Modulus (absolute value) functions and graphs (WACE Year 12 Mathematics Specialist Unit 3)

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.

The piecewise definition behind the bars

Every modulus expression can be written piecewise, and this is what makes the casework rigorous. By definition $|u| = u$ when $u \ge 0$ and $|u| = -u$ when $u < 0$. So $y = |2x - 6|$ is $2x - 6$ for $x \ge 3$ and $6 - 2x$ for $x < 3$, the change of rule happening exactly where the inside expression is zero. Writing the piecewise form first, then sketching each linear or curved piece on its own interval, prevents the most common sketching errors and makes the V-shaped vertices land in the right place.

Solving modulus equations graphically and algebraically

There are two reliable strategies. Algebraically, split on the sign of each expression inside bars and solve a separate equation on each interval, discarding any solution that violates its interval condition. Graphically, sketch each side of the equation and read off the $x$-coordinates of the intersections; this is especially useful when one side is itself a modulus. The two approaches agree, and using both is a strong way to check an answer in the exam.

Inequalities and the positive-right-side condition

A modulus inequality such as $|g(x)| < h(x)$ is only solvable where $h(x) > 0$, because a modulus is never negative. After imposing that, rewrite $|g(x)| < h(x)$ as the double inequality $-h(x) < g(x) < h(x)$ and solve both parts, intersecting all the conditions. For $|g(x)| > h(x)$ the solution is $g(x) > h(x)$ or $g(x) < -h(x)$, a union rather than an intersection. Keeping the logical connective (and versus or) correct is where most marks are lost.

Squaring as an alternative for equations

For an equation of the form $|g(x)| = |h(x)|$, where both sides are moduli, squaring is often cleaner than casework, because $|u| = |v|$ is equivalent to $u^2 = v^2$. Squaring gives $g(x)^2 = h(x)^2$, a polynomial equation with no modulus signs, and $g^2 - h^2 = 0$ factors as $(g - h)(g + h) = 0$, recovering the two cases automatically. Squaring is only safe when both sides are known to be non-negative or when both are inside modulus bars; squaring an equation such as $|g(x)| = x - 1$ can introduce extraneous roots where $x - 1 < 0$, so every candidate must still be checked.

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.