Modulus and Reciprocal Graphs (AA HL)
Putting a modulus sign around a function - or around just the input - bends the usual transformation rules, because reflection now depends on the sign of what's inside the bars rather than a fixed rule applied everywhere. \(y=|f(x)|\) and \(y=f(|x|)\) look similar on paper but do very different things to a graph, and IB questions often ask you to sketch or reason about both from the same starting function. This page is part of the broader Transformations & Graphs topic.
19 questions on this sub-topic.
The two modulus transformations
Covered under IB syllabus reference AHL2.16: the graphs of \(y=|f(x)|\) and \(y=f(|x|)\). Neither transformation is given as a formula in the booklet - you have to reason about them directly from the shape of \(f\).
\(y=|f(x)|\)
Reflect in the \(x\)-axis wherever \(f(x)<0\).
Any part of the curve below the \(x\)-axis is flipped upward; parts already on or above it stay put. The result never dips below the \(x\)-axis, and any \(x\)-intercepts of \(f\) become sharp "kinks" (cusps) on \(y=|f(x)|\).
\(y=f(|x|)\)
Keep \(x\ge0\), mirror it onto \(x<0\).
Delete whatever \(f\) does for negative \(x\) and replace it with a mirror image, in the \(y\)-axis, of the part for \(x\ge0\). The result is always an even function, symmetric about the \(y\)-axis, regardless of what the original \(f\) looked like.
A reciprocal graph \(y=\dfrac{1}{f(x)}\) follows its own set of rules - branches approach a vertical asymptote at each zero of \(f\), and a turning point of \(f\) becomes a turning point of \(y=\dfrac1{f(x)}\) with the opposite type. For the full transformation toolkit and syllabus wording, see Transformations & Graphs.
Worked examples
State the coordinates of the vertex and the equation of the axis of symmetry of \(y = 2|x + 3| - 4.\)
(a)(i) State the coordinates of the vertex.
(a)(ii) State the equation of the axis of symmetry.
Worked solution
(a)(i) The vertex is where \(|x+3| = 0\), i.e. \(x = -3.\) M1
At \(x=-3\): \(y = -4.\) Vertex \((-3,-4).\) A1
(a)(ii) Axis of symmetry: \(x = -3.\) A1
The line \(f(x)=x-2\) is drawn.
Sketch \(y=|f(x)|\) and state its minimum value.
Worked solution
the part of the line below the \(x\)-axis (\(x < 2\)) is reflected upward; M1 the part with \(x \ge 2\) is unchanged. A1
the graph is a V with vertex where \(f\) M1 \(= 0\), i.e. \((2, 0).\) A1 Minimum value \(0.\) A1
Common mistakes
- Reflecting the whole graph instead of just the negative part. \(y=|f(x)|\) only flips the sections where \(f(x)<0\) - anything already at or above the \(x\)-axis is left exactly as it was, so a graph that's entirely positive doesn't change at all.
- Confusing which axis each modulus reflects in. \(y=|f(x)|\) acts on the output and reflects in the \(x\)-axis; \(y=f(|x|)\) acts on the input and reflects in the \(y\)-axis. Mixing these up is the single most common error on this topic.
- Drawing a smooth curve through a cusp. Where \(f\) crosses the \(x\)-axis, \(y=|f(x)|\) has a sharp corner, not a rounded turning point - the gradient changes sign abruptly there rather than passing through zero smoothly.
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17 modulus and reciprocal graph questions, marked instantly like the real exam.
Quick answers
What does y = |f(x)| do to a graph?
Any part of the graph that lies below the \(x\)-axis is reflected upward in the \(x\)-axis. The parts already on or above the \(x\)-axis are left unchanged, so the result is never negative.
What does y = f(|x|) do to a graph?
The part of the graph for \(x\ge0\) is kept, and the part for \(x<0\) is discarded and replaced with a mirror image of the kept part reflected in the \(y\)-axis. The result is always an even function.