Stretches and Compressions of Graphs, precalculus lesson

Draw a graph on a sheet of rubber and pull the sheet up and down, away from the $x$ -axis. The curve gets taller, yet every point on the $x$ -axis stays where it was. Multiplying the outputs of a function by a constant does exactly this. Multiplying the inputs pulls or squeezes the sheet sideways instead, and there the effect runs opposite to the number in the formula.

Key terms

Vertical stretch and compression

Definition Let $f$ be a function and let $a > 0$ be a constant. The function $g(x) = a f(x)$ has the same domain as $f$ , and a point $(x, y)$ lies on the graph of $f$ exactly when $(x, ay)$ lies on the graph of $g$ . If $a > 1$ the graph of $g$ is a vertical stretch of the graph of $f$ by the factor $a$ . If $0 < a < 1$ it is a vertical compression by the factor $a$ . Every distance from the $x$ -axis is multiplied by $a$ , so points on the $x$ -axis stay fixed, the zeros of $g$ are the zeros of $f$ , and the range of $g$ consists of the numbers $ay$ with $y$ in the range of $f$ .

In plain words Each point moves straight up or down until it is $a$ times as far from the $x$ -axis as before. The inputs do not change, only the outputs.

Example Let $f(x) = x^2 - 4$ , with zeros $-2$ and $2$ and lowest point $(0, -4)$ . Compare it with $3f(x) = 3x^2 - 12$ and $\frac{1}{2} f(x) = \frac{1}{2} x^2 - 2$ .

$x$	$f(x)$	$3f(x)$	$\frac{1}{2} f(x)$
$-3$	$5$	$15$	$\frac{5}{2}$
$-2$	$0$	$0$	$0$
$0$	$-4$	$-12$	$-2$
$1$	$-3$	$-9$	$-\frac{3}{2}$
$2$	$0$	$0$	$0$

The zeros stay at $\pm 2$ . The lowest point moves down to $(0, -12)$ under the stretch and up to $(0, -2)$ under the compression, and the range $[-4, \infty)$ becomes $[-12, \infty)$ and $[-2, \infty)$ .

Horizontal compression and stretch

Definition Let $f$ be a function and let $b > 0$ be a constant. The function $g(x) = f(bx)$ is defined at $x$ exactly when $bx$ is in the domain of $f$ . Since

g\left(\frac{x}{b}\right) = f\left(b \cdot \frac{x}{b}\right) = f(x),

a point $(x, y)$ lies on the graph of $f$ exactly when $\left(\frac{x}{b}, y\right)$ lies on the graph of $g$ . If $b > 1$ the graph of $g$ is a horizontal compression of the graph of $f$ toward the $y$ -axis by the factor $\frac{1}{b}$ . If $0 < b < 1$ it is a horizontal stretch away from the $y$ -axis by the factor $\frac{1}{b}$ . Points on the $y$ -axis stay fixed, and the range does not change.

In plain words The new function gives the output $f(x)$ at the input $\frac{x}{b}$ instead of at $x$ . For $b = 2$ every input is halved, so the graph is squeezed to half its width.

Example Let $f(x) = x^2 - 4x$ , with zeros $0$ and $4$ and lowest point $(2, -4)$ . Then

\begin{aligned} g(x) &= f(2x) \\[0.4em] &= (2x)^2 - 4(2x) \\[0.4em] &= 4x^2 - 8x \\[0.4em] &= 4x(x - 2), \end{aligned}

with zeros $0$ and $2$ and lowest point $(1, -4)$ . Every first coordinate is halved and every height stays the same. The point $(4, 0)$ moves to $(2, 0)$ , while $(0, 0)$ on the $y$ -axis does not move. Squeezing also makes the graph steeper. The secant line of $f$ through $(2, -4)$ and $(4, 0)$ has slope $2$ , and the matching secant line of $g$ through $(1, -4)$ and $(2, 0)$ has slope $4$ . This doubling of slopes is why the derivative of $f(2x)$ carries a factor $2$ in the chain rule.

Multiplier

Definition In the formula $g(x) = a f(bx)$ with nonzero constants $a$ and $b$ , the number $a$ is the outside multiplier and $b$ is the inside multiplier. The outside multiplier multiplies each output after $f$ has acted. The inside multiplier multiplies each input before $f$ acts. When the input of $f$ also contains a constant, as in $2x + 6$ , the inside multiplier is still the coefficient $2$ of $x$ . Factoring the input as $2(x + 3)$ shows that this $2$ multiplies the whole of $x + 3$ , which matters once stretches and shifts are combined.

In plain words Outside, a multiplier does what it says to the heights. Inside, it does the opposite to the widths, because a bigger multiplier on the input makes $f$ reach each value closer to the $y$ -axis.

Example Take $f(x) = x^3$ and the multiplier $2$ in both places. At $x = 1$ the outside version gives $2f(1) = 2$ and the inside version gives $f(2) = 8$ . The point $(2, 8)$ on the graph of $f$ goes to $(2, 16)$ on the graph of $y = 2f(x)$ and to $(1, 8)$ on the graph of $y = f(2x)$ . For this particular $f$ the identity $(2x)^3 = 8x^3$ shows that the inside multiplier $2$ has the same effect as the outside multiplier $8$ . That does not happen in general. For $f(x) = x^2 - 4x$ every graph $y = af(x)$ with $a \ne 0$ has zeros $0$ and $4$ , while $f(2x)$ has zeros $0$ and $2$ , so no outside multiplier can match it.

Scale factor

Definition A scale factor is the positive number by which a stretch or compression multiplies distances in one direction. For $g(x) = a f(x)$ with $a \ne 0$ the vertical scale factor is $\lvert a \rvert$ , with distances measured from the $x$ -axis. For $g(x) = f(bx)$ with $b \ne 0$ the horizontal scale factor is $\frac{1}{\lvert b \rvert}$ , with distances measured from the $y$ -axis. A scale factor greater than $1$ gives a stretch and one between $0$ and $1$ gives a compression. A negative multiplier adds a reflection, across the $x$ -axis when $a < 0$ and across the $y$ -axis when $b < 0$ . In every case a point $(x, y)$ on the graph of $f$ corresponds to a point on the graph of $g(x) = a f(bx)$ by the rule

(x, y) \ \longmapsto \ \left(\frac{x}{b},\ ay\right).

In plain words The scale factor says how much bigger or smaller the graph gets in one direction. Read it straight from an outside multiplier, and take the reciprocal of an inside one. The sign only decides whether the graph also flips.

Example Suppose $f$ has domain $[-3, 6]$ and its graph passes through $(6, 1)$ , and let $g(x) = -2f\left(\frac{x}{3}\right)$ . The inside multiplier $\frac{1}{3}$ gives the horizontal scale factor $3$ , a stretch. The outside multiplier $-2$ gives the vertical scale factor $2$ together with a reflection across the $x$ -axis. The point $(6, 1)$ goes to $(18, -2)$ , and indeed $g(18) = -2f(6) = -2$ . The domain of $g$ is $[-9, 18]$ , since $g(x)$ needs $\frac{x}{3}$ to lie in $[-3, 6]$ .

Dilation

Definition For a constant $k > 0$ , the dilation of the plane centered at the origin with scale factor $k$ sends each point $(x, y)$ to $(kx, ky)$ . The vertical dilation with factor $k$ sends $(x, y)$ to $(x, ky)$ and keeps the $x$ -axis fixed, and the horizontal dilation with factor $k$ sends $(x, y)$ to $(kx, y)$ and keeps the $y$ -axis fixed. These three maps carry the graph of $y = f(x)$ onto the graphs of

\begin{gathered} y = k f\left(\frac{x}{k}\right), \\[0.8em] y = k f(x), \\[0.8em] y = f\left(\frac{x}{k}\right) \end{gathered}

respectively. The word dilation covers both stretches, with $k > 1$ , and compressions, with $0 < k < 1$ .

In plain words A dilation centered at the origin is a zoom. It changes the size of the picture but not its shape. A dilation in one direction pulls the picture one way only, which in general does change the shape.

Example Zoom the parabola $y = x^2$ by the factor $2$ about the origin. The first rule gives

y = 2\left(\frac{x}{2}\right)^2 = \frac{x^2}{2}.

The point $(2, 4)$ goes to $(4, 8)$ , and indeed $\frac{4^2}{2} = 8$ . In the same way the zoom with factor $k = \frac{1}{c}$ carries $y = x^2$ onto $y = cx^2$ for any $c > 0$ , so all these parabolas have the same shape and differ only in size. Parabolas are special in this way, because the vertical dilation $y = cx^2$ of $y = x^2$ is also a zoom of it. A semicircle stretched vertically, by contrast, becomes half of an ellipse and really changes shape.

Stretches combined with shifts

Definition For constants $a \ne 0$ , $b \ne 0$ , $h$ and $k$ , consider

g(x) = a f\big(b(x - h)\big) + k.

For each point $(x, y)$ on the graph of $f$ the point $\left(\frac{x}{b} + h,\ ay + k\right)$ lies on the graph of $g$ , because

\begin{aligned} g\left(\frac{x}{b} + h\right) &= a f\left(b \cdot \frac{x}{b}\right) + k \\[0.8em] &= a f(x) + k \\[0.4em] &= ay + k. \end{aligned}

So the first coordinate is scaled by $\frac{1}{b}$ and then shifted by $h$ , and the second is scaled by $a$ and then shifted by $k$ . With $a$ and $b$ equal to $1$ or $-1$ this is the rule $(x, y) \mapsto (bx + h, ay + k)$ for shifts and reflections, since then $\frac{x}{b} = bx$ . When the input is written as $bx - c$ , factor out $b$ first,

bx - c = b\left(x - \frac{c}{b}\right),

so the horizontal shift is $\frac{c}{b}$ and not $c$ .

In plain words Rewrite the inside until it reads $b(x - h)$ . Then move each point by scaling first and shifting second, in both directions.

Example Start from $f(x) = \sqrt{x}$ and take $g(x) = 2\sqrt{3x - 6} + 1$ . Factoring gives $g(x) = 2\sqrt{3(x - 2)} + 1$ , so $a = 2$ , $b = 3$ , $h = 2$ and $k = 1$ , and each point $(x, y)$ goes to $\left(\frac{x}{3} + 2,\ 2y + 1\right)$ .

Point on $y = \sqrt{x}$	Point on $y = g(x)$
$(0, 0)$	$(2, 1)$
$(1, 1)$	$\left(\frac{7}{3}, 3\right)$
$(4, 2)$	$\left(\frac{10}{3}, 5\right)$
$(9, 3)$	$(5, 7)$

As a check, $g(5) = 2\sqrt{9} + 1 = 7$ . The domain of $g$ is $[2, \infty)$ , from $3x - 6 \ge 0$ . Reading $3x - 6$ as a shift by $6$ would put the starting point at $(6, 1)$ , but $g(6) = 2\sqrt{12} + 1$ , which is not $1$ .

Check yourself

The graph of $f$ passes through $(-2, 5)$ , $(0, 1)$ and $(4, -3)$ . Find the corresponding points on the graph of $g(x) = 3f(2x)$ .

Show the answer

Each point $(x, y)$ goes to $\left(\frac{x}{2}, 3y\right)$ , which gives $(-1, 15)$ , $(0, 3)$ and $(2, -9)$ . As a check, $g(2) = 3f(4) = -9$ .

Compress the graph of $y = \sqrt{x}$ horizontally by the factor $\frac{1}{4}$ , then stretch it vertically by the factor $3$ . Write the new function and find its value at $x = 1$ .

Show the answer

The horizontal compression by $\frac{1}{4}$ uses the inside multiplier $4$ and the vertical stretch uses the outside multiplier $3$ , so $g(x) = 3\sqrt{4x}$ and $g(1) = 3 \cdot 2 = 6$ . Since $\sqrt{4x} = 2\sqrt{x}$ for $x \ge 0$ , the same graph is also $y = 6\sqrt{x}$ .

Explain how the graph of $g(x) = (2x + 6)^2$ comes from the graph of $y = x^2$ .

Show the answer

Factor the inside as $2x + 6 = 2(x + 3)$ . Compress horizontally toward the $y$ -axis by the factor $\frac{1}{2}$ , then shift left by $3$ . The vertex $(0, 0)$ goes to $(-3, 0)$ , and the point $(2, 4)$ goes to $(1 - 3, 4) = (-2, 4)$ . Indeed $g(-3) = 0$ and $g(-2) = 2^2 = 4$ .

Stretches and Compressions of Graphs

Key terms

Vertical stretch and compression

Horizontal compression and stretch

Multiplier

Scale factor

Dilation

Stretches combined with shifts

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus

Resources for this lesson

Function transformation explorer