C3 Transformations

15: More Transformations15: More Transformations
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 2: A2 Core ModulesVol. 2: A2 Core Modules

More Transformations
Module C3
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The translations and stretches that we met in AS
can be applied to any functions.
In this presentation we will look particularly at the
effect on the trig, exponential and log functions of
combining transformations.
We’ll start with a reminder of some examples we’ve
already met.
Try not to use a calculator when doing this topic.
Graphs copied from graphical calculators look peculiar
unless the scales are chosen very carefully.
If you do use a calculator remember to mark
coordinates of all significant points and clearly show
the behaviour of the curves near the axes.

e.g. 1 The translation of the function by
the vector gives the function
3
xy =






1
2
1)2( 3
+−= xy
3
xy = 1)2( 3
+−= xy
The graph becomes

so it is a stretch of s.f. 3, parallel to the y-axis
e.g. 2 Describe the transformation of that
gives .
x
y
1
=
x
y
3
=
x
y
3
=Solution: can be written as
x
y
1
3×=
x
y
3
=
x
y
1
=
3×

xy cos=
e.g. 3 Sketch the graph of the function xy 2cos=
Solution:
xyxy 2coscos =→=
is a stretch of s.f. , parallel to the x-axis.
So,
2
1

xy 2cos=
xy cos=
e.g. 3 Sketch the graph of the function xy 2cos=
Solution:
xyxy 2coscos =→=
is a stretch of s.f. , parallel to the x-axis.
So,
2
1

General Translations and Stretches





−
b
a
• The function is a translation
of by)(xfy =
baxfy ++= )(
 Translations
 Stretches
)(kxfy =• The function is obtained from )(xfy =
by a stretch of scale factor ( s.f. ) ,
parallel to the x-axis.
k
1
• The function is obtained from)(xkfy = )(xfy =
by a stretch of scale factor ( s.f. ) k,
parallel to the y-axis.

Two more Transformations
 Reflection in the x-axis
Every y-value changes sign when we reflect in
the x-axis e.g.
So, xyxy sinsin −=→=
xy sin=
xy sin−=
x
x
→
In general, a reflection in the x-axis is given by
)()( xfyxfy −=→=

 Reflection in the y-axis
Every x-value changes sign when we reflect in
the y-axis e.g.
So, xx
eyey −
=→=
x
ey = x
ey −
=
x x
→
In general, a reflection in the y-axis is given by
)()( xfyxfy −=→=

SUMMARY
 Reflections in the axes
• Reflecting in the x-axis changes the sign of y
)()( xfyxfy −=→=
)()( xfyxfy −=→=
• Reflecting in the y-axis changes the sign of x
The examples that follow illustrate combinations of
the transformations: translations, stretches and
reflections.

Combined Transformations
e.g. 1 Describe the transformations of that
give the function . Hence sketch
the function.
x
ey =
12
+= x
ey
Solution:
• x has been replaced by 2x:
so we have a stretch of s.f.
• 1 has then been added:
xx
ee 2
→
122
+→ xx
ee
so we have a translation of
parallel to the x-axis
2
1






1
0

The point on the y-axis . . .
We do the sketch in 2 stages:
x
ey = x
ey 2
=
→x
ey = x
ey =
x
ey 2
=
doesn’t move
with a stretch parallel to the x-axis
2
1×

12
+= x
ey
x
ey 2
=
12
+= x
ey
1+
1+
1+
We do the sketch in 2 stages:
x
ey = x
ey 2
=
→x
ey = x
ey =
x
ey 2
=
2
1×

give
(a) (b)
xy ln=
1ln2 += xy )1ln(2 += xy
Solution:
(a) We have →xln
but for (b),
→xln
(a) is
• a stretch of s.f. 2
)1ln( +x2
xln2 1+
→)ln(x 1+
2 xln →

give
(a) (b)
xy ln=
1ln2 += xy )1ln(2 += xy
Solution:
(a) We have →xln
but for (b),
→xln
(a) is
• a stretch of s.f.
parallel to the y-axis
2
)1ln( +x2
xln2 1+
→)ln(x 1+
• a translation of
2 xln →

give
(a) (b)
xy ln=
1ln2 += xy )1ln(2 += xy
Solution:
(a) We have →xln
but for (b),
→xln
(a) is
2






1
0
)1ln( +x2
xln2 1+
→)ln(x 1+
(b) is
2 xln →

give
(a) (b)
xy ln=
1ln2 += xy )1ln(2 += xy
Solution:
(a) We have →xln
but for (b),
→xln
(a) is
2






1
0
(b) is





−
0
1
)1ln( +x2
xln2 1+
→)ln(x 1+
2 xln →

2 xln →
give
(a) (b)
xy ln=
1ln2 += xy )1ln(2 += xy
Solution:
(a) We have →xln
but for (b),
→xln
(a) is
2






1
0
(b) is
2





−
0
1
)1ln( +x2
xln2 1+
→)ln(x 1+

give
(a) (b)
xy ln=
1ln2 += xy )1ln(2 += xy
Solution:
(a) We have →xln
but for (b),
→xln
(a) is
2






1
0
(b) is
2





−
0
1
)1ln( +x2
xln2 1+
→)ln(x 1+
2 xln →

xy ln=
)1ln( += xy
xy ln2=
xy ln=
The graphs of the functions are:
(b)
1ln2 += xy
(a)
)1ln(2 += xy
→
→
translate stretch
stretch translate
1+2×
1−
2×

then (iii) a reflection in the x-axis
(i) a stretch of s.f. 2 parallel to the x-axis
then (ii) a translation of 





2
0
e.g.3 Find the equation of the graph which is
obtained from by the following
transformations, sketching the graph at each
stage. ( Start with ).
xy cos=
π20 ≤≤ x

→xcos
Solution:
(i) a stretch of s.f. 2 parallel to the x-axis
x2
1
cos
xy 2
1
cos=
xy cos=
2×
→
stretch

Brackets aren’t essential here but I think they make
it clearer.
(ii) a translation of :





2
0
→x2
1
cos ( ) 2cos 2
1
+x
( ) 2cos 2
1
+= xy
2+
translate

( ) 2cos 2
1
−−= xy
(ii) a translation of :





2
0
→x2
1
cos ( ) 2cos 2
1
+x
( ) 2cos 2
1
−− x( ) →+ 2cos 2
1
x
( ) 2cos 2
1
+= xy
2+
translate reflect
x
x
→
(iii) a reflection in the x-axis
( )( ) =+− 2cos 2
1
x

Exercises
1. Describe the transformations that map the
graphs of the 1st
of each function given below
onto the 2nd. Sketch the graphs at each stage.
x
ey = x
ey −
= 2(a) to
xy ln= )3ln(2 −= xy(b) to
xy sin= xy 2sin1 +=(c) to
( Draw for )xsin π20 ≤≤ x

x
ey = x
ey −
= 2(a) toSolutions:
x
ey −
=→ 2
( The order doesn’t matter )
x
ey =x
ey −
=
Stretch s.f. 2 parallel to the y-axis
x
ey −
=
x
ey −
= 2
→= x
ey Reflection in the y-axis
x
ey −
=

xy ln= )3ln(2 −= xy(b) toSolutions:
)3ln(2 −=→ xy
→= xy ln )3ln( −= xy Translation






0
3
Stretch s.f. 2 parallel to the y-axis
xy ln=
)3ln( −= xy
)3ln(2 −= xy
)3ln( −= xy
( Again the order doesn’t matter )

xy sin=Solutions: xy 2sin1 +=(c) to
→= xy sin xy 2sin=
Translation






1
0xyxy 2sin12sin +=→=
Stretch s.f. parallel to the x-axis2
1
Again the order doesn’t matter.
xy sin=
xy 2sin= xy 2sin=
xy 2sin1 +=

If a stretch and a translation are in the same
direction we have to be very careful.
x
ey =
e.g. A stretch s.f. parallel to the y-axis on3
followed by a translation of gives






1
0
→= x
ey →= x
ey 3 13 += x
ey
With the translation first, we get
→= x
ey →+= 1x
ey )1(3 += x
ey
33 += x
ey

An important example involving stretches is the
transformation of a circle into an ellipse.
122
=+ yx
e.g. Find the equation of the ellipse given by
transforming the circle by
(i) a stretch of scale factor 4 parallel to the x-axis,
and
(ii) a stretch of scale factor 2 parallel to the y-axis
Method
Rearranging the equation of the circle to y = . . .
gives a clumsy expression so we don’t do it.
This means we must change the way we handle the
stretch in the y direction.

When we had , we stretched by s.f. 2
parallel to the y-axis by writing
xy ln=
xyxy ln2ln =→=
We could equally well have divided the l.h.s. by 2,
so
x
y
xy ln
2
ln =→=
i.e. multiplying the r.h.s. by 2.
So, to find the equation of a curve which is stretched
by 2 in the y direction, we can replace y by 2
y
We are then treating both stretches in the same
way.

1
24
22
=





+




 yx
(i) a stretch of scale factor 4 parallel to the x-axis,
and
(ii) a stretch of scale factor 2 parallel to the y-axis
Returning to the example . . .
→=+ 122
yx
Solution: Replace x by and replace y by
4
x
2
y
122
=+ yx
e.g. Find the equation of the ellipse given by
transforming the circle by

122
=+ yx
The ellipse looks like this . . .
1
24
22
=





+




 yx
1
416
22
=+
yx
⇒
If we want to translate the ellipse we use a similar
technique
1
2
1
4
2
22
=




 +
+




 − yx
e.g. to translate by replace x by and





− 1
2
)2( −x
replace y by )1( +y
1
24
22
=





+




 yx
So,
The answer is usually left in this form.

x
1
2
1
4
2
22
=




 +
+




 − yx
The graphs look like this:
1
24
22
=





+




 yx
122
=+ yx

SUMMARY
we can obtain stretches of scale factor k by
When we cannot easily write equations of curves in
the form
)(xfy =
k
x
• Replacing x by and replacing y by
k
y
we can obtain a translation of by






q
p
• Replacing x by )( px −
• Replacing y by )( qx −

Exercises
422
=+ yx
1. Find the equation of the curve obtained from
with the transformations given.
(i) a stretch of s.f. 3 parallel to the x-axis and
(ii) a stretch of s.f. 5 parallel to the y-axis
(iii) followed by a translation of .






3
1
xy 42
=
(i) a stretch of s.f. 2 parallel to the x-axis and
(iii) followed by a translation of .





2
0

(iii) followed by a translation of .






3
1
4
3
2
2
=+





y
x
4
3
2
2
=+





y
x
422
=+ yx
1 (i) a stretch of s.f. 3 parallel to the x-axis
4
53
22
=





+




 yx
4
5
3
3
1
22
=




 −
+




 − yx
Solutions:
4
53
22
=





+




 yx

(iii) followed by a translation of .





2
0
xy 42
=
2 (i) a stretch of s.f. 2 parallel to the x-axis






=
2
42 x
y
xy 22
=⇒
x
y
2
5
2
=





x
y
2
5
2
=





x
y
2
5
2
2
=




 −
Solutions:
( or )045042
=+−− xyy
xy 22
= xy 502
=( or )

You may have to deal with a function shown only in a
drawing ( with no equation given ).
If you are confident about the earlier work, try this
one before you look at my solution.

(i) (ii))(xfy −= )2( xfy =
The diagram shows part of the curve with equation
.)(xfy =
Copy the diagram twice and on each diagram sketch
one of the following:
)(xfy =
x
y

Solution:
)2( xfy =(ii)
)(xfy =
)2( xfy =
x
y
)(xfy =
)(xfy −=
x
y
(i) )(xfy −=

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