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TASSpecialist MathematicsSyllabus dot point

How do we sketch and solve problems with the absolute value function?

Sketch graphs involving absolute value and solve absolute value equations and inequalities.

The absolute value function, sketching y equals the modulus of f of x, and solving absolute value equations and inequalities, with worked examples for TCE Mathematics Specialised Unit 3.

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What this dot point is asking

This dot point develops fluency with the absolute value (modulus) function, both as a graph to sketch and as something to solve. The single defining idea is that x|x| measures size without regard to sign.

Sketching the basic graph

The graph of y=xy = |x| is a V shape with its vertex at the origin, made of the line y=xy = x for x0x \ge 0 and the line y=xy = -x for x<0x < 0. Transformations work just like any function: y=x2+1y = |x - 2| + 1 shifts the V two units right and one unit up, so its vertex sits at (2,1)(2, 1).

Sketching y equals the modulus of f of x

This is different from y=f(x)y = f(|x|), which instead takes the right-hand part of ff and reflects it across the yy axis to make an even function. Read carefully which version a question wants.

Solving absolute value equations

To solve f(x)=a|f(x)| = a (with a0a \ge 0), the inside can be either +a+a or a-a:

f(x)=aorf(x)=a. f(x) = a \quad\text{or}\quad f(x) = -a.

Always check solutions in the original equation, and reject any if a<0a < 0 since an absolute value can never be negative.

Solving absolute value inequalities

Two standard patterns cover most cases:

f(x)<a    a<f(x)<a,f(x)>a    f(x)>a or f(x)<a. |f(x)| < a \iff -a < f(x) < a, \qquad |f(x)| > a \iff f(x) > a \text{ or } f(x) < -a.

The first traps the inside between a-a and aa; the second lets the inside escape on either side.

Why this matters

Absolute value graphs appear in piecewise modelling and in defining regions, and the case-split logic here is the same reasoning used for reciprocal and rational graphs. Mastering the reflection rule and the two inequality patterns makes those later graphs routine.

Exam-style practice questions

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

2022 TASC3 marksSolve the inequality |3x - 2| < 5.
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An absolute value inequality of the form |A| < k (with k > 0) is equivalent to the double inequality -k < A < k.

  1. Rewrite without the modulus: -5 < 3x - 2 < 5.

  2. Add 2 throughout: -3 < 3x < 7.

  3. Divide by 3 (positive, so the inequality directions are unchanged): -1 < x < 7/3.

The solution set is the open interval -1 < x < 7/3. Marks are awarded for setting up the two-sided inequality, the correct algebra, and stating the final interval with strict inequalities.

2024 TASC3 marksFind the real values of x for which |2 - 9x| >= 6.
Show worked answer →

An inequality |A| >= k (with k > 0) splits into two separate cases: A >= k OR A <= -k.

  1. Case 1: 2 - 9x >= 6, so -9x >= 4, and dividing by -9 (negative, so reverse the sign) gives x <= -4/9.

  2. Case 2: 2 - 9x <= -6, so -9x <= -8, and dividing by -9 (reverse the sign) gives x >= 8/9.

  3. Combine: the solution is x <= -4/9 or x >= 8/9.

Markers reward splitting into the two cases, correctly reversing the inequality when dividing by the negative coefficient, and the complete solution set. A common trap is forgetting to flip the inequality sign.