What is 2.9: Graph transformations?

Transformations outside f act on output coordinates: y=af(x) scales vertically by factor |a| and reflects in the x-axis when a<0. Transformations inside f act oppositely on inputs: f(x+a) moves left by a, while f(ax) scales horizontally by factor 1/|a| and may reflect in the y-axis.

How many marks are 2.9 questions worth?

2.9 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 2.9 appeared in?

We have not yet indexed a verified past-paper appearance for 2.9 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

2.9

Understand the effect of simple transformations on the graph of y = f(x), including sketching associated graphs: y = af(x), y = f(x) + a, y = f(x + a), y = f(ax) and combinations of these transformations.

Q: How many marks are 2.9 questions worth?

2.9 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Q: Which past exam papers has 2.9 appeared in?

We have not yet indexed a verified past-paper appearance for 2.9 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

Draft — not yet indexed

Graph transformations

Worked answers and methods for 2.9 on Edexcel A-level Maths 9MA0.

Explanation

Transformations outside $f$ act on output coordinates: $y=af(x)$ scales vertically by factor $|a|$ and reflects in the $x$ -axis when $a<0$ . Transformations inside $f$ act oppositely on inputs: $f(x+a)$ moves left by $a$ , while $f(ax)$ scales horizontally by factor $1/|a|$ and may reflect in the $y$ -axis.
The graph $y=|f(x)|$ reflects every negative part above the $x$ -axis.
For $y=|f(-x)|$ , first reflect $f$ in the $y$ -axis, then apply the output modulus.
For example, a point $(u,v)$ on $y=f(x)$ maps to $(u/2,v+3)$ on $y=f(2x)+3$ .
Map a general point through combined transformations so horizontal scale factors and shifts stay in the correct order.

Replacing

x

x-a

translates the graph of

f

right by

a

Worked example

The point $(6,-1)$ lies on $y=f(x)$ . Find the corresponding point on $y=-2f(3x)+1$ .

1.For the transformed input to equal $6$ , set $3x=6$ , giving $x=2$ .
2.The transformed output is $-2(-1)+1=3$ , so the corresponding point is $(2,3)$ .

Answer: $(2,3)$

Common mistakes

Don't describe $f(x+a)$ as a translation right by $a$ instead of left by $a$ .
Don't use horizontal scale factor $a$ for $f(ax)$ instead of the reciprocal factor $1/|a|$ .

Exam tip

Track a point $(u,v)$ through the transformation and give its new coordinates to verify every shift, stretch and reflection.

Worked practice

Tier 1 · Easy

Describe fully the transformation from

y=f(x)

y=f(x-4)+2

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Translation by vector $\begin{pmatrix}4\\2\end{pmatrix}$ .	2
Notes
Replacing $x$ by $x-4$ moves the graph $4$ units right, and adding $2$ moves it $2$ units up. Together this is translation by $\begin{pmatrix}4\\2\end{pmatrix}$ .

(2 marks)

Tier 2 · Standard

The graph of

y=2^x

is transformed to the graph of

y=2^{x+1}-3

. Describe the transformation, and state the equation of the horizontal asymptote and the

y

-intercept of the transformed graph.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Translation by vector $\begin{pmatrix}-1\\-3\end{pmatrix}$ , or equivalently a stretch parallel to the $y$ -axis of scale factor $2$ followed by a translation $\begin{pmatrix}0\\-3\end{pmatrix}$ Horizontal asymptote $y=-3$ $y$ -intercept $(0,-1)$	4
Notes
Replacing $x$ by $x+1$ translates the graph one unit left, and subtracting $3$ translates it three units down, giving the vector $(-1,-3)$ . A second description earns the same marks: because $2^{x+1}=2\times2^{x}$ , the same curve is a vertical stretch of scale factor $2$ followed by a translation three units down. Either is correct — do not discard your answer if it takes the stretch route. The asymptote $y=0$ moves to $y=-3$ . At $x=0$ , $y=2^1-3=-1$ , so the $y$ -intercept is $(0,-1)$ .

(4 marks)

Tier 3 · Hard

Let

f(x)=x^2-4x+1

. For

g(x)=-f(2x+2)+4

, find the turning point and both

x

-intercepts, then state whether it is a maximum or minimum.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Turning point $(0,7)$ , a maximum $x$ -intercepts $\left(-\dfrac{\sqrt7}{2},0\right)$ and $\left(\dfrac{\sqrt7}{2},0\right)$	5
Notes
First write $f(u)=(u-2)^2-3$ . With $u=2x+2$ , $f(2x+2)=(2x)^2-3=4x^2-3$ , so $g(x)=-(4x^2-3)+4=7-4x^2$ . Its turning point is $(0,7)$ and the negative $x^2$ coefficient makes it a maximum. Setting $7-4x^2=0$ gives $x=\pm\sqrt7/2$ .

(5 marks)

Tier 1 · Easy

Describe fully the two transformations that map the graph of

y=f(x)

onto the graph of

y=f(-3x)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	A reflection in the $y$ -axis and a stretch parallel to the $x$ -axis with scale factor $\dfrac13$ .	2
Notes
The negative sign inside $f$ reflects the graph in the $y$ -axis. The factor $3$ multiplying $x$ produces a horizontal scale factor of $1/3$ .

(2 marks)

Tier 2 · Standard

The curve

y=x^3

is translated

2

units left, stretched parallel to the

y

-axis by scale factor

3

, reflected in the

x

-axis and then translated

1

unit up. Find the equation of the transformed curve and the image of

(0,0)

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$y=-3(x+2)^3+1$ The image of $(0,0)$ is $(-2,1)$ .	4
Notes
The horizontal translation replaces $x$ by $x+2$ . The vertical stretch and reflection multiply the output by $-3$ , and the final translation adds $1$ , giving $y=-3(x+2)^3+1$ . Tracking $(0,0)$ through the same changes gives $(-2,0)$ after the horizontal translation; the vertical stretch and reflection leave its zero output unchanged, and the final translation gives $(-2,1)$ .

(4 marks)

Tier 3 · Hard

Let

f(x)=1/x

and

g(x)=-2f(3x-6)+4

. Give the coordinate mapping from

y=f(x)

y=g(x)

, state both asymptotes of

g

, and find its

x

-intercept.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$(u,v)\mapsto\left(\dfrac{u}{3}+2,-2v+4\right)$ Asymptotes $x=2$ and $y=4$ $x$ -intercept $\left(\dfrac{13}{6},0\right)$	6
Notes
The input relation $u=3x-6=3(x-2)$ gives $x=u/3+2$ , while the output becomes $-2v+4$ . Hence $(u,v)$ maps to $(u/3+2,-2v+4)$ . The asymptotes $u=0$ and $v=0$ therefore map to $x=2$ and $y=4$ . Also $g(x)=-2/(3x-6)+4$ . Setting this equal to zero gives $3x-6=1/2$ , so $x=13/6$ .

(6 marks)

Tier 2 · Standard

A function

f

has domain

-1\leq x\leq4

and range

-2\leq f(x)\leq5

. Its minimum occurs at

(1,-2)

and its maximum at

(3,5)

. For

g(x)=3-2f(2x-1)

, find the domain and range of

g

and the images of these two turning points.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Domain $0\leq x\leq\dfrac52$ ; range $-7\leq g(x)\leq7$ $(1,-2)$ maps to $(1,7)$ and $(3,5)$ maps to $(2,-7)$ .	5
Notes
Write the original input as $u=2x-1$ , so $x=(u+1)/2$ . Mapping the original domain $-1\leq u\leq4$ gives $0\leq x\leq5/2$ . The output mapping is $v\mapsto3-2v$ , so $-2\leq v\leq5$ maps to $-7\leq g(x)\leq7$ . The point $(1,-2)$ maps to $((1+1)/2,3-2(-2))=(1,7)$ , while $(3,5)$ maps to $((3+1)/2,3-2(5))=(2,-7)$ .

(5 marks)

Tier 3 · Hard

A transformation sends

y=f(x)

y=g(x)

, where

g(x)=af(bx+c)+d

and

a,b\ne0

. The point

(-2,4)

maps to

(3,-5)

, and the corresponding point

(1,-1)

maps to

\left(\dfrac32,5\right)

. Find

a

b

c

and

d

, and hence write

g(x)

in terms of

f

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$a=-2$ , $b=-2$ , $c=4$ , $d=3$ $g(x)=-2f(-2x+4)+3$	6
Notes
For a transformed point with new $x$ -coordinate $X$ , the original input is $bX+c$ . The two point correspondences therefore give $3b+c=-2$ and $(3/2)b+c=1$ . Subtracting gives $b=-2$ , and then $c=4$ . The output relation is $Y=av+d$ , so $4a+d=-5$ and $-a+d=5$ . These give $a=-2$ and $d=3$ . Hence $g(x)=-2f(-2x+4)+3$ .

(6 marks)

Tier 3 · Hard

A function

f

has domain

-4\leq x\leq6

. Within this domain,

f(x)\geq2

exactly when

-3\leq x\leq-1

4\leq x\leq6

. For

g(x)=5-2f(3x+2)

, find the domain of

g

and solve

g(x)\leq1

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	The domain of $g$ is $-2\leq x\leq\dfrac43$ . $-\dfrac53\leq x\leq-1$ or $\dfrac23\leq x\leq\dfrac43$	5
Notes
The input $u=3x+2$ must satisfy $-4\leq u\leq6$ , giving $-2\leq x\leq4/3$ . Also, $5-2f(u)\leq1$ is equivalent to $f(u)\geq2$ . Mapping $-3\leq u\leq-1$ through $x=(u-2)/3$ gives $-5/3\leq x\leq-1$ . Mapping $4\leq u\leq6$ gives $2/3\leq x\leq4/3$ . These intervals already lie in the domain of $g$ .

(5 marks)

Q10

Tier 3 · Hard

10.

Starting with

y=f(x)

, a translation

2

units right is followed by a stretch parallel to the

x

-axis with scale factor

3

. Find the resulting equation and the image of the point

(-1,4)

. Repeat when the same two transformations are applied in the reverse order, and explain why the results differ.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Translation then stretch: $y=f\left(\dfrac{x-6}{3}\right)$ and $(-1,4)\mapsto(3,4)$ . Stretch then translation: $y=f\left(\dfrac{x-2}{3}\right)$ and $(-1,4)\mapsto(-1,4)$ . The horizontal transformations do not commute.	5
Notes
For translation then stretch, an original input coordinate $u$ maps first to $u+2$ and then to $3(u+2)=3u+6$ . Thus $u=(x-6)/3$ , giving $y=f((x-6)/3)$ ; $u=-1$ maps to $x=3$ . In the reverse order, $u$ maps to $3u$ and then to $3u+2$ , so $u=(x-2)/3$ , giving $y=f((x-2)/3)$ ; $u=-1$ maps to $x=-1$ . The different coordinate maps show that changing the order changes the graph.

(5 marks)

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