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More about Graphs of Functions: Paper 2 Challenge Multiple-Choice

This is the Challenge tier of the multiple-choice set. Every question has a distractor that is the right answer to a slightly different question, so the second check is the skill being tested. Work the standard Paper 2 set first. Detailed solutions can be requested by email.

10 questions · about 20 minutes · answers below each question

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

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Question 1 — reduce, then translate: the shift gets multiplied
Question 2 — when f(−x) and −f(x) give the same graph
Question 3 — where an exponential graph meets its own translate
Question 4 — the roots of f(1 − x) = 0
Question 5 — three roots of f(2x) + 3 = k
Question 6 — reflection in the vertical line x = 2
Question 7 — y = log₂(4x) as a translation and as a reduction
Question 8 — greatest value and the mirror line of 3 − f(x)
Question 9 — two candidate equations that share a turning point
Question 10 — statements about y = 3ˣ⁺¹ − 3

Question 1 · Modelled on 2024 P1 Q19(b), as an MC item — a reduction followed by a translation, where the shift ends up multiplied inside the bracket

Trap: option A is the answer for the steps in the other order.

The graph of $y = f (x)$ is reduced to $\frac{1}{2}$ of the original along the $x$ -axis, and the resulting graph is then translated leftwards by 3 units. The equation of the new graph is

$y = f (2 x + 3)$ .
$y = f (\frac{x}{2} + 3)$ .
$y = f (\frac{x}{2} + \frac{3}{2})$ .
$y = f (2 x + 6)$ .

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D. Reducing gives y = f(2x). Translating leftwards replaces x by x + 3: f(2(x + 3)) = f(2x + 6). Option A is the trap.

Question 2 · New sub-type, no direct past-paper model — the two reflections compared for a function where they coincide

New idea: for some functions the two reflections are the same graph.

Let $f (x) = x^{3} - 3 x$ . Which of the following equations represent the same graph as $y = - f (x)$ ?

I. $y = f (- x)$
II. $y = - f (- x)$
III. $y = f (x) - 2 f (x)$

I only
I and III only
II and III only
I, II and III

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B. f(−x) = −x³ + 3x = −f(x), so I matches. Then −f(−x) = f(x), so II is the original graph, not its reflection.

Question 3 · New sub-type, no direct past-paper model — a graph and its own translate, solved as an exponential equation

Trap: a diagonal translation does not mean the two graphs never meet.

The graphs of $y = 2^{x}$ and $y = 2^{x - 1} + 1$ intersect at the point

$(1, 2)$ .
$(0, 1)$ .
$(2, 4)$ .
$(- 1, \frac{1}{2})$ .

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A. 2ˣ = 2ˣ⁻¹ + 1 gives 2ˣ⁻¹ = 1, so x = 1 and y = 2.

Question 4 · New sub-type, no direct past-paper model — a reflection and a translation together, chosen so the roots come back unchanged

Trap: the distractors are the roots of f(x − 1) = 0 and f(−x) = 0.

It is given that the roots of the equation $f (x) = 0$ are $- 2$ and $3$ only. The roots of the equation $f (1 - x) = 0$ are

$- 3$ and $2$ .
$- 1$ and $4$ .
$- 2$ and $3$ .
$- 4$ and $1$ .

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C. 1 − x = −2 gives x = 3, and 1 − x = 3 gives x = −2. The roots are symmetric about x = 1/2, the mirror line of f(1 − x).

Question 5 · New sub-type, no direct past-paper model — counting roots after a reduction and a translation, with a distractor for each wrong move

Trap: each wrong option comes from applying one of the two changes to the heights incorrectly.

The figure shows the graph of $y = f (x)$ , which has turning points at $(- 1, 2)$ and $(3, - 6)$ . Find the range of values of $k$ such that the equation $f (2 x) + 3 = k$ has three distinct real roots.

Graph of y = f(x) with turning points (−1, 2) and (3, −6)

$- 6 < k < 2$
$- 12 < k < 4$
$- 9 < k < - 1$
$- 3 < k < 5$

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D. f(2x) has the same turning heights, 2 and −6; adding 3 moves them to 5 and −3. A ignores the +3, C subtracts it, B doubles the heights.

Question 6 · New sub-type, no direct past-paper model — reflection in a vertical line other than the y-axis

Technique not in the standard set: reflection in a vertical line other than the y-axis.

The graph of $y = g (x)$ is the reflection of the graph of $y = f (x)$ in the straight line $x = 2$ . Then $g (x) =$

$f (4 - x)$ .
$f (2 - x)$ .
$f (- x) + 4$ .
$f (x - 4)$ .

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A. The point (a, b) reflects to (4 − a, b), since 2 is halfway between a and 4 − a. So g(x) = f(4 − x). Check: g(2) = f(2).

Question 7 · Modelled on 2026 P2 Q35 — statements about a transformed logarithmic graph, turned round to ask which transformations produce it

New idea: a vertical translation and a reduction along the x-axis giving the same logarithmic graph.

Which of the following transformations change the graph of $y = lo g_{2} x$ to the graph of $y = lo g_{2} (4 x)$ ?

I. Translating the graph upwards by 2 units
II. Reducing the graph to $\frac{1}{4}$ of the original along the $x$ -axis
III. Translating the graph leftwards by 4 units

I only
III only
I and II only
II and III only

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C. log₂(4x) = log₂ x + 2, so I works, and replacing x by 4x is II. III would give log₂(x + 4).

Question 8 · Modelled on 2015 P1 Q18(c), as an MC item — a graph and its reflection in a horizontal line, with the mirror line itself tested

Trap: statement III names the wrong mirror line.

Let $f (x) = 2 sin x^{\circ} + 1$ and $g (x) = 3 - f (x)$ . Which of the following are true?

I. The greatest value of $g (x)$ is $4$ .
II. $g (90) = 0$
III. The graph of $y = g (x)$ is the reflection of the graph of $y = f (x)$ in the straight line $y = 3$ .

I and II only
I and III only
II and III only
I, II and III

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A. f(x) runs from −1 to 3, so g(x) runs from 0 to 4. The mirror line is halfway between y and 3 − y, which is y = 3/2, not y = 3.

Question 9 · Modelled on 2018 P2 Q31 — matching a graph to its equation, where two options agree at the turning point and only a second point separates them

Trap: options A and C both send the turning point to the right place; only a second point decides.

The figure shows the graph of $y = f (x)$ and the graph of $y = g (x)$ . The graph of $y = f (x)$ has a turning point $(1, 3)$ and passes through $(3, 0)$ . The graph of $y = g (x)$ has a turning point $(- 1, - 1)$ and passes through $(- 3, 2)$ . Then $g (x) =$

Graphs of y = f(x) (solid) through (1, 3) and (3, 0), and y = g(x) (dashed) through (−1, −1) and (−3, 2)

$- f (x + 2) + 2$ .
$f (- x) - 2$ .
$- f (- x) + 2$ .
$- f (x) + 2$ .

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C. Both A and C send (1, 3) to (−1, −1). Only C sends (3, 0) to (−3, 2); A sends it to (1, 2).

Question 10 · Modelled on 2026 P2 Q35 — I/II/III statements about a translated exponential graph

Trap: statement III holds for large x but fails far to the left.

Let $f (x) = 3^{x}$ and $g (x) = f (x + 1) - 3$ . Which of the following are true?

I. The graph of $y = g (x)$ passes through the origin.
II. $g (x) > - 3$ for all real values of $x$ .
III. The graph of $y = g (x)$ lies above the graph of $y = f (x)$ for all real values of $x$ .

I only
I and II only
I and III only
I, II and III

Show answer

B. g(x) − f(x) = 2·3ˣ − 3, which is negative for x below log₃ 1.5, so III fails far to the left.

Continue practising this topic: Try the Paper 1 long questions for harder variants, or move to Paper 1 Challenge — also available as Paper 2 multiple-choice.

Where this sits in the syllabus

More about Graphs of Functions is senior Number and Algebra content. From 2012 to 2026, Paper 2 set a graph-transformation question in at least six papers, and Paper 1 has several times placed a translation, reflection or enlargement inside a longer question on a quadratic function, most recently in 2024 and 2025. The marks are usually lost on direction: f(x − 4) moves a graph right, not left, and in a two-step transformation each step acts on the equation you already have.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

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