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 measures size without regard to sign.
Sketching the basic graph
The graph of is a V shape with its vertex at the origin, made of the line for and the line for . Transformations work just like any function: shifts the V two units right and one unit up, so its vertex sits at .
Sketching y equals the modulus of f of x
This is different from , which instead takes the right-hand part of and reflects it across the axis to make an even function. Read carefully which version a question wants.
Solving absolute value equations
To solve (with ), the inside can be either or :
Always check solutions in the original equation, and reject any if since an absolute value can never be negative.
Solving absolute value inequalities
Two standard patterns cover most cases:
The first traps the inside between and ; 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.Show worked answer →
An absolute value inequality of the form |A| < k (with k > 0) is equivalent to the double inequality -k < A < k.
Rewrite without the modulus: -5 < 3x - 2 < 5.
Add 2 throughout: -3 < 3x < 7.
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.
Case 1: 2 - 9x >= 6, so -9x >= 4, and dividing by -9 (negative, so reverse the sign) gives x <= -4/9.
Case 2: 2 - 9x <= -6, so -9x <= -8, and dividing by -9 (reverse the sign) gives x >= 8/9.
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.