Maths Quest 12 Mathematical Methods CAS Solutions Manual

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Jenny Watson Ian Younger
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
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
Multiple choice 52
Exercise 3A — The index laws 54
Exercise 3B — Logarithm laws 55
Exercise 3C — Exponential equations 56
Exercise 3D — Logarithmic equations using
any base 58
logarithms 62
Exercise 3I — Exponential and logarithmic modelling 66
Chapter review 68
Short answer 68
Multiple choice 69
any base 72
Exercise 4C — Graphs of exponential functions with
base e 81Exercise 4D — Logarithmic graphs to base e 86
Exercise 4E — Finding equations for graphs of exponential
and logarithmic functions 90
Exercise 4G — Exponential and logarithmic functions with
absolute values 95
using graphs 96
Chapter review 98
Short answer 98
Multiple choice 99
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
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 6I — Trigonometric functions with an increasing
trend 144
Exercise 7B — Limits and differentiation from first
principles 154
Exercise 7D — The chain rule 160
Exercise 7E — The derivative of e x
164 Exercise 7F — The derivative of loge

Exercise 7G — The derivatives of sin( x), cos( x) and
tan( x) 172
Exercise 7J — Mixed problems on differentiation 183
Chapter review 190
Short answer 190
Multiple choice 192
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
function is unknown 210
Exercise 8F — Related rates 215
Exercise 8G — Linear approximation 217
Chapter review 217
Short answer 217
Multiple choice 219
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
calculus 238
Exercise 9H — Areas between two curves 260
Exercise 9I — Average value of a function 270Exercise 9J — Further applications of integration 271
Chapter review 276
Short answer 276
Multiple choice 278
Exercise 10A — Probability revision 286
Exercise 10B — Discrete random variables 289
Exercise 10C — Measures of centre of discrete random
distributions 293
distributions 297
distribution for multiple probabilities 311
Exercise 11C — Markov chains and transition
matrices 315
deviation of the binomial distribution 318
Chapter review 321 Short answer 321
Multiple choice 323
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 12E — The normal distribution 343
Exercise 12F — The standard normal distribution 347Exercise 12G — The inverse cumulative normal
distribution 350
Chapter 2 363
Chapter 12 364
G r a p h s a n d p o l y n o m i a l s M M 1 2 - 1 1
Exercise 1A — The binomial theorem
1 a ( x + 3)2 = 2 2
0 x
+ 1 × 1024
= x5 + 20 x4 + 160 x3 + 640 x2 + 1280 x + 1024
c ( x − 1)8 = 88
−

= x8 − 8 x7 + 28 x6 − 56 x5 + 70 x4 − 56 x3 + 28 x2
− 8 x + 1
0 x
e (7 − x)4 = 44 7
0
1 2 x x
f (2 − 3 x)5 = 55 2
0
− 243 x5
2 a
3 1
x x
−

= 2187 x7 − 10 206 x5 + 20 412 x3 − 22 680 x
+ 3
+ 3
1458
x +
6
729
x
d
5
2
x x
−
r
⇒ r = 1
Coefficient = 3
term = 3
term = 5 3
⇒ r = 2
term = 5
term = 5
8/21/2019 Maths Quest 12 Mathematical Methods CAS Solutions Manual
M M 1 2 - 1 2 G r a p h s a n d p o l y n o m i a l s
ii x3 is the 5th term ⇒ r = 4
term =
d
ii x3 is the 4th term ⇒ r = 3
term = 6
e
6
2
ii x3 is the 2nd term ⇒ r = 1
term =
6
1
4
3
term = 3
The answer is A.
5 When the expression for C is expanded it does not contain an x5 term. The first three terms contain x8, x6 and x4
respectively. All the other expressions contain an x5 term.
The answer is C.
80
x +
2
32
x
= 243
The answer is D.
= 20 × x3 × 27 y3
=
=
1
x
, ( x2)3
2
3
1
x
term = 5C 2( x 2)3
2
3
4
x
1
x
, ( x2)2
2
2
1
x
term = 4C 2( x 2)2
2
2
3
x
= p5 + 5 p43 + 10 p332 + 10 p233 + 5 p34 + 35
= p5 + 15 p4 + 90 p3 + 270 p2 + 405 p + 243
∴(2 p − 5)( p5 + 15 p4 + 90 p3 + 270 p2 + 405 p + 243)
⇒ 2 p6 + 30 p5 + 180 p4 + 540 p3 + 810 p2 + 486 p
− 5 p5 − 75 p4 − 450 p3 − 1350 p2 − 2025 p − 1215
Coefficient of 4th term = 180 − 75 = 105
14 (2a − 1) n
coefficient: −n × 2n − 1 = −192
n × 2n × 1
negative integer powers of x only. Not Polynomial:
ii x4 + 3 x2 − 2 x + x
iii x7 + 3 x6 − 2 xy + 5 x
vi 2 x5 + x4 − x3 + x2 + 3 x − 2
x
Polynomial:
iv 3 x8 − 2 x5 + x2 − 7
v 4 x6 − x3 + 2 x − 3
2 a P ( x) + Q( x) = 8 − 3 x + 2 x2 + x4 + x5 − 3 x4 − 4 x2 − 1
= x5 − 2 x4 − 2 x2 − 3 x + 7
b Q( x) − R( x) = x5 − 3 x4 − 4 x2 − 1 − (8 x3 + 7 x2 − 4 x)
= x5 − 3 x4 − 4 x2 − 1 − 8 x3 − 7 x2 + 4 x = x5 − 3 x4 − 8 x3 − 11 x2 + 4 x − 1
c 3 P ( x) − 2 R( x).
3 P ( x) = 3(8 − 3 x + 2 x2 + x4)
= 24 − 9 x + 6 x2 + 3 x4
2 R( x) = 2(8 x3 + 7 x2 − 4 x)
= 16 x3 + 14 x2 − 8 x
∴ 3 P ( x) − 2 R( x) = 24 − 9 x + 6 x2 + 3 x4
− (16 x3 + 14 x2 − 8 x)
= 24 − 9 x + 6 x2 + 3 x4 − 16 x3 − 14 x2 + 8 x
= 3 x4 − 16 x3 − 8 x2 − x + 24
d 2 P ( x) − Q( x) + 3 R( x)
2 P ( x) = 2(8 − 3 x + 2 x2 + x4)
= 16 − 6 x + 4 x2 + 2 x4
3 R( x) = 3(8 x3 + 7 x2 − 4 x)
= 24 x3 + 21 x2 − 12 x
2 P ( x) − Q( x) + 3 R( x)
= 16 − 6 x + 4 x2 + 2 x4− ( x5 − 3 x4 − 4 x2 − 1)
+ 24 x3 + 21 x2 − 12 x
= 16 − 6 x + 4 x2 + 2 x4 − x5 + 3 x4 + 4 x2 + 1
+ 24 x3 + 21 x2 − 12 x
= 17 − 18 x + 29 x2 + 24 x3 + 5 x4 − x5
3 a P ( x) = x6 + 2 x5 − x3 + x2
i degree = 6
= 0
= 1
i degree = 7
= −8
= 280
= −3 − 2 − 1 − 8
= −14
c P ( x) = 5 x6 + 3 x4 − 2 x3 − 6 x2 + 3
i degree = 6 ii P (0) = 5 × 06 + 3 × 04 − 2 × 03 − 6 × 02 + 3
= 3
iii P (2) = 5 × 26 + 3 × 24 − 2 × 23 − 6 × 22 + 3
= 331
− 2 × (−1)3 − 6 × (−1)2 + 3
= 5 + 3 + 2 − 6 + 3
= 7
d P ( x) = −7 + 2 x − 5 x2 + 2 x3 − 3 x4
i degree = 4
ii P (0) = −7 + 2 × 0 − 5 × 02 + 2 × 03 − 3 × 04
= −7
+ 2 × 2 3
− 3 × 2 4
+ 2 × (−1)3 − 3(−1)4
= −19
4 P ( x) = x8 − 3 x6 + 2 x4 − x2 + 3
P (−2) = (−2)8 − 3 × (−2)6 + 2 × (−2)4 − (−2)2 + 3
= 95
5 P ( x) = 2 x7 + ax5 + 3 x3 + bx − 5
P (1) = 4
∴ 4 = 2 × 17 + a × 15 + 3 × 13 + b × 1 − 5
4 = 2 + a + 3 + b − 5
4 = a + b
163 = 2 × 27 + a × 25 + 3 × 23 + b × 2 − 5
= 256 + 32a + 24 + 2b − 5
−112= 32a + 2b [2]
[2] − [3] −120 = 30a
a = −4 b = 8
6 f ( x) = ax4 + bx3 − 3 x2 − 4 x + 7
f (1) = −2
∴ −2 = a × (1)4 + b × (1)3 − 3 × 12 −4 × 1 + 7
−2 = a + b − 3 − 4 + 7
−2 = a + b
∴ −2 − b = a
∴ −5 = a × 24 + b × 23 − 3 × 22 − 4 × 2 + 7
−5 = 16a + 8b − 12 − 8 + 7
−5 = 16a + 8b − 13
1 = −4 − 2b + b
3 = a
− 3 x2 − 4 x + 7
7 Q( x) = x5 + 2 x4
+ ax3 − 6 x + b
45 = 52 + 8a + b
−7 = b [2]
a = 0
− 6 x − 7.
+ x3 − 6
then 3 P ( x) = 3(ax6 + bx4
+ x3 − 6)
+ 13 − 6)
−8 = a + b − 5
then 3 P ( x) = 3(ax6 + bx4 + x3 − 6)
102 = 3[a(−2)6 + b(−2)4
+ (−2)3 − 6)
34 = 64a + 16b
Substitute [1] into [2]
3 = 4a + (−3 − a)
3 = 4a − 3 − a
b = −3 − 2
− 5 x4 + x3
+ 3 x2 − 5
+ 3 × (1)2 − 5
−1 = a − 1 + 3 − 5
−1 = a − 3
b f ( x) = xn
− 22
Exercise 1C — Division of polynomials
1 a 2
R( x) = 50
5 4
4 3
4 3
3 2
3 2
3
x x
x x
x x
x x
x x
x x
x x
171
−

Q( x) = x4 − 3 x3 + 6 x2 − 18 x + 58 R( x) = −171
c 3
6 18
17 2
17 51
53 4
53 159
155 0
155 465
x x
x x
x x
x x
x x
R( x) = 465
d 3 2
4 3 2
3
x x
x x
x x
− =

2 a i P ( x) = x3 − 2 x2 + 5 x − 2
P (4) = 43 − 2 × 42 + 5 × 4 − 2
= 50
P (1) = 1 + 1 + 3 − 7
= −2
P (−3) = (−3)5 − 3 × (−3)3 + 4 × (−3) + 3
= −171
iv P ( x) = 2 x6 − x4 + x3 + 6 x2 − 5 x
P (−2) = 2 × (−2)6 − (−2)4 + (− 2)3 + 6 × (−2)2 − 5 ×
(−2)
= 138
v P ( x) = 6 x4 − x3 + 2 x2 − 4 x
P (3) = 6 × 34 − 33 + 2 × 32 − 4 × 3
= 465
P (2) = 24 − 13 × 22 + 36
= 0
1

viii P ( x) = x5 + 3 x3 − 4 x2 + 6 x − 8
3

b The values obtained in 2 were the same as the remainder values obtained in 1.
3 a P (3) = 33 + 9 × 32 + 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.
c P (+ 4) = 4 − 9 × + 4 + 6 × 42 − 13 × (+ 4)3
− 12 × (+ 4)4 + 3 × (+ 4)5
= −768
Since P (−4) ≠ 0 then 4 − x is not a factor.
d P 1
= 0
= 0 then 2 x + 1 is a factor.
4 a f ( x) = x4 − 4 x3 − x2 + 16 x − 12
A x + 1 ⇒ f (− 1)
= (−1)4 − 4 × (−1)3 − (−1)2 + 16 × (−1) − 12
= 1 + 4 − 1 − 16 − 12
= −24
= −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
= 44 − 4 × 43 − 42 + 16 × 4 − 12
= 36 The answer is C.
b
x x
x x
x x
x x
x x
3
= 0
2
− +
− + − − −
−
− + −
− +
− −
−

x2 − 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) = x3 + 4 x2 − 3 x − 18
Test x = ± 1 P ( x) ≠ 0
x = 2, P ( x) = 0
∴ ( x − 2) is a factor
2
= ( x − 2)( x + 3)2
b P ( x) = 3 x3 − 13 x2 − 32 x + 12
Test x = ± 1 P ( x) ≠ 0
x = ± 2
3 2
3 2
3 6
19 32
19 38
6 12
6 12
∴ ( x + 2)(3 x2 − 19 x + 6)
= ( x + 2)(3 x − 1)( x − 6) c P ( x) = x4 + 2 x3 − 7 x2 − 8 x + 12
Test x = −2, P ( x) = 0
∴ ( x + 2) is a factor 3 2
4 3 2
2
x x
x x
x x
Test x = 2, P ( x) = 0
∴ ( x − 2) is a factor 2
3 2
3 2
d P ( x) = 4 x4 + 12 x3 − 24 x2 − 32 x
Test x = −1, P ( x) = 0
∴ x + 1 is a factor 3 2
4 3 2
4 4
8 24
8 8
32 32
32 32
x x
x x
x x
x x
x x
Take out factor of 4 x.
4 x( x + 1)( x2 + 2 x − 8)
∴ 4 x( x + 1)( x − 2)( x + 4)
6 a 3 x3 + 3 x2 − 18 x = 0
Test x = 2, f ( x) = 0
2
3 6
9 18
9 18
3 x( x − 2)( x + 3) = 0
∴ x = 0, 2, or −3
b 2 x4 + 10 x3 − 4 x2 − 48 x = 0
Test x = 2, f ( x) = 0 ∴ ( x − 2) is a factor
3 2
2 4
14 4
14 28
24 48
24 48
x x
x x
x x
x x
x x
+ +
+ − − − −
−
− −
−
− −
−

∴ ( x − 2)(2 x3 + 14 x2 + 24 x) Take out common factor of 2 x:
2 x( x − 2)( x2 + 7 x + 12)
= 2 x( x − 2)( x + 3)( x + 4) = 0
∴ x = 2, −3, 0, and −4
c 2 x4 + x3 − 14 x2 − 4 x + 24 = 0
Test x = 2, f ( x) = 0
∴ ( x − 2) is a factor 3 2
4 3 2
2 4
5 14
5 10
4 4
4 8
x x
x x
x x
x x
x x
Test x = −2, f (−2) = 0
∴ ( x + 2) is a factor
2
2 4
( x − 2)( x + 2)(2 x − 3)( x + 2)
x = 2, −2, or 3
2
3 2
2
0
1
1
0
x x
x x
x x
x x
x x
∴ ( x + 1) is a factor
2
( x − 1)( x + 1)( x − 1)( x + 1) = 0
x = ±1
when x = 2, f ( x) = 0
0 = x3 + ax2 − 6 x − 4
f (2) = 0 = 23 + a22 − 6 × 2 − 4
0 = 8 + 4a − 12 − 4
0 = 4a − 8
+ x2 − ax
( x − 1) is a factor
a = 5
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 = 189 − 27a
27a = 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]
when x = 2, f ( x) = 0
f (2) = 0 = 8a − 16 + 2b − 12
0 = 8a + 2b − 28
14 = −4b − 64 + b
11 (2 x − 3) and ( x + 2) are factors of
2 x3 + ax2
+ bx + 30
+ b(−2)
+ 30 = 0
4a − 2b = −14 [1]
9a + 6b = −147
−2b = 22
b = −11
x-intercept when y
x-intercept when y = 0
x-intercept when y = 0
1 = 3 × 2 + c
b y = mx + c
y = −2 x + c
3 = −2 × −4 + c
∴ y = −2 x − 5
3 a (−3, −4), (−1, −10)
m = 10 4
3 x + y + 13 = 0
b (7, 5), (2, 0)
m = 5 0
4 2 y − 3 x − 6 = 0
A 2 × 6 − 3 × 2 − 6 = 0
12 − 6 − 6 = 0
0 + 6 − 6 = 0
6 − 0 − 6 = 0
4 − 3 − 6 ≠ 0
18 − 12 − 6 = 0
The answer is D.
1 2
∴ m
= 3
5 = 3 × 4 + c
c 2 y − x + 1 = 0
2 y = x
1
2
m = −2 gradient of perpendicular line y − y1 = m( x − x1)
Sub in (−2, 4) y − 4 = −2( x
+ 2)
2 x + y
• x-intercept when y = 0
iii y − 2 x − 2 = 0
• x-intercept when y
• x-intercept when y = 0
vi x = −2 − Graph d.
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 4 y + 3 x = 24 x ∈ [−12, 12] x-intercept
3 x = 24 x = 8 y-intercept
4 y
i domain [−12, 12]
ii range [−3, 15]
2 x = 10
x = 5 y-intercept
−5 y = 10
i domain (−∞, 5)
ii range (−∞, 0)
c 4 x − 3 y − 6 = 0 x ∈ [2, 5) x-intercept
4 x = 6
3
y = −2 x + c
9 = c
b Perpendicular ∴ m = − 1
x + 3 y − 17 = 0 10 a Parallel to
4 x − 13 = 2 y
2 x − 13
b 4 x − 2 y = 13
4 x − 13 = 2 y
2 x − 13
3 x + 2 = y
12 5 x + y − 3 = 0 bx − y − 2 = 0
y = −5 x + 3 y = bx − 2
gradient −5 gradient b = 1
5
1 b2 − 4ac =
a f ( x) = x2 − 3 x + 4 a = 1, b = −3, c = 4
= 9 − 16
b f ( x) = x2 + 5 x − 8
a = 1 b = 5 c = −8
= 25 + 32
c f ( x) = 3 x2 − 5 x + 9
a = 3 b = −5 c = 9
= 25 − 108
d f ( x) = 2 x2 + 7 x − 11
a = 2 b = 7 c = −11
= 49 + 88
e f ( x) = 1 − 6 x − x2
a = −1 b = −6 c = 1
= 36 + 4
f f ( x) = 3 + 6 x + 3 x2
a = 3 b = 6 c = 3 = 36 − 36
= 0
2 a f ( x) = x2 − 6 x + 8
y-intercept x = 0
∴ x = 4 or 2
y-intercept x = 0
x = 4 or 1
y-intercept x = 0
y-intercept x = 0
x-intercept(s) y = 0
x = − 4
3 or
= x2 − 6 x + 32 − 32 + 8
= ( x − 3)2 − 9 + 8
= x2 − 5 x +
= −( x2 − 3 x − 10)
2 2 x x
= 26 2 6
6 12 12
TP = (0, 2)
ii Domain = R
iii Range = (−∞, 2]
a = 1, h = 6, k = 0
TP = (6, 0)
c i y = −( x + 2)2
a = −1, h = −2, k = 0
TP = (−2, 0)
ii Domain = R
iii Range = (−∞, 0]
a = 2, h = −3, k = −6
TP = (−3, −6)
ii Domain = R
iii Range = [−6, ∞)
a i TP =
Assume A = 1
y = x2 − 2 x + 1 − 2
y = x2 − 2 x − 1
ii Domain = R
iii Range [−2, ∞)
∴ B = −2
C = −3
= x2 − 4 x + 1 ii Domain = [−1, ∞)
iii Range = [−3, ∞)
Assume A = −1
y = −1( x2 − 2 x + 1) + 9
y = − x2 + 2 x − 1 + 9
y = − x2 + 2 x + 8
ii Domain = [−4, 4)
iii Range = [−16, 9]
TP = (0, 3) y-intercept when x = 0
y = 3
b y = 1 − 4(2 − x)2
TP = (2, 1)
x-intercepts when y = 0
0 = 1 − 4(2 − x)2
= −4 x2 + 16 x − 15
x = 3
TP = 3
, 8 2
= 4 x2 − 12 x + 9 − 8
= 4 x2 − 12 x + 1 From the graphics calculator,
x = 2.91 and x = 0.09
7 a y = x2 − 2 x − 3
x-intercepts y = 0
(3, 0)(−1, 0)
The answer is B.
y = x2 − 2 x + 12 − 12 − 3
y = ( x − 1)2 − 4
8 f ( x) = −( x + 3)2 + 4
TP = (−3, 4)
∴ range (−2, 4]
∴ range [0, 16]
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
x-int y = 0
0 = ( x − 4)( x)
∴ x = 4 or 0
TP = (−4, 9)
0 = (3 − x − 4) (3 + x + 4)
0 = (− x − 1)(7 + x)
x = −1 or −7
y = x2 + 4 x + 4 − 4 + 3
y = ( x + 2)2 − 1
0 = ( x + 1)( x + 3)
∴ x = −1 or −3
d y = 2 x
= 2[ x2 − 2 x + 1 − 1 − 3]
= 2[( x − 1)2 − 4]
TP = (1, −8)
y-int x = 0
0 = 2( x − 3)( x + 1)
∴ x = 3 or −1
11 a y = x2 − 2 x + 2 x ∈ [−2, 2]
y = x2 − 2 x + 1 − 1 + 2
y = ( x − 1)2 + 1
but TP = (1, 1)
y = −( x2 − x + 1)
1 2 2
x ∈ [−10, 6]
2 3
−
d f ( x) = −3 x2 + 6 x + 5 x ∈ [−5, 3)
= 2 5 3 2
3 x x
When x = −5, y = −100
∴ [−100, 8]
12 V (t ) = 2t 2 − 16t + 40 t ∈ [0, 10]
V (t ) = 2(t 2 − 8t + 20)
= 2[t 2 − 8t + 16 − 16 + 20]
= 2[(t − 4)2 + 4]
When t = 10
= 80
13 h(t ) = −3t 2 + 12t + 36
h(t ) = −3[t 2 − 4t − 12]
= −3[t 2 − 4t + 4 − 4 − 12]
= −3[(t − 2)2 − 16]
b When h(t ) = 0
0 = −3(t − 6)(t + 2)
c Domain [0, 6]
Range [0, 48]
14 a h(t ) = t 2 − 12t + 48, t ∈ [0, 11] The lowest point is the y-coordinate of the turning point
h(t ) = t 2 − 12t + 36 − 36 + 48
= (t − 6)2 + 12
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
ground is 48 m.
d Domain = [0, 11]
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 = 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 = 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 = x3 + x2 − 4 x − 4
y-intercept x = 0
∴ x + 1 is a factor 2
3 2
3 2
y = ( x + 1)( x − 2)( x + 2)
If y = 0, x = −1, 2, or −2
b y = 2 x3 − 8 x2 + 2 x + 12
y-int
3 2
3 2
2 2
10 2
10 10
12 12
12 12
y = 2( x + 1)( x − 2)( x − 3)
If y = 0, then x = −1, 2 or 3
c y = −2 x3 + 26 x + 24
y-int
∴ ( x + 1) is a factor.
2
2 2
2 26
2 2
24 24
24 24
If y = 0, then x = −1, −3 or 4.
d y = − x3 + 8 x2 − 21 x + 18
y-int
3 2
3 2
= −( x − 3)( x2 − 5 x + 6)
y = −1( x − 3)( x − 3)( x − 2)
∴ x = 3 or 2
Test x = −2 so y = 0
∴ ( x + 2) is a factor 2
3 2
3 2
= ( x + 2)( x + 2)2
b In form y = a( x − t )3
a = 1, t is intercept
The answer is E.
so a = −1
The answer is C.
y = (2 − x)3 + 2
Point of inflection (−1, −3)
Graph is a positive cubic
Τhe answer is A.
x-intercept at (−3, 0)
y-intercept at (0, 6)
The answer is B.
a < 0 so,
9 y = −h( x − a)2 ( x − c)
x = 0, y = −ha2(−c)
b = ha2c
h = 2
The answer is E.
10 a f ( x) = x3 + x2 − 10 x + 8 x ∈ [2, ∞)
a > 1 positive
y = 8
= 0
i Domain [2, ∞) ii range [0, ∞)
b f ( x) = 3 x3 − 5 x2 − 4 x + 4 for
x ∈ [−2, −1]
y-intercept
+ 4
= −32
= 0
i Domain [−2, −1]
ii Range [−32, 0]
c f ( x) = −3 x3 + 4 x2 + 27 x − 36
x ∈ (0, 1]
a < 1 ∴ negative
y-intercept x = 0
When x = 1
= −8
i Domain (0, 1]
ii Range (−36, −8]
d f ( x) = − x3 − 3 x for x ∈ [−1, 2)
a < 1 negative
x = 2, y = −14 open end point
i Domain − [−1, 2)
ii Range − (−14, 4]
x ∈ [−2, −1) ∪ (0, 3]
a > 1
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) = −2 x3 − x
for x ∈ (−1, 1) ∪ [2, 3)
a < 1 negative
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) = x3 + ax2 + bx − 64
0 = (−2)3+ (−2)2a + (−2)b − 64
= −8 + 4a − 2b − 64
0 = 43 + 16a + 4b − 64
0 = 16a + 4b
0 = 4a + b
0 = 7 − a [1]
0 = 6 + (7 + b) × (−1) − 4 × 1
− (−1)
0 = −4 − b
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
reflection = (−3, 3)
domain [−4, −2]
range [0, 6]
c When f ( x) = 3.375 3.375 = 3( x − 3)3 + 3
x = 3.5
14 d (t ) = at 2(b − t ) a (2, 3) and (5, 0)
∴ 3 = 4a(b − 2) [1]
0 = 25a(b − 5) [2]
3 = 4ab − 8a × 25
0 = 25ab − 125a × 4
4
4 4
t t −
4
10 = 3t
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 = 24
x = −3, −1, 2 and 4
b y = 2 x4 + 6 x3 − 16 x2 − 24 x + 32 y-intercept when x = 0
y = 32
Test x = + 2 so y = 0
( x − 1)( x − 2) = x2 − 3 x + 2 2
4 3 2 2
2 6 4
12 20 24
12 36 24
16 48 32
16 48 32
x x x
x x x
x x x
2( x − 1)( x − 2)( x + 4)( x + 2) = y
When using N.F.L, x = 1, 2, −4, −2
then y = 0.
∴ x = ± 2
d y = −2 x4 + 15 x3 − 37 x2 + 30 x
y-intercept, when x = 0
∴ y = 0
( x − 2)( x − 3) are factors
( x − 2)( x − 3) = x2 − 5 x + 6
2
2 10 12
5 25 30
5 25 30
x x x
x x x
x x x
Using N.F.L, x = 0, 2, 3 or 5
2
e y = 6 x4 + 11 x3 − 37 x2 − 36 x + 36
Test x = −3 ∴ y = 0
Test x = 2
( x + 3)( x − 2) = x2 + x − 6 2
4 3 2 2
6 6 36
x x x
x x x
x x x
+ −
+ − − + + − −
+ −
− − −
+ −
− − + −
− − +
∴ ( x + 3)( x − 2)(3 x − 2)(2 x + 3) = y Using N.F.L
x = −3, 2, 2
Turning point (0, 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)
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)
c y = ( x − 1)2( x + 1)( x + 3)
y = 0, ( x − 1)2( x + 1)( x + 3) = 0
x = 1, −1, −3
Maximum turning point
y = 0, ( x + 2)3(1 − x) = 0
x = −2, 1
x-intercept is 1 Negative quartic (1 − x) = −( x − 1)
Maximum turning point (0.25, 8.54)
3 a
f ( x) =
= (a − 4)2
f ( x) = ( x2 − 4)2
∴ ( x − 2)2( x + 2)2
The answer is E.
y-int when x = 0
y = 16 (0, 16)
x-int when y = 0
The answer is A.
0 = x4 − 8 x2 − 9
Let x2 = a
∴ x = ± 3
The answer is D.
y = a( x + 2)( x + 1)( x − 1)( x − 3)
(0, 6)
6 = a × 6
b x = 4, 2, −1
Repeated factor at x = −2.
y = a( x − 4)( x − 2)2( x + 1)
(0, 8)
8 = −16a
a = − 1
− ( x − 4)( x − 2)2( x + 1)
5 a y = (2 − x)( x2 − 4)( x + 3) x ∈ [2, 3]
y-int when x = 0
y = 2 × −4 × 3
x = 3, y = −15 closed end point
i Domain [2, 3]
ii Range [−15, 0]
b y = 9 x4 − 30 x3 + 13 x2 + 20 x + 4 x ∈
(−2, −1]
x-int when y = 0
∴ ( x − 2) factor
9 18
12 13
12 24
11 20
11 22
2 4
x x
x x
x x
x x
x x
Try for 2nd factor of x − 2
Test x = 2, 9 x3 − 12 x2 − 11 x − 2 = 0
x − 2 is a factor. So turning point at (2, 0)
2
9 18
6 11
6 12
y = ( x − 2)2(9 x2 + 6 x + 1)
= ( x − 2)2(3 x + 1)2
1 , 0
i Domain (−2, −1]
ii Range [36, 400)
y-int when x = 0
y = −1 × 4 × 1
∴ x = −1 or 2 When x = −2, y = −16.
y = −(−2 − 2)2(−2 + 1)2
y = −(−4)2(−1)2
y = −16 × 1
x ∈ [−3, −2]
x = −2, y = 0
i Domain [−3, −2]
ii Range [−45, 0]
(2, 0):
16 + 8a − 16 + 2b + 6 = 0
6 + 8a + 2b = 0
−4a − 3 = b [1]
0 = 81 − 27a − 36 − 3b + 6
0 = 51 − 27a − 3b
0 = 17 − 9a + 4a + 3,
0 = 20 − 5a
( x − 1) is a factor
P (1) = 1 + a + b − 1 + 6
a + b = −6 [1]
P (−3) = 81 − 27a + 9b + 3 + 6 = 0
27a − 9b = 90
3a − b = 10 [2]
[1] + [2] 4a = 4
a = 1 b = −7
8 y = (a − 2b) x4 − 3 x − 2 Sub in (1, 0):
0 = (a − 2b)14 − 3 − 2
0 = a − 2b − 5
0 = 1 − 1 + (a + 5b)1 − 5 + 7
0 = a + 5b + 2
−2 = a + 5b [1]
−7 = 7b
−1 = b
∴ 3 = a
Chapter review
Short answer
= (2 y)5 + 5(2 y)4(−3 x) + 10(2 y)3(−3 x)2 +
10(2 y)2(−3 x)3 + 5(2 y)(−3 x)4 + (−3 x)5
= 32 y5 + 5 × 16 × (−3) y4 x + 10 × 8
× 9 y3 x2 + 10 × 4 × (−27) y2 x3 + 5
× 2 × 81 yx4 − 243 x5
= 32 y5 − 240 y4 x + 720 y3 x2
− 1080 y2 x3 + 810 yx4 − 243 x5
b
∴ 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 = 30 + 2a
Test x = −2.
2
− 14 x + 45) ( x + 2)( x − 9)( x − 5)
b 2 x4 + 7 x3 − 31 x2 + 0 x + 36 = y
Test x = −1 ∴ y = 0
Test x = 2 ∴ y = 0
∴ ( x + 1)( x − 2) are factors
∴ ( x2 − x − 2) is a factor
2
2 2 4
9 27 0
9 9 18
x x x
x x x
x x x
4 a (−5, 6), (1, −1)
m = 6 1
−1 = 7
7 x + 6 y − 1 = 0
b 2 x − y + 10 = 0
2 x + 10 = y
3 = 3
5 y = − x2 − 2 x + 8
y-int when x = 0
∴ x = −4 or 2
= −1[( x + 1)2 − 9]
x ∈ [−3, 0)
∴ x = 1
3 x x
3 6 36 36 x x
+ + − −

25 , 0
0 = −(−3)3 + b(−3)2 + a × (−3) − 18
0 = 27 + 9b − 3a − 18
0 = 9 + 9b − 3a
0 = 3 + 3b − a
0 = a(−3)2 + b × (−3) − 75
0 = 9a − 3b − 75
25= 3a − b [2]
Sub [1] into [2]
25 = 3(3 + 3b) − b
25 = 9 + 9b − b
y-int when x = 0
3 2
3 2
−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
4 3 2 2
2
x x x
x x x
x x x
∴ x = −1, 2, or 4.
Multiple choice
+ 10 × 12 (−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.
2nd term = 8
1
x
4 D x4 + 5 x3
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 + 6 2 4
10
5 4 3 2
3
4
3
x x
x x
x x
x x
x x
x x
x x
171
was substituted the expression would be zero.
24 − 2 × 23 − 6 × 22 − 8 × 2 + 2
16 − 16 − 24 − 16 + 2 = −38
9 2 x4 − 4 x3 − 10 x2 + 12 x
To check if ( x − 3) is a factor,
substitute in x = 3
= 162 − 108 − 90 + 36 = 0 The answer is E.
10 (4, 0), (0, 2)
m = 2 0
0 = 4
2 − + c
11 (2, b), (−3, −5)
m =
5
−5 + 10 = b
12 y = 2 x + c Sub in (2, 5)
5 = 4 + c
∴ y-int (0, 1)
The answer is D. 13 3 x2 + 4 x − 5 = 0
b2 − 4ac = 42 − 4 × 3 × −5
= 16 + 60
14 y = 2 x2 + 8 x − 10
x ∈ (−6, 2)
∴ 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
16 y = −3 x3
17 f ( x) = −2 x3 + 3 x2 + 11 x − 6
Test x = −2, f ( x) = 0
x + 2 is a factor 2
3 2
3 2
2 4
7 11
7 14
3 6
3 6
= ( x + 2)(−2 x + 1)( x − 3)
∴ x = −2, 1
and (0, −6) is the y-intercept. The answer is A.
18 TP at ( x + 1)2
Other intercept ( x − 3)
Τhe answer is E.
2
∴ 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.
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)
6.25a = 14
a = 2.24
b Domain [−5, 0] Range [4, 18]
c y = 5
x = −1.83
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 ) = at 3 + bt 2 + ct + d
t = 0, d = 0
When t = 1, d = 4
4 = a + b + c [1]
When t = 2, d = 2
2 = 8a + 4b + 2c
0 = 27a + 9b + 3c [3] Solving [1], [2] and [3]
simultaneously, then
so, d (t ) = t 3 − 6t 2 + 9t
d When t = 8,d (8) = 83 − 6 × 82 + 9 × 8

000 km from Earth. So, the satellite is closer by 40 000 km.
e 490 = t 3 − 6t 2 + 9t
t 3
− 6t 2
+ 9t − 490 = 0 When t = 10, d (10) = 490 So, the satellite will self-destruct 490 000 km from Earth, 10 days
after launching.
is 10 days.
Range is [0, 490]
(2, 0) ⇒ 0 = 4a + 3
−3 = 4a
a = −0.75
y = −0.75(1.5)2 + 3
c 1.7 – 1.3125
4
line moving forwards at a constant speed of 0.75
m/min.
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.
1.3 m the wrong way for 2.55 mins, stopping
momentarily then moving forward with increasing
speed
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.
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 = ax
a = 3
c Find y when x = 1 to obtain B.
y = x4 − 3 x3
= −2
∴ B is (1, −2) Distance from road at A to river at B is 2 km.
d Distance OD = 2 22.25 8.54+
= 5.0625 72.9316+
= 0.5625 72.9316+
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.
y = a( x + 3)( x + 1)( x − 1)( x − 3)
1 0, 2
b x = 4,
= 1
= 1
−

Range 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
y = 1
f Domain [−4, 4]

7 Since smooth landing, the graph 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 = 502 × a(50 − b)
10 = 2500a(50 − b) [1]
d
d
y
0 = 7500a − 100ab
b = 75 Substituting into [1]
⇒ 10 = 2500a(50 − 75)
6250
− − = 0.04672 km
= 46.72 m c Seems to be extremely low 2 km from touchdown.
Therefore, not very accurate.
∴ y = a( x − 4)2( x + 4)2
(0, −3) ⇒ −3 = a(0 − 4)2(0 + 4)2
−3 = 256a
a = 3
256
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 = −16ab
ab = 3
16 [1]
But zero gradient at G ⇒ y = a( x2 − 8 x + 16)( x − b)
y = a( x3 − bx2 − 8 x2 + 8bx + 16 x − 16b) d
d
y
0 = 8b + 16 since a ≠ 0
8b = −16
b = −2
16
c i y = ax2 – 3
Since strut is 1 m long, F has coordinates (2, −2).
(2, −2) ⇒ −2 = a × 4 − 3
1 = 4a
a = 1
−2 = 4a
a = 1
As the gradients are different, the connection is not smooth.
iii Lower equation is still y = 2
3 4 x −
a = − 3
As the gradients are the same, the graphs meet smoothly at (3, −0.75).
d quartic: x = 2 ⇒ y = 2 23(2 4) (2 4)
256
= 0.0875 m
32 − − +
= −0.1 m
2 3
= 0.4 m
The quartic model is the closest to the actual ramp with 1.6875 m.
F u n c t i o n s a n d t r a n s f o r m a t i o n s M M 1 2 - 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 in
the y direction.
c a = −3
⇒ Dilation by a factor of 3 in the y direction and reflection in the
x-axis
2 in
the y direction, reflection in the x-axis, translation 1 unit up.
f h = 2
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
the y direction, translation of 3
2
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 (v)
⇒ graph (i)
⇒ graph (iii)
4 y = a(k − h)2 + k a Turning point (2, 2)
h = 2, k = 2
0 = a(0 − 2)2 + 2
h = −1, k = −2
0 = a(0 + 1)2 − 2
a = 2
Equation is y = 2( x + 1)2 − 2 c Turning point (1, 3)
h = 1, k = 3
0 = a(0 − 1)2 + 3
d Turning point (−2, −4)
h = −2, k = −4
0 = a(0 + 2)2 − 4
5 Turning point (c, d )
h = c, k = d
⇒ Only alternative is
6 y = x2
a a = 1
⇒ y = ( x − 2)2 − 1
⇒ y = 3 x2 − 2
⇒ y = −( x + 3)2
7 y = a( x − h)2 + k a Turning point (3, −4)
h = 3, k = −4
0 = a(5 − 3)2 − 4
b Turning point (−1, 1)
h = −1, k = 1
Using (0, −1) −1 = a(0 + 1)2 + 1
−1 = a + 1
h = −3, k = −4
Using (0, −1)
h = 2, k = 2
0 = a(0 − 2)2 + 2
h = 1, k = 6
9 = a(0 − 1)2 + 6
h = −2, k = 8
Using (− 2 − 2, 0)
+ 8 0 = 2a + 8
M M 1 2 - 2 22 F u n c t i o n s a n d t r a n s f o r m a t i o n s
8 y = x2
⇒ y = 2 x2
⇒ y = −2 x2
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, 3 y)
⇒ (2 x, y)
f A translation of 2 units horizontally in the positive
direction ⇒ ( x + 2, y)
the y-axis ⇒ ( x, y − 1)
10 a y = a( x − h)2 + k
Turning point ( z , −8)
10 = a(0 − z )2 − 8
10 = az 2 − 8
or 0 = 2
equation 1
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 z 2 − 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 = 2
y = 2
a Range is y ≤ 3 ⇒ a < 0
and k = 3 is the y-coordinate of
turning point. h = −4
Using
x = −7 or −1
f ( x + 2) −3, −4 ≤ x ≤ 0
2 h = 2, k = −3
f ( x − 2) − 3, 0 ≤ x ≤ 4
3 h = −4
4 h = 4
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 x3
a = 7 The graph is dilated by a factor of 7 in the y direction.
b y = 2
2
reflection in the x-axis.
c y = x3 + 4
d y = 6 − x3
e y = ( x − 1)3 h = 1 Translated 1 unit right.
f y = −( x + 3)3
g y = 4(2 − x)3
a = 4 Dilated in the y direction by a factor of 4. Reflection in the y-axis.
h = 2Translated 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 in the
y direction, reflected in the x-axis, translated 1 unit right, translated 6 units up.
k y = 1
y direction, translated 5
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)
F u n c t i o n s a n d t r a n s f o r m a t i o n s M M 1 2 - 2 23
3 a y = 3
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
b y = 1 − 2 x3
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
a > 0: A positive cubic Stationary point of inflection is
(0, −6)
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
a < 0: A negative cubic.
Stationary point of inflection is (2, 0).
y-intercept: x = 0
( x − 2)3 = 0
x − 2 = 0
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&nb

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Maths Quest 12 Mathematical Methods CAS Solutions Manual