A Level H1 Mathematics Graphs Coordinate Geometry Quiz

A-Level Maths H1 Quiz - Graphs Coordinate Geometry: Answer Key

Total Marks: 50

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Section A: Short Questions (Questions 1–5, 10 marks)

Question 1

Answer: $(0, -2)$ Marks: 2

Working: The curve crosses the $y$-axis where $x = 0$. Substitute $x = 0$ into $y = e^{x} - 3$: $y = e^{0} - 3 = 1 - 3 = -2$ Therefore, the coordinates are $(0, -2)$.

Teaching Note: The $y$-intercept is always found by setting $x = 0$. Recall that $e^0 = 1$, which is a fundamental property of the exponential function.

Common Mistake: Students sometimes forget that $e^0 = 1$ and incorrectly compute $e^0 - 3 = 0 - 3 = -3$.

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Question 2

Answer: $y = 2$ Marks: 1

Working: As $x \to \infty$, $e^{-x} \to 0$. Therefore, $y = 2 + e^{-x} \to 2$. The horizontal asymptote is $y = 2$.

Teaching Note: For curves of the form $y = a + e^{kx}$ where $k < 0$, the horizontal asymptote is $y = a$ because the exponential term decays to zero as $x$ increases.

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Question 3

Answer: $x = 4$ Marks: 1

Working: The natural logarithm $\ln(x - 4)$ is defined only when $x - 4 > 0$, i.e., $x > 4$. As $x \to 4^{+}$, $x - 4 \to 0^{+}$, so $\ln(x - 4) \to -\infty$. The vertical asymptote is $x = 4$.

Teaching Note: For $y = \ln(x - a)$, the vertical asymptote is always $x = a$ because the logarithm is undefined at and below zero.

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Question 4

Answer: $(0, 2)$ Marks: 2

Working: The curve crosses the $y$-axis where $x = 0$. Substitute $x = 0$ into $y = 3 - e^{2x}$: $y = 3 - e^{0} = 3 - 1 = 2$ Therefore, the coordinates are $(0, 2)$.

Teaching Note: Again, $e^0 = 1$. The coefficient 2 in the exponent does not affect the value at $x = 0$ because $2 \times 0 = 0$.

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Question 5

Answer: $f(x) > 5$ (or $y > 5$) Marks: 2

Working: For all real $x$, $e^{x} > 0$. Therefore, $e^{x} + 5 > 5$. The range is $f(x) > 5$, i.e., $(5, \infty)$.

Teaching Note: The exponential function $e^x$ is always positive for all real $x$. Adding 5 shifts the entire range upward. Note that 5 itself is not included in the range because $e^x$ never equals 0.

Marking Note: Award 1 mark for recognising $e^x > 0$; award 1 mark for the correct range.

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Section B: Graph Sketching and Transformations (Questions 6–12, 18 marks)

Question 6

Answer: See sketch description below. Marks: 3

Expected Sketch Features:

• The curve approaches $y = -2$ as $x \to -\infty$ (horizontal asymptote).
• The curve passes through $(0, -1)$ (the $y$-intercept).
• The curve crosses the $x$-axis at $(\ln 2, 0) \approx (0.693, 0)$.
• The curve increases without bound as $x \to \infty$.

Working:

• $y$-intercept: Set $x = 0$: $y = e^0 - 2 = 1 - 2 = -1$. So $(0, -1)$.
• $x$-intercept: Set $y = 0$: $0 = e^x - 2 \Rightarrow e^x = 2 \Rightarrow x = \ln 2$. So $(\ln 2, 0)$.
• Asymptote: As $x \to -\infty$, $e^x \to 0$, so $y \to -2$.

Marking Scheme:

• 1 mark: Correct shape of exponential curve (increasing, concave up).
• 1 mark: Correct $y$-intercept and asymptote labelled.
• 1 mark: Correct $x$-intercept labelled.

Common Mistake: Students often forget the $x$-intercept or mislabel the asymptote as $y = 0$ instead of $y = -2$.

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Question 7

Answer: $y = \ln(x - 3)$; vertical asymptote $x = 3$ Marks: 2

Working: A translation 3 units to the right replaces $x$ with $x - 3$: $y = \ln(x - 3)$ The vertical asymptote of $y = \ln x$ is $x = 0$. Translating 3 units right shifts the asymptote to $x = 3$.

Teaching Note: For horizontal translations, remember: "right means subtract inside the function." The asymptote moves with the graph.

Marking Scheme:

• 1 mark: Correct equation $y = \ln(x - 3)$.
• 1 mark: Correct asymptote $x = 3$.

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Question 8

Answer: A horizontal stretch with scale factor $\frac{1}{2}$ (or a stretch parallel to the $x$-axis with scale factor $\frac{1}{2}$). Marks: 2

Working: The transformation from $y = e^x$ to $y = e^{2x}$ replaces $x$ with $2x$. Replacing $x$ with $kx$ where $k > 1$ represents a horizontal stretch with scale factor $\frac{1}{k}$. Here, $k = 2$, so the scale factor is $\frac{1}{2}$.

Teaching Note: This is a common point of confusion. Replacing $x$ with $2x$ compresses the graph horizontally (makes it "steeper"), so the scale factor is $\frac{1}{2}$, not 2.

Marking Scheme:

• 1 mark: Identifying it as a horizontal stretch.
• 1 mark: Correct scale factor $\frac{1}{2}$.

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Question 9

Answer: See sketch description below. Marks: 3

Expected Sketch Features:

• The curve approaches $x = -2$ from the right (vertical asymptote).
• The curve crosses the $x$-axis at $(-1, 0)$.
• The curve increases slowly without bound as $x \to \infty$.

Working:

• Vertical asymptote: $x + 2 = 0 \Rightarrow x = -2$.
• $x$-intercept: Set $y = 0$: $0 = \ln(x + 2) \Rightarrow x + 2 = 1 \Rightarrow x = -1$. So $(-1, 0)$.

Marking Scheme:

• 1 mark: Correct shape of logarithmic curve.
• 1 mark: Correct vertical asymptote $x = -2$ labelled.
• 1 mark: Correct $x$-intercept $(-1, 0)$ labelled.

Common Mistake: Students sometimes place the asymptote at $x = 2$ instead of $x = -2$. Remember: the asymptote is where the argument of the logarithm equals zero.

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Question 10

Answer: $y = -e^{x}$ Marks: 1

Working: Reflection in the $x$-axis multiplies the entire function by $-1$: $y = -e^{x}$

Teaching Note: Reflection in the $x$-axis changes the sign of the output ($y$-values). Reflection in the $y$-axis would replace $x$ with $-x$, giving $y = e^{-x}$.

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Question 11

Answer: (a) $y = 4$ (b) $(0, 3)$ (c) See sketch description below. Marks: 4

Working: (a) As $x \to \infty$, $e^{-x} \to 0$, so $y = 4 - e^{-x} \to 4$. The horizontal asymptote is $y = 4$.

(b) Set $x = 0$: $y = 4 - e^0 = 4 - 1 = 3$. The $y$-intercept is $(0, 3)$.

(c) Expected Sketch Features:

• The curve approaches $y = 4$ as $x \to \infty$ (horizontal asymptote).
• The curve passes through $(0, 3)$.
• As $x \to -\infty$, $e^{-x} \to \infty$, so $y \to -\infty$. The curve decreases steeply to the left.

Marking Scheme:

• 1 mark: Correct asymptote $y = 4$.
• 1 mark: Correct $y$-intercept $(0, 3)$.
• 2 marks: Correct sketch with asymptote and intercept labelled.

Teaching Note: Note that $e^{-x}$ grows without bound as $x \to -\infty$ (because $-x \to +\infty$). This is why the curve goes down to negative infinity on the left.

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Question 12

Answer: $y = 3\ln x$ Marks: 1

Working: A stretch parallel to the $y$-axis with scale factor 3 multiplies the entire function by 3: $y = 3\ln x$

Teaching Note: Vertical stretches multiply the output (the $y$-value) by the scale factor. This is distinct from a horizontal stretch, which would modify the input.

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Section C: Extended Response Questions (Questions 13–20, 22 marks)

Question 13

Answer: (a) $(\ln 4, 0)$ (b) $y = -4$ (c) See sketch description below. Marks: 4

Working: (a) Set $y = 0$: $0 = e^x - 4 \Rightarrow e^x = 4 \Rightarrow x = \ln 4$. The $x$-intercept is $(\ln 4, 0)$.

(b) As $x \to -\infty$, $e^x \to 0$, so $y \to -4$. The horizontal asymptote is $y = -4$.

(c) Expected Sketch Features:

• The curve approaches $y = -4$ as $x \to -\infty$.
• The curve passes through $(0, -3)$ (the $y$-intercept: $y = e^0 - 4 = 1 - 4 = -3$).
• The curve crosses the $x$-axis at $(\ln 4, 0) \approx (1.386, 0)$.
• The curve increases without bound as $x \to \infty$.

Marking Scheme:

• 1 mark: Correct $x$-intercept $(\ln 4, 0)$.
• 1 mark: Correct asymptote $y = -4$.
• 2 marks: Correct sketch with intercepts and asymptote labelled.

Teaching Note: The exact form $\ln 4$ is required for full marks. A decimal approximation alone may lose a mark in an "exact" question.

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Question 14

Answer: $y = \ln x + 2$; $x$-intercept at $(e^{-2}, 0)$ Marks: 3

Working: A translation 2 units upwards adds 2 to the function: $y = \ln x + 2$

To find the $x$-intercept, set $y = 0$: $0 = \ln x + 2$ $\ln x = -2$ $x = e^{-2}$

The $x$-intercept is $(e^{-2}, 0)$.

Teaching Note: To solve $\ln x = a$, exponentiate both sides: $x = e^a$. Here, $a = -2$, so $x = e^{-2}$.

Marking Scheme:

• 1 mark: Correct equation $y = \ln x + 2$.
• 2 marks: Correct $x$-intercept $(e^{-2}, 0)$ with working.

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Question 15

Answer: (a) $y = 5$ (b) $(\ln 5, 0)$ (c) $(0, 4)$ Marks: 3

Working: (a) As $x \to -\infty$, $e^x \to 0$, so $y = 5 - e^x \to 5$. The horizontal asymptote is $y = 5$.

(b) Set $y = 0$: $0 = 5 - e^x \Rightarrow e^x = 5 \Rightarrow x = \ln 5$. The $x$-intercept is $(\ln 5, 0)$.

(c) Set $x = 0$: $y = 5 - e^0 = 5 - 1 = 4$. The $y$-intercept is $(0, 4)$.

Marking Scheme:

• 1 mark: Correct asymptote.
• 1 mark: Correct $x$-intercept.
• 1 mark: Correct $y$-intercept.

Common Mistake: For part (b), students sometimes write $x = \ln(-5)$ or forget the negative sign when rearranging. Remember: $5 - e^x = 0 \Rightarrow e^x = 5$.

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Question 16

Answer: $y = -e^{x+1}$ Marks: 2

Working: Step 1: Translation 1 unit left: replace $x$ with $x + 1$. $y = e^{x+1}$

Step 2: Reflection in the $x$-axis: multiply by $-1$. $y = -e^{x+1}$

Teaching Note: Apply transformations in the order stated. A translation left by 1 unit means we add 1 inside the exponent (i.e., $x$ becomes $x + 1$).

Marking Scheme:

• 1 mark: Correct translation $y = e^{x+1}$.
• 1 mark: Correct reflection $y = -e^{x+1}$.

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Question 17

Answer: (a) $x = 0$ (b) $\left(\frac{1}{2}, 0\right)$ (c) A horizontal stretch with scale factor $\frac{1}{2}$ (or stretch parallel to the $x$-axis with scale factor $\frac{1}{2}$). Marks: 3

Working: (a) $\ln(2x)$ is defined only when $2x > 0$, i.e., $x > 0$. The vertical asymptote is $x = 0$ (the $y$-axis).

(b) Set $y = 0$: $0 = \ln(2x) \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$. The $x$-intercept is $\left(\frac{1}{2}, 0\right)$.

(c) Replacing $x$ with $2x$ in $y = \ln x$ gives $y = \ln(2x)$. This is a horizontal stretch with scale factor $\frac{1}{2}$.

Marking Scheme:

• 1 mark: Correct asymptote $x = 0$.
• 1 mark: Correct $x$-intercept $\left(\frac{1}{2}, 0\right)$.
• 1 mark: Correct transformation description.

Teaching Note: The vertical asymptote of $y = \ln(2x)$ is still $x = 0$ because the argument $2x$ approaches 0 as $x$ approaches 0.

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Question 18

Answer: $y = \frac{1}{2}e^{x} - 3$ Marks: 2

Working: Step 1: Stretch parallel to the $y$-axis with scale factor $\frac{1}{2}$: multiply the function by $\frac{1}{2}$. $y = \frac{1}{2}e^{x}$

Step 2: Translate 3 units downwards: subtract 3. $y = \frac{1}{2}e^{x} - 3$

Teaching Note: Vertical stretches multiply the output; vertical translations add/subtract a constant. Apply them in the order specified in the question.

Marking Scheme:

• 1 mark: Correct stretch $y = \frac{1}{2}e^x$.
• 1 mark: Correct translation $y = \frac{1}{2}e^x - 3$.

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Question 19

Answer: (a) $x = 0$ (b) $\left(e^{-1/2}, 0\right)$ (c) See sketch description below. Marks: 3

Working: (a) $\ln x$ is defined only when $x > 0$. The vertical asymptote is $x = 0$.

(b) Set $y = 0$: $0 = 2\ln x + 1 \Rightarrow 2\ln x = -1 \Rightarrow \ln x = -\frac{1}{2} \Rightarrow x = e^{-1/2}$. The $x$-intercept is $\left(e^{-1/2}, 0\right)$.

(c) Expected Sketch Features:

• The curve approaches $x = 0$ from the right (vertical asymptote).
• The curve crosses the $x$-axis at $\left(e^{-1/2}, 0\right) \approx (0.607, 0)$.
• The curve increases slowly without bound as $x \to \infty$.

Marking Scheme:

• 1 mark: Correct asymptote $x = 0$.
• 1 mark: Correct $x$-intercept $\left(e^{-1/2}, 0\right)$.
• 1 mark: Correct sketch with asymptote and intercept labelled.

Teaching Note: The coefficient 2 in $2\ln x$ makes the curve steeper than $y = \ln x$, but does not change the asymptote or the general shape.

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Question 20

Answer: $y = \ln(-x)$; domain: $x < 0$ Marks: 2

Working: Reflection in the $y$-axis replaces $x$ with $-x$: $y = \ln(-x)$

For $\ln(-x)$ to be defined, we need $-x > 0$, i.e., $x < 0$. The domain is $x < 0$, or in interval notation, $(-\infty, 0)$.

Teaching Note: Reflecting $y = \ln x$ in the $y$-axis produces a curve that exists only for negative $x$-values. The domain changes from $x > 0$ to $x < 0$.

Marking Scheme:

• 1 mark: Correct equation $y = \ln(-x)$.
• 1 mark: Correct domain $x < 0$.

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END OF ANSWER KEY