Transformations of graphs - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 17 exam-style questions on Transformations of graphs for Edexcel A-Level Maths (Paper 2). 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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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

3 marks

The curve with equation \(y = x^3\) is stretched parallel to the x-axis with scale factor 2. The resulting curve is then translated by the vector \(\begin{pmatrix} 0 \\ -1 \end{pmatrix}\) to give the curve C.

Find an equation of C, and the coordinates of the point where C crosses the x-axis.

Model answer

Stretch parallel to the x-axis, scale factor 2: \(y = \left(\dfrac{x}{2}\right)^3 = \dfrac{x^3}{8}\)
Translation by \(\begin{pmatrix} 0 \\ -1 \end{pmatrix}\): \(y = \dfrac{x^3}{8} - 1\)
C: \(y = \dfrac{x^3}{8} - 1\)
At the x-axis: \(\dfrac{x^3}{8} = 1 \Rightarrow x^3 = 8 \Rightarrow x = 2\)
C crosses the x-axis at \((2, 0)\).

Explanation

A stretch parallel to the x-axis with scale factor 2 replaces x with \(\dfrac{x}{2}\), and a translation by \(\begin{pmatrix} 0 \\ -1 \end{pmatrix}\) subtracts 1 from y. Setting \(y = 0\) then gives \(x^3 = 8\), so \(x = 2\). A common error is to write \((2x)^3\), which is a stretch with scale factor \(\dfrac{1}{2}\).

Mark scheme

1 mark
B1 \(y = \left(\dfrac{x}{2}\right)^3 - 1\), or \(y = \dfrac{x^3}{8} - 1\); correct final answers of \(y = \dfrac{x^3}{8} - 1\) and \((2, 0)\) with no working score 3 marks
An answer of \(y = \dfrac{x^3}{8} - 1\) and \((2, 0)\) with no working scores 3 marks. A stretch parallel to the x-axis with scale factor 2 replaces x by \(\dfrac{x}{2}\): \(y = (2x)^3 - 1\) is B0.
1 mark
M1 Sets their equation equal to 0 and solves for x
Follow through their equation of the form \(y = (kx)^3 - 1\).
1 mark
A1 \((2, 0)\)
\(y = (2x)^3 - 1\) leading to \(\left(\dfrac{1}{2}, 0\right)\) scores B0 M1 A0.

Question 2

5 marks

The point \(A(4, 5)\) lies on the curve with equation \(y = f(x)\).
The curve \(C_1\) is obtained from the curve \(y = f(x)\) by a stretch parallel to the y-axis with scale factor 3, followed by a translation by the vector \(\begin{pmatrix} 0 \\ 2 \end{pmatrix}\).
The curve \(C_2\) is obtained from the curve \(y = f(x)\) by the same two transformations in the opposite order: the translation first, then the stretch.

Write down an equation of \(C_1\) and an equation of \(C_2\), in terms of f, and find the image of A on each curve.

Describe fully the single transformation that maps \(C_1\) onto \(C_2\).

Model answer

\(C_1\): the stretch gives \(y = 3f(x)\); the translation then gives \(y = 3f(x) + 2\).
Image of A on \(C_1\): \((4, 3 \times 5 + 2) = (4, 17)\)
\(C_2\): the translation gives \(y = f(x) + 2\); the stretch then gives \(y = 3\big(f(x) + 2\big) = 3f(x) + 6\).
Image of A on \(C_2\): \((4, 3(5 + 2)) = (4, 21)\)
Every point of \(C_2\) is 4 units above the corresponding point of \(C_1\), so the translation by \(\begin{pmatrix} 0 \\ 4 \end{pmatrix}\) maps \(C_1\) onto \(C_2\).

Explanation

The order of a stretch parallel to the y-axis and a vertical translation matters: when the translation comes first, the stretch multiplies the 2 as well, giving \(3f(x) + 6\) rather than \(3f(x) + 2\). The images of A differ by 4 in the y-coordinate, and so does every pair of corresponding points, which is a translation by \(\begin{pmatrix} 0 \\ 4 \end{pmatrix}\). A common error is to treat both orders as giving \(3f(x) + 2\).

Mark scheme

1 mark
B1 \(C_1\): \(y = 3f(x) + 2\)
Correct answers with no working score 5 marks. \(y = 3\big(f(x) + 2\big)\) for \(C_1\) is B0.
1 mark
B1 Image of A on \(C_1\): \((4, 17)\)
From \(3 \times 5 + 2\).
1 mark
B1 \(C_2\): \(y = 3\big(f(x) + 2\big)\) or \(y = 3f(x) + 6\)
\(y = 3f(x) + 2\) for \(C_2\) (ignoring the order) is B0. \(y = 3f(x + 2)\) is B0.
1 mark
B1 Image of A on \(C_2\): \((4, 21)\)
From \(3(5 + 2)\).
1 mark
B1 Translation by \(\begin{pmatrix} 0 \\ 4 \end{pmatrix}\)
Accept 'translation of 4 units in the positive y-direction'. Must say 'translation' (or 'translate'): 'move up 4' or 'shift' alone is B0.

Question 3

6 marks

The function f is defined by
\[f(x) = 2|x + 3| - 4, \quad x \in \mathbb{R}\]
The graph of \(y = f(x)\) and the graph of \(y = -f(x)\) together enclose a region R.

Find the coordinates of the vertices of R and the area of R.

Find the coordinates of every point where the graph of \(y = f(x)\) meets the graph of \(y = f(-x)\).

Model answer

\(y = -f(x)\) is the reflection of \(y = f(x)\) in the x-axis, so the two graphs meet where \(f(x) = -f(x)\), i.e. where \(f(x) = 0\):
\(2|x + 3| - 4 = 0 \Rightarrow |x + 3| = 2 \Rightarrow x + 3 = \pm 2 \Rightarrow x = -1\) or \(x = -5\)
The vertex of \(y = f(x)\) is \((-3, -4)\), where \(x + 3 = 0\); the vertex of \(y = -f(x)\) is its reflection, \((-3, 4)\).
For \(-5 < x < -1\), \(f(x) < 0 < -f(x)\), so R has vertices \((-5, 0)\), \((-3, -4)\), \((-1, 0)\) and \((-3, 4)\).
R is a rhombus with diagonals of lengths 4 (from \((-5, 0)\) to \((-1, 0)\)) and 8 (from \((-3, -4)\) to \((-3, 4)\)), so
Area of R \(= \dfrac{1}{2} \times 4 \times 8 = 16\)
\(f(-x) = 2|-x + 3| - 4 = 2|x - 3| - 4\)
\(2|x + 3| - 4 = 2|x - 3| - 4 \Rightarrow |x + 3| = |x - 3|\)
Squaring: \(x^2 + 6x + 9 = x^2 - 6x + 9 \Rightarrow 12x = 0 \Rightarrow x = 0\)
\(f(0) = 2|3| - 4 = 2\), so the graphs meet only at \((0, 2)\).

Explanation

The graph of \(y = -f(x)\) is the reflection of \(y = f(x)\) in the x-axis, so the two V-shapes meet on the x-axis at \((-5, 0)\) and \((-1, 0)\), and R is a rhombus whose other vertices are \((-3, -4)\) and its image \((-3, 4)\); its area is half the product of the diagonals, \(\dfrac{1}{2} \times 4 \times 8 = 16\). The graph of \(y = f(-x)\) is the reflection in the y-axis, \(y = 2|x - 3| - 4\), and as \(|x + 3| = |x - 3|\) only when \(x = 0\), the two graphs meet only at \((0, 2)\). A common error is to confuse \(y = -f(x)\) with \(y = f(-x)\).

Mark scheme

1 mark
M1 Uses \(f(x) = -f(x)\), i.e. \(f(x) = 0\), to find where the two graphs meet: \(|x + 3| = 2\); or correct final answers of \((-5, 0)\), \((-1, 0)\), \((-3, -4)\), \((-3, 4)\), an area of 16 and \((0, 2)\) only, with no working
An answer of \((-5, 0)\), \((-1, 0)\), \((-3, -4)\) and \((-3, 4)\), an area of 16, and \((0, 2)\) as the only point, with no working scores 6 marks. Allow the M1 for \(2|x + 3| - 4 = 0\) solved on at least one branch.
1 mark
A1 \((-5, 0)\) and \((-1, 0)\)
Both needed, from \(x + 3 = \pm 2\).
1 mark
B1 \((-3, -4)\) and \((-3, 4)\)
Both needed: the vertex of \(y = f(x)\) and its image under the reflection in the x-axis, the vertex of \(y = -f(x)\).
1 mark
B1ft Area of R = 16
From \(\dfrac{1}{2} \times 4 \times 8\) (half the product of the diagonals of the rhombus), or two triangles each of area \(\dfrac{1}{2} \times 4 \times 4\). Follow through their four vertices, provided they form a rhombus with one diagonal on the x-axis.
1 mark
M1 Writes \(f(-x) = 2|-x + 3| - 4\) (or \(2|x - 3| - 4\)) and solves \(f(x) = f(-x)\), reaching \(|x + 3| = |x - 3|\) and \(x = 0\)
By squaring (\(x^2 + 6x + 9 = x^2 - 6x + 9\), so \(12x = 0\)), by considering the cases, or by noting that \(x = 0\) is the only number equidistant from \(-3\) and \(3\). Writing \(f(-x)\) as \(-f(x)\) or as \(-2|x + 3| - 4\) is M0.
1 mark
A1 \((0, 2)\) only
From \(f(0) = 2|3| - 4 = 2\). Any further points given are A0. A statement that the arms of the two graphs are parallel (so they do not meet away from the y-axis) is a good reason but is not required.

Question 4

8 marks

A shop hires out floor sanders. The charge, £C, for hiring a floor sander for t hours is modelled by
\[C = f(t) = 20 + 10\sqrt{t}, \quad 0 < t \leqslant 24\]
A second shop uses the same price list as the first shop, except that the times in its list are measured in units of 40 minutes instead of hours. It also adds a fixed delivery fee of £8 to every hire. The charge, £D, at the second shop for hiring a floor sander for t hours is modelled by \(D = g(t)\), \(0 < t \leqslant 16\).

Explain why \(g(t) = f(1.5t) + 8\), and show that \(g(t) = 28 + 5\sqrt{6t}\).

Describe fully the two transformations that map the graph of \(C = f(t)\) onto the graph of \(D = g(t)\).

Show that, for every hire period, the second shop charges more than the first shop.

Find the hire period for which the second shop charges £53, giving your answer in hours and minutes.

Model answer

1 hour = 60 minutes = 1.5 units of 40 minutes, so t hours is 1.5t units. The second shop charges the first shop's list price for 1.5t units, \(f(1.5t)\), plus £8, so \(g(t) = f(1.5t) + 8\).
\(g(t) = 20 + 10\sqrt{1.5t} + 8 = 28 + 10\sqrt{\dfrac{6t}{4}} = 28 + 10 \times \dfrac{\sqrt{6t}}{2} = 28 + 5\sqrt{6t}\)
Replacing t by 1.5t is a stretch parallel to the t-axis with scale factor \(\dfrac{1}{1.5} = \dfrac{2}{3}\); adding 8 is a translation by \(\begin{pmatrix} 0 \\ 8 \end{pmatrix}\).
\(g(t) - f(t) = 28 + 5\sqrt{6}\sqrt{t} - 20 - 10\sqrt{t} = 8 + (5\sqrt{6} - 10)\sqrt{t}\)
\(5\sqrt{6} = \sqrt{150} > \sqrt{100} = 10\), so \((5\sqrt{6} - 10)\sqrt{t} > 0\) and \(g(t) - f(t) > 8 > 0\).
So the second shop always charges more than the first.
\(28 + 5\sqrt{6t} = 53 \Rightarrow \sqrt{6t} = 5 \Rightarrow 6t = 25 \Rightarrow t = \dfrac{25}{6}\)
\(\dfrac{25}{6}\) hours = 4 hours 10 minutes

Explanation

Measuring time in 40-minute units means t hours count as 1.5t units, so the charge is \(f(1.5t) + 8\): replacing t by 1.5t is a stretch parallel to the t-axis with scale factor \(\dfrac{2}{3}\), and adding 8 is a translation parallel to the vertical axis. The difference \(g(t) - f(t)\) is 8 plus a positive multiple of \(\sqrt{t}\), so it is always positive. Common errors are giving the scale factor as 1.5 and leaving the time as 4.17 hours.

Mark scheme

1 mark
B1 Explains: t hours is \(1.5t\) units of 40 minutes (as \(60 \div 40 = 1.5\)), so the second shop's list price is \(f(1.5t)\), and the £8 fee is added. An answer with no working scores 0 marks for the show-thats.
An answer with no working scores 0 marks for the show-thats; correct transformations and 4 hours 10 minutes with no working score 4 marks (B1 B1 M1 A1). The explanation must give the reason for 1.5t, not just restate the formula.
1 mark
M1 Substitutes \(1.5t\) into f: \(g(t) = 20 + 10\sqrt{1.5t} + 8\)
1 mark
A1* \(10\sqrt{1.5t} = 10\sqrt{\dfrac{6t}{4}} = 5\sqrt{6t}\), so \(g(t) = 28 + 5\sqrt{6t}\) cso
A correct intermediate step for \(10\sqrt{1.5t}\), such as \(10 \times \dfrac{\sqrt{6t}}{2}\) or \(\sqrt{150t} = 5\sqrt{6t}\), must be seen. Decimal work (12.247...) is A0.
1 mark
B1 A stretch parallel to the t-axis with scale factor \(\dfrac{2}{3}\)
Scale factor 1.5 is B0. Accept 'horizontal stretch' or 'in the t-direction'.
1 mark
B1 A translation by \(\begin{pmatrix} 0 \\ 8 \end{pmatrix}\)
The two transformations may be given in either order. Translation by \(\begin{pmatrix} 8 \\ 0 \end{pmatrix}\) is B0.
1 mark
B1 Shows \(g(t) > f(t)\) for every t: \(g(t) - f(t) = 8 + (5\sqrt{6} - 10)\sqrt{t}\) with \(5\sqrt{6} = \sqrt{150} > 10\), so the difference is positive; or \(\sqrt{1.5t} > \sqrt{t}\), so \(f(1.5t) > f(t)\), and £8 is added; with a conclusion
A reason why the difference is positive for all t and a conclusion are both needed. Checking one or two values of t is B0.
1 mark
M1 Solves \(28 + 5\sqrt{6t} = 53\) (or \(f(1.5t) = 45\)): \(\sqrt{6t} = 5\), \(t = \dfrac{25}{6}\)
Independent of the show-that.
1 mark
A1 4 hours 10 minutes
Must be in hours and minutes: 4.17 hours or \(\dfrac{25}{6}\) hours alone is A0.

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