Resources For Teachers For Tutors For Students & Parents Pricing
Create Free Account
Year 12 Maths - Methods (Unit 3 & Unit 4) Transformations

Combinations of transformations

20 practice questions 0 video lessons Theory + worked examples
Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

Combining transformations applies a dilation, a reflection and a translation to a base function \(y=f(x)\) in one rule, \(y=a\,f\big(n(x-h)\big)+k\). The order matters — dilations and reflections act first, translations last — so tracking a key point through the sequence and reading off the final domain, range and asymptotes pins down the transformed graph.

A combination of transformations takes a standard function \(y=f(x)\) — such as \(y=x^{2}\), \(y=\sqrt{x}\) or \(y=\dfrac{1}{x}\) — and applies a dilation, a reflection and a translation together. Every combination can be written in the single form \(y=a\,f\big(n(x-h)\big)+k\), and reading off \(a,n,h,k\) tells you exactly which transformations are involved.

Each parameter has a fixed meaning. \(a\) is a dilation by factor \(|a|\) from the \(x\)-axis (a reflection in the \(x\)-axis as well when \(a<0\)); \(n\) is a dilation by factor \(\dfrac{1}{|n|}\) from the \(y\)-axis (a reflection in the \(y\)-axis when \(n<0\)); \(h\) is a horizontal translation (right when \(h>0\)); and \(k\) is a vertical translation (up when \(k>0\)).

The order of application is not free: dilations and reflections are applied first, then the translations last. A convenient way to see the effect is to track a key point — the turning point of a parabola, the endpoint of \(y=\sqrt{x}\), or the crossing of the asymptotes of \(y=\dfrac1x\). Under the combination, a point \((x,y)\) on the base graph moves to \(\left(\dfrac{x}{n}+h,\;a\,y+k\right)\), and the image's domain, range and asymptotes follow from that key feature.

Key idea. Write the combination as \(y=a\,f\big(n(x-h)\big)+k\). Apply dilations and reflections first, translations last; a point \((x,y)\) maps to \(\left(\dfrac{x}{n}+h,\;a\,y+k\right)\).
Transformed square-root curveThe curve y equals square root of x with key point at the origin, and its image y equals minus two root (x minus 1) plus 3 with key point moved to (1,3); the image opens downward to the right. x y (0,0) (1,3) image y=√x
\(y=\sqrt{x}\) becomes \(y=-2\sqrt{x-1}+3\): the endpoint \((0,0)\) tracks to \((1,3)\); domain \(x\ge 1\), range \(y\le 3\)
Transformed reciprocal curve with asymptotesThe image of y equals one over x after translating 2 right and 1 up, giving y equals one over (x minus 2) plus 1, with a vertical asymptote at x equals 2 and a horizontal asymptote at y equals 1. x y x=2 y=1
\(y=\dfrac1x\) translated \(2\) right and \(1\) up gives \(y=\dfrac{1}{x-2}+1\); the asymptotes move to \(x=2\) and \(y=1\)

The general form of a combination of transformations of \(y=f(x)\):

\[y=a\,f\big(n(x-h)\big)+k\]
y=af(n(x-h))+k

Where each point on the base graph is mapped by:

\[(x,\,y)\;\longmapsto\;\left(\frac{x}{n}+h,\;\;a\,y+k\right)\]
(x,y)↦(xn+h,ay+k)

For a transformed reciprocal, the asymptotes of \(y=\dfrac1x\) move to \(x=h\) and \(y=k\):

\[y=\frac{a}{n(x-h)}+k \quad\Rightarrow\quad x=h,\;\; y=k\]
y=an(x-h)+k
Domain and range. Horizontal transformations (\(n\) and \(h\)) act on the domain; vertical transformations (\(a\) and \(k\)) act on the range. E.g. \(y=\sqrt{x}\) (domain \(x\ge0\), range \(y\ge0\)) becomes \(y=-2\sqrt{x-1}+3\) with domain \(x\ge1\), range \(y\le3\).

How to apply a combination of transformations

  1. Identify the base function and the parameters. Match the rule to \(y=a\,f\big(n(x-h)\big)+k\), or read \(a,n,h,k\) from a described sequence of a dilation, a reflection and a translation.
  2. Apply in the correct order. Carry out the dilations and reflections first (the \(a\) and \(n\) factors), then the translations last (the \(h\) and \(k\) shifts).
  3. Track a key point. Send a distinctive feature through \((x,y)\mapsto\left(\dfrac{x}{n}+h,\;a\,y+k\right)\) — the turning point, endpoint, or crossing of the asymptotes.
  4. State domain, range and asymptotes. Read them from the image: horizontal changes adjust the domain, vertical changes adjust the range, and translations move any asymptotes to \(x=h,\;y=k\).
Order shortcut. Reading the rule \(y=a\,f\big(n(x-h)\big)+k\) directly — inside-out: subtract \(h\), scale by \(n\), then \(f\), then scale by \(a\), then add \(k\) — gives the transformations in the right order automatically.
Example 1 — dilation then translation
The graph of \(y=x^{2}\) is dilated by a factor of \(2\) from the \(x\)-axis, then translated \(3\) units in the positive direction of the \(y\)-axis. Find the image rule and its range.
Solution
Dilation by factor \(2\) from the \(x\)-axis — multiply the whole function by \(2\):
\(y\)\(=\)\(2x^{2}\)
Translation \(3\) up — add \(3\):
\(y\)\(=\)\(2x^{2}+3\)
The turning point \((0,0)\) maps to \((0,\,2(0)+3)=(0,3)\), so the range is \(y\ge3\).
\(\therefore\) image \(y=2x^{2}+3\); range \(y\ge3\)
y=2x2+3
Example 2 — reflection, dilation, translation
The graph of \(y=\sqrt{x}\) is dilated by a factor of \(3\) from the \(x\)-axis, reflected in the \(x\)-axis, then translated \(2\) units right and \(1\) unit up. Find the image, and state its domain and range.
Solution
Dilations and reflections first — factor \(3\) then reflect in the \(x\)-axis (\(a=-3\)):
\(y\)\(=\)\(-3\sqrt{x}\)
Translations last — replace \(x\) by \(x-2\) and add \(1\):
\(y\)\(=\)\(-3\sqrt{x-2}+1\)
Track the endpoint \((0,0)\): \(\left(\dfrac{0}{1}+2,\;-3(0)+1\right)=(2,1)\).
domain\(:\)\(x\ge2\)
range\(:\)\(y\le1\)
\(\therefore\) \(y=-3\sqrt{x-2}+1\); domain \(x\ge2\), range \(y\le1\)
y=-3x-2+1
Example 3 — describe the transformations
The parabola \(y=x^{2}\) is transformed to \(y=-(x+1)^{2}+4\). Describe the sequence of transformations, and state the turning point and the range.
Solution
Read \(a,n,h,k\) from the rule \(y=a(x-h)^{2}+k\):
\(a\)\(=\)\(-1\)
\(h\)\(=\)\(-1\)
\(k\)\(=\)\(4\)
So: reflect in the \(x\)-axis (\(a=-1\)), then translate \(1\) left and \(4\) up.
Turning point \((0,0)\mapsto(0+(-1),\;-1(0)+4)=(-1,4)\); the graph opens down, so the range is \(y\le4\).
\(\therefore\) reflect in \(x\)-axis, then left \(1\), up \(4\); turning point \((-1,4)\), range \(y\le4\)
Reflected and translated parabolaThe parabola y equals x squared with turning point at the origin, and its image y equals minus (x plus 1) squared plus 4 with turning point at (-1,4) opening downward. x y (0,0) (-1,4)
y=-(x+1)2+4
Example 4 — reciprocal with asymptotes
Find the image of \(y=\dfrac1x\) under a dilation of factor \(2\) from the \(x\)-axis, a reflection in the \(x\)-axis, then a translation \(3\) right and \(1\) up. State the asymptotes, domain and range.
Solution
Dilations and reflections first — factor \(2\) then reflect (\(a=-2\)):
\(y\)\(=\)\(\dfrac{2}{x}\)
\(y\)\(=\)\(-\dfrac{2}{x}\)
Translations last — replace \(x\) by \(x-3\) and add \(1\):
\(y\)\(=\)\(-\dfrac{2}{x-3}+1\)
The asymptotes \(x=0,\;y=0\) move to \(x=h,\;y=k\):
asymptotes\(:\)\(x=3,\;y=1\)
domain\(:\)\(x\neq3\)
range\(:\)\(y\neq1\)
\(\therefore\) \(y=-\dfrac{2}{x-3}+1\); asymptotes \(x=3,\,y=1\); domain \(x\neq3\), range \(y\neq1\)
y=-2x-3+1

Common pitfalls

Do the translation last, not first. A dilation applied after a translation scales the shift too. For \(y=2(x-1)^{2}\) the graph moves right \(1\) then is dilated; swapping the order gives a different image. Reading \(y=a\,f\big(n(x-h)\big)+k\) directly keeps the order right.
A factor of \(n\) inside means a dilation of \(\tfrac{1}{n}\), not \(n\). In \(y=f(2x)\) the graph is dilated by factor \(\dfrac12\) from the \(y\)-axis (squashed horizontally), not stretched by \(2\).
Match horizontal changes to the domain, vertical to the range. Reflecting \(y=\sqrt{x}\) in the \(x\)-axis flips the range to \(y\le0\); it does not change the domain \(x\ge0\). Only the \(x\)-transformations move the domain.

Frequently asked questions

What is the general form for a combination of transformations?

The image is \(y=a\,f\big(n(x-h)\big)+k\) from the base \(y=f(x)\): \(a\) is a dilation by \(|a|\) from the \(x\)-axis (reflection in the \(x\)-axis if \(a<0\)), \(n\) a dilation by \(\dfrac{1}{|n|}\) from the \(y\)-axis (reflection in the \(y\)-axis if \(n<0\)), \(h\) a horizontal translation and \(k\) a vertical translation.

In what order are combined transformations applied?

Dilations and reflections first (the \(a\) and \(n\) factors), then the translations last (the \(h\) and \(k\) shifts) — matching the rule \(y=a\,f\big(n(x-h)\big)+k\) read from the inside out.

How do you track a key point through several transformations?

A point \((x,y)\) maps to \(\left(\dfrac{x}{n}+h,\;a\,y+k\right)\). Apply this to a feature such as a turning point, an endpoint, or the crossing of the asymptotes to place it on the image.

How do transformations change the domain and range?

Horizontal transformations (\(n\) and \(h\)) change the domain; vertical transformations (\(a\) and \(k\)) change the range. E.g. \(y=\sqrt{x}\) becomes \(y=-2\sqrt{x-1}+3\) with domain \(x\ge1\) and range \(y\le3\).

What happens to the asymptotes of a reciprocal graph under a translation?

The asymptotes \(x=0\) and \(y=0\) of \(y=\dfrac1x\) are unmoved by dilations and reflections but shifted by a translation, so \(y=\dfrac{a}{n(x-h)}+k\) has asymptotes \(x=h\) and \(y=k\).

Does the order of transformations matter?

Yes — a dilation then a translation generally differs from the same translation then dilation, because a dilation scales the shift. Following dilations and reflections first, translations last, avoids the error.

More in Transformations
Translations
20 questions
Dilations and reflections
20 questions
Using transformations to sketch graphs
20 questions
Transformations of power functions
20 questions
Determining the rule for a function from its graph
20 questions