The modulus |x| is the size of x, ignoring its sign, so |x| = x when x ≥ 0 and |x| = −x when x < 0. The graph of y = |ax + b| is a V shape with its vertex on the x-axis. You need to sketch y = |f(x)| and y = f(|x|), which are different transformations, and solve modulus equations by considering both cases and rejecting any solution that does not work.
Key facts about the modulus function
- |x| = x when x ≥ 0, and |x| = −x when x < 0
- y = |ax + b|: a V shape with vertex where ax + b = 0
- y = |f(x)|: reflect the parts below the x-axis upwards
- y = f(|x|): reflect the part for x ≥ 0 in the y-axis
- |f(x)| = c (c ≥ 0): solve f(x) = c and f(x) = −c
What is the modulus of a number?
The modulus, or absolute value, |x| is the distance of x from zero, so it is never negative: |5| = 5 and |−5| = 5. To evaluate |f(k)|, work out f(k) first, then take its size.
The graph of y = |ax + b| is a V. Its vertex sits on the x-axis where ax + b = 0, and its two arms have gradients a and −a.
How do you sketch y = |f(x)|?
Keep every part of y = f(x) that is on or above the x-axis, and reflect any part below the axis upwards in the x-axis. Roots stay where they are and become sharp corners, and a minimum below the axis becomes a maximum above it. The whole graph of y = |f(x)| lies on or above the x-axis.
How is y = f(|x|) different?
For y = f(|x|), keep the part of y = f(x) where x ≥ 0 and reflect it in the y-axis to make the left-hand side. The graph is symmetric about the y-axis, and anything originally to the left of the y-axis is thrown away.
So |f(x)| reflects below-axis parts up, in the x-axis, while f(|x|) mirrors the right-hand side across, in the y-axis. They usually give quite different graphs.
Solving modulus equations
To solve |f(x)| = c, where c is a non-negative number, solve both f(x) = c and f(x) = −c. If the right-hand side involves x, as in |2x − 1| = x + 4, solve both cases and then check each answer in the original equation.
A modulus is never negative, so any solution that makes the right-hand side negative must be rejected. A sketch shows how many solutions to expect.
Modulus graphs and lines
To find where a line meets a modulus graph, work with each arm of the V separately, using the expression with its sign for x on one side of the vertex and the reversed sign on the other side. Check each solution lies on the arm you used.
For a horizontal line y = k and a V opening upwards with vertex height v, there are two crossings when k > v, one when k = v and none when k < v.
Worked examples
Example 1. Sketch y = |2x − 6|, giving the vertex and the y-intercept.
- Vertex: 2x − 6 = 0, so x = 3, giving (3, 0).
- y-intercept: x = 0 gives |−6| = 6, so (0, 6).
- The graph is a V with arms of gradient 2 and −2.
Answer: Vertex (3, 0); y-intercept (0, 6).
Marks: B1 for the vertex, B1 for the y-intercept.
Example 2. Sketch y = |x² − 4|.
- Keep the parts of y = x² − 4 for x ≤ −2 and x ≥ 2, which are already above the axis.
- Reflect the dip between x = −2 and x = 2 upwards.
- The minimum (0, −4) becomes a maximum (0, 4), with corners at (±2, 0).
Answer: A W shape with corners at (−2, 0) and (2, 0) and a peak at (0, 4).
Marks: B1 for reflecting the dip, B1 for the corners at the roots.
Example 3. Given f(x) = x² − 4x, sketch y = f(|x|).
- For x ≥ 0, the curve has roots at 0 and 4 and a minimum at (2, −4).
- Reflect that part in the y-axis to make the left-hand side.
- The result is symmetric about the y-axis, with minima at (±2, −4).
Answer: Symmetric about the y-axis with minima at (2, −4) and (−2, −4).
Marks: B1 for reflecting in the y-axis, B1 for the symmetry.
Example 4. Solve |2x − 1| = 5.
- Two cases: 2x − 1 = 5 or 2x − 1 = −5.
- x = 3 or x = −2.
- The right-hand side is positive, so both are valid.
Answer: x = 3 or x = −2.
Marks: M1 for both cases, A1 for both answers.
Example 5. The line y = k meets y = |2x − 6| at two distinct points. State the possible values of k.
- The vertex of the V is at (3, 0), at height 0.
- Two crossings need the line above the vertex.
Answer: k > 0.
Marks: B1 for the vertex height, B1 for k > 0.
Common mistakes
- Confusing |f(x)| with f(|x|). |f(x)| reflects in the x-axis; f(|x|) reflects in the y-axis.
- Solving only one case. |f(x)| = c needs both f(x) = c and f(x) = −c.
- Not checking solutions. When the right-hand side contains x, substitute each answer back and reject any that fail.
- Drawing a rounded bottom on a V. Where f(x) = 0, y = |f(x)| has a sharp corner, not a smooth turn.
- Losing the minus sign on the reflected arm. For x below the vertex of y = |ax + b|, use y = −(ax + b).
How the modulus function is examined
In Edexcel International A-Level the modulus function is part of Pure 3 (WMA13), within functions and graphs. In Edexcel A-Level (9MA0) it is second-year Pure. Expect to:
- sketch y = |ax + b| or a transformed V, labelling the vertex and intercepts
- sketch y = |f(x)| and y = f(|x|) from a given graph
- solve an equation where a modulus equals a number or a linear expression
- find where a line meets a modulus graph, or the values of k giving a set number of solutions
Modulus questions with full mark schemes appear in the Pure 3 functions and graphs exam questions. You can also work through the Pure 3 functions worksheet or generate questions in the algebra revision engine.
The reflections here follow the same rules as transformations of graphs.
Frequently asked questions
What does |x| mean?
The modulus of x: its size without the sign. |3| = 3 and |−3| = 3.
What is the difference between y = |f(x)| and y = f(|x|)?
y = |f(x)| flips any part of the graph below the x-axis upwards. y = f(|x|) keeps the right-hand half and mirrors it in the y-axis.
Why do some modulus equation solutions get rejected?
Squaring or splitting into cases can produce values that do not satisfy the original equation. A modulus can never equal a negative number, so check each answer.