IB Maths AA HL Topic 2 — Functions Paper 1 & 2 HL only ~8 min read

Modulus Transformations

Two distinct transformations apply the modulus to a function. y = |f(x)| takes the modulus of the output — anything below the x-axis flips up. y = f(|x|) takes the modulus of the input — the right side is kept and reflected across the y-axis. Don’t confuse them: the first changes negative y-values, the second forces symmetry about the y-axis.

📘 What you need to know

y = |f(x)|: keep the parts of y = f(x) on or above the x-axis; reflect parts below in the x-axis.
y = f(|x|): keep the part of y = f(x) on or to the right of the y-axis; reflect that part in the y-axis to make the left side.
Cusps appear at x-intercepts (for |f|) or at x = 0 (for f(|x|)) when the original gradient is non-zero — sketch them sharp, not smooth.
Range: |f(x)| has range y ≥ 0; f(|x|) inherits the range of f on its right side only.
Symmetry: f(|x|) is always symmetric about the y-axis; |f(x)| usually isn’t symmetric.
Composite transformations apply in order: do the inner transformation first, then the modulus, then any outer stretch or translation.
Local extrema change type: a min below the axis becomes a max after |f|; the rest of the time, the type is preserved.

y = |f(x)| — reflect parts below the x-axis

Rule parts of f on or above x-axis stay; parts below are reflected in the x-axis

y = |f(x)| — flipping the negative parts up

The reflection happens at every x-intercept of f. At those points, the gradient changes sign abruptly — creating a cusp. Maxima above the axis stay as maxima; minima below the axis become maxima after the flip.

y = f(|x|) — make symmetric about the y-axis

Rule keep the part of f on or to the right of the y-axis; reflect it in the y-axis to make the left side

y = f(|x|) — discarding the left, mirroring the right

The original left side is discarded entirely — replaced by a mirror image of the right side. Whatever was on the original left, including any features there, vanishes from the new graph.

Key differences

y = |f(x)|

no parts below x-axis
not necessarily symmetric

flips the output — affects negative y-values

y = f(|x|)

can have parts below x-axis
always symmetric about y-axis

flips the input — only the right side matters

Quick test: a graph with parts below the x-axis cannot be y = |f(x)| — those parts would have been reflected up. A graph that isn’t symmetric about the y-axis cannot be y = f(|x|).

Composite transformations — order matters

Functions like y = a|f(x)| + b or y = |af(x) + b| chain the modulus with stretches and translations. The order is dictated by the brackets:

y = a|f(x)| + b

f → |f| → a|f| → a|f| + b

take modulus first, then stretch and shift

y = |af(x) + b|

f → af → af + b → |af + b|

stretch and shift first, then take modulus

Read the brackets carefully. The modulus only affects what’s inside it — anything outside the modulus signs has already been computed before the absolute value is taken.

🧭 Recipe — sketch a modulus transformation

Identify which transformation: y = |f(x)| (modulus around f) or y = f(|x|) (modulus around x).
Sketch y = f(x) first, marking key points (intercepts, vertex, asymptotes).
Apply the rule: reflect parts below the x-axis up (for |f(x)|) or reflect the right side across the y-axis (for f(|x|)).
Mark new cusps sharply — at x-intercepts (for |f|) or at x = 0 (for f(|x|)) if the gradient changes there.
Apply outer transformations (stretches, translations) after the modulus, in their stated order.
Label key points on the final graph — both originals that remained and any new ones created by the reflection.

Worked examples

WE 1

Sketch y = |f(x)| for a quadratic

Sketch the graph of y = |x² − 4|, identifying all key features.

Step 1: Sketch f(x) = x² − 4 parabola opening up; vertex (0, −4); roots at x = ±2 Step 2: Apply |f(x)| — reflect parts below x-axis up f < 0 on −2 < x < 2 → reflect this piece vertex (0, −4) reflects to (0, 4) — now a local max Step 3: Mark cusps at x-intercepts cusps at (−2, 0) and (2, 0) y-int (0, 4) — local max; cusps at (±2, 0); range y ≥ 0 graph looks like a parabola for |x| > 2, plus an inverted parabola for −2 ≤ x ≤ 2 — joined sharply at the cusps

WE 2

Sketch y = f(|x|) for an asymmetric quadratic

Given f(x) = x² + 2x − 3, sketch y = f(|x|), identifying all key features.

Step 1: Sketch f(x) — original parabola roots at x = −3 and x = 1; vertex (−1, −4); y-int (0, −3) Step 2: Apply f(|x|) — keep right side (x ≥ 0), reflect across y-axis right side starts at (0, −3), passes through (1, 0), rises left side is mirror image — passes through (−1, 0), down to (0, −3) Step 3: Identify new key features y-int (0, −3) — now a local min (cusp from the two reflections) x-intercepts at (±1, 0); original vertex (−1, −4) is gone y-int (0, −3); x-ints (±1, 0); cusp at (0, −3); symmetric about y-axis the original vertex on the left of the y-axis is replaced — only what’s on the right (and its reflection) remain

WE 3

Compare y = |f(x)| and y = f(|x|) using key points

The function y = f(x) has a local maximum at A(2, 4) and a local minimum at B(−3, −2). Describe the position and type of these features on:
(a) y = |f(x)|
(b) y = f(|x|).

(a) y = |f(x)| A(2, 4): y > 0 → unchanged → still local max at (2, 4) B(−3, −2): y < 0 → reflected up → local max at (−3, 2) (a) max at (2, 4); max at (−3, 2) (b) y = f(|x|) A(2, 4): on right side → unchanged → still local max at (2, 4) A also reflected to (−2, 4) → another local max B(−3, −2): on left side → discarded entirely (b) max at (2, 4); max at (−2, 4); B disappears the modulus changes a min on the negative side into a max — the curvature still bends the original way, but now away from the axis

WE 4

Modulus of a cubic

Sketch the graph of y = |x³ − 3x|, identifying the cusps and any local maxima.

Step 1: Factorise and find roots of f(x) = x³ − 3x x(x² − 3) = 0 → x = 0, ±√3 Step 2: Find original local extrema f'(x) = 3x² − 3 = 0 → x = ±1 f(1) = 1 − 3 = −2 (local min) f(−1) = −1 + 3 = 2 (local max) Step 3: Apply |f(x)| f < 0 on x < −√3 and 0 < x < √3 → reflect these regions cusps form at all three roots: x = 0, ±√3 (1, −2) reflects to (1, 2) → local max (−1, 2) is on a positive region → unchanged → local max cusps at (0, 0), (±√3, 0); two local maxima at (±1, 2) |f| of an odd function is even — the result is symmetric about the y-axis even when f wasn’t

WE 5

Composite transformation — track key points

The function f has a local minimum at (3, −4) and a local maximum at (−1, 2). Find the position and type of these features on the graph of y = 2|f(x)| − 1.

Step 1: Apply transformations in order — |f| → 2|f| → 2|f| − 1 Step 2: Track (3, −4) |f|: y = |−4| = 4 → (3, 4); was min, now max ×2: (3, 8) −1: (3, 7) Step 3: Track (−1, 2) |f|: y = 2 → (−1, 2); still max ×2: (−1, 4) −1: (−1, 3) (3, 7) — local max; (−1, 3) — local max order matters: |f| first reflects negative outputs; then ×2 stretches; then −1 shifts. Doing them in a different order would give a different graph

WE 6

Double modulus — sketch y = ||x| − 3|

Sketch y = ||x| − 3|, identifying all cusps and the y-intercept.

Step 1: Sketch the inner f(x) = |x| − 3 V-shape with vertex (0, −3); x-intercepts (±3, 0) Step 2: Apply outer modulus — reflect parts below x-axis up f < 0 on −3 < x < 3 — reflect this section vertex (0, −3) reflects to (0, 3) — new local max Step 3: Identify cusps and intercepts cusps appear at the x-intercepts of f: (−3, 0) and (3, 0) cusp at the original vertex (now at (0, 3)) where reflection meets reflection cusps at (−3, 0), (0, 3), (3, 0); y-int (0, 3); range y ≥ 0 final shape is an “M” — two V’s joined at a peak, all built from straight-line segments

💡 Top tips

Sketch y = f(x) lightly first with all key features. Then apply the modulus rule on top.
Cusps are sharp, not smooth. Drawing them as smooth curves makes them look like turning points and loses marks.
Mins below the axis become maxes after |f(x)|. Don’t write them as still being mins.
Whatever’s on the left of the y-axis is gone in f(|x|). Don’t try to keep features that were originally on the negative side.
Composite order matters: y = a|f(x)| + b is different from y = |af(x) + b|. Apply the modulus exactly where the brackets put it.
Test points at the suspected cusps — substituting confirms whether the gradient changes sign there.
Symmetry check for f(|x|): the final graph must be symmetric about the y-axis. If it isn’t, you’ve made a sign mistake.

⚠ Common mistakes

Confusing the two transformations. |f(x)| flips negative y-values; f(|x|) replaces the left side with a mirror.
Drawing smooth turning points at cusps. The gradient changes sign abruptly — must look like a sharp corner.
Reflecting the wrong direction: |f| reflects in the x-axis (vertical flip), not the y-axis.
Keeping features on the original left side when sketching f(|x|). Those points get replaced by the reflected right side.
Applying composite transformations in the wrong order. The brackets dictate the order — read them carefully.
Forgetting to update the type of extremum. A min reflected up becomes a max.
Missing cusps at x = 0 for f(|x|) when the original gradient at x = 0 is non-zero.

Modulus transformations are the bridge to solving modulus equations and inequalities — the next note. With the graph in hand, you can read off solutions to |f(x)| = k by finding where horizontal lines cross the modulus graph, and inequalities by reading off intervals where one curve sits above another.

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