Horizontal and vertical dilations

Stretch a graph horizontally by replacing x with x/a, vertically by multiplying the function by b, and handle fractional and negative factors.

Two kinds of stretch

Dilation	Change to the equation	Point $(p, q)$ goes to	Stays fixed
Horizontal, factor $a$ (away from the $y$-axis)	replace $x$ with $\dfrac{x}{a}$: $y = f\!\left(\dfrac{x}{a}\right)$	$(ap, q)$	the $y$-axis
Vertical, factor $b$ (away from the $x$-axis)	replace $y$ with $\dfrac{y}{b}$: $y = b\,f(x)$	$(p, bq)$	the $x$-axis

Stretch a semicircle

Faint: the semicircle y = √(9 − x²). Orange: y = b f(x / a). With a ≠ b the circle becomes an ellipse.

Try this

Which factor moves the $x$-intercepts? Which moves the top point?
Make $b$ negative. What extra transformation do you see?

Example 1●●

Stretch $y = x^2 - 4$ horizontally by factor $2$. Find the new equation and $x$-intercepts.

$y = \left(\dfrac{x}{2}\right)^2 - 4 = \dfrac{x^2}{4} - 4$

Replace x with x/2.

$x$-intercepts move from $\pm 2$ to $\pm 4$. The $y$-intercept $-4$ is on the $y$-axis, so it stays.

The old intercepts ±2 are doubled.

Example 2●●

Describe $y = 3 \times 2^x$ as a dilation of $y = 2^x$.

A vertical stretch by factor $3$.

The $y$-intercept moves from $1$ to $3$. The asymptote $y = 0$ stays.

Every y-value triples, but 3 × 0 = 0.

Combining transformations

When you combine a dilation with a translation, the order matters. In $y = 2f(x) + 3$, stretch first, then shift. In $y = 2(f(x) + 3)$, shift first, then stretch. Section 5G covers this in detail.

Flashcards

Horizontal stretch, factor a

Replace $x$ with $\dfrac{x}{a}$. $(p, q) \to (ap, q)$.

Vertical stretch, factor b

$y = b\,f(x)$. $(p, q) \to (p, bq)$.

y = f(2x)

Horizontal compression by factor $\tfrac12$.

Negative factor

Stretch plus a reflection.

Practice

$y = 4f(x)$ is $y = f(x)$:
$y = f(3x)$ is $y = f(x)$:
$(5, 2)$ is on $y = f(x)$. After a horizontal stretch by factor $2$, what is the new $x$-coordinate?
$(1, 3)$ is on $y = f(x)$. Which $y$-coordinate does it have on $y = -2f(x)$?
Stretching $y = \dfrac{1}{x}$ vertically by factor $5$ gives:
Which dilation keeps every $x$-intercept fixed?
The circle $x^2 + y^2 = 4$ is stretched horizontally by factor $3$. What is its new positive $x$-intercept?
$y = \sqrt{x}$ becomes $y = \sqrt{4x}$. This is the same as: