Parabola with absolute value

Consider the graph of $y = |x^2 - 1|$ . It is the parabola $y = x^2 - 1$ with the part below the $x$ -axis, where $-1 < x < 1$ , folded upward across the $x$ -axis.

Writing it by cases

Range	Sign of $x^2 - 1$	$y$	Shape
$\|x\| \geq 1$	zero or positive	$x^2 - 1$	concave up
$\|x\| < 1$	negative	$1 - x^2$	concave down, a hump

The second is a downward parabola forming a hump with its peak at $(0, 1)$ . The range is $y \geq 0$ : the function never takes a negative value.

The corners

At $x = -1$ and $x = 1$ , where $x^2 - 1 = 0$ , the graph reaches the $x$ -axis and turns with a corner, and there it is not differentiable. Just to the right of $x = 1$ the slope is $2x = 2$ , while just to the left the slope of $y = 1 - x^2$ is $-2x = -2$ ; the two do not agree. Taking an absolute value plants a corner wherever the original function crosses zero.

When a corner appears

In general the graph of $y = |f(x)|$ is that of $f$ with everything below the $x$ -axis reflected above it. But at a zero where $f$ merely touches the axis without crossing it, the fold creates no corner.

Function	Behavior at the zero	Result of folding
$y = \|x^2 - 1\|$	crosses at $x = \pm 1$	corners appear
$y = \|x^2\|$	touches at $x = 0$	stays $y = x^2$ , still smooth

Whether the sign actually changes is what decides whether a corner appears.

Counting the solutions

Read the real solutions of $|x^2 - 1| = k$ as intersections with the horizontal line $y = k$ .

Range of $k$	Number of real solutions	Where they lie
$k < 0$	$0$	no intersection
$k = 0$	$2$	the corners $x = \pm 1$
$0 < k < 1$	$4$	two on the hump, two outside
$k = 1$	$3$	the top of the hump and two outside
$k > 1$	$2$	the outer branches only

The count changes as the line passes the height $1$ of the hump: it is the folded-up hump that multiplies the solutions.

Symmetry

$y = |x^2 - 1|$ is an even function, symmetric about the $y$ -axis, because $x^2 - 1$ is even and taking an absolute value preserves the symmetry. The large dots mark the corners $(-1, 0)$ and $(1, 0)$ and the peak $(0, 1)$ .