Transformations of graphs

AS · P1 · 12 min

Changing the equation of a graph in a simple way moves or stretches the graph in a predictable way. Knowing the four basic transformations lets you sketch $y = 3\sin 2x + 1$ or $y = 2(x - 1)^2 - 5$ without plotting a single point, and lets you describe a change of graph in the precise words Cambridge demands: translation, reflection, stretch. Transformation questions appear on most papers, often applied to a function you have just met, or to a graph given only by its features.

The four basic transformations

Start with any graph $y = f(x)$ . There are two places you can change the equation: outside $f$ (changing the output) or inside $f$ (changing the input).

Key result

New equation	Transformation	Point $(p, q)$ moves to
$y = f(x) + a$	translation by $\begin{pmatrix} 0 \\ a \end{pmatrix}$	$(p,\ q + a)$
$y = f(x + a)$	translation by $\begin{pmatrix} -a \\ 0 \end{pmatrix}$	$(p - a,\ q)$
$y = af(x)$	stretch parallel to the $y$ -axis, scale factor $a$	$(p,\ aq)$
$y = f(ax)$	stretch parallel to the $x$ -axis, scale factor $\dfrac{1}{a}$	$\left(\dfrac{p}{a},\ q\right)$
$y = -f(x)$	reflection in the $x$ -axis	$(p,\ -q)$
$y = f(-x)$	reflection in the $y$ -axis	$(-p,\ q)$

Outside changes act on $y$ and do what they say. Inside changes act on $x$ and do the opposite.

Outside changes: the output

$y = f(x) + 3$ takes every $y$ -value of the original and adds $3$ , so the whole graph moves up $3$ . $y = 2f(x)$ doubles every $y$ -value, so the graph is stretched away from the $x$ -axis by factor $2$ . Points on the $x$ -axis ( $y = 0$ ) stay where they are.

Inside changes: the input

$y = f(x - 2)$ moves the graph right by $2$ , not left. Here is why. The original graph has some feature (a vertex, say) where the input to $f$ is $0$ . In $f(x - 2)$ the input is $0$ when $x = 2$ . So the feature now happens at $x = 2$ : two units to the right.

Similarly $y = f(2x)$ : the input $2x$ reaches any given value at half the $x$ it used to, so the graph is squashed towards the $y$ -axis by factor $\tfrac{1}{2}$ . Points on the $y$ -axis ( $x = 0$ ) stay where they are.

y = x^2 y = (x - 2)^2 y = x^2 + 1

The graphs above are $y = x^2$ , $y = (x - 2)^2$ (translated $2$ right) and $y = x^2 + 1$ (translated $1$ up).

y = sin x y = 2 sin x y = sin 2x

Here $y = \sin x$ , $y = 2\sin x$ (stretch parallel to the $y$ -axis, factor $2$ : the peaks rise to $2$ ) and $y = \sin 2x$ (stretch parallel to the $x$ -axis, factor $\tfrac{1}{2}$ : two full waves in the space of one).

Describing transformations in words

Cambridge awards marks for the exact vocabulary. Each description needs the type and all its details.

Type	Details you must give	Example wording
Translation	the vector	translation by $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$
Stretch	the direction and the scale factor	stretch parallel to the $y$ -axis with scale factor $3$
Reflection	the mirror line	reflection in the $x$ -axis

"Stretch in the $x$ -direction" and "stretch parallel to the $x$ -axis" both mean the same and are both accepted. Words such as "shift", "move", "squash" or "enlarge" are not accepted.

A combined translation

Describe the transformation that maps $y = x^2$ onto $y = (x - 3)^2 + 2$ .

Solution

Replacing $x$ by $x - 3$ translates the graph $3$ units in the positive $x$ -direction; adding $2$ outside translates it $2$ units in the positive $y$ -direction. Together:

\text{a translation by } \begin{pmatrix} 3 \\ 2 \end{pmatrix}

The vertex moves from $(0, 0)$ to $(3, 2)$ , which agrees with the completed square form.

Images of a point

The point $P(4, -2)$ lies on the curve $y = f(x)$ . State the coordinates of the image of $P$ on each of the following curves.

(a) $y = 3f(x)$ (b) $y = f(x - 1)$ (c) $y = f(2x)$ (d) $y = -f(x) + 5$ (e) $y = f(-x)$

Solution

(a) Stretch parallel to the $y$ -axis, factor $3$ : $(4, -6)$ .

(b) Translation by $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ : $(5, -2)$ .

(d) Reflection in the $x$ -axis gives $(4, 2)$ ; then translation by $\begin{pmatrix} 0 \\ 5 \end{pmatrix}$ gives $(4, 7)$ .

(e) Reflection in the $y$ -axis: $(-4, -2)$ .

Check (c): on $y = f(2x)$ at $x = 2$ , the input is $f(4) = -2$ . Correct.

Combining transformations

When two transformations act in different directions (one on $x$ , one on $y$ ), the order does not matter. When they act in the same direction, it does.

Outside: follow the order of operations

$y = 2f(x) + 3$ means: take $f(x)$ , multiply by $2$ , then add $3$ . So the transformations are, in order:

stretch parallel to the $y$ -axis, scale factor $2$ ;
translation by $\begin{pmatrix} 0 \\ 3 \end{pmatrix}$ .

Doing them the other way round would give $2\left(f(x) + 3\right) = 2f(x) + 6$ , which is different.

Inside: reverse the order of operations on $x$

$y = f(2x - 4)$ : write the inside as $2(x - 2)$ . Then two valid descriptions are:

translation by $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ , then stretch parallel to the $x$ -axis with factor $\tfrac{1}{2}$ (check: translating gives $f(x - 4)$ ; stretching replaces $x$ by $2x$ , giving $f(2x - 4)$ );
stretch parallel to the $x$ -axis with factor $\tfrac{1}{2}$ , then translation by $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ (stretching gives $f(2x)$ ; translating replaces $x$ by $x - 2$ , giving $f(2(x - 2))$ ).

Tip

To check any sequence, apply each step to the equation in turn. A translation by $\begin{pmatrix} h \\ 0 \end{pmatrix}$ means replace $x$ by $x - h$ . A stretch parallel to the $x$ -axis with factor $k$ means replace $x$ by $\dfrac{x}{k}$ . For the $y$ -direction, apply the change to the whole right-hand side.

Finding the equation after a sequence of transformations

Start with $y = f(x)$ .
Apply the first transformation to the equation, using the replacement rules.
Apply the second transformation to the new equation.
Simplify, and check with one point (for example the vertex or an intercept).

A trigonometric combination

Describe a sequence of transformations that maps the graph of $y = \cos x$ onto the graph of $y = 1 - \cos 2x$ .

Solution

Build $1 - \cos 2x$ from $\cos x$ step by step:

\cos x \ \longrightarrow\ \cos 2x \ \longrightarrow\ -\cos 2x \ \longrightarrow\ 1 - \cos 2x

Stretch parallel to the $x$ -axis with scale factor $\tfrac{1}{2}$ .
Reflection in the $x$ -axis.
Translation by $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ .

Step 1 can be done at any point in the sequence (it is the only $x$ -direction change), but steps 2 and 3 must be in this order.

y = cos x y = 1 - cos 2x

The equation after a stretch and a translation

The function $f$ is defined by $f(x) = x^2 - 6x + 11$ for $x \in \mathbb{R}$ . The graph of $y = f(x)$ is transformed to the graph of $y = g(x)$ by a stretch parallel to the $y$ -axis with scale factor $2$ , followed by a translation by $\begin{pmatrix} 1 \\ -5 \end{pmatrix}$ . Find $g(x)$ in the form $ax^2 + bx + c$ .

Solution

Stretch: $y = 2f(x)$ . Then translate: replace $x$ by $x - 1$ and subtract $5$ :

g(x) = 2f(x - 1) - 5

Now $f(x) = (x - 3)^2 + 2$ , so $f(x - 1) = (x - 4)^2 + 2$ and

g(x) = 2\left[(x - 4)^2 + 2\right] - 5 = 2(x - 4)^2 - 1 = 2x^2 - 16x + 31

Check with the vertex: $f$ has vertex $(3, 2)$ . The stretch sends it to $(3, 4)$ and the translation to $(4, -1)$ . And $g(x) = 2(x - 4)^2 - 1$ has vertex $(4, -1)$ . Correct.

Identifying transformations from the equations

The function $f$ is defined by $f(x) = x^2 - 4x + 7$ , and $g(x) = x^2 + 2x + 2$ . Describe fully the single transformation that maps $y = f(x)$ onto $y = g(x)$ .

Solution

Complete the square on both:

f(x) = (x - 2)^2 + 3, \qquad g(x) = (x + 1)^2 + 1

The vertex moves from $(2, 3)$ to $(-1, 1)$ and the shape is unchanged (both have $x^2$ coefficient $1$ ). So the transformation is a translation by

\begin{pmatrix} -1 - 2 \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} -3 \\ -2 \end{pmatrix}

Check: $f(x + 3) - 2 = (x + 1)^2 + 3 - 2 = (x + 1)^2 + 1 = g(x)$ . Correct.

A graph given by its features

The graph of $y = f(x)$ has a maximum point at $(2, 5)$ and crosses the $x$ -axis at $(-1, 0)$ and $(4, 0)$ . The graph is transformed to $y = 3 - 2f(x)$ . Describe the transformations in order and find the images of the three given points. State whether the image of $(2, 5)$ is a maximum or minimum.

Solution

$3 - 2f(x) = -2f(x) + 3$ : multiply $f(x)$ by $-2$ , then add $3$ . So:

stretch parallel to the $y$ -axis with scale factor $2$ ;
reflection in the $x$ -axis;
translation by $\begin{pmatrix} 0 \\ 3 \end{pmatrix}$ .

(Steps 1 and 2 can be swapped, since both are multiplications; the translation must come last.)

Each point $(p, q)$ goes to $(p, 3 - 2q)$ :

$(2, 5) \to (2, -7)$ ;
$(-1, 0) \to (-1, 3)$ ;
$(4, 0) \to (4, 3)$ .

The reflection turns the curve upside down, so the maximum becomes a minimum at $(2, -7)$ .

Watch out

Inside changes the wrong way. $f(x + 3)$ moves the graph left $3$ ; $f(3x)$ squashes it by factor $\tfrac{1}{3}$ , it does not stretch it by $3$ .

Translation vector with the wrong sign or order. The vector is $\begin{pmatrix} x\text{-shift} \\ y\text{-shift} \end{pmatrix}$ . $y = f(x - 2) + 5$ is a translation by $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ .

Incomplete descriptions. "A stretch of factor 2" is missing the direction. "A translation of 3" is missing the vector. Each loses the mark.

Wrong order in the $y$ -direction. $2f(x) + 3$ is "stretch, then translate by $3$ "; translating first would need a translation by $\tfrac{3}{2}$ .

Translating before reflecting in $f(-x)$ problems. Reflecting $y = f(x)$ in the $y$ -axis and then translating by $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ gives $y = f(-(x - 2)) = f(2 - x)$ , not $f(-x - 2)$ .

Exam tip

Use the words translation, stretch and reflection, with the vector, the direction and scale factor, or the mirror line. Typically one mark per transformation, and only if every detail is correct.
If the question says "a sequence of transformations", give them in a valid order and state the order (first, then).
Questions can involve any function: quadratics, trigonometric functions, or "other graphs with given features" such as a curve described only by a few points. Track key points through each step.
Always check your answer with one point. It takes a few seconds and catches most sign errors.

Summary

$f(x) + a$ : translation $\begin{pmatrix} 0 \\ a \end{pmatrix}$ . $f(x + a)$ : translation $\begin{pmatrix} -a \\ 0 \end{pmatrix}$ .
$af(x)$ : stretch parallel to the $y$ -axis, factor $a$ . $f(ax)$ : stretch parallel to the $x$ -axis, factor $\tfrac{1}{a}$ .
$-f(x)$ : reflection in the $x$ -axis. $f(-x)$ : reflection in the $y$ -axis.
Outside changes act on $y$ as written; inside changes act on $x$ in the opposite way.
In the $y$ -direction, the order follows the arithmetic: multiply, then add.
Replacement rules: translate by $h$ in $x$ , replace $x$ by $x - h$ ; stretch by $k$ in $x$ , replace $x$ by $\tfrac{x}{k}$ .
Describe fully: type plus vector, direction and factor, or mirror line.

Practice questions

Question

Describe the transformation that maps $y = x^2$ onto $y = (x + 4)^2 - 1$ .
Describe the transformation that maps $y = f(x)$ onto $y = f\left(\tfrac{1}{3}x\right)$ .
Describe the transformations that map $y = f(x)$ onto (a) $y = -f(x)$ , (b) $y = f(-x)$ .
The point $(2, 6)$ lies on $y = f(x)$ . Find its image on (a) $y = f(x) - 4$ , (b) $y = f(x + 3)$ , (c) $y = \tfrac{1}{2}f(x)$ , (d) $y = f(2x)$ , (e) $y = f(-x)$ .
$f(x) = x^2 + 2x$ . The graph of $y = f(x)$ is translated by $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ . Find the equation of the new graph in the form $y = x^2 + bx + c$ .
Describe a sequence of transformations that maps $y = \sin x$ onto $y = 2 + \sin\left(\tfrac{1}{2}x\right)$ .
Describe, in order, transformations that map $y = f(x)$ onto $y = 2f(x) + 3$ , and find the image of $(-1, 4)$ .
The graph of $y = f(x)$ is reflected in the $y$ -axis and then translated by $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ . Find the equation of the resulting graph in terms of $f$ .
$f(x) = 2x^2 - 8x + 5$ and $g(x) = 2x^2 + 4x - 1$ . Find the translation that maps $y = f(x)$ onto $y = g(x)$ .
The graph of $y = f(x)$ is transformed to $y = 4 - f(2x)$ . (a) Describe a sequence of transformations. (b) The point $(6, 1)$ is on $y = f(x)$ . Find its image. (c) If $y = f(x)$ has an asymptote $y = 2$ , state the equation of the corresponding asymptote of $y = 4 - f(2x)$ .

Answers

Translation by $\begin{pmatrix} -4 \\ -1 \end{pmatrix}$ .
Stretch parallel to the $x$ -axis with scale factor $3$ .
(a) Reflection in the $x$ -axis. (b) Reflection in the $y$ -axis.
(a) $(2, 2)$ . (b) $(-1, 6)$ . (c) $(2, 3)$ . (d) $(1, 6)$ . (e) $(-2, 6)$ .
$f(x - 3) = (x - 3)^2 + 2(x - 3) = x^2 - 6x + 9 + 2x - 6 = x^2 - 4x + 3$ .
Stretch parallel to the $x$ -axis with scale factor $2$ , and translation by $\begin{pmatrix} 0 \\ 2 \end{pmatrix}$ (in either order, as they act in different directions).
Stretch parallel to the $y$ -axis with scale factor $2$ , then translation by $\begin{pmatrix} 0 \\ 3 \end{pmatrix}$ . Image: $(-1, 2(4) + 3) = (-1, 11)$ .
Reflection gives $y = f(-x)$ . Translating replaces $x$ by $x - 2$ : $y = f(-(x - 2)) = f(2 - x)$ .
$f(x) = 2(x - 2)^2 - 3$ , vertex $(2, -3)$ ; $g(x) = 2(x + 1)^2 - 3$ , vertex $(-1, -3)$ . Same shape, so translation by $\begin{pmatrix} -3 \\ 0 \end{pmatrix}$ . Check: $f(x + 3) = 2(x + 1)^2 - 3 = g(x)$ .
(a) Stretch parallel to the $x$ -axis with scale factor $\tfrac{1}{2}$ ; reflection in the $x$ -axis; translation by $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ (the reflection must come before the translation). (b) $(6, 1) \to (3, 1) \to (3, -1) \to (3, 3)$ . (c) Horizontal asymptotes are unchanged by the $x$ -stretch; reflection gives $y = -2$ ; translation gives $y = 2$ . So the asymptote is $y = 4 - 2 = 2$ .

The four basic transformations

Outside changes: the output

Inside changes: the input

Describing transformations in words

Combining transformations

Outside: follow the order of operations

Inside: reverse the order of operations on xxx

Practice questions

Console

Inside: reverse the order of operations on $x$