Skip to main content
SASpecialist MathematicsSyllabus dot point

How does taking the absolute value reshape a graph, and how do we solve equations and inequalities involving it?

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

How the modulus reflects negative parts of a graph above the axis, the difference between |f(x)| and f(|x|), and solving modulus equations and inequalities by cases or squaring.

Generated by Claude Opus 4.78 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Definition and key properties
  3. Sketching modulus graphs
  4. Solving modulus equations
  5. Solving modulus inequalities

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

Sketching modulus graphs

Two distinct transformations get confused, so keep them separate:

  • y=f(x)y = |f(x)|: sketch y=f(x)y = f(x), then reflect every part that is below the xx-axis up to above it. Points already at or above the axis stay put. The result never dips below the xx-axis and typically has sharp corners where ff crossed the axis.
  • y=f(x)y = f(|x|): keep the graph for x0x \ge 0, then reflect it in the yy-axis to produce the x<0x < 0 side. The result is symmetric about the yy-axis and ignores the original left half.

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.

For equations of the form A=B|A| = |B|, squaring is often faster, since A=B    A2=B2    A=±B|A| = |B| \iff A^2 = B^2 \iff A = \pm B.

Solving modulus inequalities

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

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

For f(x)<c|f(x)| < c, replace xx by f(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.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2017 SACE Stage 21 marksFor f(x) = 5/((x-2)(x+3)), draw the graph of y = |f(x)|, given the graph of y = f(x).
Show worked answer →

To graph y = |f(x)| from y = f(x), keep every part of the curve that is already at or above the x-axis unchanged, and reflect every part that lies below the x-axis up across the x-axis (since the modulus makes all outputs non-negative). [1 mark]

For f(x) = 5/((x-2)(x+3)), the curve has vertical asymptotes at x = 2 and x = -3 and is negative on the middle branch between x = -3 and x = 2 (where (x-2)(x+3) is negative). After applying the modulus, that middle branch flips above the x-axis, so the whole graph of |f(x)| lies at or above y = 0, still with vertical asymptotes at x = -3 and x = 2.

2017 SACE Stage 22 marksFor f(x) = 5/((x-2)(x+3)), draw the graph of y = |f(x)| - f(x).
Show worked answer →

Use the definition of modulus on each region:
Where f(x) >= 0, |f(x)| = f(x), so |f(x)| - f(x) = 0. The graph sits on the x-axis there.
Where f(x) < 0, |f(x)| = -f(x), so |f(x)| - f(x) = -f(x) - f(x) = -2 f(x). [1 mark]

For f(x) = 5/((x-2)(x+3)), f is negative only on the middle branch -3 < x < 2. So y = |f(x)| - f(x) is zero on the two outer branches (x < -3 and x > 2) and equals -2 f(x) = -10/((x-2)(x+3)) on -3 < x < 2, a positive curve rising to the vertical asymptotes at x = -3 and x = 2. [1 mark]