This is Part 3 of a three-part tutorial on graph transformations, and covers compound transformations. If you haven’t already worked through Part 1 and Part 2 then I recommend you look at those before trying to make sense of this one!

This tutorial goes beyond GCSE/IGCSE; in fact, compound transformations are part of the second-year content of the A-level specification.

A quick recap

Basic rules for graph transformations

The table below summarises the rules you need to remember.

Table summarising the rules for graph transformations

Remember, the GCSE spec doesn’t include stretches, though Edexcel IGCSE and A-level both do.

Skills covered so far

In Part 1 we practised the following skills:

Given an original graph and the description of a transformation, finding the equation of a transformed graph and sketching it
Given the equations of an original and a transformed graph, identifying the transformation that has taken place

In Part 2 we looked at sketching transformations of a graph with unknown equation, and how the different transformations affect the x- and y-coordinates of points on the graph.

In this instalment we’re going to do the same again, but with compound transformations.

Compound transformations

A compound transformation is what you have when a graph undergoes two or more transformations in succession, for example, it might be stretched in the x-direction with scale factor 2 and then translated with vector $\begin{pmatrix}0\\-1\end{pmatrix}$ .

You’ve already encountered one type of compound transformations when we’ve looked at the two components of a vector transformation: a translation with vector $\begin{pmatrix}1\\5\end{pmatrix}$ could also be described as a translation with vector $\begin{pmatrix}1\\0\end{pmatrix}$ followed by a translation with vector $\begin{pmatrix}0\\5\end{pmatrix}$ (in either order).

Key rules for compound transformations

The key things to remember with compound transformations are:

Transformations in the y-direction go “as expected” and transformations in x-direction go “backwards”; this applies to the order of compound transformations as well as the individual transformations themselves (see VON HIR, below).
Horizontal and vertical transformations do not affect each other – remember, horizontal transformations only affect the x-coordinates and vertical ones only the y-coordinates – so you only need to worry about the order within the vertical transformations and within the horizontal transformations.

Note: Since changing the sign is the same as multiplying by -1, reflections and stretches share equal priority.

You might find it helpful to remember the acronym VON HIR:
Vertical, Outside Normal; Horizontal, Inside, Reverse.

VON HIR for compound transformations:
Vertical transformations go on the
Outside, and are carried out in the
Normal order.
Horizontal transformations go on the
Inside, and are carried out in
Reverse order.

Identifying and describing a sequence of transformations

Examples

y = x²
→ y = 3 sin x + 4 or y = 4 + 3 sin x

The transformations are applied to the Outside of the original function so they’re Vertical.
The entire sin x has been multiplied by 3 and then had 4 added to it; the order is Normal so that equates to
A stretch in the y-direction with SF 3 and then a translation with vector $\begin{pmatrix}0\\4\end{pmatrix}$

y = sin x
→ y = sin (2x – 50°)

The transformations are applied to the Inside of the original function (changing/expanding the x) so they’re Horizontal.
Working in Reverse to “undress the x”, we add 50° and then divide by 2, so this equates to
A translation with vector $\begin{pmatrix}50^{\circ}\\0\end{pmatrix}$ and then a stretch in the x-direction with SF ½.

Putting both of these together will give a format that you might encounter in a trig modelling question:
y = sin x
→ y = 4 + 3 sin (2x – 50°)

It doesn’t matter whether we list the vertical or the horizontal transformations first, but it makes sense to combine the two vectors, so that gives us
A stretch in the y-direction with SF 3
then a translation with vector $\begin{pmatrix}50^{\circ}\\4\end{pmatrix}$
then a stretch in the x-direction with SF ½.

Your turn 1

Describe each series of transformations. For questions 4-7 there are (at least) two possible interpretations; can you find both?

$y=x^2 \hspace{0.5cm} \to \hspace{0.5cm} y=3x^2+5$
$y = \cos x \hspace{0.5cm} \to \hspace{0.5cm} y = 2 \cos(x - 30) - 1$
$y=\sqrt{x} \hspace{0.5cm} \to \hspace{0.5cm} y=\sqrt{4x-1}$
$y=x^2 \hspace{0.5cm} \to \hspace{0.5cm} y=9x^2 - 3$
$y=x^2 \hspace{0.5cm} \to \hspace{0.5cm} y = 5 - \frac{x^2}{4}$
$y=\sqrt{x} \hspace{0.5cm} \to \hspace{0.5cm} y = -\sqrt{9x} + 2$
$y=2^{x} \hspace{0.5cm} \to \hspace{0.5cm} y = 5(2^{3-x} + 7)$

Click here for the answers

Finding the equation of the resulting graph

Work out what equation you’d end up with if you applied each of the sets of transformations to the start equation.
Start by separating out the vertical and horizontal transformations, then apply each set separately.

Example

Starting with $y=\cos x$ , what would be the result of a translation by $\begin{pmatrix}10\\20\end{pmatrix}$ followed by a stretch SF 2 in the x-direction?

Vertical:
Translation $\begin{pmatrix}0\\20\end{pmatrix}$ => add 20 to the whole thing

Horizontal:
Translation $\begin{pmatrix}10\\0\end{pmatrix}$ then stretch SF 2 in the x-direction
Remember, REVERSE order for horizontal transformations,
so first divide the x by 2
then subtract 10 from the $\frac{x}{2}$

So you end up with $y = \cos (\frac{x}{2} - 10) + 20$

Your turn 2

Find the result of each set of transformations:

$y = \sin x :$ Stretch SF 3 in the y-direction then translation $\begin{pmatrix}10\\20\end{pmatrix}$
$y = 5x^2:$ Reflection in the x-axis then stretch SF 2 in the y-direction
$y = 4\sqrt{x}:$ Stretch SF 4 in the x-direction then translation $\begin{pmatrix}2\\3\end{pmatrix}$
$y = 3x + 2:$ Translation $\begin{pmatrix}5\\1\end{pmatrix}$ then reflection in the y-axis
$y = 3^{2x}:$ Translation $\begin{pmatrix}1\\-2\end{pmatrix}$ , then stretch SF 4 in the y-direction, then stretch SF 2 in the x-direction.

Click here for the answers

Transforming a graph with unknown equation

Below is the graph of y = f(x), the same as we used in Part 2 of this series.
As you can see, four coordinate points are marked on the graph. Your task is to (a) sketch each of the transformed graphs listed under “Your turn 3” below, labelling the new position of each of the four marked points, and (b) describe the sequence of transformations in each case.

Graph of y = f(x) (actual equation not provided)

Don’t forget VON HIR, and that

A transformation in the x-direction will change only the x-coordinates, and
A transformation in the y-direction will change only the y-coordinates.

Your turn 3

Now try sketching these graphs and describing the sequence of transformations in each case:

y = f(2x) + 1
y = 2f(x + 1)
y = 2f(x) + 1
y = f(2x + 1)
y = -f(x) + 1
y = f(1 – x) [Hint: think of it as -x + 1]
y = 2f(-x) + 1
y = -f(1 – 2x)

Click here for the answers

That about covers it – you should now be able to deal with just about anything that the topic of compound transformations throws at you!

If you’ve found this article helpful then please share it with anyone else who you think would benefit (use the social sharing buttons if you like). If you have any suggestions for improvement or other topics that you’d like to see covered, then please comment below or drop me a line using my contact form.

On my sister site at at mathscourses.co.uk you can find – among other things – a great-value suite of courses covering the entire GCSE (and Edexcel IGCSE) Foundation content, and the “Flying Start to A-level Maths” course for those who want to get top grades at GCSE and hit the ground running at A-level – please take a look!

If you’d like to be kept up to date with my new content then please sign up to my mailing list using the form at the bottom of this page, which will also give you access to my collection of free downloads.

Your turn 1: answers

Stretch in y-direction SF 3 then translation $\begin{pmatrix}0\\5\end{pmatrix}$
Stretch in y-direction SF 2 then translation $\begin{pmatrix}30\\1\end{pmatrix}$
Translation $\begin{pmatrix}1\\0\end{pmatrix}$ then stretch in x-direction SF $\frac{1}{4}$ .
Stretch in y-direction SF 9 THEN translation $\begin{pmatrix}0\\-3\end{pmatrix}$
OR $y = (3x)^2 - 3:$ Stretch in x-direction SF $\frac{1}{3}$ then translation $\begin{pmatrix}0\2\end{pmatrix}$
Think of it as $y = - \frac{x^2}{4} + 5:$ Stretch in y-direction SF $\frac{1}{4}$ and reflection in x-axis (either way round), THEN translation $\begin{pmatrix}0\\5\end{pmatrix}$
OR $y=-(\frac{x}{2})^2 + 5:$ Stretch in x-direction SF 2 and reflection in x-axis (either way round), THEN translation $\begin{pmatrix}0\\5\end{pmatrix}$
Stretch in the x-direction SF $\frac{1}{9}$ (could go anywhere in sequence) ; reflection in the x-axis then translation $\begin{pmatrix}0\\2\end{pmatrix}$
Could also be expressed as $y=-3\sqrt{x}+2:$ Stretch in y-direction SF 3 and reflection in x-axis (either way round), THEN translation $\begin{pmatrix}0\\2\end{pmatrix}$
OR
$y=-(\sqrt{9x} - 2):$ Stretch in the x-direction SF $\frac{1}{9}$ (OR stretch in y-direction SF 3), THEN translation $\begin{pmatrix}0\\-2\end{pmatrix}$ , THEN reflection in x-axis
H: translation $\begin{pmatrix}-3\\0\end{pmatrix}$ then reflection in y-axis; V: translation $\begin{pmatrix}0\\7\end{pmatrix}$ then stretch in y-direction SF 5 (can combine translations as long as combined translation precedes both of the other transformations)
OR $y=5(2^{-(x-3)}) + 35:$ H: reflection in y-axis then translation $\begin{pmatrix}3\\0\end{pmatrix}$ ; V: stretch in y-direction SF 5 then translation $\begin{pmatrix}0\\35\end{pmatrix}$ (can combine translations as long as the combined translation comes after both other transformations)

Click here to return to questions

Your turn 2: answers

$y = 3 \sin (x - 10) + 20$
$y = -(5x^2) \times 2 = -10x^2$
H: $x \hspace{0.5cm} \to \hspace{0.5cm} x-2 \hspace{0.5cm} \to \hspace{0.5cm} \frac{x-2}{4}$
so we get $y = 4 \sqrt{ \frac{x-2}{4}}+3 = 2\sqrt{x-2} + 3$
H: $x \hspace{0.5cm} \to \hspace{0.5cm} \frac{x}{2} \hspace{0.5cm} \to \hspace{0.5cm} \frac{x}{2} - 1 \hspace{1cm}$
V: $f(x) \hspace{0.5cm} \to \hspace{0.5cm} f(x) - 2 \hspace{0.5cm} \to \hspace{0.5cm} 4(f(x) - 2)$
so we get $y=4(3^{2(\frac{x}{2}-1)} - 2) = 4(3^{x-2}) - 8$