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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: 45 minutes
Total Marks: 40

Instructions:

Answer all 20 questions.
You are expected to use an approved Graphing Calculator (GC).
Unless otherwise specified, give non-exact numerical answers correct to 3 significant figures.
Show all necessary working clearly; unsupported answers from a calculator are generally allowed, but you must show the mathematical setup.
Sketches of graphs should be clearly labeled with intercepts, asymptotes, and stationary points where relevant.

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

[10 Marks]

1. The equation of a curve is $y = 2e^{x} - 5$ .
(i) Write down the equation of the horizontal asymptote. [1]
(ii) Find the exact coordinates of the $x$ -intercept. [1]

2. Sketch the graph of $y = \ln(x - 2) + 1$ for $x > 2$ . On your sketch, clearly indicate: [2]
(i) The equation of the vertical asymptote.
(ii) The coordinates of the $x$ -intercept.
(iii) The coordinates of one other point on the graph.

3. Determine whether the quadratic expression $3x^2 - 4x + 5$ is always positive, always negative, or changes sign. Justify your answer using the discriminant. [2]

4. Solve the inequality $x^2 - 5x + 6 < 0$ . [2]

5. The curve $y = x^3 - 3x^2$ has a stationary point at $x = 0$ .
(i) Find the $y$ -coordinate of this stationary point. [1]
(ii) Determine the nature of this stationary point (maximum, minimum, or point of inflection). [1]

Section B: Calculus and Coordinate Geometry Applications (Questions 6–12)

[14 Marks]

6. Find the equation of the tangent to the curve $y = e^{2x} + 1$ at the point where $x = 0$ . Give your answer in the form $y = mx + c$ . [3]

7. A curve has equation $y = \frac{4}{x} + x$ .
(i) Find $\frac{dy}{dx}$ . [1]
(ii) Hence, find the exact coordinates of the stationary points. [2]

8. The diagram shows the curve $y = \sqrt{x}$ and the line $y = x$ .
(i) Find the coordinates of the points of intersection of the curve and the line. [2]
(ii) Calculate the area of the finite region enclosed by the curve and the line. [2]

9. Find the exact value of $\int_{1}^{e} \left( \frac{1}{x} + 2x \right) \, dx$ . [3]

10. The normal to the curve $y = \ln(2x)$ at the point where $x = 1$ intersects the $x$ -axis at point $A$ . Find the $x$ -coordinate of $A$ . [3]

11. Solve the simultaneous equations:
$y = x + 2$
$y = x^2 - 4$
[2]

12. Given that $f(x) = e^x - 3x$ , find the range of values of $x$ for which $f'(x) > 0$ . [2]

Section C: Data Analysis and Regression (Questions 13–20)

[16 Marks]

Context: A marketing firm collects data on advertising spend ( $x$ , in $000s) and monthly sales revenue ( $y$ , in $000s) for 8 different branches.

Branch	$x$ (Ad Spend)	$y$ (Sales)
A	2.0	15.2
B	3.5	22.1
C	1.5	12.5
D	4.0	25.8
E	2.5	18.4
F	5.0	30.2
G	3.0	20.5
H	4.5	28.0

13. Using your Graphing Calculator, draw a scatter diagram of $y$ against $x$ . Label the axes appropriately. [2]

14. Calculate the product moment correlation coefficient, $r$ , for this data. [1]

15. Comment on the value of $r$ in the context of the data. [1]

16. Find the equation of the least squares regression line of $y$ on $x$ in the form $y = ax + b$ . Give the values of $a$ and $b$ correct to 3 significant figures. [2]

17. Interpret the meaning of the gradient $a$ in the context of this problem. [1]

18. Use your regression equation to estimate the sales revenue for a branch that spends $3,200 on advertising. [1]

19. Explain why it might be inappropriate to use this regression line to estimate the sales revenue for a branch that spends $15,000 on advertising. [1]

20. A new data point $(6.0, 10.0)$ is added to the dataset.
(i) Without calculating, state and explain the likely effect of this new point on the value of the correlation coefficient $r$ . [2]
(ii) Would this new point be considered an outlier? Briefly explain. [1]

Answers

A-Level Maths H1 Quiz - Graphs Coordinate Geometry (Answer Key)

Total Marks: 40

Section A: Basic Skills and Graph Sketching

1.
(i) As $x \to -\infty$ , $e^x \to 0$ . Thus, $y \to -5$ .
Answer: $y = -5$ [1]
(ii) At $x$ -intercept, $y = 0$ .
$0 = 2e^x - 5 \implies 2e^x = 5 \implies e^x = 2.5 \implies x = \ln(2.5)$ .
Answer: $(\ln 2.5, 0)$ [1]

2.
(i) Vertical asymptote occurs where argument of ln is zero: $x - 2 = 0 \implies x = 2$ .
Answer: $x = 2$ [0.5]
(ii) $x$ -intercept: $0 = \ln(x-2) + 1 \implies \ln(x-2) = -1 \implies x-2 = e^{-1} \implies x = 2 + e^{-1} \approx 2.37$ .
Answer: $(2 + e^{-1}, 0)$ or approx $(2.37, 0)$ [0.5]
(iii) e.g., if $x=3$ , $y = \ln(1)+1 = 1$ . Point $(3,1)$ .
Answer: Sketch showing correct shape (increasing, concave down), asymptote at $x=2$ , and intercepts. [1]

3.
Discriminant $\Delta = b^2 - 4ac = (-4)^2 - 4(3)(5) = 16 - 60 = -44$ .
Since $\Delta < 0$ , there are no real roots.
Since coefficient of $x^2$ ( $a=3$ ) is positive, the parabola opens upwards.
Answer: Always positive. [2] (1 for discriminant calc/conclusion, 1 for justification via $a>0$ )

4.
$x^2 - 5x + 6 < 0$
$(x-2)(x-3) < 0$
Critical values: $x=2, x=3$ .
Since inequality is $<0$ , solution is between roots.
Answer: $2 < x < 3$ [2]

5.
(i) $y = 0^3 - 3(0)^2 = 0$ .
Answer: $(0,0)$ [1]
(ii) $y' = 3x^2 - 6x$ . $y'' = 6x - 6$ .
At $x=0$ , $y'' = -6 < 0$ .
Answer: Maximum point. [1]

Section B: Calculus and Coordinate Geometry Applications

6.
$y = e^{2x} + 1$ .
$\frac{dy}{dx} = 2e^{2x}$ .
At $x=0$ , $y = e^0 + 1 = 2$ . Point $(0,2)$ .
Gradient $m = 2e^0 = 2$ .
Equation: $y - 2 = 2(x - 0) \implies y = 2x + 2$ .
Answer: $y = 2x + 2$ [3] (1 for derivative, 1 for point/gradient, 1 for equation)

7.
(i) $y = 4x^{-1} + x$ .
$\frac{dy}{dx} = -4x^{-2} + 1 = 1 - \frac{4}{x^2}$ . [1]
(ii) Stationary points when $\frac{dy}{dx} = 0$ .
$1 - \frac{4}{x^2} = 0 \implies x^2 = 4 \implies x = \pm 2$ .
If $x=2, y = 4/2 + 2 = 4$ . Point $(2,4)$ .
If $x=-2, y = 4/(-2) - 2 = -4$ . Point $(-2,-4)$ .
Answer: $(2,4)$ and $(-2,-4)$ [2]

8.
(i) Intersection: $\sqrt{x} = x \implies x = x^2 \implies x^2 - x = 0 \implies x(x-1)=0$ .
$x=0 \implies y=0$ .
$x=1 \implies y=1$ .
Answer: $(0,0)$ and $(1,1)$ [2]
(ii) Area $= \int_{0}^{1} (\sqrt{x} - x) \, dx$ .
$= [\frac{2}{3}x^{3/2} - \frac{1}{2}x^2]_0^1$
$= (\frac{2}{3}(1) - \frac{1}{2}(1)) - 0 = \frac{4-3}{6} = \frac{1}{6}$ .
Answer: $\frac{1}{6}$ units $^2$ [2]

9.
$\int_{1}^{e} (\frac{1}{x} + 2x) \, dx = [\ln|x| + x^2]_1^e$
$= (\ln e + e^2) - (\ln 1 + 1^2)$
$= (1 + e^2) - (0 + 1) = e^2$ .
Answer: $e^2$ [3]

10.
$y = \ln(2x)$ . $\frac{dy}{dx} = \frac{1}{2x} \cdot 2 = \frac{1}{x}$ .
At $x=1$ , $y = \ln 2$ . Gradient of tangent $m_t = 1/1 = 1$ .
Gradient of normal $m_n = -1$ .
Equation of normal: $y - \ln 2 = -1(x - 1) \implies y = -x + 1 + \ln 2$ .
At $x$ -axis, $y=0$ : $0 = -x + 1 + \ln 2 \implies x = 1 + \ln 2$ .
Answer: $1 + \ln 2$ [3]

11.
$x + 2 = x^2 - 4 \implies x^2 - x - 6 = 0$ .
$(x-3)(x+2) = 0$ .
$x=3 \implies y=5$ .
$x=-2 \implies y=0$ .
Answer: $(3,5)$ and $(-2,0)$ [2]

12.
$f'(x) = e^x - 3$ .
$e^x - 3 > 0 \implies e^x > 3 \implies x > \ln 3$ .
Answer: $x > \ln 3$ [2]

Section C: Data Analysis and Regression

13.
Answer: Scatter plot with $x$ -axis labeled "Ad Spend ($000s)" and $y$ -axis labeled "Sales Revenue ($000s)". Points plotted correctly according to table. [2]

14.
Using GC:
Answer: $r \approx 0.994$ (accept 0.993 - 0.995) [1]

15.
Answer: There is a strong positive linear correlation between advertising spend and sales revenue. [1]

16.
Using GC for linear regression $y = ax+b$ :
$a \approx 5.37$ , $b \approx 4.66$ .
Answer: $y = 5.37x + 4.66$ [2]

17.
Answer: For every additional $1,000 spent on advertising, sales revenue increases by approximately $5,370 on average. [1]

18.
$x = 3.2$ .
$y = 5.37(3.2) + 4.66 = 17.184 + 4.66 = 21.844$ .
Answer: $21,800 (or 21.8 in $000s) [1]

19.
Answer: $15,000 ( $x=15$ ) is outside the range of the observed data ( $x$ ranges from 1.5 to 5.0). This is extrapolation, which is unreliable as the linear relationship may not hold. [1]

20.
(i) Answer: The value of $r$ will decrease (become weaker). The point $(6,10)$ deviates significantly from the existing strong positive linear trend (high $x$ but low $y$ ). [2]
(ii) Answer: Yes, it is an outlier because it does not follow the general pattern of the rest of the data. [1]