What is A13: Transforming graphs?

For y=f(x)+a, translate the graph vertically by vector (0, a). For y=f(x-a), translate it horizontally by (a, 0); the sign inside the function appears opposite to the movement.

How many marks are A13 questions worth?

In verified past-paper appearances, A13 questions on Edexcel GCSE Maths 1MA1 have been worth 1–4 marks.

Which past exam papers has A13 appeared in?

A13 has appeared in 8 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A13

Sketch translations and reflections of a given function [Higher only]

Higher only

Transforming graphs

Worked answers, methods and verified real exam appearances for A13 on Edexcel GCSE Maths 1MA1.

Explanation

For $y=f(x)+a$ , translate the graph vertically by vector $\begin{pmatrix}0\\a\end{pmatrix}$ . For $y=f(x-a)$ , translate it horizontally by $\begin{pmatrix}a\\0\end{pmatrix}$ ; the sign inside the function appears opposite to the movement.
The graph $y=-f(x)$ is the reflection of $y=f(x)$ in the $x$ -axis, while $y=f(-x)$ is its reflection in the $y$ -axis.
Track distinctive points, including intercepts and turning points, and preserve the graph's exact shape and scale.
A point $(p,q)$ provides a reliable coordinate check after transforming.
Examiners expect a fully described transformation, including the correct axis or translation vector.

A translation moves every point by the same vector while preserving the graph's shape.

Worked example

The point $(p,q)$ lies on $y=f(x)$ . Find its image on $y=f(x-3)-2$ .

1. $f(x-3)$ translates the graph $3$ units right.
2.Subtracting $2$ outside the function translates it $2$ units down.
3.Therefore $(p,q)$ maps to $(p+3,q-2)$ .

Answer: The image is $(p+3,q-2)$ , under translation by $\begin{pmatrix}3\\-2\end{pmatrix}$ .

Common mistakes

Don't move $f(x-3)$ three units left instead of right.
Don't reflect $y=f(-x)$ in the $x$ -axis rather than the $y$ -axis.
Don't move only selected points and change the graph's shape.

Exam tip

For a translation, state the vector; for a reflection, name the mirror axis.

Worked practice

Tier 1 · Easy

Describe fully the transformation from

y=f(x)

y=f(x)+4

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	Translation by vector $\begin{pmatrix}0\\4\end{pmatrix}$	2	Adding $4$ outside the function increases every $y$ -coordinate by $4$ and leaves every $x$ -coordinate unchanged. This is translation by vector $(0,4)$ .

Tier 2 · Standard

The point

(3,-2)

lies on

y=f(x)

. Find the corresponding point on

y=f(-x)

and name the transformation.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$(-3,-2)$ Reflection in the $y$ -axis	3	Replacing $x$ by $-x$ reverses every $x$ -coordinate and leaves every $y$ -coordinate unchanged. Thus $(3,-2)$ maps to $(-3,-2)$ , a reflection in the $y$ -axis.

Tier 3 · Hard

The point

(-1,5)

lies on

y=f(x)

. Find the corresponding point on

y=f(2-x)-3

, and describe the reflection and translations that produce the new graph.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Corresponding point $(3,2)$ Reflect in the $y$ -axis, translate $2$ units right, then translate $3$ units down	4	If $(a,b)$ lies on $y=f(x)$ , then $f(2-x)=b$ when $2-x=a$ , so the new $x$ -coordinate is $2-a$ and the new $y$ -coordinate is $b-3$ . With $(a,b)=(-1,5)$ this gives $(3,2)$ . Since $f(2-x)=f(-(x-2))$ , the graph is reflected in the $y$ -axis, moved $2$ units right, and then moved $3$ units down.

Tier 1 · Easy

Describe fully the transformation from

y=f(x)

y=f(x-5)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	Translation by vector $\begin{pmatrix}5\\0\end{pmatrix}$ (or translation $5$ units right)	2	Replacing $x$ by $x-5$ moves every point $5$ units to the right without changing its height. This is translation by vector $\begin{pmatrix}5\\0\end{pmatrix}$ .

Tier 2 · Standard

The point

(-2,4)

lies on

y=f(x)

. Find the corresponding point on

y=-f(x)+1

, and describe the transformations.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	Corresponding point $(-2,-3)$ Reflect in the $x$ -axis, then translate $1$ unit up	3	The reflection $y=-f(x)$ keeps the $x$ -coordinate and changes $4$ to $-4$ . Adding $1$ then changes the height to $-3$ , so the point becomes $(-2,-3)$ . The graph is reflected in the $x$ -axis and translated $1$ unit up.

Tier 3 · Hard

The graph

y=g(x)

is obtained from

y=f(x)

by reflecting in the

x

-axis and then translating by vector

\begin{pmatrix}-4\\3\end{pmatrix}

. Write

g(x)

in terms of

f(x)

. The point

(2,-5)

lies on

y=f(x)

. Find its image on

y=g(x)

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$g(x)=-f(x+4)+3$ Image $(-2,8)$	4	Reflection in the $x$ -axis gives $y=-f(x)$ . Translating $4$ units left replaces $x$ by $x+4$ , and translating $3$ units up adds $3$ , so $g(x)=-f(x+4)+3$ . The point $(2,-5)$ reflects to $(2,5)$ and then translates to $(-2,8)$ .

Tier 2 · Standard

The graph of

y=g(x)

is a translation of the graph of

y=f(x)

. The point

(-3,2)

y=f(x)

corresponds to

(4,-1)

y=g(x)

. Write down the translation vector and write

g(x)

in terms of

f(x)

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	Translation vector $\begin{pmatrix}7\\-3\end{pmatrix}$ $g(x)=f(x-7)-3$	3	The $x$ -coordinate increases by $7$ and the $y$ -coordinate decreases by $3$ , so the vector is $\begin{pmatrix}7\\-3\end{pmatrix}$ . A shift $7$ units right replaces $x$ by $x-7$ , and a shift $3$ units down subtracts $3$ , giving $g(x)=f(x-7)-3$ .

Tier 3 · Hard

A point

(p,q)

lies on

y=f(x)

. Find the corresponding point on (i)

y=-f(x)+4

and (ii)

y=-(f(x)+4)

. Are the two transformed graphs the same? Give a reason.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	(i) $(p,4-q)$ (ii) $(p,-q-4)$ No, the first graph is the reflection in the $x$ -axis translated $4$ units up, while the second is translated $4$ units down after reflection	4	For (i), reflection in the $x$ -axis changes $q$ to $-q$ , then adding $4$ gives $4-q$ . For (ii), the bracket first adds $4$ to the output and the outside negative changes $q+4$ to $-q-4$ . No, the images differ vertically by $8$ units, so the transformed graphs are not the same.

Tier 3 · Hard

The graph

y=f(x)

has turning point

(-2,5)

and

x

-intercepts

(-5,0)

and

(1,0)

. The graph

y=g(x)

is given by

g(x)=-f(x-4)

. Describe the transformations and find the corresponding turning point and

x

-intercepts of

y=g(x)

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Translate $4$ units right, then reflect in the $x$ -axis Turning point $(2,-5)$ $x$ -intercepts $(-1,0)$ and $(5,0)$	4	Replacing $x$ by $x-4$ translates every point $4$ units right. The outside negative then reflects every height in the $x$ -axis. Thus $(-2,5)$ maps to $(2,-5)$ . The intercepts move to $(-1,0)$ and $(5,0)$ ; reflection leaves their zero heights unchanged.

Q10

Tier 3 · Hard

Higher only: The graph

y=g(x)

is obtained by reflecting

y=f(x)

in the

y

-axis and then translating it. The point

(2,5)

y=f(x)

maps to

(-7,1)

y=g(x)

. Find the translation vector and write

g(x)

in terms of

f(x)

. Find the image of the point

(-3,-2)

, which also lies on

y=f(x)

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Translation vector $\begin{pmatrix}-5\\-4\end{pmatrix}$ $g(x)=f(-x-5)-4$ The image of $(-3,-2)$ is $(-2,-6)$	5	Reflection in the $y$ -axis maps $(2,5)$ to $(-2,5)$ . Reaching $(-7,1)$ then requires translation by $\begin{pmatrix}-5\\-4\end{pmatrix}$ . After reflection the equation is $y=f(-x)$ ; translating $5$ units left and $4$ units down gives $g(x)=f(-(x+5))-4=f(-x-5)-4$ . The point $(-3,-2)$ reflects to $(3,-2)$ and then translates to $(-2,-6)$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-11	3H	Q20	1	Allowed	Higher	QP MS
2019-11	1H	Q20	3	Non-calculator	Higher	QP MS
2021-11	2H	Q18	4	Allowed	Higher	QP MS
2022-06	2H	Q21	3	Allowed	Higher	QP MS
2019-06	3H	Q15	2	Allowed	Higher	QP MS
2024-06	2H	Q21	2	Allowed	Higher	QP MS
2024-11	3H	Q21	4	Allowed	Higher	QP MS
2022-11	3H	Q23	2	Allowed	Higher	QP MS

Other points in A Algebra · equations and reasoning

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