Stretches — scaling horizontally and vertically — Mathematics

Before this

Reflections of graphs
Translations of graphs
Enlargement of a shape · coming soon

This leads to

Combined transformations
Graphs of sin, cos and tan · coming soon
Harmonics on a string

Multiply by 2, and the graph gets narrower

Take $f (x) = x^{2} - 4$ . Its graph is a U-shape that crosses the $x$ -axis at $- 2$ and $2$ . So it is 4 units wide at the $x$ -axis.

Now put a 2 inside the bracket: $y = f (2 x)$ . Most students expect the graph to get twice as wide, because they multiplied by 2.

Test it. The new function is $f (2 x) = (2 x)^{2} - 4 = 4 x^{2} - 4$ . It crosses the $x$ -axis where $4 x^{2} = 4$ , so at $x = - 1$ and $x = 1$ . The graph is now only 2 units wide. Multiplying by 2 made it half as wide.

This is the surprise you met with translations, in a new form. This lecture shows why it happens, and the other kind of stretch, which does exactly what you expect.

Two places, one new change

The translations lecture showed that there are only two places to change a function: the output, after the function has worked, or the input, before it works. That lecture added a number. The reflections lecture changed a sign. This lecture multiplies.

Multiplying the output gives $y = a f (x)$ . Multiplying the input gives $y = f (a x)$ . They give two different stretches, and they behave in two different ways.

Multiplying the output: a vertical stretch

Compare $y = f (x)$ with $y = 2 f (x)$ and $y = \frac{1}{2} f (x)$ , for $f (x) = x^{2} - 4$ .

$x$	$- 3$	$- 1$	$0$	$1$	$3$
$f (x)$	5	−3	−4	−3	5
$2 f (x)$	10	−6	−8	−6	10
$\frac{1}{2} f (x)$	2.5	−1.5	−2	−1.5	2.5

Every output has been multiplied by 2, or by $\frac{1}{2}$ . So each point $(p, q)$ moves to $(p, 2 q)$ . A point 5 above the $x$ -axis moves to 10 above it. A point 3 below moves to 6 below.

This is called a vertical stretch with scale factor 2. Some books say "a stretch parallel to the $y$ -axis", because every point moves in the direction of the $y$ -axis. With scale factor $\frac{1}{2}$ , every point moves towards the $x$ -axis, to half its distance. In everyday words the graph is squashed. In mathematics it is still called a stretch, with a scale factor between 0 and 1.

The graph of

y = f (x)

, where

f (x) = x^{2} - 4

(ink), with

y = 2 f (x)

(teal) and

y = \frac{1}{2} f (x)

(dashed teal). The arrows show

(1, - 3)

moving to

(1, - 6)

, and

(2.5, 2.25)

moving to

(2.5, 4.5)

. The roots

- 2

and

2

do not move.

Look at the two roots. They did not move, because any point on the $x$ -axis has height 0, and $2 \times 0 = 0$ . So in a vertical stretch the $x$ -axis is the fixed line. Everything else moves away from it or towards it.

Multiplying the input: a horizontal stretch

Now put the 2 inside: compare $y = f (x)$ with $y = f (2 x)$ . The function does not work on the input you give it. It works on double that input. At $x = 1$ , the new function works on 2, so it gives $f (2) = 0$ . At $x = 1.5$ , it gives $f (3) = 5$ .

$x$	$- 1.5$	$- 1$	$- 0.5$	$0$	$0.5$	$1$	$1.5$
$f (2 x)$	5	0	−3	−4	−3	0	5

Look for outputs you recognise. The old 5 at $x = 3$ now appears at $x = 1.5$ . The old 0 at $x = 2$ now appears at $x = 1$ . Every output from the first table appears again, at half the input.

This is the hard part of the lecture. Spend time here. Here is the reason, written once for any function $f$ and any number $a$ that is not zero.

Take a point $(p, q)$ on the graph of $y = f (x)$ . This means that $q = f (p)$ . Where does $y = f (a x)$ give the same output $q$ ? It needs $f (a x) = f (p)$ . That happens when $a x = p$ , so when $x = \frac{p}{a}$ .

(p, q) on y = f (x) ⟹ (\frac{p}{a}, q) on y = f (a x)

The translations lecture promised that its main sentence would come back for stretches. Here it is again. A change to the input moves the graph the opposite way, because the input must be changed back before the function sees the number it needs. For $f (2 x)$ , the change is "multiply by 2". The only way to undo that is to divide by 2. So the function sees $p$ when you give it $\frac{p}{2}$ , and every point moves to half its distance from the $y$ -axis.

So $y = f (a x)$ is a horizontal stretch with scale factor $\frac{1}{a}$ . Some books say "a stretch parallel to the $x$ -axis". With $a = 2$ the scale factor is $\frac{1}{2}$ , and the graph is half as wide. With $a = \frac{1}{2}$ the scale factor is 2, and the graph is twice as wide.

The graph of

y = f (x)

, where

f (x) = x^{2} - 4

(ink), with

y = f (2 x)

(teal) and

y = f (\frac{1}{2} x)

(dashed teal). For

f (2 x)

, the roots move from

\pm 2

\pm 1

, and

(3, 5)

moves to

(1.5, 5)

. For

f (\frac{1}{2} x)

, the roots move out to

\pm 4

. The lowest point

(0, - 4)

is on the

y

-axis, so it does not move.

The lowest point $(0, - 4)$ did not move. It lies on the $y$ -axis, where the position is 0, and $\frac{0}{2} = 0$ . So in a horizontal stretch the $y$ -axis is the fixed line. Every other point moves towards it or away from it, and keeps its height.

That is it. That is the whole idea. A number multiplying the output stretches every height by that number. A number multiplying the input stretches every position by one over that number, because the input has to be divided back.

Stretches of graphs

$y = a f (x)$ is a vertical stretch with scale factor $a$ . Every point $(p, q)$ moves to $(p, a q)$ . The $x$ -axis does not move.

$y = f (a x)$ is a horizontal stretch with scale factor $\frac{1}{a}$ . Every point $(p, q)$ moves to $(\frac{p}{a}, q)$ . The $y$ -axis does not move.

Try both stretches on four graphs.

Graph builder — a number outside, a number inside

Choose a graph. Move the first slider to multiply the output, and the second to multiply the input. Watch the two marked points, and read where each one goes.

Starting graph:

Number outside, a in y = a·f(bx): 1

Number inside, b in y = a·f(bx): 1

Try:

The two stretches by 2, written out for the four graphs of the graph builder. The number outside multiplies every height. The number inside divides every position.

y = x² − 4. 2f(x): y = 2x² − 8, roots stay at ±2, (0, −4) → (0, −8). f(2x): y = 4x² − 4, roots move to ±1, (0, −4) stays.
y = x³ − 4x. 2f(x): y = 2x³ − 8x, (1, −3) → (1, −6). f(2x): y = 8x³ − 8x, (1, −3) → (0.5, −3), roots move from ±2 to ±1.
y = √x. 2f(x): y = 2√x, (4, 2) → (4, 4). f(2x): y = √(2x), (4, 2) → (2, 2).
y = x² − 2x. 2f(x): y = 2x² − 4x, (1, −1) → (1, −2). f(2x): y = 4x² − 4x, (1, −1) → (0.5, −1), roots 0 and 2 move to 0 and 1.

What should you have seen? The number outside moved every point straight up or down, and the points on the $x$ -axis stayed. The number inside moved every point straight across, and the points on the $y$ -axis stayed. A bigger number inside always made the graph narrower. And a negative number gave a stretch and a reflection together. The section after next explains why.

One graph, two descriptions

Stop and think

For $g (x) = x^{2}$ , is the graph of $y = g (2 x)$ a horizontal stretch or a vertical stretch?

Show the answer

Both. $g (2 x) = (2 x)^{2} = 4 x^{2}$ , and $4 x^{2} = 4 g (x)$ . So the same graph is a horizontal stretch with scale factor $\frac{1}{2}$ , and also a vertical stretch with scale factor 4. Both descriptions are correct. This does not break the rules. It is a fact about $x^{2}$ : squeezing it sideways and pulling it upwards give the same curve.

This only works for some functions. For $f (x) = x^{2} - 4$ , the horizontal stretch gives $f (2 x) = 4 x^{2} - 4$ . The vertical stretch by 4 gives $4 f (x) = 4 x^{2} - 16$ . These are different graphs: the first has its lowest point at $(0, - 4)$ , and the second at $(0, - 16)$ . The constant $- 4$ is not multiplied by the inside 2, but it is multiplied by the outside 4. So always substitute, and never assume that two stretches are the same.

A reflection is a stretch with scale factor −1

Look again at the last lecture. In $y = - f (x)$ , every output is multiplied by $- 1$ . That is $y = a f (x)$ with $a = - 1$ . The point $(p, q)$ moves to $(p, - q)$ , exactly as the stretch rule says. In $y = f (- x)$ , the input is multiplied by $- 1$ , so $a = - 1$ and the scale factor is $\frac{1}{- 1} = - 1$ . The point moves to $(- p, q)$ .

So a reflection is a stretch with scale factor $- 1$ . And any negative scale factor is a stretch and a reflection together. For example, $y = - 3 f (x)$ multiplies every height by 3 and flips it below the $x$ -axis.

Worked examples

Worked Example 1 — A vertical stretch

$f (x) = x^{2} - 4$ . Describe the graph of $y = 2 f (x)$ . Find its equation, its roots and its lowest point.

The 2 multiplies the output, so this is a vertical stretch with scale factor 2. Every $y$ -coordinate doubles.

The roots $(- 2, 0)$ and $(2, 0)$ stay, because $2 \times 0 = 0$ . The lowest point $(0, - 4)$ moves to $(0, - 8)$ .

The equation multiplies the whole output, so put it in a bracket: $2 f (x) = 2 (x^{2} - 4) = 2 x^{2} - 8$ .

Check: $2 x^{2} - 8 = 0$ gives $x^{2} = 4$ , so $x = \pm 2$ . ✓ At $x = 0$ , $y = - 8$ . ✓ The graph is deeper, and it still crosses the axis in the same places.

Worked Example 2 — Two horizontal stretches

$f (x) = x^{2} - 2 x$ . Describe $y = f (2 x)$ and $y = f (\frac{1}{2} x)$ . Find the lowest point of each.

Complete the square: $f (x) = (x - 1)^{2} - 1$ . So the lowest point of $f$ is $(1, - 1)$ , and the roots are 0 and 2.

$y = f (2 x)$ is a horizontal stretch with scale factor $\frac{1}{2}$ . Every $x$ -coordinate halves, and every $y$ -coordinate stays. The lowest point moves to $(0.5, - 1)$ . The roots move to 0 and 1.

$y = f (\frac{1}{2} x)$ is a horizontal stretch with scale factor 2. The lowest point moves to $(2, - 1)$ . The roots move to 0 and 4.

Check the first by substitution: $f (2 x) = (2 x)^{2} - 2 (2 x) = 4 x^{2} - 4 x = 4 (x - 0.5)^{2} - 1$ . Its lowest point is $(0.5, - 1)$ . ✓ Notice that the lowest point moved, because it was not on the $y$ -axis. The root at 0 stayed, because it was.

Worked Example 3 — A wave that repeats twice as often

You may not have drawn the graph of $y = sin x$ yet. You only need five of its points. With $x$ in degrees, a calculator gives $sin 0° = 0$ , $sin 90° = 1$ , $sin 180° = 0$ , $sin 270° = - 1$ and $sin 360° = 0$ . The graph is a smooth wave through these points, and after 360° it repeats.

Describe $y = sin 2 x$ , and find where its highest and lowest points are between 0° and 360°.

The 2 multiplies the input, so this is a horizontal stretch with scale factor $\frac{1}{2}$ . Every $x$ -coordinate halves. The highest point $(90°, 1)$ moves to $(45°, 1)$ . The lowest point $(270°, - 1)$ moves to $(135°, - 1)$ . The heights do not change.

So one full wave now fits into 180° instead of 360°. Between 0° and 360° there are two full waves, with highest points at 45° and 225°, and lowest points at 135° and 315°.

Check: $sin (2 \times 225°) = sin 450° = sin 90° = 1$ , because the wave repeats every 360°. ✓

The graph of

y = sin x

(ink) makes one full wave between 0° and 360°. The graph of

y = sin 2 x

(teal) makes two. The highest point at 90° moves to 45°, and the lowest point at 270° moves to 135°.

A sound is a wave of air pressure. When a guitar string vibrates twice as fast, its wave is squeezed sideways in exactly this way, and the ear hears the same note one octave higher. An octave is a horizontal stretch with scale factor $\frac{1}{2}$ .

Worked Example 4 — Finding the stretch from three points

The graph of $y = f (x)$ crosses the $x$ -axis at $(- 2, 0)$ and $(2, 0)$ , and passes through $(0, - 4)$ . A stretched graph crosses the $x$ -axis at $(- 6, 0)$ and $(6, 0)$ , and passes through $(0, - 4)$ . Write its equation in terms of $f$ .

The heights did not change: the point on the $y$ -axis is still at $- 4$ . The positions were multiplied by 3: $\pm 2$ became $\pm 6$ . So this is a horizontal stretch with scale factor 3.

A scale factor of 3 comes from $f (a x)$ with $\frac{1}{a} = 3$ , so $a = \frac{1}{3}$ . The equation is $y = f (\frac{1}{3} x)$ .

Check one point: at $x = 6$ , the new function gives $f (\frac{1}{3} \times 6) = f (2) = 0$ . ✓ The input 6 is divided back to 2 before $f$ sees it, which is exactly why the root moved from 2 out to 6.

The working below makes one realistic mistake. Find the line where it goes wrong.

Spot the error — a horizontal stretch of x² − 4x

A student describes the graph of y = f(2x), where f(x) = x² − 4x. Tap the line where the working first goes wrong.

A student's working for y = f(2x), where f(x) = x² − 4x, with the mistake marked.

f(x) = (x − 2)² − 4, so the lowest point of y = f(x) is (2, −4). Correct.
In y = f(2x), the input is multiplied by 2 before f works. Correct.
So every x-coordinate is multiplied by 2, and the lowest point moves to (4, −4). This is the mistake. The input must be divided by 2 to give f the number it needs, so every x-coordinate is halved. The lowest point moves to (1, −4).
The y-coordinates do not change, so the lowest value is still −4. Correct.
The equation is f(2x) = (2x)² − 4(2x) = 4x² − 8x. Correct, and it shows the mistake: 4x² − 8x = 4(x − 1)² − 4, with its lowest point at (1, −4).

Six transformations, one pattern

You have now met every basic change to a graph. Here they are together.

Equation	What happens to $(p, q)$	What does not move
$y = f (x) + b$	$(p, q + b)$ : up $b$	no point
$y = f (x - a)$	$(p + a, q)$ : right $a$	no point
$y = - f (x)$	$(p, - q)$	the $x$ -axis
$y = f (- x)$	$(- p, q)$	the $y$ -axis
$y = a f (x)$	$(p, a q)$	the $x$ -axis
$y = f (a x)$	$(\frac{p}{a}, q)$	the $y$ -axis

Read the rows in pairs. A change outside the function acts on $y$ , and does exactly what it says: add $b$ , change the sign, multiply by $a$ . A change inside acts on $x$ , and is undone before the function works: subtract becomes add, multiply becomes divide. The sign change is its own undo, which is why the reflection in the $y$ -axis looked "normal". One reason lies behind all six rows.

Three classic mistakes

Trap 1 — Using a instead of 1/a for the horizontal stretch

$y = f (3 x)$ is not a stretch with scale factor 3. It is a stretch with scale factor $\frac{1}{3}$ , and the graph gets narrower. If you are unsure, take one point, such as a root at $x = 6$ , and ask: what input now gives the function 6? The answer is $x = 2$ , so the root moved inwards.

Trap 2 — Multiplying only the first term

$2 f (x)$ means twice the whole output. For $f (x) = x^{2} - 4$ , it is $2 (x^{2} - 4) = 2 x^{2} - 8$ . Writing $2 x^{2} - 4$ doubles only the first term, and it is a different graph: its roots are $\pm 2$ , not $\pm 2$ . Always put $f (x)$ in a bracket first.

Trap 3 — Writing 2x² for (2x)²

In $f (2 x)$ , every $x$ is replaced by $(2 x)$ , with its bracket. $(2 x)^{2} = 4 x^{2}$ , because the 2 is squared too. Without the bracket, $f (2 x)$ for $f (x) = x^{2}$ would wrongly become $2 x^{2}$ , a vertical stretch with scale factor 2. The correct graph, $4 x^{2}$ , is narrower than that.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

$f (x) = x^{2} - 6 x + 5$ . Find the roots and the lowest point of $y = 3 f (x)$ and of $y = f (3 x)$ .

Hint 1
Find the roots and lowest point of $f$ first. Then decide which coordinate each stretch changes.

Hint 2
$f (x) = (x - 1) (x - 5) = (x - 3)^{2} - 4$ . For $3 f (x)$ , multiply the heights by 3. For $f (3 x)$ , divide the positions by 3.

Full answer
$f$ has roots 1 and 5, and lowest point $(3, - 4)$ . For $y = 3 f (x)$ , a vertical stretch with scale factor 3: the roots stay at 1 and 5, and the lowest point moves to $(3, - 12)$ . For $y = f (3 x)$ , a horizontal stretch with scale factor $\frac{1}{3}$ : the roots move to $\frac{1}{3}$ and $\frac{5}{3}$ , and the lowest point moves to $(1, - 4)$ . Check: $f (3 x) = 9 x^{2} - 18 x + 5$ . At $x = 1$ , this is $9 - 18 + 5 = - 4$ . ✓ At $x = \frac{1}{3}$ , it is $1 - 6 + 5 = 0$ . ✓
The graph of $y = f (x)$ has a highest point at $(4, 6)$ , crosses the $x$ -axis at $(- 2, 0)$ and $(8, 0)$ , and crosses the $y$ -axis at $(0, 2)$ . Describe the graphs of $y = \frac{1}{2} f (x)$ and $y = f (2 x)$ in the same way.

Hint 1
For each stretch, decide which coordinate changes, and which axis does not move.

Hint 2
$\frac{1}{2} f (x)$ halves every height. $f (2 x)$ halves every position.

Full answer
$y = \frac{1}{2} f (x)$ : every height halves. The highest point moves to $(4, 3)$ . The roots are on the $x$ -axis, so they stay at $(- 2, 0)$ and $(8, 0)$ . The $y$ -intercept moves to $(0, 1)$ . $y = f (2 x)$ : every position halves. The highest point moves to $(2, 6)$ . The roots move to $(- 1, 0)$ and $(4, 0)$ . The $y$ -intercept stays at $(0, 2)$ . Check one: $f (2 x)$ at $x = 4$ is $f (8) = 0$ . ✓
$f (x) = x^{2} - 4$ . (a) Find the value of $a$ for which $f (a x) = 9 x^{2} - 4$ . (b) Can $9 x^{2} - 4$ also be written as $k f (x)$ for some number $k$ ? (c) For $g (x) = x^{2}$ , can $9 x^{2}$ be written both as $g (a x)$ and as $k g (x)$ ?

Hint 1
Substitute $a x$ into $f$ , and compare with $9 x^{2} - 4$ .

Hint 2
$k f (x) = k x^{2} - 4 k$ . Both the $x^{2}$ term and the constant must match.

Full answer
(a) $f (a x) = a^{2} x^{2} - 4$ . This equals $9 x^{2} - 4$ when $a^{2} = 9$ , so $a = 3$ (or $a = - 3$ , which gives the same graph because $f$ is even). It is a horizontal stretch with scale factor $\frac{1}{3}$ . (b) No. $k f (x) = k x^{2} - 4 k$ . Matching $x^{2}$ needs $k = 9$ , but then the constant is $- 36$ , not $- 4$ . (c) Yes. $g (3 x) = 9 x^{2}$ and $9 g (x) = 9 x^{2}$ . For $x^{2}$ alone, a horizontal stretch with scale factor $\frac{1}{3}$ is the same as a vertical stretch with scale factor 9. The constant $- 4$ is what stops this for $f$ . Check (a): $f (3 \times 1) = 5$ , and $9 - 4 = 5$ . ✓
A ball is thrown upwards. Its height in metres after $t$ seconds is $h (t) = 20 t - 5 t^{2}$ . (a) When does it land, and what is its greatest height? (b) A video of the throw is played at double speed. Explain why the height shown at time $t$ of the video is $h (2 t)$ . When does the ball land in the video, and how high does it go? (c) A second video shows the same throw, filmed so that every height appears half as large. Write this as a stretch, and give its greatest height.

Hint 1
For (a), factorise $h (t)$ and use the symmetry of the graph.

Hint 2
For (b), after 1 second of video, how many seconds of the real throw have been shown? For (c), the heights change, and the times do not.

Full answer
(a) $h (t) = 5 t (4 - t)$ , so the ball is at height 0 when $t = 0$ and $t = 4$ . It lands after 4 seconds. The highest point is halfway, at $t = 2$ : $h (2) = 40 - 20 = 20$ m. (b) At double speed, 1 second of video shows 2 seconds of the real throw. So the video at time $t$ shows the real height at time $2 t$ , which is $h (2 t)$ . This is a horizontal stretch with scale factor $\frac{1}{2}$ : every time halves, and every height stays. The ball lands at $t = 2$ seconds of video, and still reaches 20 m, at $t = 1$ . Check: $h (2 \times 2) = h (4) = 0$ . ✓ (c) Every height halves and every time stays, so the graph is $y = \frac{1}{2} h (t)$ , a vertical stretch with scale factor $\frac{1}{2}$ . The greatest height shown is 10 m, still at $t = 2$ . Check: $\frac{1}{2} h (2) = 10$ . ✓

The bigger idea

Changing the units of a measurement is a stretch. If you measure time in minutes instead of seconds, every time on a graph is divided by 60, and the graph is squeezed sideways. If you measure height in centimetres instead of metres, every height is multiplied by 100, and the graph is stretched upwards. The shape of the relationship has not changed. Only the scale has.

So a slow wave and a fast one, or a small pendulum and a large one, are often the same curve underneath, stretched one way or the other. Asking "is this just a stretch of a graph I already know?" turns many new graphs into old ones.

Key concepts in this lesson

The graph of $y = a f (x)$ is a vertical stretch with scale factor $a$ : every point $(p, q)$ moves to $(p, a q)$ , and the $x$ -axis does not move.
The graph of $y = f (a x)$ is a horizontal stretch with scale factor $\frac{1}{a}$ , because the input must be divided by $a$ to give the function the number it needs: $(p, q)$ moves to $(\frac{p}{a}, q)$ .
In a horizontal stretch the $y$ -axis does not move, so points with $x = 0$ , such as the $y$ -intercept, stay where they are.
A scale factor between 0 and 1 is still called a stretch, and a negative scale factor is a stretch together with a reflection.
For some functions, such as $x^{2}$ , one graph is both a horizontal and a vertical stretch, but this is never true in general, so always substitute with brackets and check.