a from the x-axis results in y = ax2(x – b) a (50, 10) ⇒ 10 = 50 2 × a(50 − b) 10 = 2500 a(50 − b) [1] If gradient is 0 at x = 50, d dy x = 3 ax2 − 2 abx 0 = 3 a × 50 2 − 2 ab × 50 0 = 7500 a − 100 ab 100 ab = 7500 a (divide by 100 a, since a ≠ 0) b = 75 Substituting into [1] ⇒ 10 = 2500 a(50 − 75) 10 = − 6250 a a = 1 6250 − ∴ y = 2(75) 6250xx − − b x = 2, y = 22(2 75) 6250− − = 0.04672 km = 46.72 m c Seems to be extremely low 2 km from touchdown. Therefore, not very accurate. 8 a x-intercepts and turning points at x = ± 4 ∴ y = a(x − 4) 2(x + 4) 2 (0, −3) ⇒ − 3 = a(0 − 4) 2(0 + 4) 2 −3 = 256 a a = 3 256 − ∴ y = 22 3( 4) ( 4) 256xx−+ −

Graphs and polynomials MM12-1 19 b If there is a smooth connection to the platform then x = 4 could be a turning point and an intercept. So, ( x – 4) 2 could be a factor. The other turning point at (0, −3) suggests another x intercept to the left at x = b, where b is negative. So, in factor form: y = a(x − 4) 2(x − b), where a is a dilation constant. (0, −3) ⇒ − 3 = a(0 − 4) 2(0 − b) −3 = − 16 ab ab = 3 16 [1] But zero gradient at G ⇒ y = a(x2 − 8 x + 16)( x − b) y = a(x3 − bx2 − 8 x2 + 8 bx + 16 x − 16 b) d dy x = a(3x2 − 2 bx − 16 x + 8 b + 16) Gradient = 0 at x = 0: 0 = a(8b + 16) 0 = 8 b + 16 since a ≠ 0 8 b = − 16 b = − 2 Substitute in [1]: a × − 2 = 3 16 a = 3 32 − ∴ y = 2 3 (4)(2) 32 xx −− + c i y = ax2 – 3 Since strut is 1 m long, F has coordinates (2, −2). (2, −2) ⇒ −2 = a × 4 − 3 1 = 4 a a = 1 4 ∴ y = 2 3 4 x − ii y = a(x − 4) 2 (2, −2) ⇒ −2 = a(2 − 4) 2 −2 = 4 a a = 1 2 − ∴ y = 2 1 (4) 2x −− = 2 1 ( 8 16) 2xx −−+ = 2 48 2 x x−+− Lower parabola d dy x = 2 x Upper parabola d dy x= − x + 4 x = 2 ⇒ d dy x = 1 d dy x = 2 As the gradients are different, the connection is not smooth. iii Lower equation is still y = 2 3 4 x − Upper y = 2 (4)ax − (3, −0.75) ⇒ −0.75 = 2 (3 4)a − a = − 3 4 ∴ y = 2 3 (4) 4x −− ∴ y = − 2 3 ( 8 16) 4xx −+ Lower parabola d dy x = 2 x Upper parabola d dy x = 3 (2 8) 4 x −− x = 3 ⇒ d dy x = 3 2 d dy x = 3 (2) 4 −− = 3 2

MM12-1 20 Graphs and polynomials As the gradients are the same, the graphs meet smoothly at (3, −0.75). d quartic: x = 2 ⇒ y = 22 3( 2 4) ( 2 4) 256 −+ − y = 3436 256 ×× − = − 1.6875 m Difference from strut = − 1.6 – ( −1.6875) = 0.0875 m cubic: x = 2 ⇒ y = 2 3 (2 4) (2 2) 32 −− + y = 3 44 32 −×× y = 1.5 m Difference from strut = − 1.6 – ( −1.5) = − 0.1 m quadratic: x = 2 ⇒ y = 22 3 4 − y = − 2 m Difference from strut = − 1.6 – ( −2) = 0.4 m The quartic model is the closest to the actual ramp with 1.6875 m.

Functions and transformations MM12-2 21 Exercise 2A — Transformations and the parabola 1 y = a (x − h ) 2 + k a a = 2 ⇒ Dilation by a factor of 2 in the y direction. b a = 1 3 ⇒ Dilation by a factor of 1 3 in the y direction. c a = − 3 ⇒ Dilation by a factor of 3 in the y direction and reflection in the x -axis d k = − 6 ⇒ Translation 6 units down. e a = 1 2 −, k = 1 ⇒ Dilation by a factor of 1 2 in the y direction, reflection in the x -axis, translation 1 unit up. f h = 2 ⇒ Translation of 2 units to the right. g a = − 1, h = − 3 ⇒ Reflection in the x-axis and translation of 3 units to the left. h a = 2, h = 3 ⇒ Dilation by a factor of 2 in the y direction and translation of 3 units to the right. i h = − 2, k = − 1 ⇒ Translation of 2 units to the left and translation of 1 unit down. j h = 0.5, k = 2 ⇒ Translation of 0.5 units to the right and translation of 2 units up. k a = − 2, h = 3, k = 1 ⇒ Dilation by a factor of 2 in the y direction, reflection in the x -axis, translation of 3 units to the left and translation of 1 unit up. l y = 12 23 2 x  −   − 1 4 a = 12, h = 3 2, k = − 1 4 ⇒ Dilation by a factor of 12 in the y direction, translation of 3 2 (1.5) units to the right and translation of 1 4 (0.25) units down. 2 Increasing m ⇒ a decreases. Therefore the graph will be wider. The answer is D. 3 a k = 2 ⇒ graph (ii) b a = − 2, h = 2 ⇒ graph (v) c a = − 1, h = − 2, k = 2 ⇒ graph (i) d a = 1 2, h = 2 ⇒ graph (iv) e a = 2, h = − 2, k = 2 ⇒ graph (iii) 4 y = a (k − h ) 2 + k a Turning point (2, 2) h = 2, k = 2 y = a (x − 2) 2 + 2 Using (0, 0): 0 = a (0 − 2) 2 + 2 0 = 4 a + 2 4 a = − 2 a = 1 2 − Equation is y = 1 2 −( x − 2) 2 + 2 b Turning point ( −1, −2) h = − 1, k = − 2 y = a (x + 1) 2 − 2 Using (0, 0): 0 = a (0 + 1) 2 − 2 0 = a − 2 a = 2 Equation is y = 2( x + 1) 2 − 2 c Turning point (1, 3) h = 1, k = 3 y = a (x − 1) 2 + 3 Using (0, 0) 0 = a (0 − 1) 2 + 3 0 = a + 3 a = − 3 Equation is y = − 3( x − 1) 2 + 3 d Turning point ( −2, −4) h = − 2, k = − 4 y = a (x + 2) 2 − 4 Using (0, 0): 0 = a (0 + 2) 2 − 4 0 = 4 a − 4 4 a = 4 a = 1 Equation is y = ( x + 2) 2 − 4 5 Turning point ( c, d ) h = c , k = d y = a (x − c ) 2 + d Parabola is negative ⇒ a < 0 ⇒ Only alternative is y = d − ( x − c ) 2 The answer is E. 6 y = x 2 a a = 1 2 ⇒ y = 1 2x2 b a = −1 ⇒ y = − x2 c h = 2, k = −1 ⇒ y = (x − 2)2 − 1 d a = 3, k = −2 ⇒ y = 3x2 − 2 e a = −1, h = 3 ⇒ y = −(x + 3)2 7 y = a(x − h)2 + k a Turning point (3, −4) h = 3, k = −4 y = a(x − 3)2 − 4 Using (5, 0) 0 = a(5 − 3)2 − 4 0 = 4a − 4 4 a = 4 a = 1 Equation is y = (x − 3)2 − 4 b Turning point ( −1, 1) h = −1, k = 1 y = a(x + 1)2 + 1 Using (0, −1) −1 = a(0 + 1)2 + 1 −1 = a + 1 a = −2 ∴ y = −2(x + 1)2 + 1 c Turning point ( −3, −4) h = −3, k = −4 y = a(x + 3)2 − 4 Using (0, −1) −1 = a(0 + 3)2 − 4 −1 = 9a − 4 9 a = 3 a = 1 3 ∴ y = 1 3(x + 3)2 − 4 d Turning point (2, 2) h = 2, k = 2 y = a(x − 2)2 + 2 Using (0, 0) 0 = a(0 − 2)2 + 2 0 = 4a + 2 4 a = −2 a = 1 2 − ∴ y = 1 2 − (x − 2) 2 + 2 e Turning point (1, 6) h = 1, k = 6 y = a(x − 1) 2 + 6 Using (0, 9) 9 = a(0 − 1) 2 + 6 9 = a + 6 a = 3 ∴ y = 3( x − 1) 2 + 6 f Turning point ( −2, 8) h = − 2, k = 8 y = a(x + 2) 2 + 8 Using ( − 2− 2, 0) 0 = a(− 2− 2 + 2) 2 + 8 0 = 2 a + 8 2 a = − 8 a = − 4 ∴ y = − 4( x + 2) 2 + 8 Chapter 2 — Functions and transformations

MM12-2 22 Functions and transformations 8 y = x2 a dilation of 2 from the x-axis ⇒ y = 2 x2 b reflection in the x-axis ⇒ y = − 2 x2 c translation of −1 parallel to the x-axis ⇒ y = − 2( x + 1) 2 d translation of 3 parallel to the y-axis ⇒ y = − 2( x + 1) 2 + 3 9 point ( x, y) a Reflection in the y-axis ⇒ ( − x, y) b Reflection in the x-axis ⇒ (x, −y) c Dilation of 3 from the x-axis ⇒ (x, 3y) d Dilation of 2 from the y-axis ⇒ (2x, y) e Dilation of 1 3 from the y-axis ⇒ 1 , 3x y    f A translation of 2 units horizontally in the positive direction ⇒ (x + 2, y) g A translation of −1 unit parallel to the y-axis ⇒ (x, y − 1) 10 a y = a(x − h)2 + k Turning point ( z, − 8) y = a(x − z)2 − 8 Using (0, 10): 10 = a(0 − z)2 − 8 10 = az2 − 8 az2 = 18 ⇒ a = 218 z 1 Using (5, 0): 0 = a(5 − z)2 − 8 or 0 = 218 z (5 − z)2 − 8 (using equation 1 ) 8 = 218 z (25 − 10 z + z 2) 8 z 2 = 18(25 − 10 z + z 2) 8 z 2 = 450 − 180 z + 18 z 2 ⇒ 10 z 2 − 180 z + 450 = 0 or z2 − 18 z + 45 = 0 ( z − 3)( z − 15) = 0 ⇒ z = 3 or z = 15 b Substitute z = 3 or z = 15 into equation 1 a = 218 3 or 218 15 a = 2 or 2 25 ∴ y = 2( x − 3) 2 − 8 or y = 2 25( x − 15) 2 − 8 11 y = a (x − h ) 2 + k a Range is y ≤ 3 ⇒ a < 0 and k = 3 is the y-coordinate of turning point. h = − 4 b y = a (x + 4) 2 + 3 Using 1 0, 2 3  −   −2 1 3 = a (0 + 4) 2 + 3 7 3 − = 16 a + 3 16 a = 16 3− a = 1 3− ∴ y = 1 3− ( x + 4) 2 + 3 c x-intercepts: y = 0 1 3− ( x + 4) 2 + 3 = 0 1 3− ( x + 4) 2 = − 3 ( x + 4) 2 = 9 x + 4 = − 3 or 3 x = − 7 or −1 12 1 h = − 2, k = − 3 f( x + 2) −3, −4 ≤ x ≤ 0 2 h = 2, k = − 3 f ( x − 2) − 3, 0 ≤ x ≤ 4 3 h = − 4 f ( x + 4), −6 ≤ x ≤ − 2 4 h = 4 f ( x − 4), 2 ≤ x ≤ 6 5 a < 0, k = 6 − f( x ) + 6, −2 ≤ x ≤ 2 6 a < 0, h = − 4, k = 6 − f( x + 4) + 6, −6 ≤ x ≤ − 2 7 a < 0, h = 4, k = 6 − f( x − 4) + 6, 2 ≤ x ≤ 6 8 a < 0, h = − 2, k = 9 − f( x + 2) + 9, −4 ≤ x ≤ 0 9 a < 0, h = 2, k = 9 − f( x − 2) + 9, 0 ≤ x ≤ 4 Exercise 2B — The cubic function in power form 1 a y = 7 x 3 a = 7 The graph is dilated by a factor of 7 in the y direction. b y = 2 3− x3 a = 2 3− The graph is dilated by a factor of 2 3 in the y direction and also a reflection in the x-axis. c y = x 3 + 4 k = 4 The graph is translated 4 units up. d y = 6 − x 3 a = − 1 Reflected in the x-axis. k = 6 Translated 6 units up. e y = ( x − 1) 3 h = 1 Translated 1 unit right. f y = − (x + 3) 3 a = − 1 Reflected in the x-axis. h = − 3 Translated 3 units left. g y = 4(2 − x ) 3 a = 4 Dilated in the y direction by a factor of 4. Reflection in the y-axis. h = 2 Translated 2 units right. h y = − 6(7 − x ) 3 a = − 6 Dilated by a factor of 6 in the y direction. Reflected in the x-axis, reflected in the y-axis h = 7 Translated 7 units right. i y = 3( x + 3) 3 − 2 a = 3, h = − 3, k = − 2 Dilated by a factor of 3 in the y direction, translated 3 units left, translated 2 units down. j y = 6 − 1 2( x − 1) 3 a = 1 2− , h = 1, k = 6 Dilated by a factor of 1 2 in the y direction, reflected in the x-axis, translated 1 unit right, translated 6 units up. k y = 1 4(2 x + 5) 3 = 3 15 2 42 x   +     = 35 2 2 x+   a = 2, h = 5 2− Dilated by a factor of 2 in the y direction, translated 5 2 units left. l y = 3 − 2 3 1 4 2x +   = 3 − 2 () 3 1 8 2 x  +   = 3 − 1 4(8 + x ) 3 a = 1 4 −, h = − 8, k = 3 Dilated by a factor of 1 4 in the y direction, reflected in the x-axis, translated 8 units left, translated 3 units up. 2 a (i), ( iv) b (iii), ( v) c (ii) d (i), ( ii), ( iv) e (ii), ( v) f ( iii), ( iv)

Functions and transformations MM12-2 23 3 a y = 3 4x3 a = 3 4 , h = 0, k = 0 Dilated by a factor of 3 4 in the y direction. a > 0: A positive cubic. Stationary point of inflection is (0, 0) y-intercept: x = 0 y = 3 4(0) 3 = 0 x-intercept: y = 0 3 4 x 3 = 0 x 3 = 0 x = 0 b y = 1 − 2 x 3 a = − 2, h = 0, k = 1 Dilated by a factor of 2 in the y direction, reflected in the x-axis. a < 0: A negative cubic Stationary point of inflection (0, 1) y-intercept: x = 0 y = 1 − 2(0) 3 y = 1 x-intercept: y = 0 1 − 2 x 3 = 0 2 x3 = 1 x 3 = 1 2 x = 31 2  0.8 c y = 2 3x3 − 6 a = 2 3, h = 0, k = − 6 a > 0: A positive cubic Stationary point of inflection is (0, −6) y-intercept: x = 0 y = 2 3(0) 3 − 6 y = − 6 x-intercept: y = 0 2 3x3 − 6 = 0 2 3x3 = 6 x 3 = 9 x = 39 2.08 d y = 2( x − 4) 3 a = 2, h = 4, k = 0 a > 0: A positive cubic Stationary point of inflection is (4, 0) y-intercept: x = 0 y = 2(0 − 4) 3 y = − 128 x-intercept: y = 0 2( x − 4) 3 = 0 ( x − 4)3 = 0 x − 4 = 0 x = 4 e y = 1 2 −( x − 2) 3 a = 1 2 −, h = 2, k = 0 a < 0: A negative cubic. Stationary point of inflection is (2, 0). y-intercept: x = 0 y = 1 2 −(0 − 2) 3 = 1 2 − × − 8 = 4 x-intercept: y = 0 1 2 −( x − 2) 3 = 0 ( x − 2) 3 = 0 x − 2 = 0 x = 2 f y = 4(1 − x ) 3 = − 4( x − 1)3 a = − 4, h = 1, k = 0 a < 0: A negative cubic. Stationary point of inflection is (1, 0) y-intercept: x = 0 y = 4(1 − 0) 3 = 4 × 1 = 4 x-intercept: y = 0 4(1 − x ) 3 = 0 (1 − x )3 = 0 1 − x = 0 x = 1 g y = ( x − 1) 3 + 2 a = 1, h = 1, k = 2 a > 0: A positive cubic. Stationary point of inflection is (1, 2) y-intercept: x = 0 y = (0 − 1) 3 + 2 = − 1 + 2 = 1 x-intercept: y = 0 ( x − 1) 3 + 2 = 0 ( x − 1) 3 = − 2 x − 1 = − 32 x = 1 − 32  − 0.26 h y = 3 − ( x + 2) 3 a = − 1, h = − 2, k = 3 a < 0: A negative cubic. Stationary point of inflection is (− 2, 3) y-intercept: x = 0

MM12-2 24 Functions and transformations y = 3 − (0 + 2) 3 = 3 − 8 = − 5 x-intercept: y = 0 3 − ( x + 2) 3 = 0 ( x + 2)3 = 3 x + 2 = 33 x = − 2 + 33  − 0.56 i y = 2( x + 1) 3 − 6 a = 2, h = − 1, k = − 6 a > 0: A positive cubic Stationary point of inflection is (− 1, 6) y-intercept: x = 0 y = 2(0 + 1) 3 − 6 = 2 − 6 = − 4 x-intercept: y = 0 2( x + 1) 3 − 6 = 0 2( x + 1)3 = 6 ( x + 1)3 = 3 x + 1 = 33 x = − 1 + 33  0.44 Question s 4 to 6 y = 2( mx − 4) 3 − 3 = 2 3 4 mx m   −     − 3 = 2 m 3 3 4 x m  −   − 3 4 h = 4 m, k = − 3 4 ,3 m −    The answer is E. 5 a = 2 m 3 The answer is C. 6 If m > 1, h = 4 m decreases and a = 2 m 3 increases ⇒ Graph is thinner and not shifted as far to the right. The answer is B. 7 a a = 1 2 y = 1 2x3 b a = − 1, h = − 5 y = − (x + 5) 3 c h = 3, k = − 1 y = ( x − 3) 3 − 1 d a = 2, k = 3 y = 2 x 3 + 3 e a = − 1, h = − 1, k = − 1 y = − (x + 1) 3 − 1 8 y = x 3 a dilation by a factor of 2 from the x -axis ⇒ y = 2 x 3 b a reflection in the y-axis ⇒ y = 2( −x) 3 y = − 2x 3 c a translation of 2 in the positive direction parallel to the x-axis ⇒ y = − 2( x − 2) 3 d a translation of 1 in the negative direction parallel to the y-axis ⇒ y = − 2( x − 2) 3 − 1 9 a h = 0, k = 4 y = ax 3 + 4 Using (2, 0): 0 = a × 2 3 + 4 0 = 8 a + 4 8 a = − 4 a = 1 2− ∴ y = 1 2− x3 + 4 b h = 1, k = 2 y = a (x − 1) 3 + 2 Using (0, 0): 0 = a (0 − 1) 3 + 2 0 = − a + 2 a = 2 ∴ y = 2( x − 1) 3 + 2 c h = − 1, k = 1 y = a (x + 1) 3 + 1 Using (0, −2) −2 = a (0 + 1) 3 + 1 −2 = a + 1 a = − 3 ∴ y = − 3( x + 1) 3 + 1 d h = 3, k = 0 y = a (x − 3) 3 Using (0, 9): 9 = a (0 − 3) 3 9 = − 27 a a = 1 3− ∴ y = 1 3 −( x − 3) 3 e h = − 1, k = 1 2 − y = a (x + 1) 3 −1 2 Using 1 ,0 2 −    : 0 = a 3 1 1 2 −  +   − 1 2 = 8 a − 1 2 8 a = 1 2 a = 4 ∴ y = 4( x + 1) 3 − 1 2 10 y = 2( x + 3) 3 + 1 Reflected in x-axis: ⇒ y = − 2( x + 3) 3 − 1 Shifted 3 units to the right: ⇒ y = − 2( x + 3 − 3) 3 − 1 y = − 2x 3 − 1 Shifted 1 unit up: y = − 2x 3 − 1 + 1 y = − 2x3 The answer is E. 11 y = a (x − h ) 3 + k h = − 1, k = − 4 y = a (x + 1) 3 − 4 Using (0, −2) −2 = a (0 + 1) 3 − 4 −2 = a − 4 a = 2 ∴ y = 2( x + 1) 3 − 4 12 a y = a (h − x ) 3 + k h = 2, k = 1 y = a (2 − x ) 3 + 1 Using 1 1, 2    1 2 = a (2 − 1) 3 + 1 1 2 = a + 1 a = − 1 2 ∴ y = 1 2 − (2 − x ) 3 + 1 b y = − 1 2[ − (x − 2)] 3 + 1 y = 1 2 ( x − 2) 3 + 1 ⇒ a = 1 2 Therefore it is a positive cubic. 13 a y = a (x − h ) 3 + k a = 1 y = ( x − h ) 3 + k Using (0, 28): 28 = (0 − h ) 3 + k −h 3 + k = 28 Using ( −4, 0): 0 = ( −4 − h ) 3 + k

Functions and transformations MM12-2 25 0 = − 64 − 48 h − 12 h 2 − h 3 + k 2 From 1 k = 28 + h 3 Substitute into 2 : 0 = − 64 − 48 h − 12 h 2 3h − + 28 3h + 12 h 2 + 48 h + 36 = 0 12( h2 + 4 h + 3) = 0 12( h + 3)( h + 1) = 0 ⇒ h = − 3 or h = − 1 Substitute h into 1 h = − 3, k = 28 + ( −3)3 = 28 − 27 = 1 h = − 1, k = 28 + ( −1) 3 = 28 − 1 = 27 Therefore the stationary points are ( −3, 1) or ( −1, 27). b For y = ( x + 3) 3 + 1 y-intercept: x = 0 y = (0 + 3) 3 + 1 = 27 + 1 = 28 x-intercept: y = 0 ( x + 3) 3 + 1 = 0 ( x + 3) 3 = − 1 x + 3 = − 1 x = − 4 For y = ( x + 1) 3 + 27 y-intercept: x = 0 y = (0 + 1) 3 + 27 = 1 + 27 y = 28 x-intercept: y = 0 ( x + 1) 3 + 27 = 0 ( x + 1) 3 = − 27 x + 1 = − 3 x = − 4 Exercise 2C — The power function (the hyperbola) 1 y = a x h − + k a a = 2, h = 0, k = 0 Dilation in the y direction by a factor of 2. b a = − 3 Dilation in the y direction by a factor of 3, reflection in the x -axis. c a = 1, h = 6, Translation 6 units right. d a = 2, h = − 4 Dilation in the y direction by a factor of 2, translation 4 units left. e a = 1, k = 7 Translation 7 units up. f a = 2, k = − 5 Dilation by a factor of 2 in the y direction, translation 5 units down. g a = 1, h = − 4, k = − 3 Translation 4 units left, translation 3 units down. h a = 2, h = 3, k = 6 Dilation by a factor of 2 in the y direction, translation 3 units right, translation 6 units up. i a = − 4, h = 1, k = − 4 Dilation in the y direction by a factor of 4, reflection in the x -axis, translation 1 unit right, translation 4 units down. 2 a (v) b (iii) c (i) d (v), ( iii) e (v), ( ii), ( iii) f ( i), ( iii) g (v), ( i), ( iv) h (ii), ( iv) 3 y = a x h − + k a i h = 0, k = 0 Horizontal asymptote: y = 0 Vertical asymptote: x = 0 ii Domain: R\{0} iii Range: R\{0} b i h = − 6, k = 0 Horizontal asymptote: y = 0 Vertical asymptote: x = − 6 ii Domain: R\{ −6} iii Range: R\{0} c i h = 2, k = 0 x = 2 Vertical asymptote y = 0 Horizontal asymptote ii Domain: R\{2} iii Range: R\{0} d i h = 3, k = 0 x = 3 Vertical asymptote y = 0 Horizontal asymptote ii Domain: R\{3} iii Range: R\{0} e i h = 0, k = 4 x = 0 Vertical asymptote y = 4 Horizontal asymptote ii Domain: R\{0} iii Range: R\{4} f i h = 0, k = − 5 x = 0 Vertical asymptote y = − 5 Horizontal asymptote ii Domain: R\{0} iii Range: R\{ −5} g i h = − 6, k = − 2 x = − 6 Vertical asymptote y = − 2 Horizontal asymptote ii Domain: R\{ −6} iii Range: R\{ −2} h i h = 2, k = 1 x = 2 Vertical asymptote y = 1 Horizontal asymptote ii Domain: R\{2} iii Range: R\{1} i i h = − n, k = − m x = − n Vertical asymptote y = − m Horizontal asymptote ii Domain: R\{ −n} iii Range: R\{ −m } 4 a i h = 4, k = 0 x = 4 y = 0 ii Domain: R\{4} iii Range: R\{0} b i h = 0, k = 2 x = 0 y = 2 ii Domain: R\{0} iii Range: R\{2} c i h = 3, k = 2 x = 3 y = 2 ii Domain: R\{3} iii Range: R\{2} d i h = − 1, k = − 1 x = − 1 y = − 1 ii Domain: R\{ −1} iii Range: R\{ −1} e i h = m , k = n x = m y = n ii Domain: R\{ m} iii Range: R\{ n} f i h = b , k = a x = b y = a ii Domain: R\{ b} iii Range: R\{ a} 5 y = 2 3 x is dilated by a factor 2 3 in the y direction. y = 3 x is dilated by a factor 3 in the y direction. y = 4 3 x− is dilated by a factor 4 3 in the y direction and reflected through the x-axis. 6 y = a x h − + k a a = 1, h = − 3, k = 0 Asymptotes: x = − 3 y = 0

MM12-2 26 Functions and transformations y-intercept: x = 0 y = 1 03 + y = 1 3 x-intercept: y = 0 1 3 x + = 0 No solution. ⇒ No x-intercept. b a = 1, h = − 2, k = − 1 Asymptotes: x = − 2 y = − 1 y-intercept: x = 0 y = 1 02 + − 1 = − 1 2 x-intercept: y = 0 1 2 x + − 1 = 0 1 2 x + = 1 x + 2 = 1 x = − 1 c a = 3, h = 1, k = − 3 4 Asymptotes: x = 1 y = − 3 4 y-intercept: x = 0 y = 3 01 − − 3 4 = − 3 − 3 4 = − 3 3 4 x-intercept: y = 0 3 1 x − − 3 4 = 0 3 1 x − = 3 4 3( x − 1) = 12 3 x − 3 = 12 3 x = 15 x = 5 d a = − 2, h = − 5, k = 0 Asymptotes: x = − 5 y = 0 y-intercept: x = 0 y = − 2 05 + = − 2 5 x-intercept: y = 0 − 2 5 x + = 0 No solution ⇒ No x-intercept. e a = − 6, h = 1, k = − 3 Asymptotes: x = 1 y = − 3 y-intercept: x = 0 y = 6 10 − − 3 = 6 − 3 = 3 x-intercept: y = 0 6 1 x− − 3 = 0 6 1 x− = 3 6 = 3(1 − x ) 6 = 3 − 3 x −3x = 3 x = − 1 f a = − 3, h = 2, k = 6 Asymptotes: x = 2 y = 6 y-intercept: x = 0 y = 3 02 − − + 6 = 3 2 + 6 = 7 1 2 x-intercept: y = 0 3 2 x − − + 6 = 0 3 2 x − = 6 3 = 6( x − 2) = 6 x − 12 6 x = 15 x = 2 1 2 g a = + 1, h = 2, k = 1 Asymptotes: x = 2 y = 1 y-intercept: x = 0 y = 1 − 1 20 − = 1 − 1 2 = 1 2 x-intercept: y = 0 1 − 1 2 x− = 0 1 2 x− = 1 1 = 2 − x

Functions and transformations MM12-2 27 −x = − 1 x = 1 h a = 4, h = − 1, k = 2 5 Asymptotes: x = − 1 y = 2 5 y-intercept: x = 0 y = 2 5 + 4 10 + = 2 5 + 4 = 4 2 5 x-intercept: y = 0 2 5 + 4 1 x+ = 0 4 1 x+ = 2 5− 2(1 + x ) = − 20 2 + 2 x = − 20 2 x = − 22 x = − 11 i y = 1 23 x + + 4 = 1 3 2 2 x  +   + 4 a = 1 2, h = − 3 2, y = 4 Asymptotes: x = − 3 2 y = 4 y-intercept: x = 0 y = 1 20 3 ×+ + 4 = 1 3 + 4 = 4 1 3 x-intercept: y = 0 1 23 x + + 4 = 0 1 23 x + = − 4 −4(2 x + 3) = 1 −8x − 12 = 1 −8x = 13 x = − 13 8 j y = 2 34 x − − 1 = 2 3 4 4 x  −−   − 1 a = − 1 2, h = 3 4, k = − 1 Asymptotes: x = 3 4 y = − 1 y-intercept: x = 0 y = 2 340 −× − 1 = 2 3 − 1 = − 1 3 x-intercept: y = 0 2 34 x − − 1 = 0 2 34 x − = 1 2 = 3 − 4 x −4x = − 1 x = 1 4 k y = 3 2 x x + − = (2)5 2 x x −+ − = 25 22 x xx − + −− = 5 1 2 x + − x-asymptote: x = 2 y-asymptote: y = 1 x-intercept, y = 0 0 = 3 2 x x + − 0 = x + 3 x = − 3 y-intercept, x = 0 y = 3 2 − l y = 43 1 x x + − = 4( 1) 7 1 x x−+ − = 4( 1) 7 11 x xx − + −− = 7 4 1 x + − x -asymptote: x = 1 y-asymptote: y = 4 y-intercept, x = 0 y = 3 1 − = − 3 x-intercept, y = 0 0 = 43 1 x x + − 0 = 4 x + 3 −3 = 4 x x = 3 4− 7 a < 0, h = 4, k = 3 y = 1 4 x − − + 3 The answer is E. 8 Asymptotes: x = − 1 (vertical) y = − 3 (horizontal) Domain: R\{ −1} Range: R\{ −3} The answer is C.

MM12-2 28 Functions and transformations 9 a h = 2, k = 0 y = 2 a x − Using (0, −1) −1 = 02 a − = 2 a − a = 2 ∴ y = 2 2 x − b h = 0, k = 1 y = a x + 1 Using (3, 0): 0 = 3 a + 1 3 a = − 1 a = − 3 ∴ y = 3 x − + 1 c h = − 4, k = 0 y = 4 a x + Using 3 0, 4 −    3 4 − = 04 a + = 4 a −3 = a ∴ y = 3 4 x − + d h = 0, k = − 1 y = a x − 1 Using ( −4, 0) 0 = 4 a − − 1 4 a − = 1 a = − 4 ∴ y = 4 x − − 1 e h = 4, k = 2 y = 4 a x − + 2 Using 1 0, 1 2    1 1 2 = 04 a − + 2 − 1 2 = 4 a − a = 2 ∴ y = 2 4 x − + 2 f h = − 1, k = − 1 y = 1 a x + − 1 Using (0, 5) 5 = 01 a + − 1 = a − 1 a = 6 ∴ y = 6 1 x + − 1 10 f( x ) = 1 x Asymptotes: x = 0 y = 0 a f( x + 2) = 1 2 x + Asymptotes: x = − 2 y = 0 y-intercept: x = 0 y = 1 02 + = 1 2 x-intercept: y = 0 0 = 1 2 x + ⇒ No x-intercept. b f( x ) −1 = 1 x − 1 Asymptotes: x = 0 y = − 1 x-intercept: y = 0 0 = 1 x − 1 1 = 1 x x = 1 y-intercept: x = 0 y = 1 0 − 1 ⇒ No y-intercept. c −f( x ) − 2 = − 1 x − 2 Asymptotes: x = 0 y = − 2 y-intercept: x = 0 y = − 1 0 − 2 ⇒ No y-intercept. x-intercept: y = 0 1 2 x − − = 0 1 x − = 2 2 x = − 1 x = 1 2 − d f(1 − x ) + 2 = 1 1 x− + 2 Asymptotes: x = 1 y = 2 y-intercept: x = 0 y = 1 10 − + 2 = 1 + 2 = 3 x-intercept: y = 0 0 = 1 1 x− + 2 −2 = 1 1 x− −2(1 − x ) = 1 −2 + 2 x = 1 2 x = 3 x = 3 2 e −f( x − 1) − 1 = − 1 1 x − − 1 Asymptotes: x = 1 y = − 1

Functions and transformations MM12-2 29 y-intercept: x = 0 y = − 1 01 − − 1 = 1 − 1 y = 0 So the x-intercept is at the origin. f 1 − f ( x − 2) = 1 − 1 2 x − Asymptotes: x = 2 y = 1 y-intercept: x = 0 y = 1 − 1 02 − = 1 − − 1 2 = 1 1 2 x-intercept: y = 0 0 = 1 − 1 2 x − 1 = 1 2 x − x − 2 = 1 x = 3 11 yx − 3 x + 1 = 0 ⇒ y − 3 + 1 x = 0 y = 3 − 1 x Asymptotes: x = 0 y = 3 y-intercept: x = 0 y = 3 − 1 0 ⇒ No y-intercept x-intercept: y = 0 0 = 3 − 1 x 1 x = 3 3 x = 1 x = 1 3 Domain: R\{0} Range: R\{3} Exercise 2D — The power function (the truncus) 1 y = () 2 a x h − + k a a = 2, h = 0, k = 0 Dilation in the y direction by a factor of 2. b a = − 3, h = 0, k = 0 Dilation in the y direction by a factor of 3, reflection in the x -axis. c a = 1, h = − 2, k = 0 Translation 2 units left. d a = 2, h = 3 Dilation by a factor of 2 in the y direction, translation 3 units right. e a = − 5, h = − 4 Dilation in the y direction by a factor of 5, reflection in the x -axis, translation 4 units left. f a = 2, k = 6 Dilation in the y direction by a factor of 2, translation 6 units up. g a = − 1, k = 3 Reflection in the x-axis, translation 3 units up. h a = 4, h = 3, k = 1 Dilation in the y direction by a factor of 4, translation 3 units right, translation 1 unit up. i a = − 1, h = − 2, k = 5 Reflection in the x-axis, translation 2 units left, translation 5 units up. 2 a < 0 ⇒ Reflected in the x-axis h = 2 ⇒ Translated 2 units right. The answer is D. 3 Translated m units left, translated n units up The answer is C. 4 y = 2 () ax h − + k a i h = 0, k = 0 Asymptotes: x = 0 y = 0 ii Domain: R\{0} iii a = 2( > 0) Range: y > 0 b i h = 0, k = 0 Asymptotes: x = 0 y = 0 ii Domain: R\{0} iii a = 4 3 − Range: y < 0 c i h = 2, k = 0 Asymptotes: x = 2 y = 0 ii Domain: R\{2} iii a = 1 Range: y > 0 d i h = − 1, k = 0 Asymptotes: x = − 1 y = 0 ii Domain: R\{ −1} iii a = 2 Range: y > 0 e i h = − 4, k = 0 Asymptotes: x = − 4 y = 0 ii Domain: R\{ −4} iii a = − 5 Range: y < 0 f i h = 0, k = − 3 Asymptotes: x = 0 y = − 3 ii Domain: R\{0} iii a = 2 Range: y > − 3 g i h = 0, k = 4 5 Asymptotes: x = 0 y = 4 5 ii Domain: R\{0} iii a = 1 Range: y > 4 5 h i h = 0, k = 1 2 Asymptotes: x = 0 y = 1 2 ii Domain: R\{0} iii a = − 3 Range: y < 1 2 i i h = 1, k = 4 Asymptotes: x = 1 y = 4 ii Domain: R\{1} iii a = 2 Range: y > 4 5 Vertical asymptote at x = − 3 in graphs ( ii) & ( iii) The answer is C. 6 Horizontal asymptote y = − 3 and a < 0 in graphs ( iii), ( v) & ( vi) The answer is B. 7 Asymptotes y = 0 and x = − 3 in graph ( ii) The answer is B.

MM12-2 30 Functions and transformations 8 y = 2 () ax h − + k a a = − 2 5, h = 0, k = 0 Reflected in the x-axis Asymptotes: x = 0 y = 0 x-intercept: y = 0 − 22 (5 ) x = 0 No solution. ⇒ No x-intercept. y-intercept: x = 0 y = − 22 50 × Undefined ⇒ No y-intercept. b a = 1, h = 3, k = 0 Asymptotes: x = 3 y = 0 ⇒ No x-intercepts. y-intercept: x = 0 y = 2 (0 3) 1 − = 1 9 c a = 2, h = − 4, k = 0 Asymptotes: x = − 4 y = 0 ⇒ No x-intercepts. y-intercept: x = 0 y = 2 2 (4 0) + = 2 16 = 1 8 d a = − 1, h = 1, k = 0 Reflected in the x-axis Asymptotes: x = 1 y = 0 ⇒ No x intercepts. y-intercept: x = 0 y = − 2 1 (0 1) − = − 1 e a = − 4, h = 0, k = 1 Reflected in the x-axis Asymptotes: x = 0 y = 1 ⇒ No y-intercept x-intercept: y = 0 1 − 24 x = 0 24 x = 1 x 2 = 4 x = ± 2 f a = 1 2 , h = 0, k = − 3 Asymptotes: x = 0 y = − 3 ⇒ No y-intercept. x-intercepts: y = 0 21 2 x − 3 = 0 21 2 x = 3 6 x 2 = 1 x 2 = 1 6 x = ± 1 6 ≈ ± 0.4 g a = 2, h = 1, k = − 2 Asymptotes: x = 1 y = − 2 x-intercept: y = 0 2 2 (1) x − − 2 = 0 2 2 (1) x − = 2 2( x − 1) 2 = 2 ( x − 1)2 = 1 x − 1 = 1 or −1 x = 2 or 0 y-intercept: x = 0 y = 2 2 (0 1) − − 2 = 2 − 2 = 0 h a = − 2, h = − 3, k = 4 Reflected in the x-axis. Asymptotes: x = − 3 y = 4 x-intercepts: y = 0 4 − 2 2 (3 ) x+ = 0 2 2 (3 ) x+ = 4 4(3 + x ) 2 = 2 (3 + x ) 2 = 1 2 3 + x = ± 1 2 x = − 3 + 1 2 or

Functions and transformations MM12-2 31 −3 − 1 2 x ≈ − 2.3 or −3.7 y-intercept: x = 0 y = 4 − 2 2 (3 0) + = 4 − 2 9 = 3 7 9 i a = 1, h = 2, k = 2 3 Asymptotes: x = 2 y = 2 3 x-intercept: y = 0 2 3 + 2 1 (2) x − = 0 2 1 (2) x − = − 2 3 ( x − 2) 2 = − 3 2 No solution ⇒ No x-intercepts. y-intercept: x = 0 y = 2 3 + 2 1 (0 2) − = 2 3 + 1 4 = 11 12 j a = 3 4 , h = − 1, k = − 1 4 Asymptotes: x = − 1 y = 1 4− x-intercepts: y = 0 2 3 4( 1) x+ − 1 4 = 0 2 3 4( 1) x+ = 1 4 2 3 (1) x + = 1 ( x + 1) 2 = 3 x + 1 = ± 3 x = − 1 + 3, −1 − 3 ≈ 0.7, −2.7 y-intercept: x = 0 y = 2 3 4(0 1) + − 1 4 = 3 4 − 1 4 = 1 2 k a = 1, h = 2, k = 3 Asymptotes: x = 2 y = 3 x-intercepts: 2 1 (2 ) x− + 3 = 0 2 1 (2 ) x− = − 3 (2 − x ) 2 = − 1 3 No solution ⇒ No x-intercepts. y-intercept: x = 0 y = 2 1 (2 0) − + 3 = 1 4 + 3 = 3 1 4 l y = 2 4 [2( 2)] x− − 1 = 2 4 4( 2) x− − 1 = 2 1 (2) x − − 1 a = 1, h = 2, k = − 1 Asymptotes: x = 2 y = − 1 x-intercepts: y = 0 2 1 (2) x − − 1 = 0 2 1 (2) x − = 1 ( x − 2) 2 = 1 x − 2 = ± 1 x − 2 = 1 or −1 x = 3 or 1 y-intercept: x = 0 y = 2 4 [2( 2)] − − 1 = 4 16 − 1 = − 3 4 9 y = 2 () ax h − + k where a = 1 or −1. a a = 1, h = m , k = n ⇒ y = 2 1 () x m − + n b a = − 1, h = − q, k = p ⇒ y = 2 1 () x q − + + p c a = − 1, h = r , k = 0 ⇒ y = 21 () x r − − d a = 1, h = 0, k = t ⇒ y = 21 x + t e a = 1, h = − a, k = − b ⇒ y = 2 1 () x a + − b f a = 1, h = g , k = − e ⇒ y = 2 1 () x g − − e g a = 1, h = 0, k = − k ⇒ y = 21 x − k

MM12-2 32 Functions and transformations h a = − 1, h = 0, k = − c ⇒ y = 21 x − − c 10 y = 2 () ax h − + k a h = 0, k = − 2 y = 2a x − 2 Using (1, 0): 0 = 21 a − 2 2 = a ⇒ y = 22 x − 2 b h = 2, k = 0 y = 2 (2) a x − Using 3 0, 4  −   : − 3 4 = 2 (0 2) a − − 3 4 = 4 a a = − 3 ⇒ y = 2 3 (2) x − − c h = − 2, k = 1 y = 2 (2) a x + + 1 Using ( −5, 0): 0 = 2 (5 2) a −+ + 1 0 = 9 a + 1 9 a = − 1 a = − 9 ⇒ y = 2 9 (2) x − + + 1 d h = − 1, k = 4 y = 2 (1) a x + + 4 Using (0, 7): 7 = 2 (0 1) a + + 4 7 = a + 4 a = 3 ⇒ y = 2 3 (1) x + + 4 e h = 4, k = − 3 y = 2 (4) a x − − 3 Using (0, −2.5): −2.5 = 2 (0 4) a − − 3 0.5 = 16a a = 8 ⇒ y = 2 8 (4) x − − 3 f h = 1, k = − 2 y = 2 (1) a x − − 2 Using (0, −7): −7 = 2 (0 1) a − − 2 −5 = 1 a a = − 5 ⇒ y = 25 (1) x − − − 2 11 h = − 2, k = − 3, a > 0 y = 2 (2) a x + − 3 Using ( −1, 0): 0 = 2 (1 2) a −+ − 3 3 = 1 a a = 3 ⇒ y = 2 3 (2) x + − 3 12 h = 1, k = 2, a > 0 y = 2 (1) a x − + 2 Using (0, 5): 5 = 2 (0 1) a − + 2 3 = 1 a a = 3 ⇒ y = 2 3 (1) x − + 2 Exercise 2E — The square root function in power form 1 y = a x h − + k a a = 2, h = 0, k = 0 Dilated by a factor of 2 in the y direction. b a = −1 3, h = 0, k = 0 Dilated by a factor of 1 3 in the y direction, reflected in the x-axis. c a = 3, h = 1, k = 0 Dilated in the y direction by a factor of 3, translated 1 unit right. d a = −2, h = −4, k = 0 Dilated in the y direction by a factor of 2, reflected in the x-axis, translated 4 units left. e a = 1, h = 0, k = −1 Translated 1 unit down. f a = −3, h = 0, k = 2 Dilated in the y-direction by a factor of 3, reflected in the x-axis, translated 2 units up. g a = 1, h = 4, k = 3 Translated 4 units right, translated 3 units up. h a = −2, h = −3, k = 6 Dilated in the y direction by a factor of 2, reflected in the x-axis, translated 3 units left, translated 6 units up. i a = −1 , 2 h = 2, k = 2 3 Dilated by a factor of 1 2 in the y direction, reflected in the x-axis, reflected in the y-axis, translated 2 units right, translated 2 3 units up. 2 (h, k) are the coordinates of the end point. a (0, 0) b (0, 0) c (1, 0) d (−4, 0) e (0, −1) f (0, 2) g (4, 3) h (−3, 6) i 2 2, 3    3 The graph is translated 3 units up. The answer is E. 4 The graph is reflected in the y-axis, translated 2 units right, translated 4 units up. The answer is D. 5 y = a x h − + k a y = 1 x + h = −1, k = 0 Domain: x ≥ −1 Range: y ≥ 0 b y = 3 x − h = 3, k = 0 Domain: x ≥ 3 Range: y ≥ 0 c y = x − 3 h = 0, k = − 3 Domain: x ≥ 0 Range: y ≥ − 3 d y = 4 + 2 x h = 0, k = 4 Domain: x ≥ 0 Range: y ≥ 4 e y = 5 − x a < 0, h = 0, k = 5 Domain: x ≥ 0 Range: y ≤ 5 f y = 1 x − + 3 h = 1, k = 3 Domain: x ≥ 1 Range: y ≥ 3 g y = 2 x+ − 1 h = − 2, k = − 1 Domain: x ≥ − 2 Range: y ≥ − 1

Functions and transformations MM12-2 33 h y = 4 − 2 21 x + 2 x + 1 ≥ 0 2 x ≥ − 1 x ≥ − 1 2 Domain: x ≥ − 1 2 k = 4, a < 0 Range: y ≤ 4 i y = − 3 34 5 x− + 2 3 x − 4 ≥ 0 3 x ≥ 4 x ≥ 4 3 Domain: x ≥ 4 3 k = 2, a < 0 Range: y ≤ 2 j y = 3 x− − 7 h = 3, k = − 7 Domain: x ≤ 3 Range: y ≥ − 7 k y = 6 + 42 x − 4 − 2 x ≥ 0 −2x ≥ − 4 x ≤ 2 Domain: x ≤ 2 k = 6, Range: y ≥ 6 l y = 1 − 2 x− a < 0, h = 2, k = 1 Domain: x ≤ 2 Range: y ≤ 1 6 Since x ≤ − 2 and y ≤ 2, graph is of the form y = a hx − + k The answer is D. 7 Domain: ( −∞, − 2] Range: ( −∞, 2] The answer is D. 8 a y = a x h − + k a = 1, h = − 2, k = 0 End point: ( −2, 0) x-intercept: y = 0 0 = 2 x + x = − 2 y-intercept: x = 0 y = 02 + = 2 b a = 1 3, h = 0, k = 3 End point: (0, 3) x-intercept: None since y ≥ 3 y-intercept: x = 0 y = 1 0 3 + 3 y = 3 c a = − 1, h = 0, k = 2 End point: (0, 2) x-intercept: y = 0 2 − x = 0 x = 2 x = 4 y-intercept: x = 0 y = 2 d a = 1, h = 6, k = 1 End point: (6, 1) x-intercept: None since y ≥ 1 y-intercept: None since x ≥ 6 e a = 1, h = − 3, k = 2 End point: ( −3, 2) x-intercept: None since y ≥ 2 y-intercept: x = 0 y = 30 + + 2 = 3 + 2 ≈ 3.7 f a = − 1, h = − 4, k = 1 2 End point: 1 4, 2 −  x-intercept: y = 0 1 2 − 4 x+ = 0 4 x+ = 1 2 4 + x = 1 4 x = − 3 3 4 y-intercept: x = 0 y = 1 2 − 40 + = 1 2 − 2 = − 1 1 2 g y = 23 x − = 3 2 2 x  −   = 3 2 2 x  −   a = 2, h = 3 2, k = 0 End point: 3 ,0 2   x-intercept: y = 0 23 x − = 0 2 x − 3 = 0 2 x = 3

MM12-2 34 Functions and transformations x = 3 2 y-intercept: None since x ≥ 3 2 h y = 63 x + + 2 = 3( 2 ) x+ + 2 = 32 x+ + 2 a = 3, h = − 2, k = 2 End point: ( −2, 2) x-intercept: None since y ≥ 2 y-intercept: x = 0 y = 630 +× + 2 = 6 + 2 ≈ 4.4 i h = 2, k = − 1 End point: (2, −1) x-intercept: y = 0 2 x− − 1 = 0 2 x− = 1 2 − x = 1 x = 1 y-intercept: x = 0 y = 20 − − 1 = 2 − 1 ≈ 0.4 9 y = a x h − + k h = 1, k = 2 y = a 1 x − + 2 Using (2, 0) 0 = a 21 − + 2 0 = a + 2 a = − 2 ⇒ y = − 2 1 x − + 2 The answer is E. 10 a y = a x h − + k a = 2, h = m , k = − 4 y = 2 x m − − 4 Using (5, 0) 0 = 2 5 m − − 4 2 5 m − = + 4 5 m − = + 2 5 − m = 4 m = 1 b y = 2 1 x − − 4 11 Since x ≤ 4 ⇒ y = a hx − + k h = 4, k = 3 y = a 4 x− + 3 Using (0, 9) 9 = a 40 − + 3 6 = a 4 = 2 a a = 3 ⇒ y = 3 4 x− + 3 12 a y = a x h − + k a = − 4, h = − 1, k = p y = − 4 1 x + + p Using (0, 4): 4 = − 4 01 + + p 4 = − 4 + p 8 = p ⇒ p = 8 b y = − 4 1 x + + 8 c x-intercept: y = 0 −4 1 x + + 8 = 0 4 1 x + = 8 1 x + = 2 x + 1 = 4 x = 3 d Domain: x ≥ − 1 e Range: y ≤ 8 f Exercise 2F — The absolute value function 1 a Let y = 2 x Let x = 0: y = 2 × 0 = 0 (0, 0) Let x = 1: y = 2 × 1 = 2 (1, 2) b Let y = x − 1 x-intercept: y = 0 x − 1 = 0 x = 1 y-intercept: x = 0 y = 0 − 1 = − 1 c Let y = 3 − 6 x x-intercept: y = 0 3 − 6 x = 0 6 x = 3 x = 1 2 y-intercept: x = 0 y = 3 − 6 × 0 = 3

Functions and transformations MM12-2 35 d Let y = x 2 − 6 Shape: Positive parabola translated 6 units down. Turning point: (0, −6) x-intercept: y = 0 x 2 − 6 = 0 x 2 = 6 x = ± 6 e Let y = 4 − x 2 Shape: Negative parabola translated 4 units up. Turning point: (0, 4) x-intercept: y = 0 4 − x 2 = 0 x2 = 4 x = ± 4 x = 2 or −2 f Let y = ( x − 3) 2 − 4 Shape: Positive parabola translated 3 units right and 4 units down. Turning point: (3, −4) x-intercept: y = 0 ( x − 3) 2 − 4 = 0 ( x − 3) 2 = 4 x − 3 = ± 4 = 2 or −2 x = 5 or 1 y-intercept: x = 0 y = (0 − 3) 2 − 4 = 9 − 4 = 5 g Let y = 3 x 3 Shape: Positive cubic dilated by a factor of 3 in y direction. Stationary point of inflection: (0, 0) h Let y = ( x + 2) 3 − 1 Shape: Positive cubic translated 2 units left and 1 unit down. Stationary point of inflection: (− 2, −1) x-intercept: y = 0 ( x + 2) 3 − 1 = 0 ( x + 2) 3 = 1 x + 2 = 1 x = − 1 y-intercept: x = 0 y = (0 + 2) 3 − 1 = 8 − 1 = 7 i Let y = 2 1 x − Asymptotes: x = 1 y = 0 y-intercept: x = 0 y = 2 01 − = − 2 2 For x ≥ − 1: Positive cubic with stationary point of inflection (0, 1) ⇒ y = x 3 + 1 So graph is y = |x 3 + 1| The answer is C. 3 a Let y = 2 x Domain: R Range: R y = 2| x| ⇒ Domain: R Range: y ≥ 0 b Let y = x + 1 Domain: R Range: R y = |x| + 1 Domain: R Range: y ≥ 1 c Let y = 4 − 3 x Domain: R Range: R y = 4 − 3| x| Domain: R Range: y ≤ 4 d Let y = |x 2 − 3| Domain: R Range: y ≥ 0 y = |x 2 − 3| − 2 (shifted down 2 units) Domain: R Range: y ≥ − 2 e Let y = 1 1 x+ Domain: R\{ −1} Range: y > 0 y = 1 1 x+ + 1 (shifted up 1 unit) Domain: R\{ −1} Range: y > 1 f Let y = 2 − 21 x Asymptotes: x = 0 y = 2 Domain: R\{0} Range: y ≤ 2 y = 21 2 x− Domain: R\{0} Range: y ≥ 0 4 a Let y = |x| Dilated in the y direction by a factor of 2 and reflected through x -axis.

MM12-2 36 Functions and transformations b The graph of y = |x + 5| is shifted down 6 units. Let y = x + 5 y-intercept: x = 0 y = 0 + 5 y = 5 x-intercept: y = 0 x + 5 = 0 x = − 5 c The graph of y = |3 − x | is dilated in the y direction by a factor of 4 and shifted 1 unit up. Let y = 3 − x x-intercept: y = 0 3 − x = 0 x = 3 y-intercept: x = 0 y = 3 − 0 = 3 d The graph of y = |x 2 − 1| is shifted up 1 unit. Let y = x 2 − 1 Turning point: (0, −1) x-intercept: x 2 − 1 = 0 x2 = 1 x = 1 or −1 e The graph of y = |x 2 − 2| is reflected through the x-axis and translated 2 units up Let y = x 2 − 2 Turning point: (0, −2) x-intercept: y = 0 x 2 − 2 = 0 x 2 = 2 x = ± 2 f The graph of y = |( x + 1) 2 − 1| is shifted down 2 units Let y = ( x + 1) 2 − 1 Turning point: ( −1, −1) x-intercept: y = 0 ( x + 1) 2 − 1 = 0 ( x + 1) 2 = 1 x + 1 = 1 or −1 x = 0 or −2 y-intercept: x = 0 y = (0 + 1) 2 − 1 = 0 g The graph of y = 1 x is translated down 3 4 units Let y = 1 x Asymptotes: x = 0 y = 0 h The graph of y = 2 6 x− is shifted up 3 units Let y = 2 6 x− Asymptotes: x = 6 y = 0 y-intercept: x = 0 y = 2 60 − = 1 3 i The graph of y = 21 x is dilated in the y direction by a factor of 1 4 and shifted down 4 units. Let y = 21 x Asymptotes: x = 0 y = 0 j The graph of y = 21 1 x − is reflected through the x-axis. Let y = 21 x − 1 Asymptotes: x = 0

Functions and transformations MM12-2 37 y = − 1 x-intercept: y = 0 21 x − 1 = 0 21 x = 1 x 2 = 1 x = ± 1 k The graph of y = 22 x −− is shifted up 3 units Let y = 2 x− − 2 End point: (2, −2) x-intercept: y = 0 2 x−− 2 = 0 2 x− = 2 2 − x = 4 x = − 2 y-intercept: x = 0 y = 20 − − 2 = 2 − 2 ≈ − 0.6 l The graph of y = 18 x +− is reflected through the x-axis and shifted 2 units up. Let y = 1 x + − 8 End point: ( −1, −8) x-intercept: y = 0 1 x +− 8 = 0 1 x + = 8 x + 1 = 64 x = 63 y-intercept: x = 0 y = 01 + − 8 = 1 − 8 = − 7 5 f( x ) = 31x − a f( x ) = 31 where310 (3 1) where 3 1 0 xx xx −−≥  −− −

MM12-2 38 Functions and transformations Yellow: Reflection through x-axis and shifted 6 units up. ⇒ y = 6 − 3 2x, −2 ≤ x ≤ 2. Exercise 2G — Transformations with matrices 1 i reflection in the y-axis, dilation by factor 2 from the x-axis. ii dilation factor 1 2 from the y-axis, reflection in the x-axis, dilation factor 4 from the x-axis. iii reflection in the y-axis, dilation by factor 2 from the y-axis, dilation by factor 3 from the x -axis. iv reflection in the x-axis, dilation by factor 1 2 from the x-axis. 2 ( −3, 5) i x y\b   \b = 10 3 02 5 −−       = 3 10   ⇒ (3, 10) ii x y \b   \b = 1 3 0 2 5 04  −      −   = 3 2 20  −   −   ⇒ 3 ,20 2  −−   iii x y\b   \b = 20 3 03 5 −−       = 6 15   ⇒ (6, 15 ) iv x y\b   \b = 10 3 1 5 0 2  −     −    = 3 52 −    −   ⇒ 5 3, 2  −−   3 a y = 21 x i reflection in the y-axis ⇒ y = 21 () x− = 21 x dilation by factor 2 from the x-axis ⇒ y = 22 x ii dilation factor 1 2 from the y-axis ⇒ y = 21 (2 ) x y = 21 4 x reflection in the x-axis ⇒ y = 21 4 x − dilation factor 4 from the x-axis ⇒ y = 24 4 x − = 21 x− b y = x3 – 5 i reflection in the y-axis ⇒ y = ( − x)3 – 5 = − x3 − 5 dilation by factor 2 from the x-axis ⇒ y = 2 [ − x3 − 5] y = − 2 x3 – 10 ii dilation factor 1 2 from the y-axis ⇒ y = (2 x)3 – 5 = 8 x3 – 5 reflection in the x-axis ⇒ y = − [8 x3 − 5] = − 8 x3 + 5 dilation factor 4 from the x-axis ⇒ y = 4[ −8 x3 + 5] = − 32 x3 + 20 c y = x i reflection in the y-axis ⇒ y = x− dilation by factor 2 from the x-axis ⇒ y = 2 x− ii dilation factor 1 2 from the y-axis ⇒ y = 2x reflection in the x-axis ⇒ y = 2x − dilation factor 4 from the x-axis ⇒ y = 42 x − 4 i 3 2 ⇒ translation of 3 units in the positive x-direction, translation of 2 units in the positive y-direction. ii 2 2   −  ⇒ translation of 2 units in the positive x-direction, translation of 2 units in the negative y-direction iii 1 5 0   −         ⇒ translation of 1 5 units in the negative x-direction. 5 i x y \b  \b = 13 22 + − = 4 0   ⇒ (4, 0) ii x y \b  \b = 12 22 + −− = 3 4  − ⇒ (3, −4) iii x y \b  \b = 1 1 5 2 0  −   +   −    = 4 5 2    −   ⇒ 4 ,2 5  −   6 a y = | x| i translation of 3 units in the positive x-direction ⇒ y = | x − 3| translation of 2 units in the positive y-direction ⇒ y = | x − 3| + 2 ii translation of 2 units in the positive x-direction ⇒ y = | x − 2| translation of 2 units in the negative y-direction ⇒ y = | x − 2| − 2 b y = x2 – 3 x i translation of 3 units in the positive x-direction ⇒ y = ( x − 3) 2 – 3( x − 3) = x2 – 6 x + 9 – 3 x + 9 = x2 – 9 x + 18 translation of 2 units in the positive y-direction ⇒ y = x2 – 9 x + 18 +2 = x2 − 9 x + 20 ii translation of 2 units in the positive x-direction ⇒ y = ( x − 2) 2 – 3( x − 2) = x2 – 4 x + 4 – 3 x + 6 = x2 – 7 x + 10 translation of 2 units in the negative y-direction ⇒ y = x2 – 7 x + 10 − 2 = x2 – 7 x + 8 7 y = x3 ⇒ y = (3 x − 6) 3 + 1 = (3( x − 2)) 3 + 1 ⇒ dilation factor 1 3 from the y-axis, then translation 2 units right, then translation 1 unit up ∴ C. 8 f(x) = 1 x a translation 6 units in the positive x-direction ⇒ y = 1 6 x −

Functions and transformations MM12-2 39 translation 1 unit in the positive y-direction ⇒ y = 1 1 6 x + − reflection in the y-axis ⇒ y = 1 1 6 x + −− = 1 1 6 x −+ + dilation factor 2 from the x-axis ⇒ y = 1 21 6 x −+  +  = 2 2 6 x −+ + b dilation factor 2 from the y-axis ⇒ y = 1 1 2 x    = 2 x reflection in the x-axis ⇒ y = 2 x− dilation factor 1 2 from the x-axis ⇒ y = 12 2 x  −   = 1 x− translation 3 units in the positive x-direction ⇒ y = 1 (3) x − − translation 1 unit in the negative y-direction ⇒ y = 1 1 (3) x −− − c translation 1 unit in the positive x-direction ⇒ y = 1 1 x − translation 2 units in the negative y-direction ⇒ y = 1 2 1 x − − dilation factor 1 3 from the y-axis ⇒ y = 1 2 31 x − − reflection in the x-axis ⇒ y = 1 2 31 x  −−  −  = 1 2 31 x − + − 9 g(x) ⇒ h(x) h(x) = − g(4( x + 1)) + 3 ⇒ reflection in x-axis ⇒ dilation factor 1 4 from the y-axis ⇒ translation 1 unit in the negative x-direction ⇒ translation 3 units in the positive y-direction x T y     = 1 1 0 4 3 01 x y  −    +    −   10 y = x3 – 4 x a 1st two transformations ⇒ 10 02−      x T y   = x y \b   \b  = 10 02 x y −     = 2 x y −      translation ⇒ 0 1   −  x T y     = x y \b   \b = 10 0 02 1 x y−  +   −   = 0 21 x y − +   −   = 21 x y −   −   Therefore, x\b = − x and y\b = 2 y – 1 x = − x\b and y = 1 2 y\b + y = x3 – 4 x the resultant equation: 1 2 y\b + = ( − x\b)3 – 4( − x\b) 1 2 y+ = − x3 + 4 x y + 1 = 2( − x3 + 4 x) = − 2 x3 + 8 x y = − 2 x3 + 8 x – 1 b x y \b  \b = 10 2 0 020 1−  +   −   = 20 01−  +    −    = 2 1−   −  ⇒ (−2, −1) x = − 2 ⇒ y = − 2( −2) 3 + 8 ( −2) – 1 = 16 − 16 − 1 = − 1 ∴ the points lies on the curve 11 f(x) = 21 x ⇒ g(x) = 23 1 (2) x − + − ⇒ reflection in the x-axis ⇒ dilation factor 3 from the x-axis ⇒ translation 2 units in the positive x-direction ⇒ translation 1 unit in the positive y-direction x T y     = x y \b   \b  = 10 2 03 1 x y  +   −   12 f(x) = − g(2( x + 1)) + 1 g(x) = x a → reflection in the x-axis → x − → dilation factor 1 2 from the y-axis ⇒ 2x − ⇒ translation 1 unit in the negative x-direction ⇒ 2( 1)x −+ ⇒ translation 1 unit in the positive y-direction ⇒ 2( 1) 1x −++ ∴ f(x) = 2( 1) 1x −−+ b T x y    = x y \b  \b = 1 1 0 2 1 01 x y    +  −   13 f( x ) = 2 g(x − 1) – 2 g(x ) = x 2 – 3 x a → dilation factor 2 from the x -axis ⇒ 2( x 2 – 3 x) = 2 x 2 – 6 x → translation 1 unit in the positive x-direction ⇒ 2( x − 1) 2 – 6( x − 1) = 2( x 2 – 2 x + 1) – 6 x + 6 = 2 x2 – 4 x + 2 − 6 x + 6 = 2 x2 – 10 x + 8 → translation 2 units in the negative y-direction ⇒ 2x 2 – 10 x + 8 – 2 = 2 x2 – 10 x + 6 f( x ) = 2 x 2 − 10 x + 6 b x T y     = x y \b  \b = 10 1 02 2 x y   +    −    14 1 (2)1 2 hx −++ = 3 2 1 36 22 x xx −− −− h(x ) = ? ∴ → reflection in the x-axis → dilation factor 1 2 from the x-axis → translation 2 units in the negative x-direction → translation 1 unit in the positive y-direction x T y     = x y \b  \b = 10 2 1 1 0 2 x y  −    +   −    2 1 x y \b −  −   \b   = 10 1 0 2 x y     −   x y = 1 2 0 2 2 1 01 x y  \b  − −     −−     \b       = 10 2 02 1 x y \b +    \b −−   = 2 2( 1) x y \b+   \b −−  x = x\b + 2 and y = − 2( y\b − 1) x\b = x − 2 y\b − 1 = 2y − y\b = 1 2 y− +

MM12-2 40 Functions and transformations y\b = 3 2 1 36 22 x xx−−−− 1 2 y − + = 3 2 (2) 13( 2) 6( 2) 22x xx−−−−−−− = 3.5 – 0.5 x3 (using CAS) 2 y − = 2.5 – 0.5 x3 − y = 5 – x3 y = x3 − 5 Exercise 2H — Sum, difference and product functions 1 a Domain g = R Domain h = R Domain f = domain g ∩ domain h ∴ Domain f: R b Domain g = R Domain h = [0, ∞) Domain f = domain g ∩ domain h ∴ Domain f: [0, ∞) c Domain g = R Domain h = R Domain f = domain g ∩ domain h ∴ Domain f: R d Domain g = R\{0} Domain h = R Domain f = domain g ∩ domain h ∴ Domain f: R\{0} 2 a g(x) – h(x) = 1 x x +− b g(x) − h(x) = | x| − | x + 1| + 2 3 a Domain f = R Domain g = ( −∞ , 3] Domain f(x)g(x) = domain f ∩ domain g ∴ Domain f(x)g(x): ( −∞ , 3] b Domain f = R Domain g = R Domain f(x)g(x) = domain f ∩ domain g ∴ Domain f(x)g(x): R c Domain f = [0, ∞) Domain g = ( −∞ , 1] Domain f(x)g(x) = domain f ∩ domain g ∴ Domain f(x)g(x): [0, 1] d Domain f = R Domain g = R Domain f(x)g(x) = domain f ∩ domain g ∴ Domain f(x)g(x): R e Domain f = R Domain g = [ −2, ∞) Domain f(x)g(x) = domain f ∩ domain g ∴ Domain f(x)g(x): [ −2, ∞) 4 5 a b Domain f = [ −2, 2] Domain g = [0, 2] Domain h(x) = domain f ∩ domain g ∴ Domain h(x) = [0, 2] ∴ a = 0 c f(0) = 0 g(0) = 0 ∴ h(0) = 0 + 0 = 0 d f(1) = − 1 g(1) = 1 ∴ h(1) = − 1 + 1 = 0 e f(2) = − 4 g(2) = 2 ∴ h(2) = − 4 + 2 = 2 − 4 f range h(x) = [0.47, 2 4] − or [0.47, −2.59] 6 a b f( − 2) = 8 g(− 2) = 2 ∴ h(− 2) = 8 + 2 = 10 c f(0) = 0 g(0) = 0 ∴ h(0) = 0 + 0 = 0 d f(1) = − 1 g(1) = 1 ∴ h(1) = − 1 + 1 = 0 e f(2) = − 8 g(2) =2 ∴ h(2) = − 8 + 2 = − 6 f range h(x) = [ −6, 10] g 7 a Domain f = R Domain g = [0, ∞) Domain h(x) = domain f ∩ domain g ∴ Domain h(x) = [0, ∞) b c f(0) = − 3 g(0) = 0 ∴ h(0) = − 3 × 0 = 0

Functions and transformations MM12-2 41 d f(1) = − 2 g(1) = 1 ∴ h(1) = − 2 × 1 = − 2 e f(2) = − 1 g(2) = 2 ∴ h(2) = − 1 × 2 = 2 − f range h(x) = [ −2, ∞) g 8 Domain f= [ −5, ∞) Domain g = ( −∞ , 8] Domain h(x) = domain f ∩ domain g ∴ Domain h(x) = [ −5, 8] 9 Exercise 2 I — Composite functions and functional equations 1 a f(x) = 2 x – 1, g(x) = 3 x+ i f(x) g(x) Domain R [ − 3, ∞) Range R [0, ∞) for f (g(x)) ran g(x) ⊆ dom f(x) [0, ∞) ⊂ R ∴ f(g(x)) is defined ii f(g(x)) = 231 x +− dom f(g(x)) = dom g(x) = [ −3, ∞) b f(x) = 1 2 x+ , g(x) = | x| + 1 i f(x) g(x) Domain R\{ −2} R Range R\{0} [1, ∞) ran g(x) ⊆ dom f(x) [1, ∞) ⊂ R\{ −2} ∴ f(g(x)) is defined. ii f(g(x)) = 1 12 x ++ domain f( g (x )) = R = 1 3 x + c f( x ) = 3( x − 2) 3, g (x ) = x 2 i f( x ) g(x ) Domain R R Range [0, ∞)[0, ∞) ran g(x ) ⊆ dom f( x ) [0, ∞) ⊂ R ∴ f ( g (x )) is defined. ii f( g (x )) = 3( x 2 − 2) 3 domain f( g (x )) = R d f( x ) = |x|, g(x ) = x 3 i f( x ) g(x ) Domain R R Range [0, ∞) R ran g(x ) ⊆ dom f( x ) R ⊆ R ∴ f ( g (x )) is defined. ii f( g (x )) = |x 3| domain f( g (x )) = R e f( x ) = ( x + 1)( x + 3), g(x ) = x 2 i f( x ) g(x ) Domain R R Range [–1, ∞)[0, ∞) ran g(x ) ⊆ dom f( x ) [0, ∞) < R ∴ f ( g (x )) is defined. ii f( g (x )) = ( x 2 + 1)( x 2 + 3) dom f( g (x )) = R 2 f( x ) = 3 x 3 x y f +    = 3 3 x y +    = x + y () ( ) 3fxfy+ = 33 3x y + = x + y ∴ f ( x ) = 3 x satisfies the equation: 3 x y f +    = () ( ) 3fxfy+ 3 f( x ) = 2 x () ( ) ()fxfyfxy + = 22 2 x y xy + = 22 2 yx xy xy + = 22 2 yx + = y + x L.H.S = R.H.S ∴ f ( x ) = 2 x satisfies the equation () ( ) ()fxfyfxy + = x + y 4 a f( x ) = x i f( x − y ) = f ( x ) – f( y ) xy − ≠ xy − L.H.S ≠ R.H.S ii f( x − y ) = () ()fx fy x y − ≠ x y L.H.S ≠ R.H.S iii f( x ) + f ( y ) = ( x 2 + y 2) f( xy ) xy + = 22()xy xy +− L.H.S ≠ R.H.S iv x f y   = () ()fx fy x y = x y L.H.S = R.H.S v f( xy ) = f ( x ) × f ( y ) xy = x y × L.H.S = R.H.S ∴ Ans: iv, v hold true b f( x ) = |x| i f( x − y ) = f ( x ) – f( y ) | x − y | ≠ |x| − |y| L.H.S ≠ R.H.S ii f( x − y ) = () ()fx fy | x − y | ≠ x y L.H.S ≠ R.H.S iii f( x ) + f ( y ) = ( x 2 + y 2) f( xy ) | x| + |y | ≠ ( x 2 + y 2)| xy | L.H.S ≠ R.H.S iv x f y   = () ()fx fy x y = x y L.H.S = R.H.S v f( xy ) = f ( x ) × f ( y ) | xy| = |x| × |y| L.H.S = R.H.S ∴ Ans: iv, v hold true c f( x ) = 1 x i f( x − y ) = f ( x ) − f ( y ) 1 x y − ≠ 11 x y − L.H.S ≠ R.H.S ii f( x − y ) = () ()fx fy 1 x y − ≠ 1 1 x y = y x L.H.S ≠ R.H.S

MM12-2 42 Functions and transformations iii f( x ) + f ( y ) = ( x 2 + y 2) f( xy ) 11 x y + = 22 1 () xy xy + yx xy + ≠ 22x y xy + L.H.S ≠ R.H.S iv x f y   = () ()fx fy 1 x y = 1 1 x y y x = y x L.H.S = R.H.S v f( xy ) =f( x ) × f ( y ) 1 xy = 11 x y × L.H.S = R.H.S ∴ Ans: iv, v hold true. d f( x ) = 21 x i f( x – y) = f ( x ) – f( y ) 2 1 () x y − ≠ 2211 x y − L.H.S ≠ R.H.S ii f( x − y ) = () ()fx fy 2 1 () x y − ≠ 2 21 1 x y = 2 2y x L.H.S ≠ R.H.S iii f( x ) + f ( y ) = ( x2 + y2) f ( xy ) 2211 x y+ = ( x2 + y2) 21 () xy 22 22yx xy + = 22 22x y xy + L.H.S = R. H.S iv x f y   = () ()fx fy 21 x y    = 2 21 1 x y 2 2y x = 2 2y x L.H.S = R.H.S v f( xy ) = f ( x ) × f ( y ) 21 () xy = 2211 x y× L.H.S = R.H.S Ans: iii, iv, v hold true e f( x ) = x2 i f( x − y ) = f ( x ) – f( y ) ( x – y) 2 ≠ x2 – y 2 L.H.S ≠ R.H.S ii f( x − y ) = () ()fx fy ( x − y )2 ≠ 2 2x y L.H.S ≠ R.H.S iii f( x ) + f ( y ) = ( x2 + y2) f ( xy ) x 2 + y2 ≠ ( x2 + y2)( xy ) 2 L.H.S ≠ R.H.S iv x f y   = () ()fx fy 2x y    = 2 2x y L.H.S = R.H.S v f(xy) = f(x) × f(y) ( xy)2 = x2 × y2 L.H.S = R.H.S ∴ Ans: iv, v hold true. f f(x) = 2x i f(x − y) = f(x) – f(y) 2 x – y ≠ 2x – 2 y L.H.S ≠ R.H.S ii f(x − y) = () ()fx fy 2 x – y = 2 2x y L.H.S = R.H.S iii f(x) + f(y) = (x2 + y2)f(xy) 2 x +2y ≠ (x2 +y2)2xy L.H.S ≠ R.H.S iv x f y   = () ()fx fy 2 xy ≠ 2 2x y L.H.S ≠ R.H.S v f(xy) = f(x) × f(y) 2 xy ≠ 2x × 2y L.H.S ≠ R.H.S ∴ Ans: ii holds true 5 f(x) = 2 1 () x a+ , g(x) = x for f(g(x)) to exist: range g(x) ⊆ dom f [0, ∞) ⊆ dom f dom f ≥ [0, ∞) dom f: R\{ −a} ∴ a > 0 6 f(x) = 2 x −, g(x) = 1 2 1 x + + f(x) g(x) Domain [2, ∞)R\{ −1} range [0, ∞)R\{2} f (g(x)) ⇒ range g(x) ⊆ dom f(x) R\{2} ⊆ [2, ∞) ∴ f (g(x)) is not defined. g(f(x)) ⇒ range f(x) ⊆ dom g(x) [0, ∞) ⊆ R\{−1} ∴ g(f (x)) exists g(f (x)) = 1 2 21 x + −+ 7 f(x) = 3 x − , g(x) = x2 − 1 f(g(x)) ⇒ range g(x) ⊆ dom f(x) [ −1, ∞) ⊆ [0, ∞) ∴ f(g(x)) is note defined. want range g(x) = [0, ∞) x2 – 1 ≥ 0 ∴ x ∈ R\(−1, 1) ∴ h(x) = x2 – 1, x ∈ R\(−1, 1) 8 w(x) = x + 3, x > −3, v(x) = |x| − 2, x ∈ R+ w(x) v(x) Domain ( −3, ∞) R+ range (0, ∞) ( −2, ∞) w(v (x)) ⇒ range v (x) ⊆ domain w(x) ( −2, ∞) ⊆ (−3, ∞) ∴ w(v(x)) is defined. w (v(x)) = |x| − 2 + 3 = |x| + 1, x ∈ R+ v (w(x)) ⇒ range w(x) ⊆ domain v (x) (0, ∞) ⊆ R+ ∴ v (w(x)) exists v(w(x)) = |x + 3| − 2, x ∈ (−3, ∞) 9 g(x) = x3 g(−x) = −g(x) ( −x)3 = −x3 −x3 = −x3 L.H.S = R.H.S ∴ g(x) = x3 satisfies the equation g(−x) = −g(x) g(x) = xn, when n = odd natural number g(−x) = −g(x) ( −x)n = −xn −xn = −xn ∴ g(x) = xn satisfies the equation g(−x) = −g(x) 10 g(x) = x4 g(xy) = g(x )g(y) ( xy)4 = x4 × y4 x4y4 = x4y4 L.H.S = R.H.S ∴ g(x) = x4 satisfies the equation g(xy) = g(x)g(y) g(x) = xn, n is a natural number g(xy) = g(x) g(y) ( xy)n = xnyn xnyn = xnyn L.H.S = R.H.S ∴ g(x) = xn satisfies the equation g(xy) = g(x)g(y) 11 f: [0, b] → R, f(x) = 2 1 (6) 4xx − g: [0, b] → R, g(x) = −x2 + bx | f(x) – g(x)| = 22 1 (6)( 6) 4xxxx−−−+ = 22 1 (1236) 6 4xxx xx −++− = 32 2 1 39 6 4x xxxx −++−

Functions and transformations MM12-2 43 = 32 1 23 4x xx −+ Maximum: (4.43, 4.23) ∴ max value is 4.23 and this occurs when x = 4.43 12 f: [4, ∞) → R, f(x) = 4 x − g: R → R, g(x) = 1 – x f(g(x)) = 14 x −− = 3 x −− = (3)x−+ f(x) → f(g(x)) → reflection in the y-axis → translation 1 unit in the positive x-direction Exercise 2J — Modelling 1 a y = a x + b ⇒ (iii) b y = ax3 + b ⇒ (ii) c y = 2a x + b ⇒ (iv) d y = ax2 + b ⇒ (i) e y = a x + b ⇒ (v) 2 a Assume y = ax3 Using (1, 0.3): 0.3 = a × 13 a = 0.3 ⇒ y = 0.3 x3 Verifying; ( −3, −8.1): y = 0.3 × (−3)3 = −8.1 ( −2, −2.4): y = 0.3 × (−2)3 = −2.4 ( −1, −0.3): y = 0.3 × (−1)3 = −0.3 (0, 0): y = 0.3 × 03 = 0 (1, 0.3): y = 0.3 × 13 = 0.3 (2, 2.4): y = 0.3 × 23 = 2.4 (3, 8.1): y = 0.3 × 33 = 8.1 The rule that fits the data is y = ax3 where a = 0.3. b Assume that y = ax2 Using (1, −6): −6 = a × 12 = a a = −6 ⇒ y = −6x2 Verifying, ( −2, −24): y = −6 × (−2)2 = −24 ( −1, −6): y = −6 × (−1)2 = −6 (0, 0): y = −6 × 02 = 0 (1, −6): y = −6 × 12 = −6 (2, −24): y = −6 × 22 = −24 (3, −54): y = −6 × 32 = −54 The rule is y = ax2 where a = −6. c Assume y = 2a x Using (1, 2): 2 = 21 a = 1 a a = 2 ⇒ y = 22 x Verifying, ( −5, 0.08): y = 22 (5) − = 2 25 = 0.08 ( −2, 0.5): y = 22 (2) −

MM12-2 44 Functions and transformations = 2 4 = 0.5 ( −1, 2): y = 22 (1) − = 2 (1, 2): y = 22 1 = 2 (2, 0.5): y = 22 2 = 0.5 (5, 0.08): y = 22 5 = 2 25 = 0.08 The rule is y = 2a x where a = 2. d Assume y = a x Using (1, 1.6): 1.6 = a 1 a = 1.6 ⇒ y = 1.6 x Verifying, (0, 0): y = 1.6 0 = 0 (0.5, 1.13): y = 1.6 × 0.5 = 1.13 (1.5, 1.96): y = 1.6 1.5 = 1.96 (2, 2.26): y = 1.6 2 = 2.26 The rule is y = a x where a = 1.6 e Assume y = a x Using (1, 5): 5 = 1 a a = 5 ⇒ y = 5 x Verifying, (2, 2.5): y = 5 2 = 2.5 (4, 1.25): y = 5 4 = 1.25 (5, 1): y = 5 5 = 1 (10, 0.5): y = 5 10 = 0.5 The rule is y = a x where a = 5. f Assuming y = ax3 Using (1, −1.5): −1.5 = a × 13 a = −1.5 Verifying, ( −3, 40.5): y = −1.5 × (−3)3 = 40.5 ( −2, 12): y = −1.5 × (−2)3 = 12 ( −1, 1.5): y = −1.5 × (−1)3 = 1.5 (0, 0): y = −1.5 × 03 = 0 (2, −12): y = −1.5 × 23 = −12 The rule is y = ax3 where a = −1.5. 3 i, ii, and iv as they have clear vertical asymptotes along the y-axis. The answer is D. 4 a b x2 0 1 4 9 16 25 y −3.2 −1 4.9 14.5 29 46.8 c a = m = 21 21yy x x − − Using (0, −3.2) and (25, 46.8) a = 46.8 3.2 25 0 −−− = 50 25 = 2 y = 2x2 + b If x = 0, y = −3.2 ∴ b = −3.2 5 x3 −64 −8 0 8 64 216 y −28 −13.5 −12.5 −104.3 41 Given y = ax3 + b a = m = 21 21yy x x − − = 41 28 216 64 −− −− = 69 280 a  1 4 Using (0, −12.5) b = −12.5 b  −12 ⇒ y = 1 4x3 − 12. 6 a b Assume f = a λ

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