Transformations of Graphs – A Level Maths Year 1

1 The Four Transformations

DfE content area B (Algebra and functions) asks you to understand the effect of simple transformations on the graph of $y = \mathrm{f}(x)$, including sketching associated graphs: $y = a\mathrm{f}(x)$, $y = \mathrm{f}(x) + a$, $y = \mathrm{f}(x + a)$ and $y = \mathrm{f}(ax)$, and simple combinations of these.

The golden rule. A change made outside the function — to the output $\mathrm{f}(x)$ — affects the $y$-values and does what you expect. A change made inside the bracket — to the input $x$ — affects the $x$-values and does the opposite of what you might expect.

New graph	Transformation	Point $(p, q)$ goes to
$y = \mathrm{f}(x) + a$	Translation by $\begin{pmatrix}0\\a\end{pmatrix}$ — up by $a$	$(p,\; q + a)$
$y = \mathrm{f}(x + a)$	Translation by $\begin{pmatrix}-a\\0\end{pmatrix}$ — left by $a$	$(p - a,\; q)$
$y = a\mathrm{f}(x)$	Stretch parallel to the $y$-axis, scale factor $a$	$(p,\; aq)$
$y = \mathrm{f}(ax)$	Stretch parallel to the $x$-axis, scale factor $\dfrac1a$	$\left(\dfrac{p}{a},\; q\right)$
$y = -\mathrm{f}(x)$	Reflection in the $x$-axis	$(p,\; -q)$
$y = \mathrm{f}(-x)$	Reflection in the $y$-axis	$(-p,\; q)$

Use the correct vocabulary. Full marks need "translation by the vector $\begin{pmatrix}3\\0\end{pmatrix}$" or "stretch parallel to the $x$-axis, scale factor $\tfrac12$". Words like "move", "shift" or "squash" on their own will not score.

2 Translations: $y = \mathrm{f}(x) + a$ and $y = \mathrm{f}(x + a)$

Why $\mathrm{f}(x + a)$ moves left. On $y = \mathrm{f}(x + 3)$, the height that $\mathrm{f}$ used to produce at $x = 5$ is now produced when $x + 3 = 5$, i.e. at $x = 2$. Every feature happens $3$ units earlier — further left.

Worked Example 1 — Translating a parabola

$\mathrm{f}(x) = x^2$. Sketch $y = \mathrm{f}(x - 3) + 2$ and write down its equation and vertex.

①$\mathrm{f}(x - 3)$: translation by $\begin{pmatrix}3\\0\end{pmatrix}$ — right $3$ (because of the minus sign inside).

②$+2$ outside: translation by $\begin{pmatrix}0\\2\end{pmatrix}$ — up $2$.

③Combined: translation by $\begin{pmatrix}3\\2\end{pmatrix}$. The vertex $(0, 0)$ moves to $(3, 2)$.

④Equation $y = (x - 3)^2 + 2 = x^2 - 6x + 11$; $y$-intercept $(0, 11)$.

This is why completing the square (see Quadratic Functions) gives the vertex: $y = (x + p)^2 + q$ is $y = x^2$ translated by $\begin{pmatrix}-p\\q\end{pmatrix}$.

3 Stretches and Reflections: $y = a\mathrm{f}(x)$ and $y = \mathrm{f}(ax)$

What stays fixed

$y = a\mathrm{f}(x)$: points on the $x$-axis stay put (roots unchanged)
$y = \mathrm{f}(ax)$: points on the $y$-axis stay put ($y$-intercept unchanged)

Worked Example 2 — A vertical stretch

$\mathrm{f}(x) = x(x - 2)(x + 2)$. Describe and sketch $y = 2\mathrm{f}(x)$.

①Stretch parallel to the $y$-axis, scale factor $2$: every $y$-coordinate doubles.

②Roots $-2, 0, 2$ are unchanged (twice zero is still zero).

③The turning points become twice as high and twice as deep.

Worked Example 3 — A horizontal stretch

For the same $\mathrm{f}$, describe $y = \mathrm{f}(2x)$ and find where it crosses the $x$-axis.

①Stretch parallel to the $x$-axis, scale factor $\tfrac12$: every $x$-coordinate halves.

②Roots $-2, 0, 2$ become $-1, 0, 1$. Check: $\mathrm{f}(2x) = 2x(2x - 2)(2x + 2) = 8x(x - 1)(x + 1)$ ✓

③The heights of the turning points do not change; they move halfway towards the $y$-axis.

The inside factor works backwards. $\mathrm{f}(2x)$ is a stretch with scale factor $\tfrac12$, not $2$. And $\mathrm{f}\!\left(\tfrac{x}{3}\right)$ is a stretch with scale factor $3$.

Reflections are stretches with $a = -1$. $y = -\mathrm{f}(x)$ flips the graph in the $x$-axis; $y = \mathrm{f}(-x)$ flips it in the $y$-axis. For example, if $\mathrm{f}(x) = x^3 + 1$, then $\mathrm{f}(-x) = -x^3 + 1$ — a negative cubic with the same $y$-intercept.

4 Tracking Points and Asymptotes

Most exam questions give a sketch of $y = \mathrm{f}(x)$ with a few labelled points and ask for the images of those points. Apply the rule from the table to each one.

Worked Example 4 — Images of key points

The curve $y = \mathrm{f}(x)$ crosses the $x$-axis at $(-1, 0)$ and $(4, 0)$ and has a maximum at $(2, 5)$. Find the images of these three points on each graph below.

Graph	$(-1, 0)$	$(4, 0)$	Maximum $(2, 5)$
$y = \mathrm{f}(x) + 3$	$(-1, 3)$	$(4, 3)$	max $(2, 8)$
$y = \mathrm{f}(x + 3)$	$(-4, 0)$	$(1, 0)$	max $(-1, 5)$
$y = 2\mathrm{f}(x)$	$(-1, 0)$	$(4, 0)$	max $(2, 10)$
$y = \mathrm{f}(2x)$	$\left(-\tfrac12, 0\right)$	$(2, 0)$	max $(1, 5)$
$y = -\mathrm{f}(x)$	$(-1, 0)$	$(4, 0)$	min $(2, -5)$
$y = \mathrm{f}(-x)$	$(1, 0)$	$(-4, 0)$	max $(-2, 5)$

A reflection in the $x$-axis turns a maximum into a minimum. Say so — "minimum point" is part of the answer.

Asymptotes move too. A vertical asymptote $x = k$ behaves like an $x$-coordinate; a horizontal asymptote $y = k$ behaves like a $y$-coordinate. So $y = \dfrac{1}{x - 2} + 3$ — the graph of $y = \dfrac1x$ translated by $\begin{pmatrix}2\\3\end{pmatrix}$ — has asymptotes $x = 2$ and $y = 3$.

Worked Example 5 — A translated reciprocal graph

Sketch $y = \dfrac{1}{x - 2} + 3$, stating its asymptotes and where it meets the axes.

①Translation of $y = \dfrac1x$ by $\begin{pmatrix}2\\3\end{pmatrix}$: asymptotes $x = 2$ and $y = 3$.

②$y$-intercept: $x = 0$ gives $y = -\tfrac12 + 3 = \tfrac52$, so $\left(0, \tfrac52\right)$.

③$x$-intercept: $\dfrac{1}{x - 2} = -3 \Rightarrow x - 2 = -\tfrac13 \Rightarrow x = \tfrac53$, so $\left(\tfrac53, 0\right)$.

5 Combining Transformations

When two transformations act in the same direction, order matters. For $y = a\mathrm{f}(x) + b$, think "work out $\mathrm{f}(x)$, multiply by $a$, then add $b$": stretch first, then translate. When one acts on $x$ and the other on $y$, as in $y = a\mathrm{f}(x + c)$, the order does not matter.

Worked Example 6 — Order matters

The point $(3, 4)$ lies on $y = \mathrm{f}(x)$. Find its image on (a) $y = 2\mathrm{f}(x) + 1$ (b) $y = 2\left(\mathrm{f}(x) + 1\right)$ (c) $y = 3\mathrm{f}(x - 1)$.

①(a) Stretch: $y = 8$; then translate up $1$: $(3, 9)$.

②(b) Up $1$ first: $y = 5$; then stretch: $(3, 10)$. A different answer.

③(c) Right $1$ and $y$-values tripled: $(4, 12)$.

Worked Example 7 — Recognising a transformation

Describe a transformation that maps $y = x^2$ onto $y = x^2 + 6x + 11$.

①Complete the square: $x^2 + 6x + 11 = (x + 3)^2 + 2$.

②This is $\mathrm{f}(x + 3) + 2$ with $\mathrm{f}(x) = x^2$.

③Translation by the vector $\begin{pmatrix}-3\\2\end{pmatrix}$.

Worked Example 8 — Writing down a new equation

The graph of $y = \dfrac{1}{x^2}$ is reflected in the $x$-axis and then translated by $\begin{pmatrix}-1\\0\end{pmatrix}$. Write down the equation of the new graph and its asymptotes.

①Reflection in the $x$-axis: $y = -\dfrac{1}{x^2}$.

②Left $1$: replace $x$ by $x + 1$, giving $y = -\dfrac{1}{(x + 1)^2}$.

③Asymptotes: $x = -1$ and $y = 0$. The whole curve lies below the $x$-axis.

Transforming an equation is always "replace $x$ by …" or "replace $y$ by …". Replacing $x$ by $x + 1$ moves the graph left $1$.

Some stretches can be done two ways. $y = (2x)^2 = 4x^2$, so for $y = x^2$ a stretch parallel to the $x$-axis, scale factor $\tfrac12$, gives the same graph as a stretch parallel to the $y$-axis, scale factor $4$. Either description earns the marks.

6 Quick Reference

$\mathrm{f}(x) + a$

Translation $\begin{pmatrix}0\\a\end{pmatrix}$: up $a$.

$\mathrm{f}(x + a)$

Translation $\begin{pmatrix}-a\\0\end{pmatrix}$: left $a$.

$a\mathrm{f}(x)$

Stretch parallel to $y$-axis, s.f. $a$. Roots fixed.

$\mathrm{f}(ax)$

Stretch parallel to $x$-axis, s.f. $\tfrac1a$. $y$-intercept fixed.

$-\mathrm{f}(x)$

Reflect in the $x$-axis; max $\leftrightarrow$ min.

$\mathrm{f}(-x)$

Reflect in the $y$-axis.

Inside vs outside

Outside acts on $y$ as expected; inside acts on $x$ the opposite way.

Asymptotes

Transform them like coordinates.

$a\mathrm{f}(x) + b$

Stretch first, then translate.

7 Practice Questions

Question 1

Describe fully the single transformation that maps $y = x^3$ onto (a) $y = x^3 - 5$ (b) $y = (x + 4)^3$.

▶ Show solution

(a) $\mathrm{f}(x) - 5$: translation by $\begin{pmatrix}0\\-5\end{pmatrix}$.

(b) $\mathrm{f}(x + 4)$: translation by $\begin{pmatrix}-4\\0\end{pmatrix}$.

Question 2

The point $(4, -2)$ lies on $y = \mathrm{f}(x)$. Write down its image on (a) $y = \mathrm{f}(x - 3)$ (b) $y = \mathrm{f}(x) + 3$ (c) $y = 3\mathrm{f}(x)$ (d) $y = \mathrm{f}(3x)$ (e) $y = \mathrm{f}(-x)$ (f) $y = -\mathrm{f}(x)$.

▶ Show solution

(a) $(7, -2)$ (b) $(4, 1)$ (c) $(4, -6)$

(d) $\left(\tfrac43, -2\right)$ (e) $(-4, -2)$ (f) $(4, 2)$

Question 3

$\mathrm{f}(x) = x^2 - 4x$. Find and simplify $\mathrm{f}(x + 2)$, and explain the result in terms of the vertex of each graph.

▶ Show solution

$\mathrm{f}(x + 2) = (x + 2)^2 - 4(x + 2) = x^2 + 4x + 4 - 4x - 8 = x^2 - 4$.

$\mathrm{f}(x) = (x - 2)^2 - 4$ has vertex $(2, -4)$. Translating left $2$ moves it to $(0, -4)$, the vertex of $y = x^2 - 4$ ✓

Question 4

The graph of $y = \dfrac1x$ is translated by $\begin{pmatrix}-1\\2\end{pmatrix}$. Write down its new equation and asymptotes, and find where it crosses the axes.

▶ Show solution

$y = \dfrac{1}{x + 1} + 2$; asymptotes $x = -1$ and $y = 2$.

$y$-intercept: $\dfrac11 + 2 = 3$, so $(0, 3)$.

$x$-intercept: $\dfrac{1}{x + 1} = -2 \Rightarrow x + 1 = -\tfrac12 \Rightarrow x = -\tfrac32$, so $\left(-\tfrac32, 0\right)$.

Question 5

$\mathrm{f}(x) = (x - 1)(x - 4)$. On the same axes sketch $y = \mathrm{f}(x)$ and $y = \mathrm{f}\!\left(\tfrac12 x\right)$, labelling where each meets the axes.

▶ Show solution

$y = \mathrm{f}(x)$: $\cup$-shaped parabola through $(1, 0)$, $(4, 0)$ and $(0, 4)$.

$y = \mathrm{f}\!\left(\tfrac12 x\right)$ is a stretch parallel to the $x$-axis, scale factor $2$: roots at $(2, 0)$ and $(8, 0)$; the $y$-intercept stays at $(0, 4)$.

Check: $\left(\tfrac x2 - 1\right)\left(\tfrac x2 - 4\right)$ is zero at $x = 2$ and $x = 8$ ✓

Question 6

Sketch $y = \dfrac{3}{(x - 2)^2} - 1$, stating the equations of the asymptotes and the exact coordinates of any points where it meets the axes.

▶ Show solution

This is $y = \dfrac{3}{x^2}$ translated by $\begin{pmatrix}2\\-1\end{pmatrix}$. Asymptotes $x = 2$ and $y = -1$.

$y$-intercept: $\dfrac34 - 1 = -\dfrac14$, so $\left(0, -\tfrac14\right)$.

$x$-intercepts: $\dfrac{3}{(x - 2)^2} = 1 \Rightarrow (x - 2)^2 = 3 \Rightarrow x = 2 \pm \sqrt3$.

Both branches lie above $y = -1$, rising steeply towards $x = 2$.

Question 7

Describe two different single transformations that map $y = x^2$ onto $y = 9x^2$.

▶ Show solution

$9x^2 = 9\mathrm{f}(x)$: stretch parallel to the $y$-axis, scale factor $9$.

$9x^2 = (3x)^2 = \mathrm{f}(3x)$: stretch parallel to the $x$-axis, scale factor $\tfrac13$.

Question 8

The curve $y = \mathrm{f}(x)$ has a maximum point at $(-1, 4)$. Write down the coordinates and nature of the corresponding turning point on (a) $y = 2\mathrm{f}(x) - 3$ (b) $y = \mathrm{f}(x + 2) + 1$ (c) $y = -\mathrm{f}(x)$ (d) $y = \mathrm{f}(-x)$.

▶ Show solution

(a) Stretch then translate: $y = 2(4) - 3 = 5$. Maximum at $(-1, 5)$.

(b) Left $2$, up $1$: maximum at $(-3, 5)$.

(c) Reflection in the $x$-axis: minimum at $(-1, -4)$.

(d) Reflection in the $y$-axis: maximum at $(1, 4)$.

Question 9

The curve $y = x(x - 3)^2$ is translated by $\begin{pmatrix}2\\0\end{pmatrix}$. Write down the equation of the new curve, its $y$-intercept, and where it meets the $x$-axis, saying whether it crosses or touches.

▶ Show solution

Replace $x$ by $x - 2$: $y = (x - 2)(x - 5)^2$.

$y$-intercept: $(-2)(-5)^2 = -50$, so $(0, -50)$.

The original roots $0$ (cross) and $3$ (touch) move right $2$: crosses at $(2, 0)$, touches at $(5, 0)$.

Question 10

$\mathrm{f}(x) = x^2 - 6x + 5$.

(a) Write $\mathrm{f}(x)$ in the form $(x - p)^2 + q$. (b) Describe the transformation that maps $y = x^2$ onto $y = \mathrm{f}(x)$. (c) The graph of $y = \mathrm{f}(x) + k$ touches the $x$-axis. Find $k$. (d) Sketch $y = \mathrm{f}(2x)$, giving the coordinates of its vertex and of the points where it meets the axes.

▶ Show solution

(a) $x^2 - 6x + 5 = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4$, so $p = 3$, $q = -4$.

(b) Translation by $\begin{pmatrix}3\\-4\end{pmatrix}$.

(c) The vertex is $(3, -4)$. To touch the $x$-axis the vertex must move up to $(3, 0)$, so $k = 4$. (Check: $\mathrm{f}(x) + 4 = (x - 3)^2$ ✓)

(d) $\mathrm{f}(x) = (x - 1)(x - 5)$ has roots $1$ and $5$. Stretch parallel to the $x$-axis, scale factor $\tfrac12$:

roots at $\left(\tfrac12, 0\right)$ and $\left(\tfrac52, 0\right)$; vertex $\left(\tfrac32, -4\right)$; $y$-intercept unchanged at $(0, 5)$.

Check: $\mathrm{f}(2x) = 4x^2 - 12x + 5 = (2x - 1)(2x - 5)$ ✓

Transformations of Graphs · A Level Maths Year 1 · Created with MathJax