Combining transformations

Apply several transformations in order, write the resulting equation, and know which pairs of transformations commute and which do not.

Every transformation in one table

Transformation	Replace	New function
Shift right $h$	$x$ by $x - h$	$y = f(x - h)$
Shift up $k$	$y$ by $y - k$	$y = f(x) + k$
Reflect in the $y$-axis	$x$ by $-x$	$y = f(-x)$
Reflect in the $x$-axis	$y$ by $-y$	$y = -f(x)$
Stretch horizontally, factor $a$	$x$ by $\dfrac{x}{a}$	$y = f\!\left(\dfrac{x}{a}\right)$
Stretch vertically, factor $b$	$y$ by $\dfrac{y}{b}$	$y = b\,f(x)$

Does the order matter?

Two transformations commute if doing them in either order gives the same graph. A horizontal change and a vertical change always commute. Two changes in the same direction may not.

Example 1: order matters●●

Start with $y = x^2$. (a) Shift up $1$, then stretch vertically by $3$. (b) Stretch vertically by $3$, then shift up $1$.

(a) $y = 3(x^2 + 1) = 3x^2 + 3$

Shift: y = x² + 1. Stretch: multiply the whole function by 3.

(b) $y = 3x^2 + 1$

Stretch: y = 3x². Shift: add 1.

Different graphs, so these two do not commute.

Stretch then shift, or shift then stretch?

The right shift commutes with everything vertical, but the up shift and the vertical stretch do not.

Try this

When are the two graphs the same? (Try $k = 0$, and try $b = 1$.)

Example 2●●●

$y = \sqrt{x}$ is reflected in the $y$-axis, then shifted right $2$. Find the equation and domain.

Reflect: $y = \sqrt{-x}$.

Shift: $y = \sqrt{-(x - 2)} = \sqrt{2 - x}$.

Replace x with x − 2 everywhere, including inside the minus sign.

Domain: $2 - x \ge 0$, so $x \le 2$.

Flashcards

Commute

Give the same result in either order.

Horizontal and vertical changes

Always commute.

y = b f(x − h) + k

Shift right $h$, stretch vertically by $b$, shift up $k$.

Practice

$y = x^2$ is stretched vertically by $2$, then shifted down $5$. The result is:
$y = x^2$ is shifted down $5$, then stretched vertically by $2$. The result is:
Which pair always commutes?
$(1, 2)$ is on $y = f(x)$. It is mapped onto $y = 3f(x) + 1$. Find the new $y$-coordinate.
$y = 2^x$ is reflected in the $x$-axis, then shifted up $3$. The new asymptote is:
$y = -|x + 1| + 4$ is $y = |x|$:
$y = \dfrac{1}{x}$ is shifted right $3$, then stretched vertically by $2$. Where is the vertical asymptote?
Find the $y$-intercept of $y = x^2$ shifted left $1$ and then stretched vertically by $9$.