A Level H1 Mathematics Graphs Coordinate Geometry Quiz

<!-- TuitionGoWhere generation metadata: stage=3-0; model=openrouter/owl-alpha; model_label=LongCat 2.0 LLM; generated=2026-06-06; Sources: Stage 2-1 real exam-derived templates and Stage 2-2 exam-enriched syllabus. -->

A-Level Maths H1 Quiz - Graphs Coordinate Geometry

Name: ____________________
Class: ____________________
Date: ____________________
Score: _____ / 40

Duration: 60 minutes
Total Marks: 40

---

Instructions

• Answer all 20 questions in the spaces provided.
• Show all working clearly. Answers without working may not receive full marks.
• A graphing calculator may be used where appropriate.
• Give non-exact answers correct to 3 significant figures unless otherwise stated.
• The number of marks for each question is shown in brackets [ ].

---

Section A: Graph Sketching and Properties (Questions 1–5)

---

1.
The curve $C$ has equation $y = \frac{2x + 1}{x - 3}$, where $x \neq 3$.

(a) Write down the equations of the vertical and horizontal asymptotes of $C$.
[2]

(b) Find the coordinates of the point where $C$ crosses the $y$-axis.
[1]

(c) Sketch the curve $C$, clearly labelling all asymptotes and intercepts.
[2]

---

2.
The graph of $y = f(x)$ is shown below.

Generated graph for Q2.

(a) State the number of real solutions to $f(x) = 0$.
[1]

(b) State the number of real solutions to $f(x) = 2$.
[1]

(c) State the number of real solutions to $f(x) = -4$.
[1]

---

3.
The diagram below shows the graph of $y = |2x - 4|$.

Generated graph for Q3.

(a) Write down the coordinates of the vertex of the graph.
[1]

(b) Solve the equation $|2x - 4| = 6$.
[2]

(c) State the range of values of $x$ for which $|2x - 4| \leq 6$.
[2]

---

4.
The curve $C$ has equation $y = x^3 - 6x^2 + 9x + 1$.

(a) Find $\frac{dy}{dx}$.
[2]

(b) Find the coordinates of the stationary points of $C$ and determine their nature.
[4]

---

5.
The graph of $y = a^x$ passes through the point $(3, 8)$.

(a) Find the value of $a$.
[2]

(b) Write down the equation of the asymptote of the graph of $y = a^x$.
[1]

(c) Sketch the graph of $y = a^x$, clearly showing the asymptote and the point $(3, 8)$.
[2]

---

Section B: Coordinate Geometry (Questions 6–10)

---

6.
The points $A$ and $B$ have coordinates $(2, 5)$ and $(8, 11)$ respectively.

(a) Find the gradient of the line $AB$.
[1]

(b) Find the equation of the line $AB$ in the form $y = mx + c$.
[2]

(c) Find the coordinates of the midpoint of $AB$.
[2]

---

7.
The line $L_1$ has equation $3x + 4y = 12$.
The line $L_2$ is perpendicular to $L_1$ and passes through the point $(6, -2)$.

(a) Find the gradient of $L_1$.
[1]

(b) Find the equation of $L_2$.
[3]

(c) Find the coordinates of the point of intersection of $L_1$ and $L_2$.
[3]

---

8.
The circle $C$ has centre $(4, -3)$ and radius $5$.

(a) Write down the equation of $C$ in the form $(x - a)^2 + (y - b)^2 = r^2$.
[1]

(b) Show that the point $(7, 1)$ lies on $C$.
[2]

(c) Find the equation of the tangent to $C$ at the point $(7, 1)$.
[3]

---

9.
The points $P$, $Q$, and $R$ have coordinates $(1, 2)$, $(4, 6)$, and $(k, 4)$ respectively.

(a) Find the value of $k$ for which $P$, $Q$, and $R$ are collinear.
[3]

(b) For the value of $k$ found in part (a), find the ratio in which $R$ divides $PQ$.
[2]

---

10.
The diagram shows a triangle with vertices $A(0, 0)$, $B(6, 0)$, and $C(3, 4)$.

Generated diagram for Q10.

(a) Find the area of triangle $ABC$.
[2]

(b) Find the length of $AC$.
[2]

(c) Find the equation of the perpendicular bisector of $AB$.
[2]

---

Section C: Applications and Transformations (Questions 11–15)

---

11.
The graph of $y = f(x)$ undergoes the following transformations in order:

• A stretch parallel to the $y$-axis by a scale factor of 3
• A translation of 2 units in the negative $x$-direction

The resulting graph has equation $y = 3(x + 2)^2$.

(a) Write down the equation of the original graph $y = f(x)$.
[2]

(b) Describe a single transformation that maps $y = f(x)$ onto $y = 3(x + 2)^2$.
[2]

---

12.
A rectangular garden has length $2x$ metres and width $x$ metres. The perimeter of the garden is 48 metres.

(a) Write down an equation in $x$ and solve it.
[2]

(b) The area of the garden is $A$ square metres. Express $A$ in terms of $x$.
[1]

(c) By completing the square, find the maximum possible area of the garden.
[3]

---

13.
The diagram below shows the graph of $y = \sqrt{x}$ and the line $y = mx$, where $m > 0$.

Generated graph for Q13.

The line $y = mx$ is tangent to the curve $y = \sqrt{x}$ at the point $(4, 2)$.

(a) Find the gradient of the curve $y = \sqrt{x}$ at $x = 4$.
[2]

(b) Hence find the value of $m$.
[1]

(c) Find the area of the region bounded by $y = \sqrt{x}$, $y = mx$, and the $x$-axis.
[3]

---

14.
The curve $C$ has equation $y = x^2 - 4x + 7$.

(a) Express $x^2 - 4x + 7$ in the form $(x - a)^2 + b$.
[2]

(b) Write down the coordinates of the minimum point of $C$.
[1]

(c) The line $y = k$ intersects $C$ at two distinct points. State the range of possible values of $k$.
[2]

---

15.
The diagram shows the graph of $y = \frac{1}{x}$ for $x > 0$ and the tangent to the curve at the point $(2, \frac{1}{2})$.

Generated graph for Q15.

(a) Find $\frac{dy}{dx}$ for the curve $y = \frac{1}{x}$.
[1]

(b) Find the equation of the tangent at $(2, \frac{1}{2})$.
[3]

(c) Find the area of the triangle formed by the tangent and the coordinate axes.
[2]

---

Section D: Mixed Applications (Questions 16–20)

---

16.
The population $P$ (in thousands) of a town is modelled by the equation $P = 50 + 10e^{0.05t}$, where $t$ is the time in years after 2020.

(a) Find the population in 2020.
[1]

(b) Find the rate of change of the population when $t = 10$.
[2]

(c) Explain what happens to the population as $t \to \infty$ in the context of the model.
[1]

---

17.
The points $A(1, 3)$ and $B(5, 7)$ are given. The point $P(x, y)$ moves such that $PA : PB = 2 : 1$.

(a) Show that the locus of $P$ is a circle and find its centre and radius.
[5]

---

18.
The diagram shows the graph of $y = x^3 - 3x$.

Generated graph for Q18.

(a) Find the coordinates of the stationary points and determine their nature.
[4]

(b) Sketch the graph of $y = x^3 - 3x$, clearly showing all stationary points and intercepts.
[2]

---

19.
The line $y = 2x + c$ is a tangent to the curve $y = x^2 + 3x + 1$.

(a) Find the value of $c$.
[3]

(b) Find the coordinates of the point of contact.
[2]

---

20.
A company's profit $P$ (in thousands of dollars) from selling $x$ units of a product is given by $P = -x^2 + 40x - 300$.

(a) Find the number of units that must be sold to break even (i.e., $P = 0$).
[2]

(b) Find the maximum profit and the number of units that must be sold to achieve it.
[3]

(c) Sketch the graph of $P$ against $x$, clearly showing the break-even points and the maximum point.
[2]

---

End of Quiz

<!-- TuitionGoWhere generation metadata: stage=3-0; model=openrouter/owl-alpha; model_label=LongCat 2.0 LLM; generated=2026-06-06; Sources: Stage 2-1 real exam-derived templates and Stage 2-2 exam-enriched syllabus. -->

A-Level Maths H1 Quiz - Graphs Coordinate Geometry

Answer Key

---

Question 1 [5 marks]

(a)
The vertical asymptote occurs where the denominator is zero:
$x - 3 = 0 \Rightarrow x = 3$

The horizontal asymptote: as $x \to \infty$, $y \to \frac{2x}{x} = 2$, so $y = 2$.

Answer: Vertical asymptote: $x = 3$; Horizontal asymptote: $y = 2$. [2]

(b)
At the $y$-axis, $x = 0$:
$y = \frac{2(0) + 1}{0 - 3} = \frac{1}{-3} = -\frac{1}{3}$

Answer: $\left(0, -\frac{1}{3}\right)$ [1]

(c)
The sketch should show:

• Vertical asymptote at $x = 3$ (dashed line)
• Horizontal asymptote at $y = 2$ (dashed line)
• $y$-intercept at $\left(0, -\frac{1}{3}\right)$
• $x$-intercept at $\left(-\frac{1}{2}, 0\right)$ (found by setting $y = 0$)
• The curve approaching asymptotes correctly in all four regions

[2 marks for correct shape and asymptotes, 1 mark for intercepts]

---

Question 2 [3 marks]

(a) From the graph, $y = f(x)$ crosses the $x$-axis at three points (the origin and two other points).
Answer: 3 real solutions [1]

(b) The horizontal line $y = 2$ intersects the cubic curve at two points (one on the rising left branch between the maximum and the left side, and one on the right rising branch).
Answer: 2 real solutions [1]

(c) The horizontal line $y = -4$ lies below the local minimum at $(2, -3)$, so it intersects the curve at exactly one point (on the far right branch).
Answer: 1 real solution [1]

---

Question 3 [5 marks]

(a) The vertex of $y = |2x - 4|$ occurs where $2x - 4 = 0$, i.e., $x = 2$. At this point, $y = 0$.
Answer: Vertex at $(2, 0)$ [1]

(b) Solve $|2x - 4| = 6$:
$2x - 4 = 6 \Rightarrow 2x = 10 \Rightarrow x = 5$
$2x - 4 = -6 \Rightarrow 2x = -2 \Rightarrow x = -1$

Answer: $x = -1$ or $x = 5$ [2]

(c) From part (b), $|2x - 4| \leq 6$ means $-6 \leq 2x - 4 \leq 6$.
Adding 4: $-2 \leq 2x \leq 10$
Dividing by 2: $-1 \leq x \leq 5$

Answer: $-1 \leq x \leq 5$ [2]

---

Question 4 [6 marks]

(a)
$\frac{dy}{dx} = 3x^2 - 12x + 9$ [2]

(b) At stationary points, $\frac{dy}{dx} = 0$:
$3x^2 - 12x + 9 = 0$
$x^2 - 4x + 3 = 0$
$(x - 1)(x - 3) = 0$
$x = 1$ or $x = 3$

When $x = 1$: $y = 1 - 6 + 9 + 1 = 5$, so point is $(1, 5)$.
When $x = 3$: $y = 27 - 54 + 27 + 1 = 1$, so point is $(3, 1)$.

Second derivative: $\frac{d^2y}{dx^2} = 6x - 12$

At $x = 1$: $\frac{d^2y}{dx^2} = 6 - 12 = -6 < 0$ → local maximum at $(1, 5)$.
At $x = 3$: $\frac{d^2y}{dx^2} = 18 - 12 = 6 > 0$ → local minimum at $(3, 1)$.

Answer: Local maximum at $(1, 5)$; Local minimum at $(3, 1)$. [4]

---

Question 5 [5 marks]

(a) Substituting $(3, 8)$ into $y = a^x$:
$8 = a^3$
$a = \sqrt[3]{8} = 2$

Answer: $a = 2$ [2]

(b) For $y = 2^x$, as $x \to -\infty$, $y \to 0$.
Answer: $y = 0$ (the $x$-axis) [1]

(c) The sketch should show:

• An exponential curve passing through $(3, 8)$, $(0, 1)$, $(1, 2)$, $(2, 4)$
• Asymptote at $y = 0$ (dashed line)
• The curve increasing and concave up
• Point $(3, 8)$ clearly labelled

[2 marks for correct shape, 1 mark for asymptote, 1 mark for labelled point]

---

Question 6 [5 marks]

(a) Gradient $m = \frac{11 - 5}{8 - 2} = \frac{6}{6} = 1$
Answer: $m = 1$ [1]

(b) Using point-slope form with point $(2, 5)$:
$y - 5 = 1(x - 2)$
$y = x + 3$

Answer: $y = x + 3$ [2]

(c) Midpoint $= \left(\frac{2 + 8}{2}, \frac{5 + 11}{2}\right) = (5, 8)$
Answer: $(5, 8)$ [2]

---

Question 7 [7 marks]

(a) Rearranging $3x + 4y = 12$:
$4y = -3x + 12$
$y = -\frac{3}{4}x + 3$

Answer: Gradient of $L_1 = -\frac{3}{4}$ [1]

(b) Since $L_2 \perp L_1$, gradient of $L_2 = \frac{4}{3}$ (negative reciprocal).

Using point-slope form with $(6, -2)$:
$y - (-2) = \frac{4}{3}(x - 6)$
$y + 2 = \frac{4}{3}x - 8$
$y = \frac{4}{3}x - 10$

Answer: $y = \frac{4}{3}x - 10$ or $4x - 3y = 30$ [3]

(c) Substituting $y = \frac{4}{3}x - 10$ into $3x + 4y = 12$:
$3x + 4\left(\frac{4}{3}x - 10\right) = 12$
$3x + \frac{16}{3}x - 40 = 12$
$\frac{25}{3}x = 52$
$x = \frac{156}{25} = 6.24$

$y = \frac{4}{3}\left(\frac{156}{25}\right) - 10 = \frac{208}{25} - 10 = \frac{208 - 250}{25} = -\frac{42}{25} = -1.68$

Answer: $\left(\frac{156}{25}, -\frac{42}{25}\right)$ or $(6.24, -1.68)$ [3]

---

Question 8 [6 marks]

(a)
$(x - 4)^2 + (y + 3)^2 = 25$ [1]

(b) Substituting $(7, 1)$:
$(7 - 4)^2 + (1 + 3)^2 = 3^2 + 4^2 = 9 + 16 = 25$ ✓

Since this equals $r^2 = 25$, the point lies on $C$. [2]

(c) The radius to $(7, 1)$ has gradient $\frac{1 - (-3)}{7 - 4} = \frac{4}{3}$.
The tangent is perpendicular to the radius, so its gradient is $-\frac{3}{4}$.

Equation of tangent:
$y - 1 = -\frac{3}{4}(x - 7)$
$4y - 4 = -3x + 21$
$3x + 4y = 25$

Answer: $3x + 4y = 25$ [3]

---

Question 9 [5 marks]

(a) For collinearity, the gradient of $PQ$ equals the gradient of $PR$.

Gradient of $PQ = \frac{6 - 2}{4 - 1} = \frac{4}{3}$

Gradient of $PR = \frac{4 - 2}{k - 1} = \frac{2}{k - 1}$

Setting equal: $\frac{2}{k - 1} = \frac{4}{3}$
$6 = 4(k - 1)$
$6 = 4k - 4$
$4k = 10$
$k = \frac{5}{2} = 2.5$

Answer: $k = \frac{5}{2}$ [3]

(b) With $k = \frac{5}{2}$, $R = \left(\frac{5}{2}, 4\right)$.

Let $R$ divide $PQ$ in ratio $\lambda : 1$. Using section formula:
$\frac{5}{2} = \frac{\lambda \cdot 4 + 1 \cdot 1}{\lambda + 1}$
$\frac{5}{2}(\lambda + 1) = 4\lambda + 1$
$\frac{5\lambda}{2} + \frac{5}{2} = 4\lambda + 1$
$\frac{5}{2} - 1 = 4\lambda - \frac{5\lambda}{2}$
$\frac{3}{2} = \frac{3\lambda}{2}$
$\lambda = 1$

Answer: $R$ divides $PQ$ in the ratio $1 : 1$ (i.e., $R$ is the midpoint). [2]

---

Question 10 [6 marks]

(a) Base $AB = 6$, height $= 4$.
Area $= \frac{1}{2} \times 6 \times 4 = 12$ square units. [2]

(b) $AC = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units. [2]

(c) Midpoint of $AB = (3, 0)$. The perpendicular bisector is vertical (since $AB$ is horizontal).
Answer: $x = 3$ [2]

---

Question 11 [4 marks]

(a) Working backwards from $y = 3(x + 2)^2$:

• Undo translation: replace $x$ with $x - 2$ → $y = 3x^2$
• Undo stretch: divide by 3 → $y = x^2$

Answer: $f(x) = x^2$ [2]

(b) The transformation from $y = x^2$ to $y = 3(x + 2)^2$ can be described as:

• A translation of 2 units in the negative $x$-direction, followed by
• A stretch parallel to the $y$-axis by scale factor 3

Or as a single transformation: a translation by $\begin{pmatrix} -2 \\ 0 \end{pmatrix}$ followed by a stretch parallel to the $y$-axis by scale factor 3.

Note: These cannot be combined into a single elementary transformation. The question asks for the two-step description. [2]

---

Question 12 [6 marks]

(a) Perimeter $= 2(2x) + 2(x) = 6x = 48$
$x = 8$

Answer: $x = 8$ [2]

(b) $A = 2x \cdot x = 2x^2$
Answer: $A = 2x^2$ [1]

(c) With $x = 8$: $A = 2(8)^2 = 128$ square metres.

However, if the question intends a general optimisation (perhaps with a fixed perimeter constraint):
Given $6x = 48$, we have $x = 8$ fixed, so the area is fixed at $128$ m².

Alternatively, if the question allows variable dimensions with fixed perimeter:
$A = 2x^2$ where $x = 8$ is fixed by the perimeter constraint.

Answer: Maximum area $= 128$ m² (achieved when $x = 8$, giving dimensions $16$ m by $8$ m). [3]

---

Question 13 [6 marks]

(a) $y = x^{1/2}$, so $\frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

At $x = 4$: $\frac{dy}{dx} = \frac{1}{2\sqrt{4}} = \frac{1}{4}$.

Answer: Gradient $= \frac{1}{4}$ [2]

(b) Since the line is tangent to the curve at $(4, 2)$, the gradient of the line equals the gradient of the curve at that point.
Answer: $m = \frac{1}{4}$ [1]

(c) The region is bounded by $y = \sqrt{x}$, $y = \frac{1}{4}x$, and the $x$-axis.

The curves intersect at $(0, 0)$ and $(4, 2)$.

Area $= \int_0^4 \sqrt{x}\,dx - \int_0^4 \frac{1}{4}x\,dx$
$= \left[\frac{2}{3}x^{3/2}\right]_0^4 - \left[\frac{1}{8}x^2\right]_0^4$
$= \frac{2}{3}(8) - \frac{1}{8}(16)$
$= \frac{16}{3} - 2 = \frac{16 - 6}{3} = \frac{10}{3}$

Answer: Area $= \frac{10}{3}$ square units [3]

---

Question 14 [5 marks]

(a) $x^2 - 4x + 7 = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$
Answer: $(x - 2)^2 + 3$ [2]

(b) The minimum occurs when $(x - 2)^2 = 0$, i.e., $x = 2$.
At $x = 2$: $y = 3$.

Answer: Minimum point at $(2, 3)$ [1]

(c) For two distinct intersections, $k > 3$ (the line $y = k$ must be above the minimum).
Answer: $k > 3$ [2]

---

Question 15 [6 marks]

(a) $y = x^{-1}$, so $\frac{dy}{dx} = -x^{-2} = -\frac{1}{x^2}$. [1]

(b) At $(2, \frac{1}{2})$: $\frac{dy}{dx} = -\frac{1}{4}$.

Equation of tangent:
$y - \frac{1}{2} = -\frac{1}{4}(x - 2)$
$y = -\frac{1}{4}x + \frac{1}{2} + \frac{1}{2}$
$y = -\frac{1}{4}x + 1$

Answer: $y = -\frac{1}{4}x + 1$ or $x + 4y = 4$ [3]

(c) The tangent has $x$-intercept at $(4, 0)$ and $y$-intercept at $(0, 1)$.
Area of triangle $= \frac{1}{2} \times 4 \times 1 = 2$ square units. [2]

---

Question 16 [4 marks]

(a) In 2020, $t = 0$:
$P = 50 + 10e^0 = 50 + 10 = 60$ thousand.

Answer: 60,000 [1]

(b) $\frac{dP}{dt} = 10 \cdot 0.05e^{0.05t} = 0.5e^{0.05t}$

At $t = 10$: $\frac{dP}{dt} = 0.5e^{0.5} \approx 0.5 \times 1.6487 \approx 0.824$ thousand per year.

Answer: Approximately 824 people per year [2]

(c) As $t \to \infty$, $e^{0.05t} \to \infty$, so $P \to \infty$.
In context: The population grows without bound according to this model, which is unrealistic in the long term due to resource constraints. [1]

---

Question 17 [5 marks]

Given $PA : PB = 2 : 1$, so $PA = 2PB$, which gives $PA^2 = 4PB^2$.

Let $P = (x, y)$. Then:
$(x - 1)^2 + (y - 3)^2 = 4[(x - 5)^2 + (y - 7)^2]$

Expanding:
$x^2 - 2x + 1 + y^2 - 6y + 9 = 4[x^2 - 10x + 25 + y^2 - 14y + 49]$
$x^2 + y^2 - 2x - 6y + 10 = 4x^2 + 4y^2 - 40x - 56y + 296$

$0 = 3x^2 + 3y^2 - 38x - 50y + 286$

$x^2 + y^2 - \frac{38}{3}x - \frac{50}{3}y + \frac{286}{3} = 0$

Completing the square:
$\left(x - \frac{19}{3}\right)^2 - \frac{361}{9} + \left(y - \frac{25}{3}\right)^2 - \frac{625}{9} + \frac{286}{3} = 0$

$\left(x - \frac{19}{3}\right)^2 + \left(y - \frac{25}{3}\right)^2 = \frac{361 + 625 - 858}{9} = \frac{128}{9}$

Answer: Centre $\left(\frac{19}{3}, \frac{25}{3}\right)$, radius $= \frac{\sqrt{128}}{3} = \frac{8\sqrt{2}}{3}$ [5]

---

Question 18 [6 marks]

(a) $\frac{dy}{dx} = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)$

Stationary points at $x = 1$ and $x = -1$.

When $x = 1$: $y = 1 - 3 = -2$, point $(1, -2)$.
When $x = -1$: $y = -1 + 3 = 2$, point $(-1, 2)$.

Second derivative: $\frac{d^2y}{dx^2} = 6x$

At $x = 1$: $\frac{d^2y}{dx^2} = 6 > 0$ → local minimum at $(1, -2)$.
At $x = -1$: $\frac{d^2y}{dx^2} = -6 < 0$ → local maximum at $(-1, 2)$. [4]

(b) Sketch should show:

• Local maximum at $(-1, 2)$
• Local minimum at $(1, -2)$
• $x$-intercepts at $(-\sqrt{3}, 0)$, $(0, 0)$, $(\sqrt{3}, 0)$
• Curve passing through origin with correct shape (cubic with positive leading coefficient) [2]

---

Question 19 [5 marks]

(a) For the line to be tangent to the curve, the equation $x^2 + 3x + 1 = 2x + c$ must have exactly one solution.

$x^2 + x + (1 - c) = 0$

For a repeated root, discriminant $= 0$:
$1^2 - 4(1)(1 - c) = 0$
$1 - 4 + 4c = 0$
$4c = 3$
$c = \frac{3}{4}$

Answer: $c = \frac{3}{4}$ [3]

(b) Substituting $c = \frac{3}{4}$:
$x^2 + x + \frac{1}{4} = 0$
$(x + \frac{1}{2})^2 = 0$
$x = -\frac{1}{2}$

$y = 2(-\frac{1}{2}) + \frac{3}{4} = -1 + \frac{3}{4} = -\frac{1}{4}$

Answer: $\left(-\frac{1}{2}, -\frac{1}{4}\right)$ [2]

---

Question 20 [7 marks]

(a) Break even when $P = 0$:
$-x^2 + 40x - 300 = 0$
$x^2 - 40x + 300 = 0$
$(x - 10)(x - 30) = 0$
$x = 10$ or $x = 30$

Answer: 10 units or 30 units [2]

(b) Completing the square:
$P = -(x^2 - 40x) - 300$
$P = -(x - 20)^2 + 400 - 300$
$P = -(x - 20)^2 + 100$

Maximum profit $= 100$ (thousand dollars) when $x = 20$ units. [3]

(c) Sketch should show:

• Downward-opening parabola
• $x$-intercepts at $(10, 0)$ and $(30, 0)$
• Maximum point at $(20, 100)$
• $y$-intercept at $(0, -300)$ [2]

---

Total: 40 marks