Secondary 1 Mathematics Algebra Functions Quiz

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Secondary 1 Mathematics Quiz - Algebra Functions

Name: ________________________________________
Class: ________________________________________
Date: ________________________________________
Score: ____ / 40

Duration: 50 minutes
Total Marks: 40

Instructions:

• Answer ALL questions.
• Show your working clearly in the space provided.
• Non-exact answers should be given correct to 2 decimal places unless otherwise stated.
• The use of calculators is NOT allowed in Section A but is allowed in Section B.
• This quiz focuses on Algebra Functions — substitution, function notation, evaluating functions, and solving for unknowns using function equations.

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Section A: Short Answer Questions (20 marks)

Questions 1–10. Each question carries 2 marks. Show your working clearly.

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1. Given that $f(x) = 3x - 5$, find the value of $f(4)$.

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

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3. If $h(x) = 7 - 2x$, find the value of $x$ when $h(x) = 1$.

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4. Given that $f(x) = 4x + 3$ and $g(x) = x - 2$, find the value of $f(2) + g(5)$.

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5. If $p(x) = x^2 - 4x + 4$, find the value of $p(0)$ and $p(4)$.

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6. Given that $f(x) = \dfrac{2x + 6}{3}$, find the value of $f(6)$.

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7. If $f(x) = 5x - a$ and $f(3) = 7$, find the value of $a$.

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8. Given that $g(x) = x^2 - 9$, find the values of $x$ for which $g(x) = 0$.

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9. If $f(x) = 2x + 1$ and $f(k) = 11$, find the value of $k$.

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10. Given that $h(x) = 3x^2 - x + 5$, find the value of $h(-1)$.

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Section B: Structured Response Questions (20 marks)

Questions 11–20. Show all working clearly. Marks are indicated for each part.

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11. The function $f$ is defined by $f(x) = 2x^2 - 3x + 1$.

    (a) Find $f(3)$.
        (1 mark)

    (b) Find $f(-2)$.
        (1 mark)

    (c) Find the value of $x$ such that $f(x) = 1$.
        (2 marks)

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12. A function $g$ is defined by $g(x) = ax + b$, where $a$ and $b$ are constants.

It is given that $g(1) = 5$ and $g(3) = 13$.

    (a) Write down two simultaneous equations in $a$ and $b$.
        (1 mark)

    (b) Solve the equations to find the values of $a$ and $b$.
        (2 marks)

    (c) Hence find $g(0)$.
        (1 mark)

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13. Given that $f(x) = x^2 - 6x + 8$,

    (a) Find $f(1)$.
        (1 mark)

    (b) Find the value(s) of $x$ for which $f(x) = 0$.
        (2 marks)

    (c) State the least value of $f(x)$ and the corresponding value of $x$.
        (1 mark)

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14. The function $h$ is defined by $h(x) = \dfrac{4x - 8}{x - 2}$, where $x \ne 2$.

    (a) Find the value of $h(5)$.
        (1 mark)

    (b) Find the value of $h(100)$.
        (1 mark)

    (c) Explain what value $h(x)$ approaches as $x$ gets very large.
        (2 marks)

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15. Given $f(x) = 3x - 2$ and $g(x) = x^2 + 1$,

    (a) Find $f(1) \times g(1)$.
        (1 mark)

    (b) Find $f(g(2))$.
        (2 marks)

    (c) Find $g(f(2))$.
        (1 mark)

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16. The cost $C$ dollars of printing $n$ booklets is given by the function $C(n) = 2.5n + 20$.

    (a) Find the cost of printing 40 booklets.
        (1 mark)

    (b) Find the number of booklets that can be printed for $120.
        (2 marks)

    (c) State what the value 20 represents in this context.
        (1 mark)

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17. Given that $f(x) = x^2 + bx + c$, and it is known that $f(1) = 0$ and $f(4) = 0$,

    (a) Write down two equations in $b$ and $c$.
        (1 mark)

    (b) Solve to find the values of $b$ and $c$.
        (2 marks)

    (c) Hence write $f(x)$ in factorised form.
        (1 mark)

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18. The function $f$ is defined by $f(x) = 2x^2 - 8x + 6$.

    (a) Find $f(0)$ and $f(4)$.
        (1 mark)

    (b) Solve $f(x) = 0$.
        (2 marks)

    (c) Write $f(x)$ in the form $a(x - h)^2 + k$ by completing the square.
        (1 mark)

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19. A taxi fare $F$ in dollars for a journey of $d$ kilometres is given by $F(d) = 3.20 + 0.85d$.

    (a) Find the fare for a 12 km journey.
        (1 mark)

    (b) Jane paid $13.40 for a taxi ride. How far did she travel?
        (2 marks)

    (c) What is the fixed starting fare?
        (1 mark)

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20. Given $f(x) = ax^2 + bx + 3$, and it is known that $f(1) = 8$ and $f(-1) = 6$,

    (a) Form two equations in $a$ and $b$.
        (1 mark)

    (b) Solve to find $a$ and $b$.
        (2 marks)

    (c) Hence find $f(2)$.
        (1 mark)

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Secondary 1 Mathematics Quiz - Algebra Functions

Answer Key

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Section A: Short Answer Questions (20 marks)

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1. $f(x) = 3x - 5$
$f(4) = 3(4) - 5 = 12 - 5 = \boxed{7}$
[2 marks]
Marking: 1 mark for correct substitution, 1 mark for correct answer.

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2. $g(x) = x^2 + 2x - 1$
$g(-3) = (-3)^2 + 2(-3) - 1 = 9 - 6 - 1 = \boxed{2}$
[2 marks]
Marking: 1 mark for correct substitution, 1 mark for correct answer.
Common mistake: $(-3)^2 = -9$ (incorrect). Correct: $(-3)^2 = 9$.

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3. $h(x) = 7 - 2x$
$7 - 2x = 1$
$-2x = 1 - 7 = -6$
$x = \dfrac{-6}{-2} = \boxed{3}$
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.

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4. $f(x) = 4x + 3$, $g(x) = x - 2$
$f(2) = 4(2) + 3 = 8 + 3 = 11$
$g(5) = 5 - 2 = 3$
$f(2) + g(5) = 11 + 3 = \boxed{14}$
[2 marks]
Marking: 1 mark for each correct evaluation, final answer must be correct for full marks.

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5. $p(x) = x^2 - 4x + 4$
$p(0) = 0^2 - 4(0) + 4 = \boxed{4}$
$p(4) = 4^2 - 4(4) + 4 = 16 - 16 + 4 = \boxed{4}$
[2 marks]
Marking: 1 mark for each correct value.

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6. $f(x) = \dfrac{2x + 6}{3}$
$f(6) = \dfrac{2(6) + 6}{3} = \dfrac{12 + 6}{3} = \dfrac{18}{3} = \boxed{6}$
[2 marks]
Marking: 1 mark for correct substitution, 1 mark for correct simplification.

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7. $f(x) = 5x - a$ and $f(3) = 7$
$f(3) = 5(3) - a = 7$
$15 - a = 7$
$a = 15 - 7 = \boxed{8}$
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.

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8. $g(x) = x^2 - 9$
$x^2 - 9 = 0$
$x^2 = 9$
$x = \boxed{3}$ or $x = \boxed{-3}$
[2 marks]
Marking: 1 mark for factorising or rearranging, 1 mark for both correct values.
Common mistake: giving only $x = 3$ and omitting $x = -3$.

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9. $f(x) = 2x + 1$ and $f(k) = 11$
$2k + 1 = 11$
$2k = 10$
$k = \boxed{5}$
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.

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10. $h(x) = 3x^2 - x + 5$
$h(-1) = 3(-1)^2 - (-1) + 5 = 3(1) + 1 + 5 = 3 + 1 + 5 = \boxed{9}$
[2 marks]
Marking: 1 mark for correct substitution, 1 mark for correct answer.
Common mistake: $-(-1) = -1$ (incorrect). Correct: $-(-1) = +1$.

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Section B: Structured Response Questions (20 marks)

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11. $f(x) = 2x^2 - 3x + 1$

(a) $f(3) = 2(3)^2 - 3(3) + 1 = 2(9) - 9 + 1 = 18 - 9 + 1 = \boxed{10}$
[1 mark]

(b) $f(-2) = 2(-2)^2 - 3(-2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = \boxed{15}$
[1 mark]

(c) $f(x) = 1$
$2x^2 - 3x + 1 = 1$
$2x^2 - 3x = 0$
$x(2x - 3) = 0$
$x = 0$ or $2x - 3 = 0 \Rightarrow x = \dfrac{3}{2}$
$\boxed{x = 0}$ or $\boxed{x = 1.5}$
[2 marks]
Marking: 1 mark for setting up and simplifying equation, 1 mark for both correct solutions.

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12. $g(x) = ax + b$

(a) $g(1) = a(1) + b = 5 \Rightarrow \boxed{a + b = 5}$
$g(3) = a(3) + b = 13 \Rightarrow \boxed{3a + b = 13}$
[1 mark]
Mark for both equations correct.

(b) Subtracting: $(3a + b) - (a + b) = 13 - 5$
$2a = 8$
$a = 4$
Substitute into $a + b = 5$: $4 + b = 5 \Rightarrow b = 1$
$\boxed{a = 4}$, $\boxed{b = 1}$
[2 marks]
Marking: 1 mark for correct method, 1 mark for both correct values.

(c) $g(0) = 4(0) + 1 = \boxed{1}$
[1 mark]

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13. $f(x) = x^2 - 6x + 8$

(a) $f(1) = (1)^2 - 6(1) + 8 = 1 - 6 + 8 = \boxed{3}$
[1 mark]

(b) $x^2 - 6x + 8 = 0$
$(x - 2)(x - 4) = 0$
$\boxed{x = 2}$ or $\boxed{x = 4}$
[2 marks]
Marking: 1 mark for correct factorisation, 1 mark for both correct values.

(c) Completing the square: $f(x) = (x - 3)^2 - 1$
Least value occurs at $x = 3$, and the least value is $\boxed{-1}$
[1 mark]
Accept: least value is $-1$ when $x = 3$.

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14. $h(x) = \dfrac{4x - 8}{x - 2}$, $x \ne 2$

(a) $h(5) = \dfrac{4(5) - 8}{5 - 2} = \dfrac{20 - 8}{3} = \dfrac{12}{3} = \boxed{4}$
[1 mark]

(b) $h(100) = \dfrac{4(100) - 8}{100 - 2} = \dfrac{400 - 8}{98} = \dfrac{392}{98} = \boxed{4}$
[1 mark]

(c) $h(x) = \dfrac{4x - 8}{x - 2} = \dfrac{4(x - 2)}{x - 2} = 4$ (for $x \ne 2$)
As $x$ gets very large, $h(x) = \boxed{4}$.
The value of $h(x)$ is always 4 (for all $x \ne 2$), so as $x$ gets very large, $h(x)$ remains 4.
[2 marks]
Marking: 1 mark for simplifying/factorising, 1 mark for stating the value is 4 with explanation.

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15. $f(x) = 3x - 2$, $g(x) = x^2 + 1$

(a) $f(1) = 3(1) - 2 = 1$
$g(1) = (1)^2 + 1 = 2$
$f(1) \times g(1) = 1 \times 2 = \boxed{2}$
[1 mark]

(b) $g(2) = (2)^2 + 1 = 4 + 1 = 5$
$f(g(2)) = f(5) = 3(5) - 2 = 15 - 2 = \boxed{13}$
[2 marks]
Marking: 1 mark for finding $g(2)$ correctly, 1 mark for correct final answer.

(c) $f(2) = 3(2) - 2 = 6 - 2 = 4$
$g(f(2)) = g(4) = (4)^2 + 1 = 16 + 1 = \boxed{17}$
[1 mark]

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16. $C(n) = 2.5n + 20$

(a) C(40) = 2.5(40) + 20 = 100 + 20 = \boxed{\120}$
[1 mark]

(b) $2.5n + 20 = 120$
$2.5n = 100$
$n = \dfrac{100}{2.5} = \boxed{40}$ booklets
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.

(c) The value $\boxed{20}$ represents the fixed cost (or setup cost) of printing, i.e., the cost when zero booklets are printed.
[1 mark]
Accept: fixed charge / base cost / setup fee.

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17. $f(x) = x^2 + bx + c$, with $f(1) = 0$ and $f(4) = 0$

(a) $f(1) = 1 + b + c = 0 \Rightarrow \boxed{b + c = -1}$
$f(4) = 16 + 4b + c = 0 \Rightarrow \boxed{4b + c = -16}$
[1 mark]
Mark for both equations correct.

(b) Subtracting: $(4b + c) - (b + c) = -16 - (-1)$
$3b = -15$
$b = -5$
Substitute: $-5 + c = -1 \Rightarrow c = 4$
$\boxed{b = -5}$, $\boxed{c = 4}$
[2 marks]
Marking: 1 mark for correct method, 1 mark for both correct values.

(c) $f(x) = x^2 - 5x + 4 = \boxed{(x - 1)(x - 4)}$
[1 mark]

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18. $f(x) = 2x^2 - 8x + 6$

(a) $f(0) = 2(0)^2 - 8(0) + 6 = \boxed{6}$
$f(4) = 2(16) - 8(4) + 6 = 32 - 32 + 6 = \boxed{6}$
[1 mark]

(b) $2x^2 - 8x + 6 = 0$
$2(x^2 - 4x + 3) = 0$
$2(x - 1)(x - 3) = 0$
$\boxed{x = 1}$ or $\boxed{x = 3}$
[2 marks]
Marking: 1 mark for correct factorisation, 1 mark for both correct values.

(c) $f(x) = 2x^2 - 8x + 6$
$= 2(x^2 - 4x) + 6$
$= 2(x^2 - 4x + 4 - 4) + 6$
$= 2[(x - 2)^2 - 4] + 6$
$= 2(x - 2)^2 - 8 + 6$
$= \boxed{2(x - 2)^2 - 2}$
[1 mark]

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19. $F(d) = 3.20 + 0.85d$

(a) F(12) = 3.20 + 0.85(12) = 3.20 + 10.20 = \boxed{\13.40}$
[1 mark]

(b) $3.20 + 0.85d = 13.40$
$0.85d = 13.40 - 3.20 = 10.20$
$d = \dfrac{10.20}{0.85} = \boxed{12}$ km
[2 marks]
Marking: 1 mark for setting up equation, 1 mark for correct answer.

(c) The fixed starting fare is \boxed{\3.20}$.
[1 mark]

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20. $f(x) = ax^2 + bx + 3$, with $f(1) = 8$ and $f(-1) = 6$

(a) $f(1) = a + b + 3 = 8 \Rightarrow \boxed{a + b = 5}$
$f(-1) = a - b + 3 = 6 \Rightarrow \boxed{a - b = 3}$
[1 mark]
Mark for both equations correct.

(b) Adding: $(a + b) + (a - b) = 5 + 3$
$2a = 8 \Rightarrow a = 4$
Substitute: $4 + b = 5 \Rightarrow b = 1$
$\boxed{a = 4}$, $\boxed{b = 1}$
[2 marks]
Marking: 1 mark for correct method, 1 mark for both correct values.

(c) $f(x) = 4x^2 + x + 3$
$f(2) = 4(4) + 2 + 3 = 16 + 2 + 3 = \boxed{21}$
[1 mark]

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