Sketch $y = |f(x)|$ by reflecting the part of $y = f(x)$ below the $x$-axis upward, and sketch $y = f(|x|)$ by keeping the part right of the $y$-axis and mirroring it across the $y$-axis, recognising that $y = f(|x|)$ is always even and that the two transformations coincide only in special cases

What this dot point is asking

NESA gives you the graph of $y = f(x)$ and asks you to sketch two new graphs from it: $y = |f(x)|$ and $y = f(|x|)$. These look almost the same on the page, but they are two completely different transformations, and the single most common mistake in this topic is to muddle them. The whole skill is knowing which one acts on the output and which acts on the input, and reading the picture off accordingly.

The two rules in one sentence each:

• $y = |f(x)|$ takes the absolute value of the height, so it reflects the part of the curve below the $x$-axis up above it and leaves the rest alone. The output can never be negative.
• $y = f(|x|)$ takes the absolute value of the input, so it throws away the left half and replaces it with a mirror image of the right half across the $y$-axis. The result is always even.

Everything else on this page is a careful working-out of those two rules, the cases where they happen to give the same picture, and how to combine them. The same "act on the picture, feature by feature" habit drives sketching the reciprocal and adding ordinates; here the features are the $x$-intercepts and the $y$-axis.

The answer

Two rules, two different parts of the graph

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

The outer modulus $y = |f(x)|$ acts on the output ($y$). For any $x$, the new height is $|f(x)|$. Where $f(x) \ge 0$ the height is unchanged, because the absolute value of a non-negative number is itself. Where $f(x) < 0$ the new height is $-f(x)$, the positive version, which is exactly the old point reflected in the $x$-axis. So the recipe is:

Keep everything on or above the $x$-axis; reflect everything below the $x$-axis up.

The $x$-intercepts are the hinge points: a height of $0$ stays $0$, so the curve is pinned to the axis there, and it bounces off, creating a sharp corner at each intercept.

The inner modulus $y = f(|x|)$ acts on the input ($x$). For $x \ge 0$ we have $|x| = x$, so $f(|x|) = f(x)$ and the right half is unchanged. For $x < 0$ we have $|x| = -x$, so $f(|x|) = f(-x)$, which is the value of $f$ at the matching positive input: the left half is the mirror image of the right half across the $y$-axis. So the recipe is:

Keep the part right of the $y$-axis; replace the left part with its mirror image across the $y$-axis.

The original left half of $y = f(x)$ is thrown away entirely and plays no part. Because the two sides are now mirror images, $y = f(|x|)$ is always an even function.

$y = |f(x)|$: flip the lower lobe up (a parabola)

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 $(0, -4)$.

The single lobe between the intercepts is below the axis, so it flips up. The vertex $(0, -4)$ becomes the peak $(0, 4)$, a reflection straight up through twice $4 = 8$ units. The arms for $|x| > 2$ are already above the axis and do not move. The intercepts $(-2, 0)$ and $(2, 0)$ are fixed, and the smooth turning point of the parabola becomes two sharp corners there, because the curve arrives going down and leaves going up. A quick table confirms every height is just made positive:

$y = f(|x|)$: discard the left, mirror the right (a cubic)

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 right half ($x \ge 0$) is kept exactly: it passes through the origin, dips to a minimum at $\left(2/\sqrt{3}, -3.079\right)$ and crosses up through $(2, 0)$. The original left half is deleted and replaced by the mirror image of that right half, giving a matching dip at $\left(-2/\sqrt{3}, -3.079\right)$ and an intercept at $(-2, 0)$. Where the two halves meet at the origin there is a corner, because the right half arrives with a negative slope and the mirror leaves with the opposite slope. The finished graph has line symmetry in the $y$-axis: it is even, which you can confirm because $h(|{-x}|) = h(|x|)$ for every $x$.

Compare this with what $y = |h(x)|$ does to the same cubic: there we reflect the below-axis pieces up instead of mirroring, and the picture is different.

For $|h(x)|$ both below-axis pieces (the far left for $x < -2$ and the dip between $0$ and $2$) flip up, giving three corners at $-2, 0, 2$ and turning the original minimum $\left(2/\sqrt{3}, -3.079\right)$ into a maximum $\left(2/\sqrt{3}, 3.079\right)$. This graph is not even. Putting the two cubic graphs side by side is the quickest way to feel that $|f(x)|$ and $f(|x|)$ really are different operations.

When the two transformations agree (and when they cannot)

Because the rules act on different parts, they usually give different graphs, but two clean cases make them coincide or trivial.

An even function is unchanged by $f(|x|)$. 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.

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

The exponential $f(x) = 2^x$ is the standard example: $2^x > 0$ for all $x$, so $|2^x| = 2^x$ and the graph does not change at all. (Be careful: this is special to $|f(x)|$. The other transformation, $f(|x|) = 2^{|x|}$, the muted curve above, does change $2^x$, into the even V-bottomed exponential that agrees with $2^x$ only for $x \ge 0$.) The contrast matters: $y = |2^x - 1|$, where the function does dip below the axis for $x < 0$, is genuinely altered, so do not assume every exponential is immune.

Combining the two, stage by stage

The richest exam questions ask for both transformations at once, usually $y = |f(|x|)|$: first make the function even with $f(|x|)$, then make it non-negative with the outer $|\,\cdot\,|$. Order matters, so work from the inside out. Take $k(x) = (x - 1)^2 - 2$, a parabola with vertex $(1, -2)$ and $x$-intercepts at $x = 1 \pm \sqrt{2}$ (about $-0.41$ and $2.41$). It is deliberately not even, so every stage does visible work.

Stage 1, plot the original $y = k(x)$. Read off the landmarks of $k(x) = (x - 1)^2 - 2$: vertex $(1, -2)$, $y$-intercept $(0, -1)$, and $x$-intercepts at $x = 1 - \sqrt{2} \approx -0.41$ and $x = 1 + \sqrt{2} \approx 2.41$. The graph is not symmetric in the $y$-axis.

Stage 2, apply the inner modulus $y = k(|x|)$. Keep the right half ($x \ge 0$) and mirror it across the $y$-axis. The right half starts at $(0, -1)$, dips to the vertex $(1, -2)$ and rises through the intercept at $x \approx 2.41$. Mirroring gives a matching dip at $(-1, -2)$ and an intercept at $x \approx -2.41$, with a corner at $(0, -1)$ where the halves meet. The graph is now even, but it still dips below the axis.

Stage 3, apply the outer modulus $y = |k(|x|)|$. Now reflect the part of the stage-2 graph below the $x$-axis up. The whole central region between $x \approx -2.41$ and $x \approx 2.41$ is below the axis, so it flips up: the two dips at $(\pm 1, -2)$ become two humps peaking at $(\pm 1, 2)$, and the corner at $(0, -1)$ becomes a notch at $(0, 1)$.

Stage 4, finish and read off the features. The completed graph $y = |k(|x|)|$ is even and never negative (range $y \ge 0$). It has $x$-intercepts at $x \approx \pm 2.41$ (each a sharp corner where the curve touches the axis), two peaks at $(\pm 1, 2)$, and a central notch at $(0, 1)$, then rises on the outside. Every feature was produced by applying the inner modulus first, then the outer one.

A reciprocal under the modulus

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.

Under $y = |g(x)|$ the right branch is unchanged (already positive), and the left branch flips up. Instead of the left branch plunging to $-\infty$ as $x \to 1^-$, it now rises to $+\infty$; instead of approaching the $x$-axis from below as $x \to -\infty$, it approaches from above. So both arms of $y = \left|\dfrac{1}{x - 1}\right|$ shoot up beside the asymptote $x = 1$, and the whole graph lies above the $x$-axis with horizontal asymptote $y = 0$. The fixed point $(2, 1)$ on the right branch does not move; the point $(0, -1)$ on the left branch reflects up to $(0, 1)$.

A note on even and odd

The two transformations relate to symmetry in opposite ways, and this is worth pinning down because it explains every "they agree / they differ" case above.

$y = f(|x|)$ is always even, whatever $f$ was. 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.

$y = |f(x)|$ keeps the symmetry $f$ already had. 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 $f$.

How exam questions ask about absolute-value graphs

The wording tells you which rule to reach for.

• "Sketch $y = |f(x)|$ from the graph of $y = f(x)$." Reflect the part below the $x$-axis up; keep the rest. Mark the $x$-intercepts as corners and flip any below-axis minimum into a maximum.
• "Sketch $y = f(|x|)$." Keep the right half ($x \ge 0$) and mirror it across the $y$-axis. State that the result is even.
• "Sketch $y = |f(|x|)|$" (or any combination). Apply the inner modulus first ($f(|x|)$, making it even), then the outer modulus (reflecting below-axis parts up). Show both stages.
• "Explain why $y = |f(x)|$ is the same as $y = f(x)$." Argue that $f(x) \ge 0$ for all $x$ (no part below the axis to reflect), as for $y = 2^x$.
• "Explain why $y = f(|x|)$ is the same as $y = f(x)$." Argue that $f$ is already even (its graph already has $y$-axis symmetry), as for $y = x^2 - 4$.
• "Is the transformed graph even?" $f(|x|)$ is always even. $|f(x)|$ is even exactly when $f$ is even or odd; otherwise check directly.
• "State the range of $y = |f(x)|$." It is a subset of $y \ge 0$; read the maximum from the reflected peak if the function is bounded.