Reflections in the axes

Reflect a graph in the y-axis by replacing x with −x, in the x-axis by replacing y with −y, and in both axes at once (a rotation of 180° about the origin).

The two reflections

Reflection	Change to the equation	Point $(a, b)$ goes to
In the $y$-axis	replace $x$ with $-x$: $y = f(-x)$	$(-a, b)$
In the $x$-axis	replace $y$ with $-y$: $y = -f(x)$	$(a, -b)$
In both (half-turn about $O$)	both: $y = -f(-x)$	$(-a, -b)$

Reflect y = 2ˣ

Faint: y = 2ˣ. Orange: the reflected graph. Set each slider to −1 to reflect.

Try this

Which reflection keeps the $y$-intercept fixed?
Which one keeps the horizontal asymptote $y = 0$ fixed?

Example 1●

Write the equation of $y = x^2 - 4x$ after reflection in (a) the $y$-axis, (b) the $x$-axis.

(a) $y = (-x)^2 - 4(-x) = x^2 + 4x$

Replace every x with (−x).

(b) $-y = x^2 - 4x$, so $y = -x^2 + 4x$.

Replace y with −y, then multiply through by −1.

Example 2●●

$y = \sqrt{x - 1}$ is reflected in the $y$-axis. State the new equation, domain and range.

$y = \sqrt{-x - 1}$

$-x - 1 \ge 0$, so the domain is $x \le -1$.

The inside must be non-negative.

Range is still $y \ge 0$.

A vertical reflection would change the range; a horizontal one doesn't.

Flashcards

Reflect in the y-axis

Replace $x$ with $-x$. Point $(a, b) \to (-a, b)$.

Reflect in the x-axis

Replace $y$ with $-y$, so $y = -f(x)$. Point $(a, b) \to (a, -b)$.

Reflect in both axes

$y = -f(-x)$: a half-turn about the origin.

Practice

Reflecting $y = 3^x$ in the $y$-axis gives:
The point $(4, -2)$ is on $y = f(x)$. Which point is on $y = -f(x)$?
$y = x^2 + 5x - 7$ is reflected in the $y$-axis. What is the constant term of the new equation?
$y = \dfrac{1}{x - 2}$ is reflected in the $y$-axis. Its vertical asymptote becomes:
Reflecting $y = x^3$ in both axes gives:
$y = f(x)$ has an $x$-intercept at $x = 5$. Where is the $x$-intercept of $y = f(-x)$?
Which reflection changes the range of $y = 2^x$?
$y = -(x + 1)^2$ is $y = (x + 1)^2$ reflected in: