Sketch the modulus of a function
Build y = |f(x)| by keeping non-negative parts of y = f(x) and reflecting negative parts above the x-axis.
Section 1 · Functions
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which parts of y = f(x) change for y = |f(x)|?
What you will learn
- Relate the graphs of y = f(x) and y = |f(x)|
- Identify unchanged and reflected intervals
- Use intercepts and key points to sketch a modulus graph
The modulus changes output signs, not input positions
For each x, |f(x)| equals f(x) when f(x) ≥ 0 and equals −f(x) when f(x) < 0. The x-intercepts stay in the same places because zero is unchanged.
To sketch y = |f(x)|, keep every part of y = f(x) above the x-axis and reflect each part below it across the x-axis. This rule applies to linear, quadratic, cubic and the stated trigonometric graphs.
Use sign intervals rather than guessing the new shape
For f(x) = x² − 4, the original graph is negative between −2 and 2. On that interval, |f(x)| = 4 − x²; outside it, |f(x)| = x² − 4.
The zeros at x = −2 and x = 2 do not move. The reflected middle section has its highest value at x = 0, where it reaches 4.
The same sign-interval method applies when f is one of the stated trigonometric functions. For f(x) = 2 sin(x/2) on 0 ≤ x ≤ 4π, f is non-negative from 0 to 2π and negative from 2π to 4π, so y = |f(x)| reflects only the second half-cycle upward; its zeros stay at 0, 2π and 4π.
In the 0606 forms y = a sin bx + c, y = a cos bx + c and y = a tan bx + c, a is a positive integer, b is an integer or simple fraction, and c is an integer. If b is fractional its denominator is 2, 3, 4, 6 or 8. A vertical shift c can change where f changes sign, so locate the zeros before reflecting.
Pause and explain
Where are the x-intercepts of y = |f(x)| compared with y = f(x)?
Worked example
Describe the key changes from y = x² − 9 to y = |x² − 9|.
Show the worked solution
- Find the zeros: x² − 9 = 0 gives x = −3 and x = 3.
- The original parabola is below the x-axis for −3 < x < 3.
- Reflect only that central arc; keep the outside branches and the two intercepts fixed.
Answer The central part is reflected above the axis, creating a peak at (0, 9); the zeros remain (−3, 0) and (3, 0).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Sketch y = |x − 2| using its relationship with y = x − 2.
Give me a hint
Find where the original line changes sign.
Compare my reasoning
- The line crosses the x-axis at x = 2.
- For x ≥ 2, keep y = x − 2.
- For x < 2, reflect the negative part to y = 2 − x.
A V-shape with vertex (2, 0), opening upward.
Look for these in your work
- Intercept located
- Positive part retained
- Negative part reflected
For y = |x² − 1|, state the interval reflected and the maximum of the reflected part.
Give me a hint
Find the roots, then test between them.
Compare my reasoning
- The roots are x = −1 and x = 1.
- x² − 1 is negative for −1 < x < 1.
- The lowest original value is −1 at x = 0, so the reflected peak is 1.
Reflect −1 < x < 1; the reflected section reaches 1 at x = 0.
Look for these in your work
- Correct interval
- Sign checked
- New peak found
For y = |2 sin(x/2)| on 0 ≤ x ≤ 4π, state the interval reflected, the zeros, and the range.
Give me a hint
Find one sine cycle's sign before taking the modulus.
Compare my reasoning
- The period of sin(x/2) is 4π, so on this domain sin(x/2) is positive from 0 to 2π and negative from 2π to 4π.
- The second interval is reflected above the x-axis; the zeros remain x = 0, 2π and 4π.
- The original sine range is [−1,1]; multiplication by 2 and taking modulus gives range [0,2].
Reflect 2π < x < 4π; zeros at 0, 2π and 4π; range 0 ≤ y ≤ 2.
Look for these in your work
- Sign interval found
- Reflection leaves zeros fixed
- Range includes only non-negative outputs
Common mistakes
- Reflecting the whole graph
- Moving the x-intercepts
- Reflecting negative outputs in y = x instead of across the x-axis
What can you explain now?
At a point where f(x) = −6, what is |f(x)|?
Compare with the explanation
6
The modulus changes a negative output to its positive distance from zero.
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Make it stick
Describe the graph transformation that converts y = f(x) into y = |f(x)|.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.