A Level H1 Mathematics Graphs Coordinate Geometry Quiz

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A-Level Maths H1 Quiz - Graphs Coordinate Geometry

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 40

Duration: 50 minutes
Total Marks: 40
Topic: Graphs & Coordinate Geometry (graphs-coordinate-geometry)
Version: 1 of 5 (Practice Quiz)

Instructions:

• Answer all 20 questions.
• Show your working clearly where required.
• Use a graphing calculator if helpful.
• Write answers in the spaces provided.

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Section A: Lines and Basic Coordinate Geometry (Questions 1–5)

1. [2 marks] Find the gradient of the line passing through the points $(2, -3)$ and $(5, 6)$.

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2. [2 marks] Find the equation of the line that passes through $(4, 1)$ and is parallel to the line $y = 3x - 2$. Give your answer in the form $y = mx + c$.

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3. [2 marks] Determine whether the points $A(1, 2)$, $B(3, 5)$, and $C(5, 8)$ are collinear. Justify your answer.

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4. [2 marks] Find the coordinates of the midpoint of the line segment joining $(-2, 4)$ and $(6, -2)$.

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5. [2 marks] A line has equation $2x - 3y = 6$. Find the $x$-intercept and $y$-intercept.

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Section B: Curves, Graphs and Transformations (Questions 6–10)

6. [2 marks] Sketch the graph of $y = e^x$ for $-2 \le x \le 2$. State the coordinates of the $y$-intercept and the equation of any asymptote.

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7. [2 marks] The graph of $y = \ln x$ is transformed to $y = \ln(x - 3)$. Describe the transformation and state the new vertical asymptote.

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8. [2 marks] Solve the equation $e^{2x} = 5$ exactly, giving your answer in terms of $\ln$.

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9. [2 marks] Using a graphing calculator, find the approximate $x$-coordinate of the point where the curves $y = x^2$ and $y = 4 - x$ intersect in the first quadrant. Give your answer to 2 decimal places.

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10. [2 marks] State the range of values of $k$ for which the quadratic equation $x^2 + kx + 4 = 0$ has no real roots.

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Section C: Simultaneous Equations and Inequalities (Questions 11–15)

11. [2 marks] Solve the simultaneous equations $y = x + 1$ and $y = x^2 - 3x + 1$ by substitution.

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12. [2 marks] Solve the inequality $x^2 - 5x + 6 < 0$. Show your answer as an interval.

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13. [2 marks] Use a graphing calculator to solve $x^3 - 2x = 1$ approximately. Give all real roots to 2 decimal places.

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14. [2 marks] Find the equation of the perpendicular bisector of the line segment joining $(0, 0)$ and $(4, 6)$.

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15. [2 marks] The curve $y = \frac{1}{x}$ and the line $y = 2x$ intersect. Find the positive $x$-coordinate of intersection exactly.

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Section D: Applied and Interpretative Graph Questions (Questions 16–20)

16. [2 marks] A scatter diagram shows the data below. Sketch the scatter diagram on the provided axes.

$x$	1	2	3	4	5
$y$	2	4	5	7	8

Image pending generation: graph for Q16.

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17. [2 marks] The graph below shows $y = f(x)$. State the equation of the horizontal asymptote and the $y$-intercept.

Image pending generation: graph for Q17.

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18. [2 marks] A line $L$ passes through $(1, 2)$ and is perpendicular to $y = -\frac{1}{2}x + 3$. Find the equation of $L$.

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19. [2 marks] The quadratic $y = x^2 - 4x + 3$ crosses the $x$-axis at two points. Find them and state the line of symmetry.

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20. [2 marks] A business models profit $P$ (in $ thousands) against time $t$ (months) as $P = 2e^{0.1t}$. Find the profit after 10 months, to the nearest $ hundred.

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

Topic: Graphs & Coordinate Geometry
Version: 1 of 5
Total Marks: 40 (2 marks per question)

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Section A: Lines and Basic Coordinate Geometry

Q1. [2 marks]
Gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-3)}{5 - 2} = \frac{9}{3} = 3$.
Answer: Gradient = 3.
Teaching note: Gradient measures steepness; subtract y's over x's in the same order. Common mistake: reversing order inconsistently.

Q2. [2 marks]
Parallel line has same gradient $m = 3$. Through $(4,1)$: $y - 1 = 3(x - 4) \Rightarrow y = 3x - 12 + 1 = 3x - 11$.
Answer: $y = 3x - 11$.
Marking: 1 mark gradient, 1 mark equation.

Q3. [2 marks]
Gradient $AB = \frac{5-2}{3-1} = \frac{3}{2}$; gradient $BC = \frac{8-5}{5-3} = \frac{3}{2}$. Same gradient and shared point B ⇒ collinear.
Answer: Yes, collinear because gradients equal.
Note: Also check via area of triangle = 0.

Q4. [2 marks]
Midpoint $= \left(\frac{-2+6}{2}, \frac{4+(-2)}{2}\right) = (2, 1)$.
Answer: $(2, 1)$.

Q5. [2 marks]
$x$-intercept: set $y=0 \Rightarrow 2x=6 \Rightarrow x=3$.
$y$-intercept: set $x=0 \Rightarrow -3y=6 \Rightarrow y=-2$.
Answer: $(3,0)$ and $(0,-2)$.

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Section B: Curves, Graphs and Transformations

Q6. [2 marks]
$y$-intercept at $x=0$: $y=e^0=1 \Rightarrow (0,1)$. Asymptote: $y=0$ (x-axis). Curve increasing, passes through (0,1).
Answer: y-int (0,1), asymptote $y=0$.
Sketch: rising curve left to right, never touching x-axis.

Q7. [2 marks]
$y=\ln(x-3)$ is translation of $y=\ln x$ by 3 units right. Vertical asymptote moves from $x=0$ to $x=3$.
Answer: Translation right by 3; new asymptote $x=3$.

Q8. [2 marks]
$e^{2x}=5 \Rightarrow 2x = \ln 5 \Rightarrow x = \frac{1}{2}\ln 5$.
Answer: $x = \frac{\ln 5}{2}$.

Q9. [2 marks]
Solve $x^2 = 4 - x \Rightarrow x^2 + x - 4 = 0$. GC gives positive root $x \approx 1.56$.
Answer: $x \approx 1.56$.

Q10. [2 marks]
No real roots ⇒ discriminant $<0$: $k^2 - 16 < 0 \Rightarrow -4 < k < 4$.
Answer: $-4 < k < 4$.

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Section C: Simultaneous Equations and Inequalities

Q11. [2 marks]
Substitute: $x+1 = x^2 - 3x + 1 \Rightarrow x^2 - 4x = 0 \Rightarrow x(x-4)=0 \Rightarrow x=0,4$.
Then $y=1,5$. Points: $(0,1), (4,5)$.
Answer: $(0,1)$ and $(4,5)$.

Q12. [2 marks]
$x^2 -5x+6 = (x-2)(x-3) < 0 \Rightarrow 2 < x < 3$.
Answer: $(2,3)$.

Q13. [2 marks]
GC root solve: $x^3-2x-1=0$ gives $x \approx -1.00, -0.62, 1.62$.
Answer: $-1.00, -0.62, 1.62$ (2 d.p.).

Q14. [2 marks]
Midpoint $(2,3)$, gradient of segment $= \frac{6}{4}=1.5$, perpendicular gradient $= -\frac{2}{3}$.
Equation: $y-3 = -\frac{2}{3}(x-2) \Rightarrow y = -\frac{2}{3}x + \frac{13}{3}$.
Answer: $y = -\frac{2}{3}x + \frac{13}{3}$.

Q15. [2 marks]
$\frac{1}{x}=2x \Rightarrow 1=2x^2 \Rightarrow x^2=\frac{1}{2} \Rightarrow x=\frac{1}{\sqrt{2}}$ (positive).
Answer: $x = \frac{1}{\sqrt{2}}$.

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Section D: Applied and Interpretative

Q16. [2 marks]
Plot points (1,2),(2,4),(3,5),(4,7),(5,8) as crosses. Positive linear trend.
Answer: Scatter as described; axes labelled 0–6 and 0–10.

Q17. [2 marks]
From image: y-int (0,3); horizontal asymptote $y=0$.
Answer: Asymptote $y=0$, y-int 3.

Q18. [2 marks]
Perpendicular gradient = 2 (negative reciprocal of $-\frac{1}{2}$). Through (1,2): $y-2=2(x-1) \Rightarrow y=2x$.
Answer: $y=2x$.

Q19. [2 marks]
$x^2-4x+3=0 \Rightarrow (x-1)(x-3)=0 \Rightarrow x=1,3$. Line of symmetry $x=2$.
Answer: (1,0),(3,0); symmetry $x=2$.

Q20. [2 marks]
$P=2e^{0.1\times10}=2e^1 \approx 2\times2.718 = 5.436$ ($ thousand) = $5436 ≈ $5400 (nearest hundred).
Answer: $5400.