Modulus functions (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Sketch graphs involving the modulus (absolute value) function and solve equations and inequalities containing modulus expressions.”

8 min readUpdated 2 Jun 2026

Quick answer

The modulus $|x|$ equals $x$ when $x \ge 0$ and $-x$ when $x < 0$ , so it is never negative. To sketch $y = |f(x)|$ , draw $y = f(x)$ then reflect any part below the $x$ -axis up over it. Solve modulus equations and inequalities either by splitting into cases ( $\ge 0$ and $< 0$ ), or - for $|A| = |B|$ type problems - by squaring both sides to remove the modulus.

Jump to a section

What this dot point is asking
Definition and key properties
Sketching modulus graphs
Solving modulus equations
The triangle inequality
Solving modulus inequalities
Exam-style questions

What this dot point is asking

You need to sketch graphs involving the modulus function and to solve equations and inequalities that contain modulus expressions.

Definition and key properties

The modulus function

|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}

It returns the distance of

x

from

0

, so

|x| \ge 0

always. Useful facts:

|x|^2 = x^2

|ab| = |a||b|

, and

|x| = c

(for

c > 0

) means

x = c

x = -c

. Also

|x - a|

is the distance between

x

and

a

on the number line.

Sketching modulus graphs

Two distinct transformations get confused, so keep them separate:

$y = |f(x)|$ : sketch $y = f(x)$ , then reflect every part that is below the $x$ -axis up to above it. Points already at or above the axis stay put. The result never dips below the $x$ -axis and typically has sharp corners where $f$ crossed the axis.
$y = f(|x|)$ : keep the graph for $x \ge 0$ , then reflect it in the $y$ -axis to produce the $x < 0$ side. The result is symmetric about the $y$ -axis and ignores the original left half.

Corners, not curves

The modulus produces sharp corners (non-differentiable points) wherever the inside expression equals zero. At those points the graph changes direction abruptly; a smooth curve there is a sketching error.

Solving modulus equations

The cleanest general method is to split at the points where each modulus argument changes sign, solve on each interval, and discard solutions that fall outside the interval assumed.

Solving a modulus equation by cases

Solve $|x - 3| = 2x - 1$

Step 1: Identify the sign-change point and set up cases

The expression $x - 3$ inside the modulus changes sign at $x = 3$ . This splits the number line into two regions where $|x-3|$ has a different algebraic form, so we treat each region separately.

The modulus argument $x - 3$ changes sign at $x = 3$ .

Step 2: Solve Case 1 and check

For $x \ge 3$ , the expression $x-3$ is non-negative, so $|x-3|=x-3$ . Solve the resulting linear equation and verify the solution lies in this region.

Case 1 ( $x \ge 3$ ): here $|x - 3| = x - 3$ , so

x - 3 = 2x - 1 \implies -2 = x.

But

x = -2

does not satisfy

x \ge 3

, so reject it.

Step 3: Solve Case 2 and check

For $x < 3$ , the expression $x-3$ is negative, so $|x-3|=-(x-3)=3-x$ . Solve the new equation, confirm the solution is in this region, and also verify that the right-hand side is non-negative (since a modulus cannot equal a negative number).

Case 2 ( $x < 3$ ): here $|x - 3| = -(x - 3) = 3 - x$ , so

3 - x = 2x - 1 \implies 4 = 3x \implies x = \tfrac{4}{3}.

Since

\tfrac{4}{3} < 3

, this is valid. Check the right side is non-negative:

2(\tfrac{4}{3}) - 1 = \tfrac{5}{3} > 0

. Good.

Final answer: $x = \dfrac{4}{3}$ .

For equations of the form $|A| = |B|$ , squaring is often faster, since $|A| = |B| \iff A^2 = B^2 \iff A = \pm B$ .

Solving an equation with two moduli

Solve $|x + 1| = |2x - 4|$

Step 1: Square both sides

When both sides of an equation are moduli (and hence non-negative), squaring is valid because $|A|=|B|$ is equivalent to $|A|^2=|B|^2$ , and squaring removes the modulus signs on both sides without introducing extraneous solutions from sign changes.

Both sides are moduli, so square: $(x + 1)^2 = (2x - 4)^2$ .

Step 2: Expand and simplify

Expand each side using the standard binomial identity and collect all terms on one side.

Expanding gives $x^2 + 2x + 1 = 4x^2 - 16x + 16$ , so $3x^2 - 18x + 15 = 0$ , that is $x^2 - 6x + 5 = 0$ .

Step 3: Factorise, solve and verify

Factorise the quadratic to read off the solutions, then confirm each by substituting back into the original equation.

The equation factorises to $(x - 1)(x - 5) = 0$ . The solutions are $x = 1$ and $x = 5$ , both of which can be confirmed by substitution into the original equation.

Final answer: $x = 1$ and $x = 5$ .

The triangle inequality

A property worth knowing is the triangle inequality, $|a + b| \le |a| + |b|$ , with equality only when $a$ and $b$ have the same sign. It captures the idea that the distance of a sum is never more than the sum of the distances, and it underlies many bounds in later mathematics. While SACE rarely asks you to prove it, recognising the modulus as a measure of distance - $|x - a|$ being the gap between $x$ and $a$ on the number line - is the intuition that makes both the triangle inequality and the geometric reading of modulus equations natural.

Solving modulus inequalities

Two standard equivalences handle most cases, for $c > 0$ :

|x| < c \iff -c < x < c, \qquad |x| > c \iff x < -c \ \text{or} \ x > c.

For

|f(x)| < c

, replace

x

f(x)

and solve the resulting double inequality. A reliable alternative is to square (valid because both sides are non-negative) and solve the polynomial inequality, then sketch to read off the solution set.

A modulus inequality

Solve $|2x + 1| \le 5$

Step 1: Convert the modulus inequality to a double inequality

The rule $|A| \le c$ (for $c > 0$ ) is equivalent to $-c \le A \le c$ . This captures the idea that the expression $A$ must lie within distance $c$ of zero on the number line. Applying the rule to $|2x+1| \le 5$ gives a compound inequality with no modulus.

-5 \le 2x + 1 \le 5.

Step 2: Solve for $x$

Perform the same arithmetic operation across all three parts of the compound inequality to isolate $x$ .

Subtract $1$ : $-6 \le 2x \le 4$ . Divide by $2$ : $-3 \le x \le 2$ .

Final answer: $-3 \le x \le 2$ .

Common errors

Confusing $|f(x)|$ with $f(|x|)$: The first reflects below-axis parts upward; the second reflects the right half across the $y$ -axis. They give different graphs.
Forgetting to check solutions against the case: A value found in the " $x \ge a$ " case must actually satisfy $x \ge a$ ; otherwise discard it. This is the most common lost mark.
Dropping the negative branch: $|x| = c$ gives both $x = c$ and $x = -c$ . Solving only the positive case loses half the answers.
Squaring when a side can be negative: Squaring is safe for $|A| = |B|$ , but for $|A| = g(x)$ you must also require $g(x) \ge 0$ , or check each solution.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style2 marks

Calculator-assumed. For

f(x) = \dfrac{5}{(x-2)(x+3)}

, describe how to obtain the graph of

y = |f(x)|

from the graph of

y = f(x)

, and state the asymptotes.

Show worked answer →

Keep every part of $y = f(x)$ that is already at or above the $x$ -axis, and reflect every part below the $x$ -axis up across it, since the modulus makes all outputs non-negative. [1 mark]

For $f(x) = \dfrac{5}{(x-2)(x+3)}$ the curve has vertical asymptotes at $x = 2$ and $x = -3$ , and is negative on the middle branch $-3 < x < 2$ . After the modulus, that middle branch flips above the axis, so $y = |f(x)|$ lies at or above $y = 0$ , still with vertical asymptotes at $x = -3$ and $x = 2$ and horizontal asymptote $y = 0$ . [1 mark]

SACE 2023-style3 marks

Calculator-free. Solve

|2x - 3| = x + 4

Show worked answer →

The argument $2x - 3$ changes sign at $x = \tfrac{3}{2}$ .

Case 1 ( $x \ge \tfrac{3}{2}$ ): $2x - 3 = x + 4$ , so $x = 7$ . Since $7 \ge \tfrac{3}{2}$ , valid. [1 mark]

Case 2 ( $x < \tfrac{3}{2}$ ): $-(2x - 3) = x + 4$ , so $3 - 2x = x + 4$ , giving $-1 = 3x$ and $x = -\tfrac{1}{3}$ . Since $-\tfrac{1}{3} < \tfrac{3}{2}$ , valid. [1 mark]

Both make the right side $x + 4$ non-negative, so the solutions are $x = 7$ and $x = -\tfrac{1}{3}$ . [1 mark]

SACE 2021-style2 marks

Calculator-free. Solve the inequality

|3x + 2| < 8

Show worked answer →

Use $|A| < c \iff -c < A < c$ :

-8 < 3x + 2 < 8.

Subtract $2$ : $-10 < 3x < 6$ . Divide by $3$ : $-\tfrac{10}{3} < x < 2$ .

Marks: one for the double inequality, one for the solution $-\tfrac{10}{3} < x < 2$ .

Practise this

Sources & how we know this

Stage 2 Specialist Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 Specialist Mathematics: external assessment — SACE Board

What this dot point is asking

Definition and key properties

Sketching modulus graphs

Solving modulus equations

Solve ∣x−3∣=2x−1|x - 3| = 2x - 1∣x−3∣=2x−1

Step 1: Identify the sign-change point and set up cases

Step 2: Solve Case 1 and check

Step 3: Solve Case 2 and check

Solve ∣x+1∣=∣2x−4∣|x + 1| = |2x - 4|∣x+1∣=∣2x−4∣

Step 1: Square both sides

Step 2: Expand and simplify

Step 3: Factorise, solve and verify

The triangle inequality

Solving modulus inequalities

Solve ∣2x+1∣≤5|2x + 1| \le 5∣2x+1∣≤5

Step 1: Convert the modulus inequality to a double inequality

Step 2: Solve for xxx

Exam-style questions

Practise this

Sources & how we know this

Solve $|x - 3| = 2x - 1$

Solve $|x + 1| = |2x - 4|$

Solve $|2x + 1| \le 5$

Step 2: Solve for $x$