Additional Mathematics questions, chapter 2

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Quadratic Functions

45 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working for Section B.

Section A: objective questions

11 marks

Choose the correct answer. Each question carries 1 mark.

Question 1

Solve x² + 4x − 7 = 0. Give the answers correct to 3 decimal places.

x = −5.317 or x = 1.317
x = −3.317 or x = 3.317
x = −1.317 or x = 5.317
x = −2 or x = 11

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(x + 2)² = 11, so x = −2 ± 3.317, which gives −5.317 and 1.317. Forgetting the −2 gives ±3.317.

For the teacher: Applying, low difficulty, standard 2.1.1

Textbook p. 38

Question 2

What are the sum and the product of the roots of 2x² − 7x + 3 = 0?

sum = 72, product = 32
sum = −72, product = 32
sum = 72, product = 3
sum = 7, product = 3

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Sum = −b/a = −(−7)/2 = 7/2 and product = c/a = 3/2. Forgetting to divide by a gives the last two options.

For the teacher: Understanding, low difficulty, standard 2.1.2

Textbook p. 39

Question 3

Which quadratic equation has roots −2 and 5?

x² + 3x − 10 = 0
x² − 3x + 10 = 0
x² − 3x − 10 = 0
x² + 3x + 10 = 0

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Sum = −2 + 5 = 3, product = −10. The equation is x² − (3)x + (−10) = 0.

For the teacher: Applying, low difficulty, standard 2.1.2

Textbook p. 39

Question 4

Find the range of values of x for (x + 3)(x − 5) < 0.

x < −3 or x > 5
−5 < x < 3
x < −5 or x > 3
−3 < x < 5

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The roots are −3 and 5. The graph opens upwards, so it is below the x-axis (negative) between the roots.

For the teacher: Applying, medium difficulty, standard 2.1.3

Textbook p. 42

Question 5

The quadratic equation 2x² − 4x + 5 = 0 has

two different real roots, because b² − 4ac = 24
two equal real roots, because b² − 4ac = 0
no real roots, because b² − 4ac = −24
two different real roots, because b² − 4ac = −24

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D = (−4)² − 4(2)(5) = 16 − 40 = −24, which is negative, so there are no real roots.

For the teacher: Applying, low difficulty, standard 2.2.1

Textbook p. 46

Question 6

The equation x² + kx + 9 = 0 has two equal real roots. The possible values of k are

6 only
−6 only
3 or −3
6 or −6

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D = 0: k² − 36 = 0, so k = ±6. Taking the square root gives both signs.

For the teacher: Applying, medium difficulty, standard 2.2.2

Textbook p. 47

Question 7

The graph of f(x) = x² + 2x + 3 is changed to f(x) = x² + 2x + 7. What happens to the graph?

It becomes narrower
It moves 4 units upwards with the same shape
It moves 4 units to the right
It opens downwards

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Only c changes (3 to 7). Changing c moves the graph vertically; the shape is unchanged.

For the teacher: Understanding, low difficulty, standard 2.3.1

Textbook p. 50

Question 8

Which statement about the graph of f(x) = −x² + 2x − 1 is correct?

It has a minimum point and touches the x-axis at one point only
It has a maximum point and cuts the x-axis at two points
It has a maximum point and touches the x-axis at one point only
It has a minimum point and does not meet the x-axis

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a = −1 < 0, so it has a maximum point. D = 4 − 4(−1)(−1) = 0, so it touches the x-axis once. The whole graph lies on or below the axis.

For the teacher: Understanding, medium difficulty, standard 2.3.2

Textbook p. 52

Question 9

Express f(x) = x² − 6x + 5 in the form (x − h)² + k. What is the minimum point?

(3, −4)
(−3, −4)
(3, 4)
(−3, 4)

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x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4. The vertex is (h, k) = (3, −4).

For the teacher: Applying, medium difficulty, standard 2.3.3

Textbook p. 55

Question 10

The graph of f(x) = 2(x − 3)² + 1 is changed to f(x) = 2(x − 5)² + 1. Which statement is true?

The graph moves 2 units to the left and the minimum value stays 1
The graph moves 2 units upwards and the axis stays x = 3
The graph becomes narrower and the axis stays x = 3
The graph moves 2 units to the right and the minimum value stays 1

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h changes from 3 to 5, so the graph moves 2 units to the right. The axis changes from x = 3 to x = 5; the minimum value k = 1 is unchanged.

For the teacher: Understanding, medium difficulty, standard 2.3.4

Textbook p. 58

Question 11

The height of a stone is h(t) = −5t² + 20t + 1 metres after t seconds. What is the maximum height?

11 m
21 m
41 m
81 m

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h(t) = −5(t − 2)² + 21, so the maximum height is 21 m when t = 2 s. (Check: −20 + 40 + 1 = 21.)

For the teacher: Applying, high difficulty, standard 2.3.6

Textbook p. 62

Section B: structured questions

34 marks

Answer all questions. Show your working.

Question 12

α and β are the roots of the equation 2x² + 3x − 4 = 0.

[2 marks] (a)Find α + β and αβ. Hence form the quadratic equation with roots α + 2 and β + 2.

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α + β = −32, αβ = −2 [1]1
New sum = 52, new product = αβ + 2(α + β) + 4 = −1; 2x² − 5x − 2 = 0 [1]1

Tick each point you got.

[2 marks] (b)Form the quadratic equation with roots 1α and 1β.

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Sum = α + βαβ = 34, product = 1αβ = −12 [1]1
4x² − 3x − 2 = 0 [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 2.1.2

Textbook p. 40

Question 13

Study the inequality 2x² > x + 6.

[3 marks] (a)Find the range of values of x.

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2x² − x − 6 > 0 and (2x + 3)(x − 2) > 0 [1]1
Roots −32 and 2 [1]1
x < −32 or x > 2 [1]1

Tick each point you got.

[2 marks] (b)Write down a quadratic inequality of the form x² + bx + c < 0 whose solution is −2 < x < 5.

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(x + 2)(x − 5) < 0 [1]1
x² − 3x − 10 < 0 [1]1

Tick each point you got.

For the teacher: Creating, medium difficulty, standard 2.1.3

Textbook p. 42

Question 14

The equation x² + 2x + 5 = p(x + 1), where p is a constant, is given.

[2 marks] (a)Show that the equation can be written x² + (2 − p)x + 5 − p = 0 and find the discriminant in terms of p.

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Expand and move: x² + 2x − px + 5 − p = 0 [1]1
D = (2 − p)² − 4(5 − p) = p² − 16 [1]1

Tick each point you got.

[3 marks] (b)Find the range of values of p for which the equation has two real and different roots.

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D > 0: p² − 16 > 0, (p + 4)(p − 4) > 0 [1]1
Roots of the boundary −4 and 4 [1]1
p < −4 or p > 4 [1]1

Tick each point you got.

For the teacher: Analysing, medium difficulty, standard 2.2.2

Textbook p. 47

Question 15

Diagram 1 shows three graphs P, Q and R. They have the form y = ax² + 1, where a is 2, 1 or ½ (not in that order).

Three parabolas P, Q and R of the form y = ax² + 1, all crossing the y-axis at 1; P is the narrowest, R is in the middle and Q is the widest — Diagram 1

[1 mark] (a)State the y-intercept shared by the three graphs.

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1, so (0, 1) [1]1

Tick each point you got.

[1 mark] (b)Which graph has a = 2? Give a reason.

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P, it is the narrowest, and a bigger a gives a narrower graph [1]1

Tick each point you got.

[1 mark] (c)Arrange the three values of a in ascending order, matching them with the graphs.

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Q (½), R (1), P (2) [1]1

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For the teacher: Understanding, low difficulty, standard 2.3.1

Textbook p. 50

Question 16

The quadratic function f(x) = 2x² − 4x + m, where m is a constant, is given.

[2 marks] (a)The graph of f(x) touches the x-axis at one point only. Find the value of m.

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D = 0: 16 − 8m = 0 [1]1
m = 2 [1]1

Tick each point you got.

[2 marks] (b)Find the range of values of m for which the graph does not cut the x-axis.

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D < 0: 16 − 8m < 0 [1]1
m > 2 [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 2.3.2

Textbook p. 53

Question 17

Diagram 2 shows the graph of the quadratic function f(x) = −2x² + 8x + 10. The curve cuts the x-axis at A and B and the y-axis at C. V is the maximum point.

Downward parabola with maximum point V on the dashed axis, cutting the x-axis at A on the left and B on the right and the y-axis at C above the origin — Diagram 2

[3 marks] (a)Express f(x) in the form a(x − h)² + k and state the coordinates of V.

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f(x) = −2(x² − 4x) + 10 = −2[(x − 2)² − 4] + 10 [1]1
f(x) = −2(x − 2)² + 18 [1]1
V = (2, 18) [1]1

Tick each point you got.

[2 marks] (b)Express f(x) in the form a(x − p)(x − q). Hence state the x-coordinates of A and B.

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f(x) = −2(x² − 4x − 5) = −2(x + 1)(x − 5) [1]1
A (−1, 0) and B (5, 0) [1]1

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[1 mark] (c)State the coordinates of C.

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f(0) = 10, so C = (0, 10) [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 2.3.3

Textbook p. 55

Question 18

Diagram 3 shows the height h, in metres, of a ball kicked from the ground, t seconds after the kick. The height is given by h(t) = −5t² + 20t.

Parabola of height h in metres against time t in seconds, starting at the origin, rising to a maximum and returning to the t-axis at t = 4 — Diagram 3

[3 marks] (a)Express h(t) in the form a(t − p)² + q. Hence find the maximum height and when it is reached.

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h(t) = −5(t² − 4t) = −5[(t − 2)² − 4] [1]1
h(t) = −5(t − 2)² + 20 [1]1
Maximum height 20 m when t = 2 s [1]1

Tick each point you got.

[1 mark] (b)Find the time the ball hits the ground again.

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−5t(t − 4) = 0, so t = 4 s (t = 0 is the kick) [1]1

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[3 marks] (c)Aiman says the ball is higher than 15 m for more than 2 seconds. Is he correct? Show your working.

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h(t) > 15: −5t² + 20t − 15 > 0, so t² − 4t + 3 < 0 [1]1
(t − 1)(t − 3) < 0, so 1 < t < 3 [1]1
Time = 3 − 1 = 2 s, exactly 2 seconds, so he is not correct [1]1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 2.3.6

Textbook p. 62

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