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Secondary 3 Elementary Mathematics Algebra Functions Quiz

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Questions

Secondary 3 Elementary Mathematics Quiz - Algebra Functions

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 40

Duration: 45 minutes Total Marks: 40

Instructions:

Answer ALL questions in the spaces provided.
Show all working clearly. Marks are awarded for method.
Calculators are allowed unless otherwise stated.
Give non-exact answers correct to 3 significant figures unless stated otherwise.

Section A: Short Answer (10 marks)

Answer all questions in this section.

1. Given the function $f(x) = 2x^2 - 3x + 1$ , find the value of $f(-2)$ .

[2 marks]

2. The graph of a quadratic function has a minimum point at $(3, -4)$ and passes through the point $(1, 4)$ . Express the function in the form $y = a(x - p)^2 + q$ , where $a$ , $p$ , and $q$ are constants.

[2 marks]

3. Factorise completely: $3x^2 - 12x + 9$

[2 marks]

4. Solve the equation $x^2 - 5x - 14 = 0$ by factorisation.

[2 marks]

5. The function $g(x) = (x + 2)^2 - 9$ is given. Write down the coordinates of the vertex of the graph of $y = g(x)$ .

[2 marks]

Section B: Structured Questions (18 marks)

Answer all questions in this section. Show all working clearly.

6. A quadratic function is given by $y = x^2 - 6x + 5$ .

(a) Express $x^2 - 6x + 5$ in the form $(x - p)^2 + q$ , where $p$ and $q$ are integers. [2 marks]

(b) Hence, or otherwise, write down the equation of the line of symmetry of the graph. [1 mark]

(d) Sketch the graph of $y = x^2 - 6x + 5$ , showing clearly the turning point and the points where the graph crosses the axes. [3 marks]

7. The function $h(x) = -2(x + 1)(x - 3)$ is given.

(a) State the coordinates of the points where the graph of $y = h(x)$ cuts the $x$ -axis. [2 marks]

(b) Find the coordinates of the maximum point of the graph. [3 marks]

(d) State the maximum value of $h(x)$ . [1 mark]

8. Solve the equation $\frac{2x}{x+1} = \frac{3}{x-2}$ .

[3 marks]

9. Given the function $f(x) = x^2 + 4x - 5$ , find the coordinates of the points where the graph of $y = f(x)$ cuts the $y$ -axis.

[2 marks]

10. The function $p(x) = (x - 1)^2 + 3$ is given. Write down the minimum value of $p(x)$ .

[1 mark]

Section C: Problem Solving (12 marks)

Answer all questions in this section. Show all working clearly.

11. A ball is thrown upwards from a platform. Its height, $h$ metres, above the ground after $t$ seconds is given by $h = -5t^2 + 20t + 25$ .

(a) Express $-5t^2 + 20t + 25$ in the form $a(t - p)^2 + q$ , where $a$ , $p$ , and $q$ are constants. [3 marks]

(b) Hence, find the maximum height reached by the ball. [1 mark]

12. The graph of $y = x^2 + bx + c$ has a minimum point at $(2, -1)$ .

(a) Find the values of $b$ and $c$ . [3 marks]

(b) Find the coordinates of the points where the graph cuts the $y$ -axis. [1 mark]

13. Solve the equation $2x^2 - 3x - 5 = 0$ using the quadratic formula.

[3 marks]

14. The function $q(x) = -x^2 + 6x - 8$ is given. Find the coordinates of the maximum point of the graph of $y = q(x)$ .

[3 marks]

15. Factorise completely: $4x^2 - 25$

[2 marks]

Section D: Applications and Analysis (10 marks)

Answer all questions in this section. Show all working clearly.

16. The product of two consecutive positive integers is 72. Form a quadratic equation and solve it to find the two integers.

[3 marks]

17. The graph of $y = 2x^2 - 8x + k$ has a minimum value of $-2$ . Find the value of $k$ .

[3 marks]

18. Given the function $f(x) = \frac{1}{x-3}$ , state the value of $x$ for which $f(x)$ is undefined.

[1 mark]

19. The function $g(x) = x^2 - 2x - 8$ is given. Find the coordinates of the points where the graph of $y = g(x)$ cuts the $x$ -axis.

[2 marks]

20. Solve the equation $\frac{x}{x+2} = \frac{2}{x-1}$ .

[3 marks]

END OF QUIZ

Check your work carefully.

Answers

Secondary 3 Elementary Mathematics Quiz - Algebra Functions

ANSWER KEY AND MARKING SCHEME

Total Marks: 40

Section A: Short Answer (10 marks)

1. $f(-2) = 2(-2)^2 - 3(-2) + 1$ $= 2(4) + 6 + 1$ $= 8 + 6 + 1$ $= 15$ ✓

[M2] Award M1 for correct substitution, A1 for correct answer.

2. Vertex form: $y = a(x - p)^2 + q$ with vertex $(p, q) = (3, -4)$ So $y = a(x - 3)^2 - 4$

Substitute $(1, 4)$ : $4 = a(1 - 3)^2 - 4$ $4 = a(4) - 4$ $8 = 4a$ $a = 2$

Therefore $y = 2(x - 3)^2 - 4$ ✓

[M2] Award M1 for correct vertex form with unknown $a$ , A1 for correct $a$ and final expression.

3. $3x^2 - 12x + 9$ $= 3(x^2 - 4x + 3)$ $= 3(x - 1)(x - 3)$ ✓

[M2] Award M1 for factorising out 3, A1 for complete factorisation. Accept $3(x - 3)(x - 1)$ .

4. $x^2 - 5x - 14 = 0$ $(x - 7)(x + 2) = 0$ $x - 7 = 0$ or $x + 2 = 0$ $x = 7$ or $x = -2$ ✓

[M2] Award M1 for correct factorisation, A1 for both solutions.

5. $g(x) = (x + 2)^2 - 9$ Vertex form: $y = (x - (-2))^2 + (-9)$ Vertex: $(-2, -9)$ ✓

[M2] Award M1 for identifying vertex form, A1 for correct coordinates.

Section B: Structured Questions (18 marks)

6. (a) $y = x^2 - 6x + 5$ $= (x^2 - 6x + 9) - 9 + 5$ $= (x - 3)^2 - 4$ ✓

[M2] Award M1 for completing the square correctly, A1 for correct expression. $p = 3$ , $q = -4$ .

(b) Line of symmetry: $x = 3$ ✓

[B1] Follow through from part (a).

(c) When $y = 0$ : $(x - 3)^2 - 4 = 0$ $(x - 3)^2 = 4$ $x - 3 = \pm 2$ $x = 5$ or $x = 1$ Coordinates: $(1, 0)$ and $(5, 0)$ ✓

[M2] Award M1 for setting $y = 0$ and solving, A1 for both coordinates.

(d) Sketch should show:

U-shaped parabola (positive coefficient of $x^2$ )
Vertex at $(3, -4)$
$x$ -intercepts at $(1, 0)$ and $(5, 0)$
$y$ -intercept at $(0, 5)$
Line of symmetry $x = 3$

[M3] Award M1 for correct shape, M1 for correct vertex and intercepts, M1 for smooth curve with symmetry.

7. (a) $h(x) = -2(x + 1)(x - 3)$ When $h(x) = 0$ : $-2(x + 1)(x - 3) = 0$ $x + 1 = 0$ or $x - 3 = 0$ $x = -1$ or $x = 3$ Coordinates: $(-1, 0)$ and $(3, 0)$ ✓

[M2] Award M1 for setting $h(x) = 0$ , A1 for both coordinates.

(b) $h(x) = -2(x + 1)(x - 3)$ $= -2(x^2 - 2x - 3)$ $= -2x^2 + 4x + 6$

$x$ -coordinate of vertex: $x = \frac{-1 + 3}{2} = 1$ (midpoint of roots) $h(1) = -2(1 + 1)(1 - 3) = -2(2)(-2) = 8$ Maximum point: $(1, 8)$ ✓

[M3] Award M1 for finding $x$ -coordinate of vertex, M1 for substituting to find $y$ , A1 for correct coordinates.

[B1]

(d) Maximum value of $h(x) = 8$ ✓

[B1]

8. $\frac{2x}{x+1} = \frac{3}{x-2}$

Cross-multiply: $2x(x - 2) = 3(x + 1)$ $2x^2 - 4x = 3x + 3$ $2x^2 - 7x - 3 = 0$

Using quadratic formula: $x = \frac{7 \pm \sqrt{49 + 24}}{4} = \frac{7 \pm \sqrt{73}}{4}$ $x = \frac{7 + \sqrt{73}}{4}$ or $x = \frac{7 - \sqrt{73}}{4}$

Check denominators: $x \neq -1$ and $x \neq 2$ . Both solutions are valid. $x \approx 3.89$ or $x \approx -0.386$ (3 s.f.) ✓

[M3] Award M1 for correct cross-multiplication, M1 for rearranging to quadratic and solving, A1 for both solutions (exact or 3 s.f.).

9. $f(x) = x^2 + 4x - 5$ $y$ -intercept occurs when $x = 0$ : $f(0) = 0^2 + 4(0) - 5 = -5$ Coordinates: $(0, -5)$ ✓

[M2] Award M1 for substituting $x = 0$ , A1 for correct coordinates.

10. $p(x) = (x - 1)^2 + 3$ Since $(x - 1)^2 \geq 0$ , the minimum value occurs when $(x - 1)^2 = 0$ . Minimum value of $p(x) = 3$ ✓

[B1]

Section C: Problem Solving (12 marks)

11. (a) $h = -5t^2 + 20t + 25$ $= -5(t^2 - 4t) + 25$ $= -5(t^2 - 4t + 4 - 4) + 25$ $= -5((t - 2)^2 - 4) + 25$ $= -5(t - 2)^2 + 20 + 25$ $= -5(t - 2)^2 + 45$ ✓

[M3] Award M1 for factorising out $-5$ , M1 for completing the square, A1 for correct expression. $a = -5$ , $p = 2$ , $q = 45$ .

(b) Maximum height occurs at $t = 2$ seconds. Maximum height $= 45$ metres ✓

[B1]

(c) When ball hits ground, $h = 0$ : $-5(t - 2)^2 + 45 = 0$ $-5(t - 2)^2 = -45$ $(t - 2)^2 = 9$ $t - 2 = \pm 3$ $t = 5$ or $t = -1$ (reject negative time) Time $= 5$ seconds ✓

[M2] Award M1 for setting $h = 0$ and solving, A1 for correct time with rejection of invalid solution.

12. (a) $y = x^2 + bx + c$ has minimum at $(2, -1)$ . Vertex form: $y = (x - 2)^2 - 1$ $= x^2 - 4x + 4 - 1$ $= x^2 - 4x + 3$

Therefore $b = -4$ and $c = 3$ ✓

[M3] Award M1 for writing vertex form, M1 for expanding, A1 for both values.

(b) $y$ -intercept: when $x = 0$ , $y = 0^2 - 4(0) + 3 = 3$ Coordinates: $(0, 3)$ ✓

[B1]

(c) Discriminant: $b^2 - 4ac = (-4)^2 - 4(1)(3) = 16 - 12 = 4$ Since discriminant $> 0$ , the graph cuts the $x$ -axis at two distinct points. ✓

[M2] Award M1 for calculating discriminant, A1 for correct conclusion with reasoning.

13. $2x^2 - 3x - 5 = 0$ $a = 2$ , $b = -3$ , $c = -5$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}$ $x = \frac{3 \pm \sqrt{9 + 40}}{4}$ $x = \frac{3 \pm \sqrt{49}}{4}$ $x = \frac{3 \pm 7}{4}$ $x = \frac{10}{4} = 2.5$ or $x = \frac{-4}{4} = -1$ ✓

[M3] Award M1 for correct substitution into formula, M1 for correct simplification, A1 for both solutions.

14. $q(x) = -x^2 + 6x - 8$ Complete the square: $q(x) = -(x^2 - 6x) - 8$ $= -(x^2 - 6x + 9 - 9) - 8$ $= -((x - 3)^2 - 9) - 8$ $= -(x - 3)^2 + 9 - 8$ $= -(x - 3)^2 + 1$

Maximum point occurs at $x = 3$ , $q(3) = 1$ . Coordinates: $(3, 1)$ ✓

[M3] Award M1 for factorising out $-1$ , M1 for completing the square, A1 for correct coordinates.

15. $4x^2 - 25$ $= (2x)^2 - 5^2$ $= (2x - 5)(2x + 5)$ ✓

[M2] Award M1 for recognising difference of squares, A1 for correct factorisation.

Section D: Applications and Analysis (10 marks)

16. Let the two consecutive positive integers be $n$ and $n + 1$ . Product: $n(n + 1) = 72$ $n^2 + n - 72 = 0$ $(n + 9)(n - 8) = 0$ $n = -9$ (reject, not positive) or $n = 8$ The integers are 8 and 9. ✓

[M3] Award M1 for forming equation, M1 for solving, A1 for correct integers with rejection of invalid solution.

17. $y = 2x^2 - 8x + k$ Complete the square: $y = 2(x^2 - 4x) + k$ $= 2(x^2 - 4x + 4 - 4) + k$ $= 2((x - 2)^2 - 4) + k$ $= 2(x - 2)^2 - 8 + k$

Minimum value is $-8 + k$ . Given minimum value is $-2$ : $-8 + k = -2$ $k = 6$ ✓

[M3] Award M1 for completing the square, M1 for setting minimum equal to $-2$ , A1 for correct $k$ .

18. $f(x) = \frac{1}{x-3}$ $f(x)$ is undefined when denominator is zero: $x - 3 = 0$ $x = 3$ ✓

[B1]

19. $g(x) = x^2 - 2x - 8$ When $g(x) = 0$ : $x^2 - 2x - 8 = 0$ $(x - 4)(x + 2) = 0$ $x = 4$ or $x = -2$ Coordinates: $(4, 0)$ and $(-2, 0)$ ✓

[M2] Award M1 for setting $g(x) = 0$ and factorising, A1 for both coordinates.

20. $\frac{x}{x+2} = \frac{2}{x-1}$

Cross-multiply: $x(x - 1) = 2(x + 2)$ $x^2 - x = 2x + 4$ $x^2 - 3x - 4 = 0$ $(x - 4)(x + 1) = 0$ $x = 4$ or $x = -1$

Check denominators: $x \neq -2$ and $x \neq 1$ . Both solutions are valid. $x = 4$ or $x = -1$ ✓

[M3] Award M1 for correct cross-multiplication, M1 for rearranging and solving, A1 for both solutions with check.

END OF ANSWER KEY

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