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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · Transforming Functions

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.13
2.x=4x = 4
3.x∈R,  x≠3x \in \mathbb{R},\; x \neq 3
4.x≥1x \geq 1
5.6
6.Yes — passes the vertical line test.
7.R\mathbb{R}
8.f(−2)=−8f(-2) = -8
9.x>4x > 4
10.f(n)=3n+1f(n) = 3n + 1
11.x=3x = 3 or x=4x = 4
12.x=±4x = \pm 4
13.x=−2x = -2 or x=−3x = -3
14.Vertex (3,−4)(3, -4); minimum
15.(0,7)(0, 7)
16.x=−12x = -\dfrac{1}{2} or x=3x = 3
17.x=3x = 3
18.x=−1x = -1 or x=5x = 5
19.Downwards (because the leading coefficient is negative).
20.x=2x = 2 and x=−5x = -5
21.Translation 5 units left
22.Translation 4 units up
23.Reflection in the xx-axis
24.Vertical stretch (dilation) by factor 3
25.(0,1)(0, 1)
26.(6,−3)(6, -3)
27.(−2,5)(-2, 5)
28.(2,5)(2, 5)
29.Horizontal compression by factor 12\dfrac{1}{2}
30.Translation 3 units up
Silver
31.17
32.2x2−12x^2 - 1
33.f−1(x)=x−52f^{-1}(x) = \dfrac{x - 5}{2}
34.Domain x≠2x \neq 2; range y≠0y \neq 0
35.x=1x = 1
36.Domain R\mathbb{R}; range y≥0y \geq 0
37.x≥3x \geq 3
38.x=−2x = -2 or x=3x = 3
39.3
40.f(−3)=−5f(-3) = -5; f(2)=4f(2) = 4
41.(x+4)2−6(x + 4)^2 - 6
42.(3,2)(3, 2)
43.x=−32x = -\dfrac{3}{2} or x=2x = 2
44.x2+2x−15x^2 + 2x - 15
45.y=2x2−12x+13y = 2x^2 - 12x + 13
46.x=−4±6x = -4 \pm \sqrt{6}
47.x=5x = 5 or x=−6x = -6
48.(a) (3,−4)(3, -4); (b) x=1x = 1 and x=5x = 5
49.y=2(x−2)2−3y = 2(x - 2)^2 - 3
50.x=3x = 3 cm
51.y=(x−3)2+2y = (x - 3)^2 + 2
52.y=−x2+4y = -x^2 + 4
53.Vertical stretch by factor 3
54.(−1,−1)(-1, -1)
55.Turning point (−1,−4)(-1, -4) — maximum; yy-intercept (0,−7)(0, -7)
56.(0,4)(0, 4) and (4,−2)(4, -2)
57.y=−(x+2)2y = -(x + 2)^2
58.(3,8)(3, 8)
59.Max at (3,8)(3, 8)
60.x=3x = 3
Gold
61.f−1(x)=x−52f^{-1}(x) = \dfrac{x - 5}{2}; domain R\mathbb{R}
62.(f∘g)(x)=x−4(f \circ g)(x) = \sqrt{x - 4}; domain x≥4x \geq 4
63.y≥2y \geq 2
64.f(2.7)=2f(2.7) = 2; f(−1.4)=−2f(-1.4) = -2
65.Restrict to x≥0x \geq 0; f−1(x)=xf^{-1}(x) = \sqrt{x}
66.Yes — exponential decay with horizontal asymptote y=0y = 0 and f(0)=4f(0) = 4.
67.(g∘f)(x)=12x+1(g \circ f)(x) = \dfrac{1}{2x + 1}; domain x≠−12x \neq -\dfrac{1}{2}
68.x=−1x = -1 or x=5x = 5
69.(a) CHF 0.50 cost per extra bottle; (b) CHF 200 fixed cost.
70.y=xy = x
71.(x+4)2−6(x + 4)^2 - 6; range y≥−6y \geq -6
72.b=−6b = -6
73.No real roots (Δ=16−24=−8<0\Delta = 16 - 24 = -8 < 0).
74.k=2k = 2
75.2≤x≤32 \leq x \leq 3
76.x2−3x−10x^2 - 3x - 10
77.(a) 21 m; (b) at t=2t = 2 s
78.Sum −5-5; product −6-6
79.(a) 9 m; (b) 1≤x≤71 \leq x \leq 7
80.−6<k<6-6 < k < 6
81.y=−(x−2)2+4y = -(x - 2)^2 + 4; vertex (2,4)(2, 4)
82.Vertical stretch ×2, then translate 3 right and 1 down.
83.x=−1x = -1 or x=3x = 3
84.(2,1)(2, 1)
85.Feature at x=13x = \dfrac{1}{3} with yy-value 6.
86.y=2(x−2)2−1y = 2(x - 2)^2 - 1
87.xx-intercepts at x=1x = 1 and x=5x = 5 (unchanged); yy-intercept (0,5)(0, 5).
88.g(x)=f(−x)−2g(x) = f(-x) - 2
89.y=2(x+3)2−5y = 2(x + 3)^2 - 5; vertical stretch ×2, translate 3 left, then 5 down.
90.(3,4)(3, 4)
Platinum
91.(g∘f)(x)=2x+12x(g \circ f)(x) = \dfrac{2x + 1}{2x}; domain x≠0x \neq 0
92.(f∘f)(x)=x−12−x(f \circ f)(x) = \dfrac{x - 1}{2 - x}; defined for x≠1,2x \neq 1, 2
93.f−1(x)=x+3x−2f^{-1}(x) = \dfrac{x + 3}{x - 2}; domain x≠2x \neq 2
94.x<−2x < -2 or x>2x > 2
95.a=−1a = -1
96.y≥3y \geq 3
97.f−1(x)=x−13f^{-1}(x) = \dfrac{x - 1}{3}
98.x=3.5x = 3.5
99.f−1(x)=2+xf^{-1}(x) = 2 + \sqrt{x}; domain x≥0x \geq 0
100.Not a function: a vertical line e.g. x=0x = 0 meets the circle at (0,5)(0, 5) and (0,−5)(0, -5). The two function pieces are f1(x)=25−x2f_1(x) = \sqrt{25 - x^2} (top) and f2(x)=−25−x2f_2(x) = -\sqrt{25 - x^2} (bottom).
101.5 cm by 8 cm
102.k>−14k > -\dfrac{1}{4}
103.a=−52a = -\dfrac{5}{2}, b=112b = \dfrac{11}{2}, c=5c = 5
104.y=(x−3)2+2y = (x - 3)^2 + 2; translation 3 right and 2 up.
105.y=3(x−2)2+3y = 3(x - 2)^2 + 3
106.(−4,−3)(-4, -3) and (1,2)(1, 2)
107.−5-5
108.Δ=(2k)2−4(k2+1)=−4<0\Delta = (2k)^2 - 4(k^2 + 1) = -4 < 0.
109.20 m parallel to wall, 10 m perpendicular; area 200 m2^2
110.a=−0.5a = -0.5, b=4b = 4, c=1.5c = 1.5; h(3)=9h(3) = 9
111.Maximum at (5,8)(5, 8)
112.Compress vertically by factor 12\dfrac{1}{2}, then translate 3 left.
113.72
114.Translate 3 left, then 4 down.
115.y=3sin⁡(2x)+2y = 3\sin(2x) + 2
116.Vertex (−2,−1)(-2, -1); g(x)=x2+4x+3g(x) = x^2 + 4x + 3
117.(3,7)(3, 7) and (5,13)(5, 13); average rate of change 3 (unchanged from ff).
118.h=4h = 4, k=−7k = -7
119.Corresponds to (5,2)(5, 2); reflection of ff in the line y=xy = x.
120.ff is odd. After translation, g(x)=(x−2)3−4(x−2)g(x) = (x - 2)^3 - 4(x - 2) is neither.

Pack B — Answers

Bronze
1.11
2.x=4x = 4
3.x∈R,  x≠−2x \in \mathbb{R},\; x \neq -2
4.x≥5x \geq 5
5.10
6.No — a vertical line, e.g. x=4x = 4, meets x=y2x = y^2 at (4,2)(4, 2) and (4,−2)(4, -2).
7.R\mathbb{R}
8.f(3)=−1f(3) = -1
9.x>−1x > -1
10.f(n)=3n−1f(n) = 3n - 1
11.x=4x = 4 or x=5x = 5
12.x=±5x = \pm 5
13.x=−2x = -2 or x=−5x = -5
14.Vertex (−2,5)(-2, 5); minimum
15.(0,4)(0, 4)
16.x=−1x = -1 or x=23x = \dfrac{2}{3}
17.x=2x = 2
18.x=−4x = -4 or x=2x = 2
19.Upwards.
20.x=4x = 4 and x=−3x = -3
21.Translation 3 units right
22.Translation 5 units down
23.Reflection in the yy-axis
24.Vertical dilation by factor 12\dfrac{1}{2} (compression)
25.(0,6)(0, 6)
26.(1,−3)(1, -3)
27.(−3,7)(-3, 7)
28.(−1,−4)(-1, -4)
29.Horizontal stretch by factor 3
30.Translation 4 units right
Silver
31.31
32.3x2+53x^2 + 5
33.f−1(x)=x+13f^{-1}(x) = \dfrac{x + 1}{3}
34.Domain x≠−3x \neq -3; range y≠0y \neq 0
35.x=4x = 4
36.Domain R\mathbb{R}; range y≥0y \geq 0
37.x≤5x \leq 5
38.x=−3x = -3 or x=2x = 2
39.−2-2
40.f(−1)=−1f(-1) = -1; f(3)=9f(3) = 9
41.(x+3)2−4(x + 3)^2 - 4
42.(−2,−5)(-2, -5)
43.x=−2x = -2 or x=13x = \dfrac{1}{3}
44.2x2+7x−42x^2 + 7x - 4
45.y=−3x2−6x+1y = -3x^2 - 6x + 1
46.x=3±7x = 3 \pm \sqrt{7}
47.x=7x = 7 or x=−8x = -8
48.(a) (−1,−9)(-1, -9); (b) x=−4x = -4 and x=2x = 2
49.y=3(x+2)2−13y = 3(x + 2)^2 - 13
50.x=3x = 3 cm
51.y=(x+4)2−1y = (x + 4)^2 - 1
52.y=−(x+3)2y = -(x + 3)^2
53.Vertical compression by factor 12\frac{1}{2}
54.(−1,7)(-1, 7)
55.Turning point (−2,4)(-2, 4) — still a minimum; yy-intercept (0,7)(0, 7)
56.(0,5)(0, 5) and (4,2)(4, 2)
57.y=−(x−3)2y = -(x - 3)^2
58.(18,8)(18, 8)
59.Now a minimum at (2,−3)(2, -3)
60.x=−2x = -2
Gold
61.f−1(x)=x+34f^{-1}(x) = \dfrac{x + 3}{4}; domain R\mathbb{R}
62.(f∘g)(x)=2x+1(f \circ g)(x) = \sqrt{2x + 1}; domain x≥−12x \geq -\frac{1}{2}
63.y≤5y \leq 5
64.f(3.9)=3f(3.9) = 3; f(−2.1)=−3f(-2.1) = -3
65.Restrict to x≤0x \leq 0; f−1(x)=−xf^{-1}(x) = -\sqrt{x}
66.Yes — f(0)=1+3=4f(0) = 1 + 3 = 4, decreasing to y=1y = 1.
67.(f∘g)(x)=2x+1(f \circ g)(x) = \dfrac{2}{x} + 1; domain x≠0x \neq 0
68.x=−5x = -5 or x=1x = 1
69.(a) CHF 0.80 per bottle; (b) CHF 150 fixed cost.
70.Domain of f−1f^{-1} = range of ff (and range of f−1f^{-1} = domain of ff).
71.(x−3)2−7(x - 3)^2 - 7; range y≥−7y \geq -7
72.b=4b = 4
73.One repeated real root (Δ=36−36=0\Delta = 36 - 36 = 0).
74.k=1k = 1
75.x<−3x < -3 or x>4x > 4
76.x2+x−12x^2 + x - 12
77.(a) 47 m; (b) at t=3t = 3 s
78.Sum 32\dfrac{3}{2}; product −2-2
79.(a) 16 m; (b) −1≤x≤7-1 \leq x \leq 7
80.−4<k<4-4 < k < 4
81.y=−[(x+1)2−3]=−(x+1)2+3y = -[(x + 1)^2 - 3] = -(x + 1)^2 + 3; vertex (−1,3)(-1, 3)
82.Reflect in xx-axis, then translate 2 left and 5 up.
83.x=−5x = -5 or x=1x = 1
84.(2,3)(2, 3)
85.Feature at x=2x = 2 with yy-value 32\dfrac{3}{2}.
86.y=−2(x+1)2+3y = -2(x + 1)^2 + 3
87.xx-intercepts at x=−1x = -1 and x=−5x = -5; yy-intercept (0,−5)(0, -5) unchanged.
88.g(x)=−f(x−3)g(x) = -f(x - 3)
89.y=−(x−2)2+3y = -(x - 2)^2 + 3; reflect in xx-axis, translate 2 right, then 3 up.
90.Maximum at (−1,10)(-1, 10)
Platinum
91.(f∘g)(x)=3x−1x−1(f \circ g)(x) = \dfrac{3x - 1}{x - 1}; domain x≠1x \neq 1
92.(f∘f)(x)=2−x3−2x(f \circ f)(x) = \dfrac{2 - x}{3 - 2x}
93.f−1(x)=2x+13−xf^{-1}(x) = \dfrac{2x + 1}{3 - x}; domain x≠3x \neq 3
94.x≥1x \geq 1 with x≠3x \neq 3
95.a=−1a = -1
96.y≤4y \leq 4
97.f−1(x)=x+13f^{-1}(x) = \dfrac{x + 1}{3}
98.x=69x = \dfrac{6}{9} (i.e. 23\frac{2}{3})
99.f−1(x)=−1−xf^{-1}(x) = -1 - \sqrt{x}; domain x≥0x \geq 0
100.Not a function. Top: f1(x)=21−x2/9f_1(x) = 2\sqrt{1 - x^2/9}; bottom: f2(x)=−21−x2/9f_2(x) = -2\sqrt{1 - x^2/9}.
101.3 cm by 8 cm
102.k>0k > 0
103.a=3a = 3, b=0b = 0, c=−2c = -2
104.y=(x+2)2−5y = (x + 2)^2 - 5; translation 2 left and 5 down.
105.y=3(x+1)2−4y = 3(x + 1)^2 - 4
106.(0,−1)(0, -1) and (3,2)(3, 2)
107.−11-11
108.Δ=(−2k)2−4(k2+4)=−16<0\Delta = (-2k)^2 - 4(k^2 + 4) = -16 < 0.
109.30 m parallel, 15 m perpendicular; area 450 m2^2
110.a=−1a = -1, b=5b = 5, c=2c = 2; h(3)=8h(3) = 8
111.Minimum at (3,−2)(3, -2)
112.Translate down 4, then reflect in the xx-axis.
113.12
114.Vertical stretch ×2, reflect in xx-axis, translate 1 right, then 5 up.
115.y=12sin⁡ ⁣(x2)−1y = \dfrac{1}{2}\sin\!\left(\dfrac{x}{2}\right) - 1
116.Vertex (3,3)(3, 3) — now a maximum; g(x)=−(x−3)2+3g(x) = -(x - 3)^2 + 3
117.(1,4)(1, 4) and (3,16)(3, 16); average rate of change 6 (doubled).
118.h=−3h = -3, k=2k = 2
119.Corresponds to (3,−1)(3, -1); same reflection.
120.ff is even. g(x)=(x−3)2+1g(x) = (x - 3)^2 + 1 is neither (translation in xx breaks yy-axis symmetry).

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 13. (b) 1. (c) x=4x = 4.
Full working
(a) f(5)=2(5)+3=13f(5) = 2(5) + 3 = 13. (b) f(−1)=2(−1)+3=1f(-1) = 2(-1) + 3 = 1. (c) 2x+3=11⇒x=42x + 3 = 11 \Rightarrow x = 4.
2Problem 2
Answer
(a) x∈R,x≠3x \in \mathbb{R}, x \neq 3. (b) x≥1x \geq 1. (c) x<4x < 4.
Full working
(a) Denominator ≠0\neq 0. (b) Radicand ≥0\geq 0. (c) Radicand must be **strictly** positive (also in denominator).
3Problem 3
Answer
(a) Yes. (b) No. (c) No. (d) Yes.
Full working
(a) Each xx gives exactly one yy. (b) x=4x = 4 gives y=2y = 2 or −2-2 — fails VLT. (c) Circle — vertical lines −3<x<3-3 < x < 3 give two yy. (d) Each xx gives exactly one non-negative yy.
4Problem 4
Answer
(a) 17. (b) 2x2−12x^2 - 1. (c) f−1(x)=x−52f^{-1}(x) = \frac{x - 5}{2}; domain R\mathbb{R}.
Full working
(a) g(3)=6g(3) = 6; f(6)=17f(6) = 17. (b) f(g(x))=2(x2−3)+5=2x2−1f(g(x)) = 2(x^2 - 3) + 5 = 2x^2 - 1. (c) y=2x+5⇒x=y−52y = 2x + 5 \Rightarrow x = \frac{y - 5}{2}, so f−1(x)=x−52f^{-1}(x) = \frac{x - 5}{2}. Domain of f−1f^{-1} = range of ff = R\mathbb{R}.
5Problem 5
Answer
(a) 5−x2\sqrt{5 - x^2}. (b) −5≤x≤5-\sqrt{5} \leq x \leq \sqrt{5}. (c) 0≤y≤50 \leq y \leq \sqrt{5}.
Full working
(a) f(g(x))=5−x2f(g(x)) = \sqrt{5 - x^2}. (b) Need 5−x2≥0⇒x2≤55 - x^2 \geq 0 \Rightarrow x^2 \leq 5, so −5≤x≤5-\sqrt{5} \leq x \leq \sqrt{5}. (c) Max when x=0x = 0: 5\sqrt{5}. Min at endpoints: 0. Range [0,5][0, \sqrt{5}].
6Problem 6
Answer
(a) −3,0,4,9-3, 0, 4, 9. (b) Linear piece, then upward parabola from (0,0)(0,0) to (3,9)(3,9), then horizontal at y=9y = 9. (c) Range: f≥−5f \geq -5, but with the linear piece dipping arbitrarily low as x→−∞x \to -\infty; on the stated domain the range is [−5,9][-5, 9].
Full working
(a) Pick the correct piece for each xx. (b) Three sections joined; check continuity at x=0x = 0 (yes, both give 0) and x=3x = 3 (yes, both give 9). (c) On −3≤x≤5-3 \leq x \leq 5: minimum at x=−3x = -3: f(−3)=−5f(-3) = -5. Maximum 9 reached at x=3x = 3 and beyond. Range [−5,9][-5, 9].
7Problem 7
Answer
(a) x=−2x = -2 or x=5x = 5. (b) −3<x<5-3 < x < 5, i.e. (−3,5)(-3, 5). (c) V-shape with vertex at (2,−1)(2, -1); xx-intercepts (1,0)(1, 0) and (3,0)(3, 0); yy-intercept (0,1)(0, 1).
Full working
(a) 2x−3=±7⇒x=52x - 3 = \pm 7 \Rightarrow x = 5 or x=−2x = -2. (b) ∣x−1∣<4⇔−4<x−1<4⇔−3<x<5|x - 1| < 4 \Leftrightarrow -4 < x - 1 < 4 \Leftrightarrow -3 < x < 5. (c) V-shape shifted right 2 and down 1. Vertex (2,−1)(2, -1). Set y=0y = 0: ∣x−2∣=1⇒x=1|x - 2| = 1 \Rightarrow x = 1 or 33. yy-intercept ∣0−2∣−1=1|0 - 2| - 1 = 1.
8Problem 8
Answer
(a) C(d)=cd+FC(d) = cd + F. (b) c=0.2c = 0.2, F=55F = 55. (c) CHF 155.
Full working
(b) System: 200c+F=95200c + F = 95, 350c+F=125350c + F = 125. Subtract: 150c=30⇒c=0.20150c = 30 \Rightarrow c = 0.20. Then F=95−40=55F = 95 - 40 = 55. (c) C(500)=0.20(500)+55=155C(500) = 0.20(500) + 55 = 155.
9Problem 9
Answer
(a) c=3c = 3; a+b+c=6a + b + c = 6; 4a+2b+c=134a + 2b + c = 13. (b) a=2a = 2, b=1b = 1, c=3c = 3. (c) f(−1)=4f(-1) = 4.
Full working
(a) Substitute each point. From f(0)=3f(0) = 3: c=3c = 3. Then a+b=3a + b = 3 and 4a+2b=104a + 2b = 10. (b) From 4a+2b=104a + 2b = 10: 2a+b=52a + b = 5. Subtract a+b=3a + b = 3: a=2a = 2. Then b=1b = 1. (c) f(−1)=2(1)+(−1)+3=4f(-1) = 2(1) + (-1) + 3 = 4.
10Problem 10
Answer
(a) (x−2)2+3(x - 2)^2 + 3. (b) Min value 3 at x=2x = 2. (c) Range y≥3y \geq 3.
Full working
(a) Half of 4 is 2: (x−2)2=x2−4x+4(x - 2)^2 = x^2 - 4x + 4, so f(x)=(x−2)2+3f(x) = (x - 2)^2 + 3. (b) Minimum 3 at x=2x = 2. (c) Range [3,∞)[3, \infty).
11Problem 11
Answer
(a) f(x)=12x+2f(x) = \frac{1}{2}x + 2. (b) f−1(x)=2x−4f^{-1}(x) = 2x - 4. (c) ff and f−1f^{-1} are reflections of each other in y=xy = x.
Full working
(a) Gradient =4−14−(−2)=12= \frac{4 - 1}{4 - (-2)} = \frac{1}{2}. Through (−2,1)(-2, 1): 1=12(−2)+c⇒c=21 = \frac{1}{2}(-2) + c \Rightarrow c = 2. So f(x)=x2+2f(x) = \frac{x}{2} + 2. (b) y=x2+2⇒x=2y−4⇒f−1(x)=2x−4y = \frac{x}{2} + 2 \Rightarrow x = 2y - 4 \Rightarrow f^{-1}(x) = 2x - 4. (c) The two lines are reflections of each other across y=xy = x.
12Problem 12
Answer
(a) Anya: 3, Bao: 3, Cara: 3. (b) x≠1x \neq 1. (c) Not strictly equal — they agree everywhere except at x=1x = 1. (d) Cara — she states the domain restriction explicitly.
Full working
(a) Anya: 4−12−1=3\frac{4 - 1}{2 - 1} = 3. Bao: 2+1=32 + 1 = 3. Cara: 2+1=32 + 1 = 3 (with x≠1x \neq 1, OK). (b) Anya's denominator must not be zero, so x≠1x \neq 1. (c) Bao's formula is defined at x=1x = 1 (giving 2), but Anya's gives 00\frac{0}{0} — undefined. So they have different natural domains. (d) Cara — she matches Anya's natural domain and clears up the ambiguity.
13Problem 13
Answer
(a) x=3x = 3 or x=4x = 4. (b) x=−12x = -\frac{1}{2} or x=3x = 3. (c) x=±3x = \pm 3.
Full working
(a) (x−3)(x−4)=0(x - 3)(x - 4) = 0. (b) (2x+1)(x−3)=0(2x + 1)(x - 3) = 0. (c) Difference of squares: (x−3)(x+3)=0(x - 3)(x + 3) = 0.
14Problem 14
Answer
(a) (x+4)2−6(x + 4)^2 - 6. (b) x=−4±6x = -4 \pm \sqrt{6}. (c) y≥−6y \geq -6. (d) Translation 5 units left.
Full working
(a) Half of 8 is 4. (x+4)2=x2+8x+16(x + 4)^2 = x^2 + 8x + 16, so f(x)=(x+4)2−6f(x) = (x + 4)^2 - 6. (b) (x+4)2=6⇒x=−4±6(x + 4)^2 = 6 \Rightarrow x = -4 \pm \sqrt{6}. (c) Min value −6-6, so range y≥−6y \geq -6. (d) Replacing xx with x+5x + 5 shifts the graph 5 units **left**.
15Problem 15
Answer
(a) x=2±3x = 2 \pm \sqrt{3}. (b) x=−3x = -3 or x=12x = \frac{1}{2}. (c) x=2x = 2 or x=13x = \frac{1}{3}.
Full working
(a) x=4±16−42=2±3x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}. (b) Factor: (2x−1)(x+3)=0(2x - 1)(x + 3) = 0. (c) (3x−1)(x−2)=0(3x - 1)(x - 2) = 0.
16Problem 16
Answer
(a) xx-intercepts (2,0)(2, 0) and (4,0)(4, 0); yy-intercept (0,8)(0, 8). (b) (3,−1)(3, -1). (c) x=3x = 3.
Full working
(a) x2−6x+8=(x−2)(x−4)x^2 - 6x + 8 = (x - 2)(x - 4). yy-intercept at x=0x = 0: 8. (b) Axis x=3x = 3; f(3)=9−18+8=−1f(3) = 9 - 18 + 8 = -1. (c) Stated. (d) Upward parabola with min at (3,−1)(3, -1), crossing xx-axis at x=2,4x = 2, 4 and yy-axis at 8.
17Problem 17
Answer
(a) 1.5 m — initial height (release point). (b) t=2t = 2 s. (c) 21.5 m. (d) t≈4.07t \approx 4.07 s.
Full working
(a) h(0)=1.5h(0) = 1.5 m. (b) Axis t=2010=2t = \frac{20}{10} = 2. (c) h(2)=−20+40+1.5=21.5h(2) = -20 + 40 + 1.5 = 21.5 m. (d) h(t)=0h(t) = 0: −5t2+20t+1.5=0⇒5t2−20t−1.5=0⇒t=20±400+3010=2±43010-5t^2 + 20t + 1.5 = 0 \Rightarrow 5t^2 - 20t - 1.5 = 0 \Rightarrow t = \frac{20 \pm \sqrt{400 + 30}}{10} = 2 \pm \frac{\sqrt{430}}{10}. Positive: t≈4.07t \approx 4.07.
18Problem 18
Answer
(a) 9 m at x=4x = 4 m. (b) h(0)=−7h(0) = -7 — nozzle is 7 m below ground. (c) x=1x = 1 or x=7x = 7.
Full working
(a) Vertex form: max 9 at x=4x = 4. (b) h(0)=−16+9=−7h(0) = -16 + 9 = -7 — the nozzle sits 7 m below ground level. (c) (x−4)2=9⇒x=1(x - 4)^2 = 9 \Rightarrow x = 1 or 77.
19Problem 19
Answer
(a) Length =80−2x= 80 - 2x; area A=x(80−2x)=80x−2x2A = x(80 - 2x) = 80x - 2x^2. (b) x=20x = 20. (c) Max area 800 m2^2 with length 40 m.
Full working
(a) Two widths and one length total fencing: 2x+L=80⇒L=80−2x2x + L = 80 \Rightarrow L = 80 - 2x. A=xL=80x−2x2A = xL = 80x - 2x^2. (b) Vertex of A(x)=−2x2+80xA(x) = -2x^2 + 80x: x=804=20x = \frac{80}{4} = 20. (c) A(20)=1600−800=800A(20) = 1600 - 800 = 800; L=40L = 40.
20Problem 20
Answer
(a) See working. (b) k=2k = 2. (c) (1,3)(1, 3).
Full working
(a) Equate yy's: x2+k=2x+1⇒x2−2x+(k−1)=0x^2 + k = 2x + 1 \Rightarrow x^2 - 2x + (k - 1) = 0. (b) Tangent ⇔ Δ=0\Delta = 0: 4−4(k−1)=0⇒k=24 - 4(k - 1) = 0 \Rightarrow k = 2. (c) Substitute: x2−2x+1=0⇒(x−1)2=0x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2 = 0, so x=1x = 1, y=2(1)+1=3y = 2(1) + 1 = 3.
21Problem 21
Answer
x≤−2x \leq -2 or x≥3x \geq 3; i.e. (−∞,−2]∪[3,∞)(-\infty, -2] \cup [3, \infty).
Full working
Factor: (x−3)(x+2)≥0(x - 3)(x + 2) \geq 0. The parabola opens upwards and the product is ≥0\geq 0 outside (and at) the roots: x≤−2x \leq -2 or x≥3x \geq 3.
22Problem 22
Answer
(a) c=5c = 5; a+b+c=6a + b + c = 6; 4a+2b+c=134a + 2b + c = 13. (b) a=3a = 3, b=−2b = -2, c=5c = 5. (c) Axis x=13x = \frac{1}{3}; vertex (13,143)\left(\frac{1}{3}, \frac{14}{3}\right).
Full working
(a) From f(0)=5f(0) = 5: c=5c = 5. Then a+b=1a + b = 1 and 4a+2b=8⇒2a+b=44a + 2b = 8 \Rightarrow 2a + b = 4. (b) Subtract: a=3a = 3, b=−2b = -2. (c) Axis x=−b2a=26=13x = -\frac{b}{2a} = \frac{2}{6} = \frac{1}{3}. f(1/3)=3(1/9)−2(1/3)+5=1/3−2/3+5=−1/3+5=14/3f(1/3) = 3(1/9) - 2(1/3) + 5 = 1/3 - 2/3 + 5 = -1/3 + 5 = 14/3.
23Problem 23
Answer
(a) n(n+1)=156n(n + 1) = 156. (b) 12 and 13.
Full working
(a) Consecutive integers nn and n+1n + 1 multiply to give 156. (b) n2+n−156=0⇒(n−12)(n+13)=0n^2 + n - 156 = 0 \Rightarrow (n - 12)(n + 13) = 0. Positive root n=12n = 12, so integers are 12 and 13.
24Problem 24
Answer
(a) A=(w−4)(200w−4)=200−4w−800w+16=216−4w−800wA = (w - 4)\left(\dfrac{200}{w} - 4\right) = 200 - 4w - \dfrac{800}{w} + 16 = 216 - 4w - \dfrac{800}{w}. (b) Differentiating gives w=200≈14.14w = \sqrt{200} \approx 14.14, so w=14w = 14 cm. (c) Approximately 130.3130.3 cm2^2.
Full working
(a) Printed dimensions: width w−4w - 4, height 200w−4\frac{200}{w} - 4. (b) A=(w−4)(200w−4)A = (w - 4)(\frac{200}{w} - 4). Expand: A=200−4w−800w+16A = 200 - 4w - \frac{800}{w} + 16. Taking derivative: A′(w)=−4+800w2=0⇒w2=200⇒w≈14.14A'(w) = -4 + \frac{800}{w^2} = 0 \Rightarrow w^2 = 200 \Rightarrow w \approx 14.14. Closest integer: w=14w = 14 giving height 20014≈14.29\frac{200}{14} \approx 14.29 → round to 14 to keep the "integer cm" constraint. (c) A(14)=216−56−80014≈216−56−57.14=102.86A(14) = 216 - 56 - \frac{800}{14} \approx 216 - 56 - 57.14 = 102.86 cm2^2 (approximately). [Note: the integer constraint complicates this; the unconstrained optimum is the square poster 200×200\sqrt{200} \times \sqrt{200}, which gives the largest printed area.]
25Problem 25
Answer
(a) Translation 5 units left. (b) Translation 3 units up. (c) Reflection in the xx-axis. (d) Vertical stretch by factor 4.
Full working
Replace xx with x+hx + h: left hh. Add constant outside: up. Negate outside: reflect in xx-axis. Multiply outside: vertical stretch.
26Problem 26
Answer
(a) (−1,−1)(-1, -1), (0,2)(0, 2). (b) (−4,4)(-4, 4), yy-intercept depends on f(3)f(3). (c) (0,−7)(0, -7), max at (−1,−4)(-1, -4). (d) (−2,4)(-2, 4), (0,7)(0, 7) unchanged.
Full working
(a) Vertical shift down 5. (b) Horizontal shift left 3: min →(−4,4)\to (-4, 4). yy-intercept becomes f(0+3)=f(3)f(0 + 3) = f(3) — value of original ff at x=3x = 3. (c) Reflect in xx-axis. (d) Horizontal stretch by 2.
27Problem 27
Answer
(a) y=(x−2)2−3y = (x - 2)^2 - 3. (b) y=−(x+1)2+4y = -(x + 1)^2 + 4. (c) y=2(x−3)2y = 2(x - 3)^2.
Full working
(a) Right 2 replaces xx with x−2x - 2; down 3 subtracts 3. (b) Reflect: y=−x2y = -x^2. Then translate: y=−(x+1)2+4y = -(x + 1)^2 + 4. (c) Stretch first: y=2x2y = 2x^2. Translate right: y=2(x−3)2y = 2(x - 3)^2.
28Problem 28
Answer
(a) Vertex (−3,5)(-3, 5); maximum. (b) x=−3x = -3. (c) (0,−13)(0, -13). (d) Vertical stretch ×2, reflect in xx-axis, translate 3 left and 5 up.
Full working
(a) Vertex (h,k)=(−3,5)(h, k) = (-3, 5); coefficient −2<0-2 < 0 → max. (b) x=−3x = -3. (c) f(0)=−2(9)+5=−13f(0) = -2(9) + 5 = -13. (d) Read off coefficient sign and shifts.
29Problem 29
Answer
(a) y=−3(x−4)2y = -3(x - 4)^2. (b) Reflect in the xx-axis, vertically compress by factor 13\frac{1}{3}, then translate 4 left.
Full working
(a) Translate first: y=(x−4)2y = (x - 4)^2. Stretch: y=3(x−4)2y = 3(x - 4)^2. Reflect: y=−3(x−4)2y = -3(x - 4)^2. (b) Inverse: reverse the order and invert each step.
30Problem 30
Answer
(a) (2,4)(2, 4), (3,7)(3, 7), (4,16)(4, 16). (b) g(3)=f(1)+3=7g(3) = f(1) + 3 = 7. (c) g(4)=f(2)+3=16g(4) = f(2) + 3 = 16.
Full working
Translate right 2 (add 2 to xx) and up 3 (add 3 to yy).
31Problem 31
Answer
(a) x=1,5x = 1, 5 unchanged; yy-intercept (0,−10)(0, -10). (b) x=12,52x = \frac{1}{2}, \frac{5}{2}; yy-intercept (0,−5)(0, -5) unchanged. (c) x=1,5x = 1, 5 unchanged; yy-intercept (0,5)(0, 5).
Full working
Vertical stretch preserves xx-intercepts (zeros remain zeros) and scales the yy-intercept. Horizontal compression halves xx-intercepts and leaves yy-intercept unchanged (it depends on f(0)f(0)). Reflection in xx-axis flips yy-intercept sign.
32Problem 32
Answer
(a) y=3sin⁡(2x)y = 3\sin(2x). (b) Amplitude 3, period π\pi. (c) y=3sin⁡(2x)+1y = 3\sin(2x) + 1.
Full working
(a) Multiply sin⁡\sin by 3 and replace xx with 2x2x. (b) Amplitude = coefficient of sin⁡\sin; period = 2πB=π\frac{2\pi}{B} = \pi. (c) Add 1.
33Problem 33
Answer
y=2f(x)y = 2f(x).
Full working
Each yy-coord has doubled; xx-coords unchanged. This is a vertical stretch by factor 2.
34Problem 34
Answer
(a) x=3x = 3 and y=2y = 2. (b) Translate 3 right and 2 up. (c) (0,53)(0, \frac{5}{3}).
Full working
(a) Asymptotes shift with the graph. (b) Inside: shift right 3; outside: shift up 2. (c) y=1−3+2=53y = \frac{1}{-3} + 2 = \frac{5}{3}.
35Problem 35
Answer
(a) Translate 1 right, vertical stretch ×3, reflect in xx-axis, translate 4 up. (b) (1,4)(1, 4), maximum. (c) (0,1)(0, 1).
Full working
(a) Inside: x−1x - 1 → right 1. Outside: −3⋅-3 \cdot → stretch ×3 and reflect. Then +4+ 4. (b) New vertex at (1,4)(1, 4); coefficient is −3<0-3 < 0 so max. (c) g(0)=−3(0−1)2+4=−3+4=1g(0) = -3(0 - 1)^2 + 4 = -3 + 4 = 1.
36Problem 36
Answer
(a) Chain 1: y=3f(x)+2y = 3f(x) + 2. Chain 2: y=3(f(x)+2)=3f(x)+6y = 3(f(x) + 2) = 3f(x) + 6. (b) Not the same — stretch is applied to the constant 2 as well in Chain 2. (c) Two translations (horizontal and vertical) commute, since each acts on a different coordinate.
Full working
(a) Apply each step to the function in order. (b) Stretch acts before the addition in Chain 1 but after in Chain 2 — so the constant gets stretched in Chain 2. (c) Horizontal and vertical translations commute because they act on xx and yy independently.