Quick questions on Graphing y = |f(x)| and y = f(|x|)

Start from a graph of $y = f(x)$ and ask what each modulus does.

The clearest first example is a parabola that dips below the axis. Take $f(x) = x^2 - 4$, with $x$-intercepts at $x = -2$ and $x = 2$ and vertex $(0, -4)$.

For $y = f(|x|)$ the contrast is sharpest with a function that is not already symmetric, so the left half genuinely changes. Take the cubic $h(x) = x^3 - 4x = x(x - 2)(x + 2)$, with $x$-intercepts at $x = -2$, $x = 0$ and $x = 2$.

The outer modulus is just as useful on a curve with an asymptote. Take $g(x) = \dfrac{1}{x - 1}$, with a vertical asymptote at $x = 1$: the right branch ($x > 1$) is positive and the left branch ($x < 1$) is negative.

If $f$ is already even, its graph already has $y$-axis symmetry, so "keep the right, mirror it" reproduces the same curve: $f(|x|) = f(x)$. That is why the parabola $f(x) = x^2 - 4$ above is identical under $f(|x|)$. The transformation only does visible work when $f$ is not even.

If $f(x) \ge 0$ for every $x$, no part of the graph is below the axis, so there is nothing to reflect and $|f(x)| = f(x)$.

It is built by mirroring the right half across the $y$-axis, which forces line symmetry: $f(|{-x}|) = f(|x|)$. So if a question asks you to make a function even, $f(|x|)$ is the tool.

Reflecting heights up in the $x$-axis does not impose $y$-axis symmetry; it preserves whatever symmetry $f$ has. If $f$ is even, $|f(x)|$ is even; if $f$ is odd, $|f(x)|$ becomes even (because $|{-f(x)}| = |f(x)|$, so the heights match across the axis) but only as a side effect of the heights, not the shape. The safe statement to make in an exam is the precise one: $f(|x|)$ is always even; $|f(x)|$ is even whenever the heights are symmetric, which holds for any even or odd $f$.