Secondary 1 Mathematics Algebra Functions Quiz

Secondary 1 Mathematics Quiz - Algebra Functions (Answer Key)

Total Marks: 50

---

Section A: Multiple Choice and Short Answer (Questions 1–10, 20 marks)

1. Given the function $f(x) = 3x - 5$, find $f(4)$.
[1]

Answer: $7$
Working: $f(4) = 3(4) - 5 = 12 - 5 = 7$
Marking Note: 1 mark for correct answer. Substitution must be shown or implied.

2. If $g(x) = x^2 + 2x$, evaluate $g(-3)$.
[1]

Answer: $3$
Working: $g(-3) = (-3)^2 + 2(-3) = 9 - 6 = 3$
Marking Note: 1 mark for correct answer. Common error: $(-3)^2 = -9$ (incorrect). Remind students that squaring a negative gives a positive.

3. The function $h$ is defined as $h(x) = \frac{12}{x}$ for $x \neq 0$. Find the value of $h(4)$.
[1]

Answer: $3$
Working: $h(4) = \frac{12}{4} = 3$
Marking Note: 1 mark for correct answer. Note the domain restriction $x \neq 0$.

4. A function $p$ is defined by $p(x) = 2x + 7$. If $p(k) = 19$, find the value of $k$.
[2]

Answer: $k = 6$
Working:
$2k + 7 = 19$
$2k = 12$
$k = 6$
Marking Note: 1 mark for setting up equation $2k + 7 = 19$, 1 mark for correct solution $k = 6$.

5. The function $q(x) = ax + b$ passes through the points $(1, 5)$ and $(3, 11)$. Find the values of $a$ and $b$.
[2]

Answer: $a = 3$, $b = 2$
Working:
Substitute $(1, 5)$: $a(1) + b = 5 \Rightarrow a + b = 5$
Substitute $(3, 11)$: $a(3) + b = 11 \Rightarrow 3a + b = 11$
Subtract: $(3a + b) - (a + b) = 11 - 5 \Rightarrow 2a = 6 \Rightarrow a = 3$
Substitute $a = 3$ into $a + b = 5$: $3 + b = 5 \Rightarrow b = 2$
Marking Note: 1 mark for correct $a$, 1 mark for correct $b$. Alternative: gradient $m = \frac{11-5}{3-1} = 3$, then $y = 3x + c$, substitute $(1,5)$ gives $c=2$.

6. Given $f(x) = 4 - 2x$, solve $f(x) = 10$.
[2]

Answer: $x = -3$
Working:
$4 - 2x = 10$
$-2x = 6$
$x = -3$
Marking Note: 1 mark for setting up equation, 1 mark for correct solution. Common error: forgetting to reverse inequality sign (not applicable here, but watch for sign errors when dividing by negative).

7. The function $C(d) = 50 + 15d$ represents the cost $C$ (in dollars) of hiring a bicycle for $d$ days. Find the cost of hiring the bicycle for 7 days.
[2]

Answer: $155$
Working: $C(7) = 50 + 15(7) = 50 + 105 = 155$
Marking Note: 1 mark for correct substitution, 1 mark for correct answer with units (dollars). The $50 is a fixed cost (deposit/base fee),$15 is the daily rate.

8. A function $T(n) = 3n + 2$ gives the number of matchsticks needed to form a pattern with $n$ triangles. How many matchsticks are needed for 15 triangles?
[2]

Answer: $47$
Working: $T(15) = 3(15) + 2 = 45 + 2 = 47$
Marking Note: 1 mark for substitution, 1 mark for answer. The $+2$ represents the starting matchsticks (e.g., 2 matchsticks for the first triangle's base), and each additional triangle adds 3 matchsticks.

9. The function $V(r) = \frac{4}{3}\pi r^3$ gives the volume of a sphere of radius $r$. If the radius is 3 cm, find the volume in terms of $\pi$.
[2]

Answer: $36\pi \text{ cm}^3$
Working: $V(3) = \frac{4}{3}\pi (3)^3 = \frac{4}{3}\pi (27) = 36\pi$
Marking Note: 1 mark for correct substitution and $3^3 = 27$, 1 mark for simplification to $36\pi$. Units $\text{cm}^3$ required.

10. The table below shows some values of a linear function $y = mx + c$.

$x$	0	2	4	6
$y$	3	7	11	15

Find the values of $m$ and $c$.
[2]

Answer: $m = 2$, $c = 3$
Working:
When $x = 0$, $y = 3 \Rightarrow c = 3$ (y-intercept)
Gradient $m = \frac{7 - 3}{2 - 0} = \frac{4}{2} = 2$
Check: $y = 2x + 3$ gives $x=4 \Rightarrow y=11$, $x=6 \Rightarrow y=15$ ✓
Marking Note: 1 mark for $c = 3$ (reading intercept from table), 1 mark for $m = 2$ (calculating gradient). Common error: using wrong pair of points for gradient.

---

Section B: Structured Questions (Questions 11–16, 18 marks)

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

(a) Find $f(2)$.
[1]

Answer: $3$
Working: $f(2) = 2(2)^2 - 3(2) + 1 = 2(4) - 6 + 1 = 8 - 6 + 1 = 3$

(b) Find $f(-1)$.
[1]

Answer: $6$
Working: $f(-1) = 2(-1)^2 - 3(-1) + 1 = 2(1) + 3 + 1 = 2 + 3 + 1 = 6$
Marking Note: Common error: $2(-1)^2 = -2$ (incorrect). Remind: $(-1)^2 = 1$, then $2 \times 1 = 2$.

(c) Solve $f(x) = 3$.
[3]

Answer: $x = -1$ or $x = 2.5$
Working:
$2x^2 - 3x + 1 = 3$
$2x^2 - 3x - 2 = 0$
$(2x + 1)(x - 2) = 0$
$2x + 1 = 0 \Rightarrow x = -0.5$
$x - 2 = 0 \Rightarrow x = 2$
Wait, let me recheck:
$2x^2 - 3x + 1 = 3 \Rightarrow 2x^2 - 3x - 2 = 0$
Using quadratic formula: $x = \frac{3 \pm \sqrt{9 - 4(2)(-2)}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4}$
$x = \frac{8}{4} = 2$ or $x = \frac{-2}{4} = -0.5$
Correct Answer: $x = 2$ or $x = -0.5$
Marking Note: 1 mark for setting up quadratic equation $2x^2 - 3x - 2 = 0$, 1 mark for correct factorisation/formula, 1 mark for both solutions. Accept $x = 2, -\frac{1}{2}$.

12. The cost $C$ (in dollars) of printing $n$ T-shirts is given by the function $C(n) = 20 + 8n$.

(a) State the fixed cost and the variable cost per T-shirt.
[2]

Answer: Fixed cost = $20, Variable cost =$8 per T-shirt
Marking Note: 1 mark each. Fixed cost is the constant term (cost when $n=0$). Variable cost is the coefficient of $n$.

(b) Find the cost of printing 50 T-shirts.
[1]

Answer: $420$
Working: $C(50) = 20 + 8(50) = 20 + 400 = 420$

(c) If the total cost is $180, how many T-shirts were printed?
[2]

Answer: $20$
Working:
$20 + 8n = 180$
$8n = 160$
$n = 20$
Marking Note: 1 mark for equation, 1 mark for solution. Check: $n$ must be integer (number of T-shirts).

13. A car rental company charges a flat fee of $30 plus$0.50 per kilometre driven.

(a) Write a function $C(k)$ for the total cost $C$ (in dollars) in terms of the distance $k$ (in km).
[1]

Answer: $C(k) = 30 + 0.5k$ or $C(k) = 30 + \frac{1}{2}k$
Marking Note: 1 mark for correct function. $30$ is fixed, $0.5k$ is variable.

(b) Find the cost of driving 120 km.
[1]

Answer: $90$
Working: $C(120) = 30 + 0.5(120) = 30 + 60 = 90$

(c) If a customer paid $85, how many kilometres did they drive?
[2]

Answer: $110 \text{ km}$
Working:
$30 + 0.5k = 85$
$0.5k = 55$
$k = 110$
Marking Note: 1 mark for equation, 1 mark for solution.

14. The function $A(s) = s^2$ gives the area of a square of side length $s$ cm.

(a) Find $A(5)$.
[1]

Answer: $25 \text{ cm}^2$
Working: $A(5) = 5^2 = 25$

(b) If the area is 64 cm², find the side length.
[1]

Answer: $8 \text{ cm}$
Working: $s^2 = 64 \Rightarrow s = \sqrt{64} = 8$ (reject negative root as length > 0)
Marking Note: Must reject $s = -8$ since side length cannot be negative.

(c) The side length of a square increases from 4 cm to 7 cm. Find the increase in area.
[2]

Answer: $33 \text{ cm}^2$
Working:
Original area: $A(4) = 4^2 = 16 \text{ cm}^2$
New area: $A(7) = 7^2 = 49 \text{ cm}^2$
Increase: $49 - 16 = 33 \text{ cm}^2$
Marking Note: 1 mark for both areas, 1 mark for difference. Alternative: $A(7) - A(4) = 49 - 16 = 33$.

15. A function $P(t) = 500(1.05)^t$ models the population of a town $t$ years after 2020.

(a) What was the population in 2020?
[1]

Answer: $500$
Working: In 2020, $t = 0$. $P(0) = 500(1.05)^0 = 500(1) = 500$
Marking Note: 1 mark. Any number to the power 0 is 1.

(b) Find the population in 2023 (correct to the nearest whole number).
[2]

Answer: $579$
Working: 2023 is 3 years after 2020, so $t = 3$.
$P(3) = 500(1.05)^3 = 500(1.157625) = 578.8125 \approx 579$
Marking Note: 1 mark for correct substitution $t=3$, 1 mark for correct evaluation and rounding.

(c) Explain what the number 1.05 represents in this context.
[1]

Answer: The population grows by 5% each year (or the annual growth factor is 1.05, meaning a 5% increase per year).
Marking Note: 1 mark for correct interpretation. Must mention "5% increase" or "growth factor" and "per year".

16. The diagram below shows a mapping diagram for a function $f$.

<image_placeholder> id: Q16-fig1 type: diagram linked_question: Q16 description: Mapping diagram showing domain {1, 2, 3, 4} mapping to co-domain {3, 5, 7, 9} with arrows: 1→3, 2→5, 3→7, 4→9 labels: Domain: 1, 2, 3, 4; Co-domain: 3, 5, 7, 9; Arrows showing mapping values: f(1)=3, f(2)=5, f(3)=7, f(4)=9 must_show: Clear arrows from each domain element to its image; domain and co-domain sets labelled </image_placeholder>

(a) Write down the set of ordered pairs for this function.
[1]

Answer: $\{(1, 3), (2, 5), (3, 7), (4, 9)\}$
Marking Note: 1 mark for correct set notation with all 4 pairs.

(b) Express the function in the form $f(x) = ax + b$.
[2]

Answer: $f(x) = 2x + 1$
Working:
The outputs increase by 2 each time (3, 5, 7, 9), so gradient $a = 2$.
When $x = 1$, $f(1) = 3 \Rightarrow 2(1) + b = 3 \Rightarrow b = 1$.
Check: $f(2) = 2(2) + 1 = 5$ ✓, $f(3) = 7$ ✓, $f(4) = 9$ ✓
Marking Note: 1 mark for $a = 2$, 1 mark for $b = 1$. Can also use two points to find $a$ and $b$.

(c) Find $f(10)$.
[1]

Answer: $21$
Working: $f(10) = 2(10) + 1 = 21$
Marking Note: 1 mark. Follow-through from (b) if their function is linear.

---

Section C: Application and Problem Solving (Questions 17–20, 12 marks)

17. A rectangular garden has a length that is 3 m more than its width $w$ metres.

(a) Write a function $A(w)$ for the area of the garden in terms of $w$.
[1]

Answer: $A(w) = w(w + 3) = w^2 + 3w$
Marking Note: 1 mark. Length = $w + 3$, Area = length × width = $w(w+3)$.

(b) If the area of the garden is 70 m², form an equation in $w$ and solve it to find the dimensions of the garden.
[3]

Answer: Width = 7 m, Length = 10 m
Working:
$w^2 + 3w = 70$
$w^2 + 3w - 70 = 0$
$(w + 10)(w - 7) = 0$
$w = -10$ (reject, width > 0) or $w = 7$
Width = 7 m, Length = 7 + 3 = 10 m
Marking Note: 1 mark for quadratic equation, 1 mark for solving (factorisation or formula), 1 mark for correct dimensions with rejection of negative root and units.

(c) Find the perimeter of the garden.
[1]

Answer: $34 \text{ m}$
Working: Perimeter = $2(\text{length} + \text{width}) = 2(10 + 7) = 34 \text{ m}$
Marking Note: 1 mark. Follow-through from (b).

18. A taxi company charges a flag-down fare of $3.50 for the first kilometre and$0.60 for each additional kilometre or part thereof.

(a) Write a function $F(d)$ for the fare $F$ (in dollars) for a journey of $d$ kilometres, where $d \geq 1$ and $d$ is an integer.
[2]

Answer: $F(d) = 3.50 + 0.60(d - 1)$ or $F(d) = 2.90 + 0.60d$
Working: First km costs $3.50. Remaining$(d-1)$km cost$0.60 each.
$F(d) = 3.50 + 0.60(d - 1) = 3.50 + 0.60d - 0.60 = 2.90 + 0.60d$
Marking Note: 1 mark for correct structure (flag-down + additional), 1 mark for simplified form. Must handle "first km" correctly.

(b) Calculate the fare for a journey of 8 km.
[1]

Answer: $7.70$
Working: $F(8) = 3.50 + 0.60(7) = 3.50 + 4.20 = 7.70$
Marking Note: 1 mark. 7 additional km after the first.

(c) A passenger pays $10.10. Find the distance travelled.
[2]

Answer: $13 \text{ km}$
Working:
$3.50 + 0.60(d - 1) = 10.10$
$0.60(d - 1) = 6.60$
$d - 1 = 11$
$d = 12$
Wait, let me recalculate:
$3.50 + 0.60(d-1) = 10.10$
$0.60(d-1) = 6.60$
$d-1 = 11$
$d = 12$
Correct Answer: $12 \text{ km}$
Marking Note: 1 mark for equation, 1 mark for solution. Check: $3.50 + 0.60(11) = 3.50 + 6.60 = 10.10$ ✓.

19. The function $h(t) = -5t^2 + 20t + 2$ gives the height $h$ (in metres) of a ball thrown upwards $t$ seconds after release.

(a) Find the height of the ball at $t = 0$. What does this represent?
[2]

Answer: Height = 2 m. This represents the initial height from which the ball is thrown (the height of the thrower's hand above ground).
Working: $h(0) = -5(0)^2 + 20(0) + 2 = 2$
Marking Note: 1 mark for height, 1 mark for interpretation.

(b) Find the height of the ball at $t = 2$.
[1]

Answer: $22 \text{ m}$
Working: $h(2) = -5(2)^2 + 20(2) + 2 = -5(4) + 40 + 2 = -20 + 40 + 2 = 22$
Marking Note: 1 mark.

(c) Solve $h(t) = 22$ and interpret your answer.
[3]

Answer: $t = 2$ or $t = 2$ (repeated root). The ball reaches 22 m at $t = 2$ seconds (at its maximum height).
Working:
$-5t^2 + 20t + 2 = 22$
$-5t^2 + 20t - 20 = 0$
Divide by -5: $t^2 - 4t + 4 = 0$
$(t - 2)^2 = 0$
$t = 2$ (repeated root)
Interpretation: The ball reaches a height of 22 m only once, at $t = 2$ seconds, which is its maximum height (vertex of the parabola).
Marking Note: 1 mark for equation, 1 mark for solving (showing repeated root), 1 mark for interpretation (maximum height / vertex). The discriminant is zero, indicating the line $h=22$ is tangent to the parabola at its vertex.

20. A company's profit $P$ (in thousands of dollars) from selling $x$ hundred units of a product is given by $P(x) = -2x^2 + 24x - 40$.

(a) Find the profit when 500 units are sold.
[1]

Answer: 10 \text{ thousand dollars} = \10,000$**Working:** 500 units = 5 hundred units, so$x = 5$.$P(5) = -2(5)^2 + 24(5) - 40 = -2(25) + 120 - 40 = -50 + 120 - 40 = 30$**Wait, recalculate:**$-50 + 120 = 70$,$70 - 40 = 30$. **Correct Answer:**$30 \text{ thousand dollars} = $30,000$**Marking Note:** 1 mark. Must convert 500 units to$x = 5$ (hundred units).

(b) Find the number of units that must be sold to break even (profit = 0).
[3]

Answer: 200 units or 1000 units
Working:
$-2x^2 + 24x - 40 = 0$
Divide by -2: $x^2 - 12x + 20 = 0$
$(x - 2)(x - 10) = 0$
$x = 2$ or $x = 10$
Since $x$ is in hundreds of units: 200 units or 1000 units.
Marking Note: 1 mark for equation, 1 mark for solving, 1 mark for correct units conversion (hundreds to units). Both break-even points required.

(c) What is the maximum profit, and how many units must be sold to achieve it?
[2]

Answer: Maximum profit = 32 \text{ thousand dollars} = \32,000$at 600 units. **Working:** The quadratic$P(x) = -2x^2 + 24x - 40$has a maximum at its vertex.$x$-coordinate of vertex:$x = -\frac{b}{2a} = -\frac{24}{2(-2)} = \frac{24}{4} = 6$Maximum profit:$P(6) = -2(6)^2 + 24(6) - 40 = -2(36) + 144 - 40 = -72 + 144 - 40 = 32$$x = 6$hundred units = 600 units. **Marking Note:** 1 mark for finding$x = 6$(vertex), 1 mark for maximum profit value and units conversion. Alternative: complete the square$P(x) = -2(x-6)^2 + 32$.

---

End of Answer Key