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Algebra and functions · Subtopic 1 of 9

Modulus Graphs and Equations

What ∣x∣|x| means, how to sketch y=∣ax+b∣y=|ax+b| by reflecting the negative part, and two reliable ways to solve modulus equations: squaring both sides, or splitting into cases and checking.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Understand ∣x∣|x| as distance from 00, and ∣a−b∣|a-b| as the distance between aa and bb.
  • Sketch y=∣ax+b∣y=|ax+b|, marking the vertex and intercepts.
  • Solve ∣ax+b∣=k|ax+b|=k and ∣ax+b∣=∣cx+d∣|ax+b|=|cx+d| by squaring or by cases.
  • Solve ∣ax+b∣=cx+d|ax+b|=cx+d, checking for and rejecting invalid solutions.
Pause and think

On a number line, which numbers are exactly 33 units away from 55?

Section 2 of 14

Core idea

Start hereHow far, not which way

A lift is at floor 22. It moves to floor −3-3 (car park) or floor 77. How far did it travel? 55 floors either way, regardless of direction. Distance ignores sign. The modulus ∣x∣|x| is the mathematical "distance from zero", and it creates graphs with sharp corners and equations with two answers.

The big idea

∣x∣|x| is the size of xx without its sign: ∣x∣=x|x|=x if x≥0x\ge0, ∣x∣=−x|x|=-x if x<0x<0. ∣x−a∣|x-a| is the distance between xx and aa. ∣f(x)∣=k|f(x)|=k splits into f(x)=kf(x)=k or f(x)=−kf(x)=-k; y=∣f(x)∣y=|f(x)| reflects the negative parts of y=f(x)y=f(x) upward.

1. Definition

∣x∣={x,x≥0−x,x<0|x|=\begin{cases}x,&x\ge0\\-x,&x<0\end{cases}
where
−x-x when x<0x<0
minus a negative is positive: ∣−5∣=−(−5)=5|-5|=-(-5)=5.
  • ∣5∣=5|5|=5, ∣−5∣=5|-5|=5, ∣0∣=0|0|=0.
  • ∣x−a∣|x-a| = distance between xx and aa on the number line.
  • ∣ab∣=∣a∣∣b∣|ab|=|a||b| but ∣a+b∣≤∣a∣+∣b∣|a+b|\le|a|+|b| (not always equal).

2. Graphs

Making y = |2x − 3| from y = 2x − 3
−11234−4−2240xyy = 2x − 3y = |2x − 3|vertex (1.5, 0)
Any part of the line below the xx-axis is reflected upwards, giving a V shape. The point of the V is where the inside, 2x−32x-3, equals zero.
Reflecting the negative part of a curve
−3−2−1123−4−2240xy
The part of y=x2−4y=x^2-4 between −2-2 and 22 (below the axis) flips over the axis to make y=∣x2−4∣y=|x^2-4|.

To draw y=∣f(x)∣y=|f(x)|: draw y=f(x)y=f(x), then reflect every part below the xx-axis in the axis. The vertex of y=∣2x−3∣y=|2x-3| is where the inside is zero: x=32x=\frac32.

Edge case∣f(x)∣|f(x)| versus f(∣x∣)f(|x|) y=f(∣x∣)y=f(|x|) is different: take the part for x≥0x\ge0 and reflect it in the yy-axis. ∣x∣2−4∣x∣|x|^2-4|x| is symmetric about the yy-axis; ∣x2−4x∣|x^2-4x| is not.

3. Solving equations

Worked example∣2x−3∣=7|2x-3|=7

Solve.

Goal
All xx.
Strategy
The inside is 77 or −7-7.
  1. 2x−3=7⇒x=52x-3=7\Rightarrow x=5
    Positive case.
  2. 2x−3=−7⇒x=−22x-3=-7\Rightarrow x=-2
    Negative case.
Answerx=5x=5 or x=−2x=-2.
✓ Check∣7∣=7|7|=7 ✓; ∣−7∣=7|-7|=7 ✓.
Worked example∣x−1∣=∣2x+3∣|x-1|=|2x+3| by squaring

Solve.

Goal
All xx.
Strategy
Both sides are ≥0\ge0, so squaring loses nothing and gains nothing; ∣A∣2=A2|A|^2=A^2.
  1. (x−1)2=(2x+3)2(x-1)^2=(2x+3)^2
    Square both sides.
  2. x2−2x+1=4x2+12x+9x^2-2x+1=4x^2+12x+9
    Expand.
  3. 0=3x2+14x+8=(3x+2)(x+4)0=3x^2+14x+8=(3x+2)(x+4)
    Collect; factorise.
Answerx=−23x=-\frac23 or x=−4x=-4.
✓ Checkx=−4x=-4: ∣−5∣=5|-5|=5 and ∣−5∣=5|-5|=5 ✓.
Worked example∣x−4∣=2x+1|x-4|=2x+1: checking is essential

Solve.

Goal
All valid xx.
Strategy
Try both cases; reject any where the right side would be negative.
  1. x−4=2x+1⇒x=−5x-4=2x+1\Rightarrow x=-5
    Case 1.
  2. Check: RHS =2(−5)+1=−9<0=2(-5)+1=-9<0, but a modulus can't be negative. Reject.

  3. −(x−4)=2x+1⇒3=3x⇒x=1-(x-4)=2x+1\Rightarrow3=3x\Rightarrow x=1
    Case 2.
  4. Check: ∣1−4∣=3|1-4|=3 and 2+1=32+1=3 ✓.

Answerx=1x=1 only.
✓ CheckA sketch shows the line y=2x+1y=2x+1 crossing the V once ✓.
Common misconceptionNot checking When the other side isn't a modulus, one case can give a false root. Always substitute back, or sketch first.
Pause and think

Solve ∣3x+1∣=5|3x+1|=5.

Summary

  • ∣x∣|x| = distance from 00.
  • ∣f∣=k|f|=k: two cases.
  • ∣f∣=∣g∣|f|=|g|: square.
  • ∣f∣=g|f|=g: check solutions.
Key words
Modulus
Absolute value.
HistoryTwo straight lines. The notation ∣x∣|x| was introduced by Karl Weierstrass in 1841. Before that, mathematicians wrote "the numerical value of xx". Weierstrass also made calculus rigorous, and his vertical bars are now used for size everywhere: the modulus of a complex number, the magnitude of a vector, the determinant of a matrix.
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