Quadratic Functions and Equations in One Variable: Mathematics Tingkatan 4 questions

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Which of the following is a quadratic expression in one variable?

x³ − 6
3x² + 2y
2x² − 5
4x − 1

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The highest power must be 2 and there must be one variable. x³ − 6 has power 3, 3x² + 2y has two variables and 4x − 1 has power 1.

For the teacher: Remembering, low difficulty, standard 1.1.1

Textbook p. 3

The quadratic expression 2r(r − 3) is written as ax² + bx + c form in r. What is the value of b?

−6
2
−3
6

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Expand: 2r(r − 3) = 2r² − 6r, so a = 2, b = −6 and c = 0. b = −3 forgets to multiply by 2.

For the teacher: Understanding, low difficulty, standard 1.1.1

Textbook p. 4

What type of relation is shown by a quadratic function?

One-to-one
Many-to-one
One-to-many
Many-to-many

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Two different values of x give the same value of f(x), for example f(x) = x² gives 4 for both x = 2 and x = −2. This is a many-to-one relation.

For the teacher: Remembering, low difficulty, standard 1.1.2

Textbook p. 6

Which function has a graph with a maximum point?

f(x) = x² − 5
f(x) = 2x² + x
f(x) = (x − 2)²
f(x) = 5 − 2x²

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A maximum point needs a < 0. In 5 − 2x² the coefficient of x² is −2. The other three have a > 0 and have minimum points.

For the teacher: Understanding, low difficulty, standard 1.1.2

Textbook p. 7

Diagram 1 shows part of the graph of a quadratic function. P and Q are two points at the same height. What is the equation of the axis of symmetry?

A parabola opening upward; P at (0, 1) and Q at (4, 1) are joined by a dashed horizontal line — Diagram 1

x = 4
x = 2
x = 1
y = 1

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P is at x = 0 and Q at x = 4, so the axis is halfway: x = (0 + 4) ÷ 2 = 2. y = 1 is the line PQ, not the axis. The axis is vertical, so its equation has the form x = h.

For the teacher: Applying, low difficulty, standard 1.1.2

Textbook p. 9

The graph of f(x) = x² − 3x + c passes through (−1, 3). What is the value of c?

1
3
−7
−1

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Substitute x = −1 and f(x) = 3: 3 = (−1)² − 3(−1) + c = 1 + 3 + c = 4 + c, so c = −1. c = 1 comes from forgetting that −3(−1) = +3.

For the teacher: Applying, medium difficulty, standard 1.1.3

Textbook p. 13

Which graph is the widest?

f(x) = ½x²
f(x) = x²
f(x) = 2x²
f(x) = 3x²

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For positive a, the smaller the value of a the wider the curve. ½ is the smallest, so ½x² is the widest and 3x² is the narrowest.

For the teacher: Understanding, low difficulty, standard 1.1.3

Textbook p. 11

The graph of f(x) = x² − 6x + 5 has axis of symmetry x = 3. Compared with the y-axis, where does the axis lie, and why?

On the left of the y-axis, because a > 0 and b < 0
On the right of the y-axis, because a > 0 and b > 0
On the right of the y-axis, because a > 0 and b < 0
On the y-axis, because c > 0

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Here a = 1 and b = −6. When a > 0 and b < 0 the axis is on the right of the y-axis. x = 3 is positive, which agrees.

For the teacher: Analysing, medium difficulty, standard 1.1.3

Textbook p. 12

The length of a rectangle is (x + 5) cm and its width is x cm. The area is 84 cm². Which quadratic equation is formed?

x² + 5x + 84 = 0
x² + 5 = 84
2x + 10 = 84
x² + 5x − 84 = 0

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Area = x(x + 5) = x² + 5x. Set equal to 84: x² + 5x = 84, so x² + 5x − 84 = 0. Move 84 to the left with a sign change. 2x + 10 is the perimeter.

For the teacher: Applying, low difficulty, standard 1.1.4

Textbook p. 15

Which value is a root of 2x² − 7x + 3 = 0?

x = 1
x = 3
x = −3
x = 2

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For x = 3: 18 − 21 + 3 = 0. For x = 1: 2 − 7 + 3 = −2. For x = −3: 18 + 21 + 3 = 42. For x = 2: 8 − 14 + 3 = −3. Only x = 3 gives 0.

For the teacher: Understanding, low difficulty, standard 1.1.5

Textbook p. 19

Diagram 2 shows the graph of f(x) = x² − x − 6. What are the roots of the equation x² − x − 6 = 0?

Graph of f(x) = x² − x − 6 crossing the x-axis between −2 and 3 on a grid labelled from −4 to 4 — Diagram 2

−6
2 and −3
−2 and 3
−1 and 6

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The roots are the x-coordinates where the graph meets the x-axis (f(x) = 0): x = −2 and x = 3. −6 is the y-intercept, not a root. 2 and −3 have the signs reversed.

For the teacher: Applying, low difficulty, standard 1.1.5

Textbook p. 18

Solve 2x² + 7x − 4 = 0.

x = 12 or x = −4
x = −12 or x = 4
x = 2 or x = −2
x = 12 or x = 4

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2x² + 7x − 4 = (2x − 1)(x + 4) = 0. So 2x − 1 = 0 gives x = ½ and x + 4 = 0 gives x = −4. Check: 2(¼) + 3.5 − 4 = 0.

For the teacher: Applying, medium difficulty, standard 1.1.6

Textbook p. 21

Section B: structured questions

33 marks

Answer all questions. Show your working.

Diagram 3 shows the graph of the quadratic function f(x) = −x² + 4x + 5. The line y = 5 meets the curve at A and B.

Graph of f(x) = −x² + 4x + 5 with the dashed line y = 5 meeting the curve at A on the y-axis and B — Diagram 3

[2 marks] (a)Write the coordinates of A and B.

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A (0, 5) [1]1
B (4, 5) [1]1

A is on the y-axis, so x = 0 and f(0) = c = 5. Solving −x² + 4x + 5 = 5 gives x = 0 or x = 4.

Tick each point you got.

[1 mark] (b)State the equation of the axis of symmetry.

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x = 2 [1]1

Halfway between 0 and 4. Or x = −4 ÷ (2 × −1) = 2.

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[2 marks] (c)Find the coordinates of the maximum point.

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x = 2 [1]1
f(2) = −4 + 8 + 5 = 9, so (2, 9) [1]1

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[2 marks] (d)State the roots of −x² + 4x + 5 = 0.

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x² − 4x − 5 = 0, (x − 5)(x + 1) = 0 [1]1
x = 5 or x = −1 [1]1

These are the x-intercepts of the graph.

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For the teacher: Understanding, low difficulty, standard 1.1.2

Textbook p. 9

Solve each quadratic equation by factorisation.

[2 marks] (a)x² − 3x − 10 = 0

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(x − 5)(x + 2) = 0 [1]1
x = 5 or x = −2 [1]1

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[2 marks] (b)3x² − 5x + 2 = 0

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(3x − 2)(x − 1) = 0 [1]1
x = 23 or x = 1 [1]1

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[3 marks] (c)2h + 6/h = 7, where h ≠ 0

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Multiply by h: 2h² + 6 = 7h, so 2h² − 7h + 6 = 0 [1]1
(2h − 3)(h − 2) = 0 [1]1
h = 32 or h = 2 [1]1

Remove the fraction first, then write the equation with 0 on the right.

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For the teacher: Applying, medium difficulty, standard 1.1.6

Textbook p. 21

The quadratic function f(x) = kx² + 8x + h has a maximum point at (2, 11), where k and h are constants.

[2 marks] (a)The axis of symmetry is x = 2. Using x = −b ÷ 2a, find the value of k.

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2 = −8 ÷ (2k) [1]1
k = −2 [1]1

a = k is negative, which agrees with a maximum point.

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[2 marks] (b)Find the value of h.

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f(2) = −2(4) + 8(2) + h = 11 [1]1
h = 3 [1]1

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[1 mark] (c)State the y-intercept of the graph.

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3, or (0, 3) [1]1

The y-intercept equals c, which is h = 3.

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For the teacher: Analysing, medium difficulty, standard 1.1.3

Textbook p. 14

A box has a square base of side x cm and a height of (x + 10) cm. The total surface area of the box is 1 000 cm². The box needs a frame made from rods that cost RM5 per metre. The budget is RM6.

[2 marks] (a)Show that the total surface area is 6x² + 40x cm².

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2(x)(x) + 4(x)(x + 10) [1]1
2x² + 4x² + 40x = 6x² + 40x [1]1

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[3 marks] (b)Form a quadratic equation and solve it to find x.

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6x² + 40x = 1 000, so 3x² + 20x − 500 = 0 [1]1
(3x + 50)(x − 10) = 0 [1]1
x = 10, since x = −503 is not acceptable [1]1

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[3 marks] (c)The frame uses the 12 edges of the box. Does the budget cover the cost of the rods? Show your calculation.

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Edges: 8(10) + 4(20) = 160 cm [1]1
160 cm = 1.6 m, cost = 1.6 × 5 = RM8 [1]1
RM8 > RM6, so the budget is not enough [1]1

Height = 10 + 10 = 20 cm. Convert to metres before using RM5 per metre.

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For the teacher: Analysing, high difficulty, standard 1.1.8

Textbook p. 26

Sketch information for the graph of f(x) = x² − 2x − 8.

[1 mark] (a)State the shape of the graph and give a reason.

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Opens upward (smile) because a = 1 > 0 [1]1

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[1 mark] (b)State the y-intercept.

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−8 [1]1

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[2 marks] (c)Find the x-intercepts.

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(x − 4)(x + 2) = 0 [1]1
x = 4 and x = −2 [1]1

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[2 marks] (d)Find the equation of the axis of symmetry and the coordinates of the minimum point.

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Axis x = (4 + (−2)) ÷ 2 = 1 [1]1
f(1) = 1 − 2 − 8 = −9, minimum point (1, −9) [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 1.1.7

Textbook p. 23

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Section B: structured questions

Question 13

Question 14

Question 15

Question 16

Question 17