Modulus Functions and Transformations

This lesson covers the modulus function and its graphs, solving modulus equations and inequalities, and a comprehensive review of graph transformations.

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The Modulus Function

The modulus (or absolute value) of x is written |x| and defined as:

|x| = x if x ≥ 0
|x| = −x if x < 0

In other words, |x| makes any negative value positive. It represents the distance of x from zero on the number line.

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Graph of y = |f(x)|

To sketch y = |f(x)|:

1. Sketch y = f(x).
2. Reflect any parts below the x-axis upwards (in the x-axis).

Worked Example 1

Sketch y = |2x − 3|.

First sketch y = 2x − 3 (a straight line crossing the x-axis at x = 3/2).

The part where y < 0 (x < 3/2) is reflected upward. The result is a V-shape with vertex at (3/2, 0).

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Graph of y = f(|x|)

To sketch y = f(|x|):

1. Sketch y = f(x) for x ≥ 0.
2. Reflect this in the y-axis to get the part for x < 0.

The graph is always symmetric about the y-axis.

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Solving Modulus Equations

Method 1: Algebraic (squaring or considering cases)

To solve |f(x)| = g(x):

• Either f(x) = g(x) or f(x) = −g(x), then check solutions are valid.

Worked Example 2

Solve |3x − 2| = x + 4.

Case 1: 3x − 2 = x + 4 → 2x = 6 → x = 3
Check: |7| = 7, x + 4 = 7 ✓

Case 2: 3x − 2 = −(x + 4) → 3x − 2 = −x − 4 → 4x = −2 → x = −1/2
Check: |−7/2| = 7/2, x + 4 = 7/2 ✓

Solutions: x = 3 or x = −1/2

Worked Example 3

Solve |x − 1| = |2x + 3|.

Square both sides:

(x − 1)² = (2x + 3)²
x² − 2x + 1 = 4x² + 12x + 9
3x² + 14x + 8 = 0 → (3x + 2)(x + 4) = 0

x = −2/3 or x = −4. Both check out.

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Solving Modulus Inequalities

For |f(x)| < a (where a > 0): −a < f(x) < a

For |f(x)| > a (where a > 0): f(x) > a or f(x) < −a

Worked Example 4

Solve |2x − 5| < 3.

−3 < 2x − 5 < 3 → 2 < 2x < 8 → 1 < x < 4

Answer: 1 < x < 4

Worked Example 5

Solve |x + 1| ≥ 4.

x + 1 ≥ 4 or x + 1 ≤ −4 → x ≥ 3 or x ≤ −5

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Graph Transformations

Summary of Transformations

Transformation	Effect on graph
y = f(x) + a	Translation (0, a) — up by a
y = f(x + a)	Translation (−a, 0) — left by a
y = af(x)	Vertical stretch, scale factor a
y = f(ax)	Horizontal stretch, scale factor 1/a
y = −f(x)	Reflection in the x-axis
y = f(−x)	Reflection in the y-axis

Worked Example 6

The curve y = f(x) passes through (3, 5). Find the corresponding point on:

(a) y = f(x) + 4 → (3, 9) (shift up 4) (b) y = f(x − 2) → (5, 5) (shift right 2) (c) y = 2f(x) → (3, 10) (stretch vertically by 2) (d) y = f(3x) → (1, 5) (compress horizontally by 3) (e) y = −f(x) → (3, −5) (reflect in x-axis) (f) y = f(−x) → (−3, 5) (reflect in y-axis)

Combined Transformations

Apply transformations in the correct order. For y = af(bx + c) + d:

1. Horizontal translation by −c/b.
2. Horizontal stretch by 1/b.
3. Vertical stretch by a.
4. Vertical translation by d.

Worked Example 7

Describe the sequence of transformations that maps y = x² to y = 3(x − 1)² + 2.

1. Translation (1, 0) — replace x with (x − 1).
2. Vertical stretch, scale factor 3.
3. Translation (0, 2) — add 2.

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Exam Tips

• When solving modulus equations, always check your solutions in the original equation.
• Squaring both sides is efficient for |f(x)| = |g(x)| type equations.
• For graph transformations, note whether the constant is inside or outside the function brackets.
• Inside brackets (e.g. f(x + 2)) affects x — horizontal, and the direction is opposite to the sign.
• Outside brackets (e.g. f(x) + 2) affects y — vertical, and the direction matches the sign.

A-Level Deep Dive: The Modulus Function and Transformations

Spec mapping

AQA 7357 specification, Paper 2 Pure Mathematics, Section B — Algebra and Functions (Year 2 content) covers the modulus function; sketch $y = |f(x)|$y=∣f(x)∣ and $y = f(|x|)$y=f(∣x∣) from the graph of $y = f(x)$y=f(x); combine transformations of graphs and use them to solve equations and inequalities involving the modulus function (refer to the official specification document for exact wording). This sub-strand sits within the Year 2 functions block and connects directly to Paper 1 Section D (Sequences and Series) through piecewise definitions, Paper 2 Section C (Coordinate Geometry) through reflection arguments, and Paper 3 Section M (Mechanics — Kinematics) where speed $|v|$∣v∣ is the modulus of velocity. The AQA formula booklet does not list the modulus definition or the transformation rules — they must be memorised as the piecewise formula $|x| = x$∣x∣=x for $x \geq 0$x≥0 and $|x| = -x$∣x∣=−x for $x < 0$x<0.

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Worked example with full mark scheme

Question (8 marks):

(a) Solve $|2x - 3| = x + 1$∣2x−3∣=x+1, giving all values of $x$x. (5)

(b) On the same axes, sketch $y = |2x - 3|$y=∣2x−3∣ and $y = x + 1$y=x+1, marking the points of intersection found in (a) and the coordinates of every axis crossing. (3)

Solution with mark scheme:

(a) Step 1 — note the constraint on the right-hand side.

Since $|2x - 3| \geq 0$∣2x−3∣≥0 for all real $x$x, any solution must satisfy $x + 1 \geq 0$x+1≥0, i.e. $x \geq -1$x≥−1. Solutions outside this range must be rejected at the end.

B1 — stating (or using) the constraint $x + 1 \geq 0$x+1≥0 at any point in the working. Many candidates skip this and lose the mark even when their final answers are correct, because they have not justified rejecting any extraneous root.

Step 2 — split the modulus into its two branches.

Branch 1 (positive branch): if $2x - 3 \geq 0$2x−3≥0, i.e. $x \geq \tfrac{3}{2}$x≥23​, then $|2x - 3| = 2x - 3$∣2x−3∣=2x−3. The equation becomes:

$2x - 3 = x + 1 \implies x = 4$2x−3=x+1⟹x=4

Check: $x = 4 \geq \tfrac{3}{2}$x=4≥23​, so this branch is valid; also $x + 1 = 5 \geq 0$x+1=5≥0, consistent.

M1 — setting up Branch 1 with the correct sign and stating (or implying) the domain restriction $x \geq \tfrac{3}{2}$x≥23​.

A1 — $x = 4$x=4, accompanied by a domain check.

Step 3 — Branch 2 (negative branch).

If $2x - 3 < 0$2x−3<0, i.e. $x < \tfrac{3}{2}$x<23​, then $|2x - 3| = -(2x - 3) = 3 - 2x$∣2x−3∣=−(2x−3)=3−2x. The equation becomes:

$3 - 2x = x + 1 \implies 2 = 3x \implies x = \tfrac{2}{3}$3−2x=x+1⟹2=3x⟹x=32​

Check: $x = \tfrac{2}{3} < \tfrac{3}{2}$x=32​<23​, valid for Branch 2; and $x + 1 = \tfrac{5}{3} \geq 0$x+1=35​≥0, consistent.

M1 — setting up Branch 2 with the negated bracket.

A1 — $x = \tfrac{2}{3}$x=32​, with domain consistency confirmed.

Final answer: $x = \tfrac{2}{3}$x=32​ or $x = 4$x=4.

(b) Sketch: $y = |2x - 3|$y=∣2x−3∣ is a V-shape with vertex at $(\tfrac{3}{2}, 0)$(23​,0), gradient $+2$+2 to the right and $-2$−2 to the left, crossing the $y$y-axis at $(0, 3)$(0,3). $y = x + 1$y=x+1 is a straight line of gradient $1$1 through $(-1, 0)$(−1,0) and $(0, 1)$(0,1). The intersections at $(\tfrac{2}{3}, \tfrac{5}{3})$(32​,35​) and $(4, 5)$(4,5) should be labelled.

B1 — V-shape vertex correctly placed at $(\tfrac{3}{2}, 0)$(23​,0). B1 — straight line correctly drawn with both intercepts marked. B1 — both intersections labelled with coordinates.

Total: 8 marks (B1 M2 A2 + B3, structured as shown).

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): The function $f$f is defined by $f(x) = x^2 - 4$f(x)=x2−4 for $x \in \mathbb{R}$x∈R.

(a) Sketch $y = |f(x)|$y=∣f(x)∣, labelling all axis crossings and turning points. (3)

(b) Sketch $y = f(|x|)$y=f(∣x∣) on a separate set of axes, labelling all axis crossings and turning points. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — sketching $y = x^2 - 4$y=x2−4 as the underlying parabola: vertex $(0, -4)$(0,−4), roots $x = \pm 2$x=±2.
• M1 (AO1.2) — reflecting the portion below the $x$x-axis (where $f(x) < 0$f(x)<0, i.e. $-2 < x < 2$−2<x<2) in the $x$x-axis to obtain $y = |f(x)|$y=∣f(x)∣.
• A1 (AO2.5) — final shape: a "W-like" curve with a local maximum at $(0, 4)$(0,4) (the reflected vertex) and roots at $x = \pm 2$x=±2 now appearing as cusps.

(b)

• M1 (AO1.1a) — recognising that $f(|x|) = |x|^2 - 4 = x^2 - 4$f(∣x∣)=∣x∣2−4=x2−4 (since $|x|^2 = x^2$∣x∣2=x2 for all real $x$x).
• M1 (AO1.2) — alternatively, applying the rule "delete the graph for $x < 0$x<0 and reflect the $x \geq 0$x≥0 portion in the $y$y-axis".
• A1 (AO2.5) — final answer: identical to $y = f(x)$y=f(x) in this special case because $f$f is an even function. The candidate must state this conclusion explicitly to score the A1.

Total: 6 marks split AO1 = 4, AO2 = 2. Part (b) has a built-in trap: candidates who do not notice that $f$f is even draw an incorrect graph; candidates who do notice, and say so, earn the AO2 reasoning mark.

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Synoptic links

Connects to:

1. Paper 1 Section A (Proof) and Section D (Sequences and Series): the formal piecewise definition $|x| = x$∣x∣=x for $x \geq 0$x≥0, $|x| = -x$∣x∣=−x for $x < 0$x<0 is the prototypical "proof by cases" object. Many AQA proof questions ask candidates to prove identities such as $|ab| = |a||b|$∣ab∣=∣a∣∣b∣ by considering the four sign combinations. The same case-splitting underpins convergence proofs for sequences in Year 2 work on series.

2. Paper 2 Section C (Coordinate Geometry): the graph $y = |f(x)|$y=∣f(x)∣ is obtained by reflecting in the $x$x-axis where $f(x) < 0$f(x)<0; $y = f(|x|)$y=f(∣x∣) is obtained by reflecting in the $y$y-axis the $x \geq 0$x≥0 portion. Both are special cases of the general transformation rules examined throughout coordinate geometry.

3. Paper 2 Section G (Differentiation): $|x|$∣x∣ is not differentiable at $x = 0$x=0 (the V-vertex is a cusp). This shows up in Year 2 differentiation when candidates are asked to identify points where $\frac{dy}{dx}$dxdy​ does not exist. The piecewise derivative is $+1$+1 for $x > 0$x>0 and $-1$−1 for $x < 0$x<0, with no value at $x = 0$x=0.

4. Paper 3 Section M (Mechanics — Kinematics): speed is $|v|$∣v∣, the modulus of velocity. Sketching distance–time and speed–time graphs from a velocity–time graph is exactly the Pure-Mathematics transformation $y = |f(x)|$y=∣f(x)∣ in disguise. Candidates who have internalised the V-shape produce neater Mechanics sketches.

5. Paper 2 Section H (Integration): integrating $|f(x)|$∣f(x)∣ over an interval where $f$f changes sign requires splitting the integral at each root, taking absolute values branch-by-branch, then summing. This is the "area enclosed between a curve and the $x$x-axis" technique that appears in every $\int_a^b |f(x)|\,dx$∫ab​∣f(x)∣dx question.

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Mark-scheme literacy

Modulus and transformation questions on AQA 7357 split AO marks more evenly than surd questions:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Splitting modulus into branches, applying transformation rules procedurally, sketching V-shapes and reflected parabolas
AO2 (reasoning / interpretation)	30–40%	Justifying domain constraints, rejecting extraneous roots with a stated reason, identifying when a function is even and graph-types coincide
AO3 (problem-solving)	5–15%	Modelling problems where speed = $

Examiner-rewarded phrasing: "since $|f(x)| \geq 0$∣f(x)∣≥0, the equation has no solution unless $g(x) \geq 0$g(x)≥0"; "the graph of $y = |f(x)|$y=∣f(x)∣ is obtained by reflecting in the $x$x-axis the portion of $y = f(x)$y=f(x) for which $f(x) < 0$f(x)<0"; "checking $x = \tfrac{2}{3}$x=32​ in the original equation: $|2(\tfrac{2}{3}) - 3| = |\tfrac{4}{3} - 3| = \tfrac{5}{3}$∣2(32​)−3∣=∣34​−3∣=35​, and $\tfrac{2}{3} + 1 = \tfrac{5}{3}$32​+1=35​, consistent." Phrases that lose marks: "I squared both sides" without acknowledging the introduction of extraneous roots; "the modulus just removes the negative sign" (it does not — it removes the value of the negative branch, which may itself be more complicated than a sign flip).

A specific AQA pattern: questions phrased "solve, showing that there are exactly two solutions" require candidates to prove exhaustivity, not merely list two answers. The case-split structure is the proof.

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Grade-band model answers

3-mark question

Question: Solve $|x - 5| = 3$∣x−5∣=3.

Grade C response (~150 words):

The modulus equation gives two cases: