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A Level H1 Mathematics Graphs Coordinate Geometry Quiz

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Questions

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]

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

<image_placeholder> id: Q2-fig1 type: graph linked_question: Q2 description: Graph of y = f(x) showing a cubic curve with a local maximum at approximately (-1, 4), a local minimum at approximately (2, -3), and passing through the origin (0, 0). The curve rises from the left, peaks at (-1, 4), falls through (0, 0) to a minimum at (2, -3), then rises to the right. labels: x-axis, y-axis, curve y = f(x), local maximum at (-1, 4), local minimum at (2, -3), origin (0, 0) values: x-range: -3 to 4, y-range: -5 to 6 must_show: The cubic shape with labelled turning points at (-1, 4) and (2, -3), and the curve passing through (0, 0). Axes labelled. </image_placeholder>

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

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

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

<image_placeholder> id: Q3-fig1 type: graph linked_question: Q3 description: V-shaped graph of y = |2x - 4| with vertex at (2, 0). The graph has two linear branches: for x < 2, the line has slope -2 and passes through (0, 4); for x > 2, the line has slope 2 and passes through (4, 4). The vertex is at (2, 0). labels: x-axis, y-axis, vertex at (2, 0), point (0, 4), point (4, 4) values: x-range: -1 to 5, y-range: -1 to 6 must_show: V-shape with vertex clearly marked at (2, 0), the two branches with correct slopes, and intercept at (0, 4). </image_placeholder>

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

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

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]

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]

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]

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 $C(3, 4)$ .

<image_placeholder> id: Q10-fig1 type: diagram linked_question: Q10 description: Triangle ABC with vertices at A(0, 0), B(6, 0), and C(3, 4). The base AB lies on the x-axis from (0, 0) to (6, 0). Point C is directly above the midpoint of AB at (3, 4). The triangle is isosceles with AC = BC. labels: A(0, 0), B(6, 0), C(3, 4), base AB on x-axis values: AB = 6 units, height from C to AB = 4 units must_show: Triangle with all three vertices labelled with coordinates, base on x-axis, and the perpendicular from C to AB shown as a dashed line. </image_placeholder>

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

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

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

<image_placeholder> id: Q13-fig1 type: graph linked_question: Q13 description: Graph showing the curve y = sqrt(x) starting at the origin and curving upward to the right, and a straight line y = mx passing through the origin with positive slope m. The line intersects the curve at the origin and at one other point in the first quadrant. The line is steeper than the curve initially but the curve eventually overtakes it. labels: y = sqrt(x), y = mx, origin (0, 0), intersection point (a, b) in first quadrant values: x-range: 0 to 5, y-range: 0 to 4 must_show: The curve y = sqrt(x) and the line y = mx intersecting at the origin and one other point. Both curves labelled. </image_placeholder>

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]

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]

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

<image_placeholder> id: Q15-fig1 type: graph linked_question: Q15 description: Graph of y = 1/x for x > 0 showing a decreasing hyperbolic curve in the first quadrant. The tangent line at (2, 1/2) is shown, with negative slope, intersecting the x-axis at (4, 0) and the y-axis at (0, 1). labels: y = 1/x, tangent at (2, 1/2), x-intercept at (4, 0), y-intercept at (0, 1) values: x-range: 0 to 5, y-range: 0 to 3 must_show: The curve y = 1/x, the tangent line at (2, 1/2), and the intercepts of the tangent at (4, 0) and (0, 1). </image_placeholder>

(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]

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]

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 $y = x^3 - 3x$ .

<image_placeholder> id: Q18-fig1 type: graph linked_question: Q18 description: Cubic graph of y = x^3 - 3x showing a local maximum at (-1, 2), a local minimum at (1, -2), and passing through the origin (0, 0). The curve crosses the x-axis at (-sqrt(3), 0), (0, 0), and (sqrt(3), 0). labels: y = x^3 - 3x, local maximum at (-1, 2), local minimum at (1, -2), x-intercepts at (-sqrt(3), 0), (0, 0), (sqrt(3), 0) values: x-range: -3 to 3, y-range: -4 to 4 must_show: The cubic curve with turning points at (-1, 2) and (1, -2), and x-intercepts at approximately (-1.73, 0), (0, 0), and (1.73, 0). </image_placeholder>

(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]

End of Quiz

Answers

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]

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