Stretching and Compressing Graphs, foundations for calculus lesson

A photo can be stretched tall or squeezed narrow, and the two edits are clearly different. Graphs work the same way. Multiplying the output of a function pulls the graph up and down, multiplying the input squeezes it toward or away from the $y$ -axis, and the only reliable way to tell which is which is to watch where specific points go.

Key terms

Point mapping

Definition A point mapping for a new function $g$ built from $f$ is a rule that sends each point $(u, f(u))$ of the graph of $f$ to a point of the graph of $g$ , and that reaches every point of the graph of $g$ this way. To find it, solve for the input of $g$ whose inside equals $u$ and then compute the output.

In plain words Pick a point you know on the old graph and work out exactly where it lands on the new one. A handful of such points draws the new graph for you.

Example For $f(x) = x^2$ and $g(x) = 3x^2$ the input stays the same and the output triples, so $(u, u^2)$ maps to $(u, 3u^2)$ . The point $(2, 4)$ goes to $(2, 12)$ .

Output scaling

Definition For a real number $A \ne 0$ , the function $g(x) = A f(x)$ is an output scaling of $f$ . It has the same domain as $f$ , and each point $(u, f(u))$ maps to

(u, A f(u)).

When $A > 0$ this is a vertical scaling of the graph, which multiplies every point’s distance from the $x$ -axis by $A$ . When $A < 0$ it is a vertical scaling by $\lvert A \rvert$ followed by a reflection in the $x$ -axis.

In plain words The function does its usual work and then you multiply the answer. Only heights change. Points on the $x$ -axis stay put.

Example Compare $f(x) = x^2$ with $g(x) = 2x^2$ .

$x$	$-2$	$-1$	$0$	$1$	$2$
$x^2$	$4$	$1$	$0$	$1$	$4$
$2x^2$	$8$	$2$	$0$	$2$	$8$

Each input keeps its place and each height doubles.

Input scaling

Definition For a real number $B \ne 0$ , the function $g(x) = f(Bx)$ is an input scaling of $f$ . Its domain is the set of $x$ with $Bx$ in the domain of $f$ . The inside equals $u$ when $x = u/B$ , so each point $(u, f(u))$ maps to

\left(\frac{u}{B}, f(u)\right).

When $B > 0$ this is a horizontal scaling of the graph, which multiplies every point’s distance from the $y$ -axis by $\frac{1}{B}$ . In particular, when $B > 1$ horizontal distances from the $y$ -axis shrink.

In plain words Multiplying the input by $B$ means the old output now appears at the input $u/B$ instead of $u$ , which for $B > 1$ is closer to the $y$ -axis. That is why multiplying by $2$ squeezes the graph rather than widening it. There is no mysterious opposite rule, just the question of which input gives the old output.

Example Let $h(x) = \sqrt{4x}$ , with domain $[0, \infty)$ . The inside $4x$ equals $u$ when $x = \frac{u}{4}$ .

Point on $y = \sqrt{x}$	Image on $y = \sqrt{4x}$
$(0, 0)$	$(0, 0)$
$(4, 2)$	$(1, 2)$
$(16, 4)$	$(4, 4)$

Check one of them. The value $h(1) = \sqrt{4} = 2$ matches the image $(1, 2)$ .

Scale factor

Definition The scale factor of a scaling is the number that every distance in the scaled direction is multiplied by. For $y = A f(x)$ the vertical scale factor is $\lvert A \rvert$ , measured from the $x$ -axis. For $y = f(Bx)$ the horizontal scale factor is $\frac{1}{\lvert B \rvert}$ , measured from the $y$ -axis.

In plain words The vertical factor is the number in front. The horizontal factor is one over the number multiplying $x$ inside.

Example Suppose $(6, 9)$ is on the graph of $f$ , and let $g(x) = \frac{1}{2} f(3x)$ . The vertical factor is $\frac{1}{2}$ and the horizontal factor is $\frac{1}{3}$ . Solving $3x = 6$ gives $x = 2$ , and the output is $\frac{1}{2} \cdot 9$ , so the image is $\left(2, \frac{9}{2}\right)$ .

Stretch

Definition A stretch is a scaling whose scale factor is greater than $1$ . The graph of $y = A f(x)$ is a vertical stretch when $\lvert A \rvert > 1$ , and the graph of $y = f(Bx)$ is a horizontal stretch when $0 < \lvert B \rvert < 1$ . A negative $A$ or $B$ adds a reflection, described below.

In plain words A stretch pulls points farther from the axis they are measured from, taller for a vertical stretch and wider for a horizontal one.

Example For $y = 3\sqrt{x}$ the point $(4, 2)$ goes to $(4, 6)$ , a vertical stretch by $3$ . For $y = \sqrt{x/2}$ , where $B = \frac{1}{2}$ , the inside equals $4$ when $x = 8$ , so $(4, 2)$ goes to $(8, 2)$ , a horizontal stretch by $2$ .

Compression

Definition A compression is a scaling whose scale factor lies strictly between $0$ and $1$ . The graph of $y = A f(x)$ is a vertical compression when $0 < \lvert A \rvert < 1$ , and the graph of $y = f(Bx)$ is a horizontal compression when $\lvert B \rvert > 1$ , again with a reflection added when $A$ or $B$ is negative.

In plain words A compression pulls points closer to the axis they are measured from, flatter for a vertical compression and narrower for a horizontal one.

Example For $y = \frac{1}{2}x^2$ the point $(2, 4)$ goes to $(2, 2)$ , a vertical compression by $\frac{1}{2}$ . The graph of $y = \sqrt{4x}$ from above is a horizontal compression by $\frac{1}{4}$ .

Negative factors

Definition For $B > 0$ the graph of $y = f(-Bx)$ is the horizontal scaling by $\frac{1}{B}$ of the graph of $f$ followed by a reflection in the $y$ -axis, and each point $(u, f(u))$ maps to $\left(-\frac{u}{B}, f(u)\right)$ . For $A > 0$ the graph of $y = -A f(x)$ is the vertical scaling by $A$ followed by a reflection in the $x$ -axis.

In plain words Handle the size first with a positive factor, then let the minus sign flip the graph.

Example Take $g(x) = \sqrt{-2x}$ , defined for $x \le 0$ . The inside $-2x$ equals $4$ when $x = -2$ , so $(4, 2)$ goes to $(-2, 2)$ , and indeed $g(-2) = \sqrt{4} = 2$ . The graph of $\sqrt{x}$ has been squeezed by $\frac{1}{2}$ and flipped to the left side.

When two scalings coincide

Definition For some special parent functions, an input scaling and an output scaling produce the same graph. Two examples are

\begin{gathered} \sqrt{4x} = 2\sqrt{x} \text{ for } x \ge 0, \\[0.4em] (2x)^2 = 4x^2 \text{ for every real } x. \end{gathered}

These identities come from properties of roots and powers. They do not hold for functions in general, so $f(Bx)$ and $A f(x)$ must be treated as different operations unless you have checked the equality for the specific $f$ .

In plain words Sometimes squeezing sideways happens to look exactly like stretching upward. That is a feature of the particular function, not a rule.

Example For $f(x) = x + 1$ the input scaling gives $f(2x) = 2x + 1$ , while the output scaling gives $2f(x) = 2x + 2$ . At $x = 0$ they give $1$ and $2$ , so the two graphs differ.

Check yourself

The point $(3, 5)$ lies on the graph of $f$ . Find the corresponding point on the graphs of $y = 4f(x)$ , $y = f(3x)$ and $y = f\left(\frac{x}{2}\right)$ .

Show the answer

For $y = 4f(x)$ the input stays and the output becomes $20$ , giving $(3, 20)$ . For $y = f(3x)$ solve $3x = 3$ to get $(1, 5)$ . For $y = f\left(\frac{x}{2}\right)$ solve $\frac{x}{2} = 3$ to get $(6, 5)$ .

Find the images of $(-2, 4)$ , $(0, 0)$ and $(2, 4)$ on the graph of $y = \frac{1}{2}x^2$ , and say whether it is a stretch or a compression.

Show the answer

The inputs stay and the outputs halve, so the images are $(-2, 2)$ , $(0, 0)$ and $(2, 2)$ . This is a vertical compression with scale factor $\frac{1}{2}$ .

Find the domain of $g(x) = \sqrt{9x}$ and the images of $(0, 0)$ , $(1, 1)$ and $(9, 3)$ on $y = \sqrt{x}$ .

Show the answer

The condition $9x \ge 0$ gives the domain $[0, \infty)$ . Solving $9x = u$ gives $x = \frac{u}{9}$ , so the images are $(0, 0)$ , $\left(\frac{1}{9}, 1\right)$ and $(1, 3)$ . This is a horizontal compression by $\frac{1}{9}$ , and it happens to equal the vertical stretch $3\sqrt{x}$ .

Stretching and Compressing Graphs

Key terms

Point mapping

Output scaling

Input scaling

Scale factor

Stretch

Compression

Negative factors

When two scalings coincide

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus

Resources for this lesson

Function transformation explorer