IP AMaths Notes (Upper Sec, Year 3-4): 11) Modulus Functions

Study guide08 Nov 2025, 00:00 ZUpdated 14 Oct 2026

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Marcus Pang·Managing Director (Maths)

Piecewise rewriting, graph sketching, and inequality solving with modulus expressions for IP AMaths.

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For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.

How this chapter applies

Eclat core: piecewise interpretation, graph reflection, and modulus equations form an Eclat IP extension where the current school teaches them.
School-sensitive extension: the whole chapter is outside K341 core and must not be assumed examinable for every IP student.
2027 national comparison: K341 has no standalone modulus-functions topic or outcome.
Check your school: check the current topic list before spending assessment preparation time on this chapter.
Exam-track route: use the separate O-Level and SEC G3 Additional Mathematics notes for K341 topic ownership.

Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 11) Modulus Functions cover?
A: Piecewise rewriting, graph sketching, and inequality solving with modulus expressions for IP AMaths.

Modulus expressions measure distance from zero. When the inside expressions are linear, rewrite them in piecewise linear form to analyse turning points and solve inequalities.

Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.

New to the Integrated Programme? Start with What is IP? | Browse all free IP notes.

The core idea is simple: A modulus expression measures distance from zero.

Use it as a working check: Find the critical points where the expression inside the modulus is zero, then solve each region with the correct sign.

Then go one layer deeper: Example: for $\lvert 2x - 3\rvert$ , split at $x = \dfrac{3}{2}$ . To the right it is $2x - 3$ ; to the left it is $-(2x - 3)$ .

Sign-region workflow

Use a sign-region chart before expanding a modulus expression. It keeps the algebra tied to the graph.

For y = |2x - 3| - 1:

critical point: 2x - 3 = 0 -> x = 3/2

region                 sign of 2x - 3     rewrite
x < 3/2                negative            y = -(2x - 3) - 1 = -2x + 2
x >= 3/2               non-negative        y = 2x - 3 - 1 = 2x - 4

The two line segments must meet at the critical point. If they do not meet, check the sign rewrite before sketching.

Multiple-modulus checkpoint

When an expression has more than one modulus term, collect every critical point first, then build intervals from left to right. Each interval needs its own sign decision for every modulus term.

Step	What to do	Common trap
1	Set each inside expression equal to zero.	Using only the first modulus term's critical point.
2	Sort the critical points on a number line.	Treating overlapping regions out of order.
3	Pick a test value in each interval.	Guessing signs without checking both modulus terms.
4	Rewrite every modulus term on that interval.	Changing one modulus term but leaving the other unchanged.
5	Check that adjacent pieces meet at the critical points.	Sketching disconnected line segments.

Worked check: for $y = |x + 1| + |x - 3|$ , the critical points are $x=-1$ and $x=3$ . On $x < -1$ , both inside expressions are negative, so

$y = -(x+1) -(x-3) = -2x + 2.$

On $-1 \leq x < 3$ , $x+1$ is non-negative but $x-3$ is negative, so

$y = (x+1) -(x-3) = 4.$

On $x \geq 3$ , both are non-negative, so

$y = (x+1) + (x-3) = 2x - 2.$

Misconception check: the minimum is not found by setting each modulus term to zero at the same time. Here the expression stays at $4$ for every $x$ between $-1$ and $3$ , so the minimum value is $4$ across an interval.

1 Fundamental identities

$|x| = \begin{cases} x & x \geq 0 \ -x & x \lt 0 \end{cases}$ .
$|ax + b| = |a| \cdot \bigl|x + \dfrac{b}{a}\bigr|$ for $a \neq 0$ .
For $c \geq 0$ , solve $|f(x)| = c$ by considering $f(x) = c$ or $f(x) = -c$ . For $c < 0$ , there is no real solution.

Modulus method-choice checkpoint

Before solving, decide what the modulus is doing in the question.

Question form	Best first move	What it means	Trap to avoid
$\vert f(x) \vert = c$ , where $c \geq 0$	Split into $f(x) = c$ or $f(x) = -c$ .	The inside expression is distance $c$ from zero.	If $c < 0$ , there is no solution.
$\vert f(x) \vert \leq c$ , where $c \geq 0$	Write $-c \leq f(x) \leq c$ .	The inside expression stays within a band.	Do not use "or" for the inside band.
$\vert f(x) \vert \geq c$ , where $c \geq 0$	Write $f(x) \leq -c$ or $f(x) \geq c$ .	The inside expression lies outside the band.	Do not join the two outside regions with "and".
$\vert f(x) \vert = \vert g(x) \vert$	Use piecewise regions or square both sides carefully.	Two expressions are the same distance from zero.	Both sides are non-negative, so squaring is equivalent on the common real domain. Retain the original domain restrictions.
Graph of $y = \vert f(x) \vert$	Sketch $y = f(x)$ , then reflect negative parts above the x-axis.	The output cannot be negative.	Do not move the whole graph up; only negative y-values are reflected.

Misconception check: modulus is not a command to "make x positive". It makes the value of the inside expression non-negative, so the critical points still come from the inside expression.

Graph transformation checkpoint

For graph questions, decide whether the modulus is outside the function or inside the input. These are different transformations.

Function form	What changes	Sketch move	Common trap
$y = \vert f(x) \vert$	Negative output values become positive.	Sketch $y=f(x)$ , then reflect parts below the $x$ -axis above it.	Reflecting the left half of the graph just because there is a modulus sign.
$y = f( \vert x \vert )$	Negative input values are replaced by positive inputs.	Keep the $x \geq 0$ half, then reflect that half in the $y$ -axis.	Reflecting parts below the $x$ -axis instead of using the right half.
$y = \vert f( \vert x \vert ) \vert$	Input and output are both affected.	First build $f( \vert x \vert )$ , then reflect any negative output above the $x$ -axis.	Trying to do both reflections at once without checking the intermediate graph.

Worked check: if $f(x)=x-2$ , then $y=|f(x)|=|x-2|$ has a vertex at $(2,0)$ . But $y=f(|x|)=|x|-2$ has a vertex at $(0,-2)$ . The two graphs are not the same.

Misconception check: the position of the modulus bars tells you what is being made non-negative. Bars around the whole function affect $y$ -values; bars around $x$ affect the input before the function is evaluated.

2 Worked example - Piecewise sketch

Sketch $y = |2x - 3| - 1$ .

Set $2x - 3 = 0$ → $x = \dfrac{3}{2}$ .
For $x \geq \dfrac{3}{2}$ , $y = 2x - 3 - 1 = 2x - 4$ .
For $x \lt \dfrac{3}{2}$ , $y = -(2x - 3) - 1 = -2x + 2$ .
Plot both lines; the vertex occurs at $\bigl(\dfrac{3}{2}, -1\bigr)$ .

Trap check: the vertex x-coordinate comes from the inside expression $2x - 3 = 0$ , not from solving $y = 0$ . The x-intercepts are separate points on the graph.

3 Worked example - Modulus inequality

Solve $|x - 4| \leq 3$ .

Write double inequality: $-3 \leq x - 4 \leq 3$ .
Add $4$ throughout: $1 \leq x \leq 7$ .

Now solve $|2x + 5| > 9$ .

Either $2x + 5 \gt 9$ or $2x + 5 \lt -9$ .
First case: $2x \gt 4$ → $x \gt 2$ .
Second case: $2x \lt -14$ → $x \lt -7$ .

Solution: $x \lt -7$ or $x \gt 2$ .

Common inequality pattern for $c \geq 0$ :

|expression| <= c  ->  -c <= expression <= c
|expression| >= c  ->  expression <= -c or expression >= c

The first pattern describes the values inside a band. The second pattern describes the values outside the band.

4 Worked example - Solving a modulus equation

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

Identify critical points where expressions change sign: $x = \dfrac{1}{2}$ and $x = -4$ .
Case 1 ( $x \geq \dfrac{1}{2}$ ): both brackets non-negative. Equation becomes $2x - 1 = x + 4$ giving $x = 5$ .
Case 2 ( $-4 \leq x \leq \dfrac{1}{2}$ ): $2x - 1$ is non-positive while $x + 4$ is non-negative. Solve $-(2x - 1) = x + 4$ to get $-2x + 1 = x + 4$ ⇒ $x = -1$ .
Case 3 ( $x \leq -4$ ): both brackets non-positive. Equation becomes $-(2x - 1) = -(x + 4)$ ⇒ $-2x + 1 = -x - 4$ leading to $x = 5$ , which does not lie in this region, so discard.
Verify both admissible roots in the original equation: $x = -1$ and $x = 5$ satisfy the modulus equality.

Answer: $x = -1$ or $x = 5$ .

5 Practice Quiz

Make sure piecewise rewrites, absolute inequalities, and sketching routines are automatic before layering them onto coordinate geometry.

6 Try this

Rewrite $y = |x + 1| + |x - 3|$ as piecewise linear segments and determine the minimum value of $y$ .

Sources

Add Eclat Institute as a Preferred Source on Google.