Transformations of graphs - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 17 exam-style questions on Transformations of graphs for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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2–10 marks each
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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

4 marks

The function f is given by
\[f(x) = x^3 - 2x, \quad x \in \mathbb{R}\]
Given that \(g(x) = 8x^3 - 4x\), describe fully the single transformation that maps the graph of \(y = f(x)\) onto the graph of \(y = g(x)\).

Given that \(h(x) = 2x - x^3\), describe fully two different single transformations, each of which maps the graph of \(y = f(x)\) onto the graph of \(y = h(x)\).

Model answer

\(g(x) = 8x^3 - 4x = (2x)^3 - 2(2x) = f(2x)\)
So g is a stretch parallel to the x-axis with scale factor \(\tfrac{1}{2}\).
\(h(x) = 2x - x^3 = -(x^3 - 2x) = -f(x)\): a reflection in the x-axis.
Also \(f(-x) = (-x)^3 - 2(-x) = -x^3 + 2x = h(x)\): a reflection in the y-axis.

Explanation

Writing \(8x^3 - 4x\) as \((2x)^3 - 2(2x)\) shows that g(x) = f(2x), and replacing x by 2x is a stretch parallel to the x-axis with scale factor \(\tfrac{1}{2}\), not 2. Since \(-f(x)\) and \(f(-x)\) are both equal to \(2x - x^3\), a reflection in either axis maps f onto h. Common errors are giving the scale factor as 2 and describing a stretch parallel to the y-axis.

Mark scheme

1 mark
B1 Stretch parallel to the x-axis
Because \(g(x) = (2x)^3 - 2(2x) = f(2x)\). Accept 'stretch in the x-direction' or 'horizontal stretch'. Do not accept 'squash', 'enlargement' or a stretch parallel to the y-axis (\(8x^3 - 4x\) is not a multiple of f(x)).
1 mark
B1 Scale factor \(\tfrac{1}{2}\)
This mark needs the stretch to be parallel to the x-axis. A scale factor of 2 is B0.
1 mark
B1 One correct single transformation for h: reflection in the x-axis, or reflection in the y-axis
\(h(x) = -f(x)\) (reflection in the x-axis) and also \(h(x) = f(-x)\) (reflection in the y-axis), because f is an odd function. Accept 'in the line y = 0' or 'in the line x = 0'. 'Reflection' alone, without the mirror line, is B0. A rotation through 180 degrees about the origin is B0: it maps \(y = f(x)\) onto \(y = -f(-x)\), which is \(y = f(x)\) itself because f is odd, not \(y = h(x)\).
1 mark
B1 A second, different correct single transformation from the same list
1 mark for each different correct transformation, max 2 for h. Two descriptions of the same transformation score only 1 of these 2 marks.

Question 2

5 marks

The curve \(C_1\) has equation \(y = x^2 - 4x + 1\).

\(C_1\) is translated by the vector \(\begin{pmatrix} -2 \\ 0 \end{pmatrix}\) to give the curve \(C_2\). The curve \(C_2\) is then stretched parallel to the x-axis with scale factor \(\tfrac{1}{2}\) to give the curve \(C_3\).

Find an equation for \(C_3\) in the form \(y = ax^2 + bx + c\), where a, b and c are constants.

Write down the coordinates of the turning point of \(C_3\).

Model answer

Translation by \(\begin{pmatrix} -2 \\ 0 \end{pmatrix}\): replace x with \(x + 2\)
\(C_2\): \(y = (x + 2)^2 - 4(x + 2) + 1 = x^2 + 4x + 4 - 4x - 8 + 1 = x^2 - 3\)
Stretch parallel to the x-axis with scale factor \(\tfrac{1}{2}\): replace x with 2x
\(C_3\): \(y = (2x)^2 - 3 = 4x^2 - 3\)
Turning point of \(C_3\): \((0, -3)\)

Explanation

A translation by \(\begin{pmatrix} -2 \\ 0 \end{pmatrix}\) replaces x with x + 2, and a stretch parallel to the x-axis with scale factor \(\tfrac{1}{2}\) then replaces x with 2x in the new equation, giving \(y = 4x^2 - 3\). Tracking the vertex (2, -3) gives the same turning point, (0, -3). A common error is to do the stretch first, which gives \(4x^2 + 8x + 1\), with turning point (-1, -3).

Mark scheme

1 mark
M1 Applies the translation by replacing x with \(x + 2\): \(y = (x + 2)^2 - 4(x + 2) + 1\); or a correct final answer of \(y = 4x^2 - 3\) and \((0, -3)\) with no working
An answer of \(y = 4x^2 - 3\) and (0, -3) with no working scores 5 marks. Replacing x with x - 2 is M0.
1 mark
A1 \(C_2\): \(y = x^2 - 3\)
May be implied by a correct equation for \(C_3\).
1 mark
M1 Applies the stretch to their \(C_2\) by replacing x with 2x
This mark does not depend on the previous marks. Replacing x with \(\tfrac{1}{2}x\) is M0. Applying the stretch to \(C_1\) first, giving \(y = (2x + 4)^2 - 4(2x + 4) + 1 = 4x^2 + 8x + 1\), is the transformations in the wrong order and scores M0 A0 here.
1 mark
A1 \(y = 4x^2 - 3\)
a = 4, b = 0, c = -3. Accept \(y = 4x^2 + 0x - 3\).
1 mark
B1ft \((0, -3)\)
ft the turning point of their equation for \(C_3\). It can also be found by tracking the vertex of \(C_1\): \((2, -3) \to (0, -3) \to (0, -3)\).

Question 3

5 marks

The temperature, \(\theta^\circ\mathrm{C}\), inside a greenhouse t hours after midnight on one day is modelled by the equation \(\theta = f(t)\), \(0 \leqslant t \leqslant 24\). According to the model, the greatest temperature is \(28^\circ\mathrm{C}\), when t = 14, and the least temperature is \(9^\circ\mathrm{C}\), when t = 4.

A gardener's thermometer shows the temperature in degrees Fahrenheit, F, where \(F = 1.8\theta + 32\).
Describe fully the transformations that map the graph of \(\theta = f(t)\) onto the graph of \(F = 1.8f(t) + 32\). Write down the greatest temperature in degrees Fahrenheit, according to the model, and the time t at which it occurs.

On the next day the temperature is modelled by \(\theta = g(t)\), with t now measured in hours from midnight on that day. The graph of \(\theta = g(t)\) has the same shape as the graph of \(\theta = f(t)\), but each temperature occurs 3 hours later and is \(2^\circ\mathrm{C}\) lower.
Write down g(t) in terms of f, and write down the coordinates of the lowest point on the graph of \(\theta = g(t)\).

Model answer

\(F = 1.8f(t) + 32\): first a stretch parallel to the vertical axis with scale factor 1.8, then a translation by \(\begin{pmatrix} 0 \\ 32 \end{pmatrix}\).
Greatest temperature: \(1.8 \times 28 + 32 = 82.4\) degrees Fahrenheit, at \(t = 14\).
Each temperature occurs 3 hours later (replace t with t - 3) and is 2 degrees lower (subtract 2):
\(g(t) = f(t - 3) - 2\)
The minimum point (4, 9) of \(\theta = f(t)\) moves to \((4 + 3, 9 - 2) = (7, 7)\).

Explanation

Multiplying the function by 1.8 is a vertical stretch and adding 32 is a vertical translation, applied in that order, so the maximum (14, 28) becomes (14, 82.4). A delay of 3 hours replaces t with t - 3, which translates the graph 3 units to the right, and the drop of 2 degrees is a translation 2 units down. A common error is to write f(t + 3), which moves the graph to the left.

Mark scheme

1 mark
B1 A stretch parallel to the vertical axis with scale factor 1.8
Accept 'stretch in the \(\theta\)-direction' or 'in the F-direction'. A stretch parallel to the t-axis is B0.
1 mark
B1 followed by a translation of 32 parallel to the vertical axis, vector \(\begin{pmatrix} 0 \\ 32 \end{pmatrix}\)
The order matters: a translation by \(\begin{pmatrix} 0 \\ 32 \end{pmatrix}\) before the stretch gives 1.8f(t) + 57.6, so it scores B0 unless the vector is adjusted to \(\begin{pmatrix} 0 \\ 32/1.8 \end{pmatrix}\).
1 mark
B1 82.4 degrees Fahrenheit, at t = 14
Both values needed: \(1.8 \times 28 + 32 = 82.4\). Accept 'at 2 pm'.
1 mark
B1 \(g(t) = f(t - 3) - 2\)
f(t + 3) - 2 is B0 (that would make each temperature occur 3 hours earlier).
1 mark
B1 \((7, 7)\)
The minimum (4, 9) moves 3 hours later and 2 degrees lower. Accept t = 7, \(\theta = 7\).

Question 4

7 marks

The curve C has equation
\[y = \dfrac{2x + 1}{x - 1}, \quad x ≠ 1\]
Write \(\dfrac{2x + 1}{x - 1}\) in the form \(A + \dfrac{B}{x - 1}\), where A and B are integers.

Hence describe fully a sequence of two transformations that maps the curve with equation \(y = \dfrac{1}{x}\) onto C.

Write down an equation for each asymptote of C, and find the coordinates of each point at which C crosses an axis.

Model answer

\(2x + 1 = 2(x - 1) + 3\), so \(\dfrac{2x + 1}{x - 1} = 2 + \dfrac{3}{x - 1}\): \(A = 2\), \(B = 3\)
\(y = \dfrac{1}{x} \to y = \dfrac{3}{x}\): stretch parallel to the y-axis with scale factor 3
\(y = \dfrac{3}{x} \to y = \dfrac{3}{x - 1} + 2\): translation by \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\)
The asymptotes \(x = 0\) and \(y = 0\) of \(y = \dfrac{1}{x}\) become \(x = 1\) and \(y = 2\).
\(y = 0\): \(2x + 1 = 0 \Rightarrow x = -\tfrac{1}{2}\), so \(\left(-\tfrac{1}{2}, 0\right)\)
\(x = 0\): \(y = \dfrac{1}{-1} = -1\), so \((0, -1)\)

Explanation

Writing the curve as \(y = 2 + \dfrac{3}{x - 1}\) shows it is \(y = \dfrac{3}{x}\) moved 1 right and 2 up, so the asymptotes become x = 1 and y = 2. The stretch must come before the translation (or the translation must be adjusted), because stretching after a vertical translation also stretches the translation. The intercepts come straight from the original form: y = 0 when 2x + 1 = 0, and y = -1 when x = 0.

Mark scheme

1 mark
M1 Writes \(2x + 1 = A(x - 1) + B\) and finds A or B, or divides \(2x + 1\) by \(x - 1\); or a correct final answer of \(2 + \dfrac{3}{x - 1}\) with no working
An answer of \(2 + \dfrac{3}{x - 1}\), a correct sequence of two transformations, asymptotes x = 1 and y = 2, and points (-0.5, 0) and (0, -1) with no working scores 7 marks.
1 mark
A1 \(A = 2\), \(B = 3\)
Or \(2 + \dfrac{3}{x - 1}\) written out.
1 mark
B1 One transformation described fully and correctly placed in a valid sequence, e.g. a stretch parallel to the y-axis with scale factor 3 (first), or a translation by \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\) (second)
ft their A and B. A stretch needs its direction and scale factor; a translation needs its vector (or e.g. '1 right and 2 up').
1 mark
B1 A fully correct sequence of two transformations, e.g. a stretch parallel to the y-axis with scale factor 3, followed by a translation by \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\)
Accept any correct sequence of two transformations; check it by applying it to y = 1/x. Correct examples: a stretch with scale factor 3 parallel to the y-axis or parallel to the x-axis, then the translation \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\); the translation \(\begin{pmatrix} 1 \\ 2/3 \end{pmatrix}\), then a stretch with scale factor 3 parallel to the y-axis. The translation \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\) followed by a stretch parallel to the y-axis gives \(\dfrac{3}{x - 1} + 6\) and is B0 here.
1 mark
B1 Asymptotes \(x = 1\) and \(y = 2\)
Both equations are needed for this mark.
1 mark
B1 \(\left(-\tfrac{1}{2}, 0\right)\)
From 2x + 1 = 0. Accept (-0.5, 0).
1 mark
B1 \((0, -1)\)
From y = 1/(-1).

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