Graphs with absolute value

Sketch y = |f(x)| and y = f(|x|) from the graph of y = f(x) using reflections in the axes, and see both as composites with the absolute value function.

The graph of y = |f(x)|

y = |f(x)| for f(x) = (x − a)² − 4

Blue dashed: y = f(x). Orange: y = |f(x)|. The piece of the parabola between the zeroes flips up, leaving sharp corners at the zeroes.

Try this

Which points of the blue graph do not move at all?

The graph of y = f(|x|)

y = f(|x|) for f(x) = (x − a)² − 4

Blue dashed: y = f(x). Orange: y = f(|x|). Only the part with x ≥ 0 survives, and it is copied across the y-axis.

Try this

Make $a$ negative. Why is the lowest point of $y = f(|x|)$ then at the $y$-intercept?

Example 1●●

Let $f(x) = x^2 - 2x - 3 = (x + 1)(x - 3)$. Describe $y = |f(x)|$ and $y = f(|x|)$.

$y = |f(x)|$: the same zeroes $-1$ and $3$, the vertex flips up to a local maximum $(1, 4)$, and the $y$-intercept becomes $3$.

f is negative between its zeroes −1 and 3, with vertex (1, −4).

$y = f(|x|)$: zeroes $\pm 3$, minimum points $(\pm 1, -4)$, and a sharp corner at $(0, -3)$.

Keep x ≥ 0: from (0, −3) down to (1, −4), then up through (3, 0).

As composite functions

Let $A(x) = |x|$. Then $|f(x)| = A(f(x))$: apply $f$ first, then take the absolute value of the output. And $f(|x|) = f(A(x))$: take the absolute value of the input first. That is why the first changes $y$-values (a vertical effect) and the second changes $x$-values (a horizontal effect).

Flashcards

$y = |f(x)|$

Reflect the parts below the $x$-axis up into it.

$y = f(|x|)$

Keep the part with $x \ge 0$ and copy it into $x < 0$ by reflecting in the $y$-axis.

Symmetry of $f(|x|)$

Always even.

Practice

To sketch $y = |f(x)|$ from $y = f(x)$:
To sketch $y = f(|x|)$ from $y = f(x)$:
The function $y = f(|x|)$ is always:
For $f(x) = x^2 - 2x - 3$, find the $y$-coordinate of the local maximum of $y = |f(x)|$.
For $f(x) = x^2 - 2x - 3$, how many zeroes does $y = f(|x|)$ have?
For $f(x) = x^2 - 2x - 3$, find $f(|-2|)$.
For $f(x) = x^2 - 2x - 3$, find the $y$-intercept of $y = |f(x)|$.
$|f(x)| = f(x)$ exactly when:
The range of $y = |x^2 - 4|$ is:
With $A(x) = |x|$, the function $y = f(|x|)$ is the composite: