Sketch graphs of the modulus functions y equals the absolute value of f of x and y equals f of the absolute value of x, and solve modulus equations

What this dot point is asking

SCSA wants you to apply the two distinct modulus transformations to a known graph and to solve equations and inequalities involving absolute values.

The transformation y equals the absolute value of f of x

The absolute value forces the output to be non-negative. So $y = |f(x)|$ leaves the graph of $f$ unchanged wherever $f \ge 0$, and reflects the parts where $f < 0$ up across the $x$-axis. The $x$-intercepts of $f$ become sharp corners (the curve touches the axis and bounces back up).

The transformation y equals f of the absolute value of x

Here the input is made non-negative before applying $f$. For $x \ge 0$, $|x| = x$ so the graph is identical to $f$. For $x < 0$, $f(|x|) = f(-x)$, which is the mirror image of the right-hand part across the $y$-axis. The result is always symmetric about the $y$-axis, and the behaviour for $x < 0$ in the original $f$ is discarded.

Solving modulus equations and inequalities

Split on the sign of the expression inside the bars. For $|g(x)| = c$ with $c \ge 0$, solve $g(x) = c$ and $g(x) = -c$. For inequalities, $|g(x)| < c$ means $-c < g(x) < c$, while $|g(x)| > c$ means $g(x) > c$ or $g(x) < -c$. Always check candidate solutions against the original equation, since squaring or casing can introduce extraneous roots.