How Do You Graph y = |f(x)| and y = f(|x|)?

For $y=|f(x)|$, reflect every part of $y=f(x)$ below the $x$-axis above it. For $y=f(|x|)$, keep the part for $x\geq 0$, reflect it in the $y$-axis, and discard the original part for $x<0$.

This is AHL 2.16: Further transformations and modulus in IB Mathematics: Analysis and Approaches HL.

How the transformations work

The modulus acts on different quantities in the two expressions:

For $y=|f(x)|$, outputs cannot be negative, so the transformed range is restricted to $y\geq0$. The $x$-coordinates do not change, meaning the zeros of $f$ remain in the same positions.

For $y=f(|x|)$, the input is always non-negative. Therefore, only the original graph to the right of the $y$-axis determines the result. The transformed graph is always an even function, so it is symmetric about the $y$-axis.

For example, if $f(x)=x-2$, then

$$y=|f(x)|=|x-2|$$

reflects the negative portion of the line above the $x$-axis. In contrast,

$$y=f(|x|)=|x|-2$$

copies the right-hand part of $y=x-2$ onto the left.

An algebraic check confirms the distinction. If $f(x)=x^2-4$, then $|f(x)|=|x^2-4|$ lifts the section between the roots, while leaves the graph unchanged because is already even.

Also track the domain carefully. For $f(|x|)$, a value $x$ is allowed only when (|x|) lies in the domain of $f$; this can create a symmetric restricted domain.

A common misconception is to reflect the entire graph for $f(|x|)$. You must first discard the original $x<0$ section, then mirror the $x\geq0$ section.

IB exam technique

In a sketch question, clearly label intercepts, turning points, sharp corners, and asymptotes. On Paper 1, apply the transformation analytically; on Paper 2, a GDC may check your sketch, but your submitted graph must still show the required features.