Learning Goal
Syllabus Requirement
Students need to have knowledge of the effect of
simple transformations on the graph of ??y = f(x)??, with the following points highlighted:
 Students should be able to apply one of these transformations to either a quadratic, cubic or reciprocal function and sketch the resulting graph.
 Given the graph of any function ??y = f(x)?? students should be able to sketch the graph resulting from one of these transformations:
??y = af(x)??
??y = f(x) + a??
??y = f(x + a)??
??y = f(ax)??
Study Notes
There are only 4 simple transformations to learn and the result of these can be split into to two types:

Translations of the curve of a function in either the ??x?? or ??y?? axis.

Stretching of the curve of a function in either the ??x?? or ??y?? axis.
The result of each of these transformations are listed in the information box below. The same rules can be used for either a quadratic, cubic or reciprocal functions.
The transformations resulting in a translation of the curve of a function ??f(x)?? have an addition to the function
??y = f(x + a)??:
Results in a translation of ??a?? along the ??x??axis.
??\implies?? We add ??a?? to each ??x?? coordinate of the curve, the ??y?? coordinates remain unchanged
??y = f(x) + a??:
Results in a translation of ??+a?? along the ??y??axis
??\implies?? We add ??+a?? to each ??y?? coordinate of the curve, the ??x?? coordinates remain unchanged
The transformations resulting in a stretch of the curve of a function ??f(x)?? have a multiplication to the function
??y = f(ax)??:
Results in a stretch of ??\frac {1}{a}?? along the ??x??axis
??\implies?? We multiply each ??x?? coordinate by ??\frac {1}{a}??, the ??y?? coordinates remain unchanged
??y = af(x)??:
Results in a stretch of ??a?? along the ??y??axis
??\implies?? We multiply each ??y?? coordinate by ??a??, the ??x?? coordinates remain unchanged
Note a simple way of remembering which axis the transformation will be, is if ??a?? is inside or outside the brackets.
If inside the bracket eg. ??y = f(x + a)??, then the transformation will be along the ??x?? axis
If outside the bracket eg ??y = f(ax)??, then the transformation will be along the ??y?? axis
Remember, the same rules can be used for either a quadratic, cubic or reciprocal functions.
Worked Examples
We'll now run through an example of each transformation in turn on the same function:
Figure 1  Graph of ??y = f(x)??
Figure 1 shows a sketch of the curve with equation ??y = f(x)??.
The curve has a turning point at ??(2, 4)?? and crosses the ??x??axis at the points ??(0, 0)?? and ??(3, 0)??.
Using ??y = f(x + a)??
Example 1
Question
Given the sketch of the curve with equation ??y = f(x)?? shown in Figure 1 above.
Sketch the curve with equation ??y = f(x+2)??
On the diagram, show clearly the coordinates of all the points at which the curve meets the axis.
Show Solution
Worked Solution
Recall the question asks for the curve with the equation ??y = f(x+2)?? which has the effect of
translating the curve along the ??x??axis.
For each point given we
subtract ??a??, which in this question is ??2??, from each ??x?? coordinate, the ??y?? coordinates remain unchanged
Taking each point given we can therefore subtract ??2?? from each ??x?? coordinate:
??x??axis intersection ??(0,0) \implies (2,0)??
??x??axis intersection ??(3,0) \implies (1,0)??
Turning point ??(2,4) \implies (0,4)??
We can now sketch the graph ??y=f(x+2)?? as shown below.
Graph of ??y=f(x+2)??
Using ??y = f(x) + a??
Example 2
Question
Given the sketch of the curve with equation ??y = f(x)?? shown in Figure 1 above.
Sketch the curve with equation ??y = f(x)3??
On the diagram, show clearly the coordinates of the turning point and intersection with the ??y?? axis.
Show Solution
Worked Solution
Recall the question asks for the curve with the equation ??y = f(x)3?? which has the effect of
translating the curve along the ??y??axis.
For each point given we
add ??a??, which in this question is ??3??, to each ??y?? coordinate, the ??x?? coordinates remain unchanged
Taking each point given we can therefore subtract ??3?? from each ??y?? coordinate:
??y??axis intersection ??(0,0) \implies (0,3)??
Turning point ??(2,4) \implies (2,7)??
We can now sketch the graph ??y=f(x)3?? as shown below.
Graph of ??y=f(x)3??
Using ??y = f(ax)??
Example 3
Question
Given the sketch of the curve with equation ??y = f(x)?? shown in Figure 1 above.
Sketch the curve with equation ??y = f(2x)??
On the diagram, show clearly the coordinates of all the points at which the curve meets the axis.
Show Solution
Worked Solution
Recall the question asks for the curve with the equation ??y = f(2x)?? which has the effect of
stretching the curve along the ??x??axis.
For each point given we
multiply each ??x?? coordinate by ??\frac {1}{a}??, which in this example ??a=2??. The ??y?? coordinates remain unchanged
Taking each point given we can therefore multiply the ??x?? coordinate by ??\frac {1}{2}??:
??x??axis intersection ??(0,0) \implies (0,0)??
??x??axis intersection ??(3,0) \implies (1.5,0)??
Turning point ??(2,4) \implies (1,4)??
We can now sketch the graph ??y=f(x+2)?? as shown below.
Graph of ??y=f(2x)??
Using ??y = af(x)??
Example 4
Question
Given the sketch of the curve with equation ??y = f(x)?? shown in Figure 1 above.
Sketch the curve with equation ??y = 2f(x)??
On the diagram, show clearly the coordinates of all the points at which the curve meets the axis.
Show Solution
Worked Solution
Recall the question asks for the curve with the equation ??y = 2f(x)?? which has the effect of
stretching the curve along the ??y??axis.
For each point given we
multiply each ??y?? coordinate by ??a??, which in this example ??a=2??. The ??x?? coordinates remain unchanged
Taking each point given we can therefore multiply the ??y?? coordinate by ??2??:
??x??axis intersection ??(0,0) \implies (0,0)??
??x??axis intersection ??(3,0) \implies (3,0)??
Turning point ??(2,4) \implies (2,8)??
We can now sketch the graph ??y=2f(x)?? as shown below.
Graph of ??y=2f(x)??
Exam Questions
The table below contains exam questions that have been asked on this topic, this includes normal papers from both January
and June sittings, International papers and Specimen papers.
To see an exam question and solution simply click on the load question icon (), the question will appear below the table and
the solution can be shown by clicking the "Show Solution" button that also appears.
The full exam paper and mark schemes are also available for download by clicking on the download icons in each
row of the table ().
Load Question 
Exam Board 
Subject 
Paper 
Year 
Month 
Module 
Question No. 
Parts 
Total Marks 
Exam Paper 
Mark Scheme 

Edexcel 
Maths 
Standard 
2005 
January 
Core 1 
6 
n/a 
6 



Edexcel 
Maths 
Standard 
2005 
June 
Core 1 
4 
n/a 
5 



Edexcel 
Maths 
Standard 
2006 
January 
Core 1 
6 
n/a 
9 



Edexcel 
Maths 
Standard 
2006 
June 
Core 1 
3 
(b) 
2 



Edexcel 
Maths 
Standard 
2007 
January 
Core 1 
3 
n/a 
6 



Edexcel 
Maths 
Standard 
2007 
June 
Core 1 
5 
n/a 
5 



Edexcel 
Maths 
Standard 
2008 
January 
Core 1 
6 
n/a 
7 



Edexcel 
Maths 
Standard 
2008 
June 
Core 1 
3 
n/a 
5 



Edexcel 
Maths 
Standard 
2009 
January 
Core 1 
5 
n/a 
6 


Edexcel  Maths  Standard  2005  January  Core 1  Question 6
Question
6.
Figure 1.
Figure 1 shows a sketch of the curve with equation ??y = f(x)??.
The curve crosses the ??x??axis at the points ??(2,0)?? and ??(4,0)??. The minimum point on the curve is ??P(3,2)??.
In separate diagrams sketch the curve with the equation
(a) ??y=f(x)?? [3]
(b) ??y=f(2x)?? [3]
On each diagram, give the coordinates of the points at which the curve crosses the ??x??axis, and the coordinates of the image of P under the given transformation.
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Worked Solution
6.a
Recall the question asks for the curve with the equation ??y=f(x)?? which has the effect of
mirroring the curve about the ??x??axis.
The intersections will remain unchanged (since ??y=0?? at these points), however the ??y?? value of the minimum point on the curve, ??P?? will reverse sign.
Taking each point given we can therefore say:
??x??axis intersection ??(2,0) \implies (2,0)??
??x??axis intersection ??(4,0) \implies (4,0)??
Turning Point ??P?? ??(3,2) \implies (3,2)??
We can now sketch the graph ??y=f(x)?? as shown below, clearly stating that the
Image of ??P(3,2)??.
Graph of ??y=f(x)??
6.b
Recall the question asks for the curve with the equation ??y=f(kx)?? which has the effect of
stretching the curve along the ??x??axis.
For each point given we
divide the ??x?? coordinate by the factor given, which in this question is ??k=2??. So the curve will be compressed along the ??x?? axis in this example.
Taking each point given we can therefore say:
??x??axis intersection ??(2,0) \implies (1,0)??
??x??axis intersection ??(4,0) \implies (2,0)??
Turning Point ??P?? ??(3,2) \implies (1.5,2)??
We can now sketch the graph ??y=f(2x)?? as shown below, clearly stating that the
Image of ??P(1.5,2)??.
Graph of ??y=f(2x)??
Edexcel  Maths  Standard  2005  June  Core 1  Question 4
Question
4.
Figure 1.
Figure 1 shows a sketch of the curve with equation ??y = f(x)??.
The curve passes through the origin ??O?? and through the point ??(6, 0)??. The maximum point on the curve is ??(3, 5)??.
On separate diagrams, sketch the curve with equation
(a) ??y = 3 f(x)?? [2]
(b) ??y = f(x+2)?? [3]
On each diagram, show clearly the coordinates of the maximum point and of each point at which the curve crosses the ??x??axis.
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Worked Solution
4.a
Recall the question asks for the curve with the equation ??y=3f(x)?? which has the effect of
stretching the curve along the ??y??axis.
For each point given we
multiply the ??y?? coordinate by the factor given, which in this question is ??3??. The ??x??axis intersections will remain unchanged since ??y=0?? at these points.
Taking each point given we can therefore say:
??x??axis intersection ??(0,0) \implies (0,0)??
??x??axis intersection ??(6,0) \implies (6,0)??
Turning Point ??(3,5) \implies (3,15)??
We can now sketch the graph ??y=3f(x)?? as shown below, clearly stating that the
maximum point ??(3,15)??.
Graph of ??y=3f(x)??
4.b
Recall the question asks for the curve with the equation ??y = f(x+2)?? which has the effect of
translating the curve along the ??x??axis.
For each point given we
subtract the offset, which in this question is ??2??, from each ??x?? coordinate.
Taking each point given we can therefore say:
??x??axis intersection ??(0,0) \implies (2,0)??
??x??axis intersection ??(6,0) \implies (4,0)??
Turning Point ??P?? ??(3,5) \implies (1,5)??
We can now sketch the graph ??y=f(x+2)?? as shown below, clearly stating that the
maximum point ??(1,5)??.
Graph of ??y=f(x+2)??
Edexcel  Maths  Standard  2006  January  Core 1  Question 6
Question
6.
Figure 1.
Figure 1 shows a sketch of the curve with equation ??y = f(x)??.
The curve passes through the points ??(0, 3)?? and ??(4, 0)?? and touches the ??x??axis at the point ??(1, 0)??.
On separate diagrams, sketch the curve with equation
(a) ??y = f(x+1)?? [3]
(b) ??y = 2f(x)?? [3]
(c) ??y = f(\frac{1}{2} x)?? [3]
On each diagram, show clearly the coordinates of all the points at which the curve meets the axis.
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Worked Solution
6.a
Recall the question asks for the curve with the equation ??y = f(x+k)?? which has the effect of
translating the curve along the ??x??axis.
For each point given we
subtract the offset, which in this question is ??k = 1??, from each ??x?? coordinate.
Taking each point given we can therefore say:
??x??axis intersection & turning point ??(1, 0) \implies (0,0)?? therefore passing through the origin
??x??axis intersection ??(4, 0) \implies (3, 0)??
??y??axis intersection ??(0, 3) \implies (1, 3)??, therefore no longer an intersection
We can now sketch the graph ??y=f(x+1)??.
Graph of ??y=f(x+1)??
6.b
Recall the question asks for the curve with the equation ??y=2f(x)?? which has the effect of
stretching the curve along the ??y??axis.
For each point given we
multiply the ??y?? coordinate by the factor given, which in this question is ??2??. The ??x??axis intersections will remain unchanged since ??y=0?? at these points.
Taking each point given we can therefore say:
??x??axis intersection & turning point ??(1, 0) \implies (1,0)??
??x??axis intersection ??(4, 0) \implies (4, 0)??
??y??axis intersection ??(0, 3) \implies (0, 6)??
We can now sketch the graph ??y=2f(x)??.
Graph of ??y=2f(x)??
6.c
Recall the question asks for the curve with the equation ??f(\frac{1}{2} x)?? which has the effect of
stretching the curve along the ??x??axis.
For each point given we
divide the ??x?? coordinate by the factor given, which in this question is ??\frac{1}{2}??. So the curve will be stretched along the ??x?? axis in this example.
Taking each point given we can therefore say:
??x??axis intersection & turning point ??(1, 0) \implies (2,0)??
??x??axis intersection ??(4, 0) \implies (8, 0)??
??y??axis intersection ??(0, 3) \implies (0, 3)??
We can now sketch the graph ??f(\frac{1}{2} x)??.
Graph of ??y=f(\frac{1}{2} x)??
Edexcel  Maths  Standard  2006  June  Core 1  Question 3
Question
3.b
Graph of ??y = (x + 3)^{2}??
The graph of ??y = (x + 3)^{2}?? shown above
Sketch the graph of ??y = (x + 3)^{2} + k??, where ??k?? is a positive constant. [2]
Show on the sketch the coordinates of each point at which the graph meets the axes.
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Worked Solution
3.b
Recall the question asks for the curve with the equation ??y = f(x)+k??, which has the effect of
translating the curve along the ??y??axis by ??k??.
For each point given we
add the offset, which in this question is ??k??, to each ??y?? coordinate.
From the sketch of ??y = (x + 3)^{2}?? we know the curve touches the ??x??axis at ??(3,0)?? and crosses the ??y??axis at ??(0,9)??
Taking each point given we can therefore say:
??x??axis intersection & turning point ??(3, 0) \implies (3,k)??
??y??axis intersection ??(0, 9) \implies (0, 9+k)??
We can now sketch the graph ??y = (x + 3)^{2} + k??.
Graph of ??y = (x + 3)^{2} + k??
Edexcel  Maths  Standard  2007  January  Core 1  Question 3
Question
3.
Given that ??f(x)= \frac{1}{x}, x \neq {0}??
(a) Sketch the graph of ??y = f(x) + 3?? and state the equation of the asymptotes [4]
(b) Find the coordinates of the point where ??y = f(x) + 3?? crosses a coordinate axis. [2]
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Worked Solution
3.a
Start with the curve for ??y = \frac{1}{x}??:
Recall that the functions of the form ??\frac{k}{x}?? (in this case where ??k=1??) are
reciprocals. With asymptotes at ??x=0?? (vertical) and ??y=0?? (horizontal).
Now consider the translation:
Recall the question asks for the curve with the equation ??y = f(x)+k??, which has the effect of
translating the curve along the ??y??axis by ??k??.
For each point given we
add the offset, which in this question is ??3??, to each ??y?? coordinate. Therefore we can say:
Vertical asymptote at ??x=0 \implies x=0??
Horizontal asymptote at ??y=0 \implies y=3??
We can now sketch the graph ??y = \frac{1}{x} + 3??.
Graph of ??y = \frac{1}{x} + 3??
3.b
Where ??y = \frac{1}{x} + 3?? crosses the ??x??axis, ??y = 0??
$$\implies \frac{1}{x} + 3 = 0$$
$$\implies \frac{1}{x} = 3$$
$$\implies x = \frac{1}{3}$$
Therefore the coordinate where ??y = \frac{1}{x} + 3?? crosses the ??x??axis is ??(\frac{1}{3},0)??
Graph of ??y = \frac{1}{x} + 3??
Edexcel  Maths  Standard  2007  June  Core 1  Question 5
Question
5.a
Figure 1
Figure 1 shows a sketch of the curve with equation ??y = \frac{3}{x}, x \neq {0}??
(a) On a separate diagram, sketch the curve with equation ??y = \frac{3}{x+2}, x \neq {2}??. Showing the coordinates of any point at which the curve crosses a coordinate axis. [3]
(b) Write down the equation of the asymptotes of the curve in part (a). [2]
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Worked Solution
5.
Recall the question asks for the curve with the equation ??y = f(x+k)?? which has the effect of
translating the curve along the ??x??axis.
For each point given we
subtract the offset, which in this question is ??k=2??, from each ??x?? coordinate. ie a translation of ??2?? parallel to the ??x?? axis
We can now sketch the graph ??y = \frac{3}{x+2}??.
Graph of ??y = \frac{3}{x+2}??
The curve has a single intercept on the ??y??axis, the question asks to show this coordinate
Where ??y = \frac{3}{x+2}?? crosses the ??y??axis, ??x = 0??
$$\implies y = \frac{3}{0+2}$$
$$\implies y = \frac{3}{2} = 1.5$$
Therefore the coordinate where ??y = \frac{3}{x+2}?? crosses the ??y??axis is ??(0, 1.5)??
5.b
Recall that the horizontal asymptote for a reciprocal of form ??y=\frac{k}{x}?? is ??y=0??
This will remain unchanged therefore we can say the vertical asymptote of ??y = \frac{3}{x+2}??, is ??y=0??
Recall that the vertical asymptote for a reciprocal of form ??y=\frac{k}{x}?? is ??x=0??
Therefore the horizontal asymptote of ??y = \frac{3}{x+2}??, is ??x = 0  2 = 2??
Note: There is a clue in the question as it states ??x \neq {2}??.
??y = \frac{3}{x+2}?? showing the vertical and horizontal asymptotes.
Edexcel  Maths  Standard  2008  January  Core 1  Question 6
Question
6.
Figure 1.
Figure 1 shows a sketch of the curve with equation ??y = f(x)??.
The curve crosses the ??x??axis at the points ??(1, 0)?? and ??(4, 0)??. The maximum point on the curve is ??(2, 5)??.
In separate diagrams sketch the curves with the following equations.
On each diagram show clearly the coordinates of the maximum point and of each point at which the curve crosses the ??x??axis.
(a) ??y = 2f(x)?? [3]
(b) ??y=f(x)?? [3]
The maximum point on the curve with equation ??y = f(x+a)?? is on the ??y??axis
(c) Write down the value of the constant ??a??. [1]
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Worked Solution
6.a
Recall the question asks for the curve with the equation ??y=kf(x)?? which has the effect of
stretching the curve along the ??y??axis.
For each point given we
multiply the ??y?? coordinate by the factor given, which in this question is ??k = 2??. The ??x??axis intersections will remain unchanged since ??y=0?? at these points.
Taking each point given we can therefore say:
??x??axis intersection ??(1, 0) \implies (1, 0)??
??x??axis intersection ??(4, 0) \implies (4, 0)??
Maximum point ??(2, 5) \implies (2, 10)??
We can now sketch the graph ??y=2f(x)??.
Graph of ??y = 2f(x)??
6.b
Recall the question asks for the curve with the equation ??y=f(x)?? which has the effect of
mirroring the curve about the ??y??axis.
Taking each point given we can therefore say:
??x??axis intersection ??(1, 0) \implies (1, 0)??
??x??axis intersection ??(4, 0) \implies (4, 0)??
Maximum point ??(2, 5) \implies (2, 5)??
We can now sketch the graph ??y=f(x)?? as shown below.
Graph of ??y=f(x)??
6.c
We are told in the question that the curve with equation ??y = f(x)?? has the maximum point at ??(2, 5)??
The maximum point on the curve with equation ??y = f(x+a)?? is on the ??y??axis at ??(0, 5)?? therefore we must transpose the curve by ??2?? along the ??x??axis
Recall the question asks for the curve with the equation ??y = f(x+k)?? which has the effect of
translating the curve along the ??x??axis.
For each point given we
subtract the offset, which in this question is ??k = 1??, from each ??x?? coordinate.
??\implies a = 2??, ??y = f(x + 2)??.
Edexcel  Maths  Standard  2008  June  Core 1  Question 3
Question
3.
Figure 1.
Figure 1 shows a sketch of the curve with equation ??y = f(x)??. The curve passes through the point ??(0, 7)?? and has a minimum point at ??(7, 0)??.
On separate diagrams, sketch the curve with equation
(a) ??y = f(x) + 3?? [3]
(b) ??y = f(2x)?? [2]
On each diagram, show clearly the coordinates of the minimum point and the coordinates of the point at which the curve crosses the ??y??axis.
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Worked Solution
3.a
Recall the question asks for the curve with the equation ??y = f(x)+k??, which has the effect of
translating the curve along the ??y??axis by ??k??.
For each point given we
add the offset, which in this question is ??k=3??, to each ??y?? coordinate.
Taking each point given we can therefore say:
??y??axis intersection ??(0, 7) \implies (0, 10)??
Minimum point ??(7, 0) \implies (7, 3)??
We can now sketch the graph ??y = f(x) + 3??.
Graph of ??y = f(x) + 3??
3.b
Recall the question asks for the curve with the equation ??y=f(kx)?? which has the effect of
stretching the curve along the ??x??axis.
For each point given we
divide the ??x?? coordinate by the factor given, which in this question is ??k=2??. So the curve will be compressed along the ??x?? axis in this example.
Taking each point given we can therefore say:
??y??axis intersection ??(0, 7) \implies (0, 7)??
Minimum point ??(7, 0) \implies (3.5, 0)??
We can now sketch the graph ??y = f(2x)??.
Graph of ??y = f(2x)??
Edexcel  Maths  Standard  2009  January  Core 1  Question 5
Question
5.
Figure 1.
Figure 1 shows a sketch of the curve ??C?? with equation ??y = f(x)??. There is a maximum at ??(0, 0)??, at minimum at ??(2, 1)?? and ??C?? passes through ??(3, 0)??.
On separate diagrams, sketch the curve with equation
(a) ??y = f(x + 3)?? [3]
(b) ??y = f(x)?? [3]
On each diagram show clearly the coordinates of the maximum point, the minimum point and any points of intersection with the ??x??axis.
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Worked Solution
5.
Recall the question asks for the curve with the equation ??y = f(x+k)?? which has the effect of
translating the curve along the ??x??axis.
For each point given we
subtract the offset, which in this question is ??k=3??, from each ??x?? coordinate. ie a translation of ??3?? parallel to the ??x?? axis
Taking each point given we can therefore say:
Maximum point ??(0, 0) \implies (3, 0)??
Minimum point ??(2, 1) \implies (1, 1)??
??x??axis intersection ??(3, 0) \implies (0, 0)??
We can now sketch the graph ??y = f(x + 3)??.
Graph of ??y = f(x + 3)??
5.b
Recall the question asks for the curve with the equation ??y=f(x)?? which has the effect of
mirroring the curve about the ??y??axis.
Taking each point given we can therefore say:
Maximum point ??(0, 0) \implies (0, 0)??
Minimum point ??(2, 1) \implies (2, 1)??
??x??axis intersection ??(3, 0) \implies (3, 0)??
We can now sketch the graph ??y=f(x)??, mirrored about the ??y?? axis as shown below.
Graph of ??y=f(x)??
Edexcel  Maths  Standard  2009  June  Core 1  Question 10
Question
10.
Question here. [2]
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Worked Solution
10.
Answer here
Edexcel  Maths  Standard  2010  January  Core 1  Question 8
Question
8.
Question here. [7]
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Worked Solution
8.
Answer here
Edexcel  Maths  Standard  2010  June  Core 1  Question 6
Question
6.
Question here. [7]
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Worked Solution
6.
Answer here
Edexcel  Maths  Standard  2011  January  Core 1  Question 5
Question
5.
Question here. [7]
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Worked Solution
5.
Answer here
Edexcel  Maths  Standard  2011  June  Core 1  Question 8
Question
8.
Question here. [10]
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Worked Solution
8.
Answer here
Edexcel  Maths  Standard  2012  June  Core 1  Question 10
Question
10.
Question here. [10]
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Worked Solution
10.
Answer here
Edexcel  Maths  Standard  2013  January  Core 1  Question 6
Question
6.
Question here. [7]
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Worked Solution
6.
Answer here
Edexcel  Maths  Standard  2013  June  Core 1  Question 8
Question
8.
Question here. [6]
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Worked Solution
8.
Answer here
Edexcel  Maths  Standard  2013  June  Core 1  Question 9
Question
9.
Question here. [3]
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9.
Answer here
Exam Tips

Exam questions typically take the form shown in the Worked Examples where a function is given and each transformation type tested.

The transformations can be grouped into the following two types:
Translation of the curve of a function ??f(x)?? which have an addition to the function.
??y = f(x {\color{Red} +} a)?? or ??y = f(x) {\color{Red} +} a??
Stretching of the curve of a function ??f(x)?? which have a multiplication to the function.
??y = f(ax)?? or ??y = af(x)??

A quick way to determine along which axis the the transformation will act is to inspect if ??a?? is inside or outside the bracket:
Inside the bracket the transformation will be along the ??x?? axis
??y = f(x + {\color{Red} a})?? or ??y = f({\color{Red} a}x)??
Outside the bracket the transformation will be along the ??y?? axis
??y = f(x) + {\color{Red} a}?? or ??y = {\color{Red} a}f(x)??

Remember that the same rules can be used for either a quadratic, cubic or reciprocal functions.