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Ecolint Campus des NationsMathématiques
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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · 11.4 Quadratic Functions

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.x=3x = 3 or x=4x = 4
2.x=±4x = \pm 4
3.x=−2x = -2 or x=−3x = -3
4.Vertex (3,−4)(3, -4); minimum
5.(0,7)(0, 7)
6.x=−12x = -\dfrac{1}{2} or x=3x = 3
7.x=3x = 3
8.x=−1x = -1 or x=5x = 5
9.Downwards (because the leading coefficient is negative).
10.x=2x = 2 and x=−5x = -5
Silver
11.(x+4)2−6(x + 4)^2 - 6
12.(3,2)(3, 2)
13.x=−32x = -\dfrac{3}{2} or x=2x = 2
14.x2+2x−15x^2 + 2x - 15
15.y=2x2−12x+13y = 2x^2 - 12x + 13
16.x=−4±6x = -4 \pm \sqrt{6}
17.x=5x = 5 or x=−6x = -6
18.(a) (3,−4)(3, -4); (b) x=1x = 1 and x=5x = 5
19.y=2(x−2)2−3y = 2(x - 2)^2 - 3
20.x=3x = 3 cm
Gold
21.(x+4)2−6(x + 4)^2 - 6; range y≥−6y \geq -6
22.b=−6b = -6
23.No real roots (Δ=16−24=−8<0\Delta = 16 - 24 = -8 < 0).
24.k=2k = 2
25.2≤x≤32 \leq x \leq 3
26.x2−3x−10x^2 - 3x - 10
27.(a) 21 m; (b) at t=2t = 2 s
28.Sum −5-5; product −6-6
29.(a) 9 m; (b) 1≤x≤71 \leq x \leq 7
30.−6<k<6-6 < k < 6
Platinum
31.5 cm by 8 cm
32.k>−14k > -\dfrac{1}{4}
33.a=−52a = -\dfrac{5}{2}, b=112b = \dfrac{11}{2}, c=5c = 5
34.y=(x−3)2+2y = (x - 3)^2 + 2; translation 3 right and 2 up.
35.y=3(x−2)2+3y = 3(x - 2)^2 + 3
36.(−4,−3)(-4, -3) and (1,2)(1, 2)
37.−5-5
38.Δ=(2k)2−4(k2+1)=−4<0\Delta = (2k)^2 - 4(k^2 + 1) = -4 < 0.
39.20 m parallel to wall, 10 m perpendicular; area 200 m2^2
40.a=−0.5a = -0.5, b=4b = 4, c=1.5c = 1.5; h(3)=9h(3) = 9

Pack B — Answers

Bronze
1.x=4x = 4 or x=5x = 5
2.x=±5x = \pm 5
3.x=−2x = -2 or x=−5x = -5
4.Vertex (−2,5)(-2, 5); minimum
5.(0,4)(0, 4)
6.x=−1x = -1 or x=23x = \dfrac{2}{3}
7.x=2x = 2
8.x=−4x = -4 or x=2x = 2
9.Upwards.
10.x=4x = 4 and x=−3x = -3
Silver
11.(x+3)2−4(x + 3)^2 - 4
12.(−2,−5)(-2, -5)
13.x=−2x = -2 or x=13x = \dfrac{1}{3}
14.2x2+7x−42x^2 + 7x - 4
15.y=−3x2−6x+1y = -3x^2 - 6x + 1
16.x=3±7x = 3 \pm \sqrt{7}
17.x=7x = 7 or x=−8x = -8
18.(a) (−1,−9)(-1, -9); (b) x=−4x = -4 and x=2x = 2
19.y=3(x+2)2−13y = 3(x + 2)^2 - 13
20.x=3x = 3 cm
Gold
21.(x−3)2−7(x - 3)^2 - 7; range y≥−7y \geq -7
22.b=4b = 4
23.One repeated real root (Δ=36−36=0\Delta = 36 - 36 = 0).
24.k=1k = 1
25.x<−3x < -3 or x>4x > 4
26.x2+x−12x^2 + x - 12
27.(a) 47 m; (b) at t=3t = 3 s
28.Sum 32\dfrac{3}{2}; product −2-2
29.(a) 16 m; (b) −1≤x≤7-1 \leq x \leq 7
30.−4<k<4-4 < k < 4
Platinum
31.3 cm by 8 cm
32.k>0k > 0
33.a=3a = 3, b=0b = 0, c=−2c = -2
34.y=(x+2)2−5y = (x + 2)^2 - 5; translation 2 left and 5 down.
35.y=3(x+1)2−4y = 3(x + 1)^2 - 4
36.(0,−1)(0, -1) and (3,2)(3, 2)
37.−11-11
38.Δ=(−2k)2−4(k2+4)=−16<0\Delta = (-2k)^2 - 4(k^2 + 4) = -16 < 0.
39.30 m parallel, 15 m perpendicular; area 450 m2^2
40.a=−1a = -1, b=5b = 5, c=2c = 2; h(3)=8h(3) = 8

Problem-solving — Worked Solutions

1Problem 1
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.
2Problem 2
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**.
3Problem 3
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.
4Problem 4
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.
5Problem 5
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.
6Problem 6
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.
7Problem 7
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.
8Problem 8
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.
9Problem 9
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.
10Problem 10
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.
11Problem 11
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.
12Problem 12
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.]