Transformations of graphs - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Transformations of graphs for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 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

2 marks

The curve with equation \(y = f(x)\) has a maximum point at \((2, 5)\).

Write down the coordinates of the turning point of the curve with equation \(y = -f(x)\), and state whether it is a maximum or a minimum.

Model answer

\((2, -5)\), and it is a minimum point.

Explanation

\(y = -f(x)\) is a reflection in the x-axis: every y-value changes sign and the x-values stay the same. Turning the graph upside down makes the highest point into the lowest point, so the maximum becomes a minimum.

Mark scheme

1 mark
B1 for \((2, -5)\)
correct answer with no working scores full marks
1 mark
B1 for minimum

Question 2

2 marks

Mo says,
"The graph of \(y = (x + 3)^2\) is the graph of \(y = x^2\) translated 3 units to the right."

Explain why Mo is wrong, and describe the correct translation.

Model answer

\((x + 3)^2 = 0\) when \(x = -3\), so the minimum point of \(y = (x + 3)^2\) is \((-3, 0)\), which is 3 units to the left of \((0, 0)\).
The correct translation is by the vector \(\begin{pmatrix}-3\\0\end{pmatrix}\).

Explanation

\(y = f(x + a)\) moves the graph \(a\) units to the left, which feels backwards. Checking where the bracket equals zero always gives the new position of the turning point, so you can confirm the direction.

Mark scheme

1 mark
C1 for explaining that the minimum point of \(y = (x + 3)^2\) is at \(x = -3\) (where the bracket is 0), which is to the left
oe
1 mark
B1 for translation by \(\begin{pmatrix}-3\\0\end{pmatrix}\)
accept 3 units to the left

Question 3

3 marks

The diagram shows the graph of \(y = \cos x\) and a curve \(C\), for \(0^\circ \le x \le 360^\circ\).
\(C\) is a translation of the graph of \(y = \cos x\).

Describe fully the translation.
Write down an equation for \(C\) in the form \(y = \cos(x + a)\), where \(-180 < a < 180\).
Write down another equation for \(C\), in terms of \(\sin x\).

A coordinate grid with x from 0 to 360 and y from −1.5 to 1.5; the x-axis is marked every 45 and the y-axis every 0.5, and the axes cross at the origin O. The curve labelled "y = cos x" is a cosine curve (a smooth repeating wave): it crosses the x-axis at (90, 0) and (270, 0); it crosses the y-axis at (0, 1); it has a minimum point at (180, −1). Curve C is a cosine curve (a smooth repeating wave): it crosses the x-axis at (0, 0), (180, 0) and (360, 0); it has a maximum point at (90, 1) and a minimum point at (270, −1). The curve labelled "y = cos x" and curve C cross at about (45, 0.7) and about (225, −0.7).

Model answer

The maximum of \(y = \cos x\) at \(x = 0^\circ\) has moved to \(x = 90^\circ\): translation by \(\begin{pmatrix}90\\0\end{pmatrix}\).
\(C\): \(y = \cos(x - 90^\circ)\)
\(C\) has maximum \((90, 1)\), minimum \((270, -1)\) and passes through the origin, so it is \(y = \sin x\).

Explanation

Compare matching features, such as maximum points, to find the horizontal movement. A move of 90° to the right replaces \(x\) with \(x - 90\). The sine curve is the cosine curve shifted 90° to the right, which is why \(\cos(x - 90^\circ) = \sin x\).

Mark scheme

1 mark
B1 for translation by \(\begin{pmatrix}90\\0\end{pmatrix}\)
accept 90° to the right
1 mark
B1 for \(y = \cos(x - 90^\circ)\)
accept \(a = -90\)
1 mark
B1 for \(y = \sin x\)

Question 4

4 marks

The diagram shows the graph of \(y = f(x)\).

Find the coordinates of
• the points where the graph of \(y = f(x + 2)\) crosses the x-axis
• the turning point of the graph of \(y = f(x + 2)\)
• the point where the graph of \(y = f(x + 2)\) crosses the y-axis.

A coordinate grid with x from −2 to 4 and y from −5 to 5; each grid square is 1 unit across, and the axes cross at the origin O. The curve labelled "y = f(x)" is a U-shaped parabola: it crosses the x-axis at (−1, 0) and (3, 0); it crosses the y-axis at (0, −3); it has a minimum point at (1, −4).

Model answer

\(y = f(x + 2)\) is a translation of 2 units to the left.
Roots of \(y = f(x)\): \(-1\) and 3, so the new roots are \(-3\) and 1: \((-3, 0)\), \((1, 0)\).
Turning point \((1, -4)\) moves to \((-1, -4)\).
New y-intercept: when \(x = 0\), \(y = f(2) = -3\), so \((0, -3)\).

Explanation

Adding 2 inside the bracket translates the graph by \(\begin{pmatrix}-2\\0\end{pmatrix}\), so every x-coordinate decreases by 2 and the y-coordinates stay the same. The new y-intercept is not the old one: it is the old value at \(x = 2\), because \(f(0 + 2) = f(2)\).

Mark scheme

1 mark
B1 for \((-3, 0)\)
correct answer with no working scores full marks
1 mark
B1 for \((1, 0)\)
1 mark
B1 for \((-1, -4)\)
1 mark
B1 for \((0, -3)\)
from f(2) = −3

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