The Modulus Function

Understanding absolute value, sketching modulus graphs, and solving modulus equations and inequalities

SPEC 2.7 • ALGEBRA AND FUNCTIONS

What is the Modulus Function?

Definition

The modulus (or absolute value) of a number x, written |x|, gives the magnitude of x without regard to its sign. |x| = x if x ≥ 0

|x| = −x if x < 0

The output of the modulus function is always non-negative.

Geometric Interpretation

|x| represents the distance of x from 0 on a number line.

More generally, |x − a| represents the distance between x and a.

ExpressionValueReason

|5|55 is positive, so |5| = 5

|−3|3−3 is negative, so |−3| = −(−3) = 3

|0|00 is neither positive nor negative

Graph of y = |x|

The graph of y = |x| is a V-shape:

For x ≥ 0: the graph follows y = x

For x < 0: the graph follows y = −x

Key Features

Vertex: (0, 0)

Symmetric about the y-axis

Domain: all real numbers

Range: y ≥ 0

y = |x|

Remember

The V-shape always opens upward and the vertex is the point where the function changes from one linear piece to another.

01Worked Example

Evaluate the following:

(a) |3 − 7| (b) |2(−4) + 5| (c) |2x − 5| when x = 1

(a)

|3 − 7| = |−4| = 4

•

Since 3 − 7 = −4 is negative, we take the positive value.

(b)

|2(−4) + 5| = |−8 + 5| = |−3| = 3

•

Evaluate inside first, then take the modulus.

(c)

When x = 1: |2(1) − 5| = |2 − 5| = |−3| = 3

Answers: (a) 4 (b) 3 (c) 3

Graphs of y = |f(x)|

The Reflection Rule

To sketch y = |f(x)| from y = f(x):

Keep all parts of the graph where f(x) ≥ 0 (on or above the x-axis)

Reflect all parts where f(x) < 0 in the x-axis

y = x − 2 (original)

--- original ━ y = |x − 2|

Key Result

The graph of y = |f(x)| is always on or above the x-axis.

The vertex occurs where f(x) = 0 (i.e., at the x-intercept of the original graph).

Worked Example02

Sketch the graph of y = |2x − 4|, stating the coordinates of the vertex and any intercepts.

1

First, consider y = 2x − 4 (the linear function inside the modulus)

2

Find where 2x − 4 = 0: solving gives x = 2. This is the vertex.

3

Vertex: When x = 2, y = |2(2) − 4| = |0| = 0. So vertex is (2, 0)

4

y-intercept: When x = 0, y = |2(0) − 4| = |−4| = 4. So y-intercept is (0, 4)

5

The gradient of y = 2x − 4 is 2, so:

For x ≥ 2: y = 2x − 4 (gradient +2)

For x < 2: y = −(2x − 4) = −2x + 4 (gradient −2)

--- y = 2x − 4 ━ y = |2x − 4|

Vertex: (2, 0)y-intercept: (0, 4)

Worked Example03

Sketch the graph of y = |x² − 9| showing all key points.

1

First sketch y = x² − 9:

Factorise: x² − 9 = (x − 3)(x + 3)

x-intercepts: x = −3 and x = 3

2

Vertex: At x = 0: y = 0 − 9 = −9

Vertex at (0, −9) — this is a minimum below the x-axis

3

For |x² − 9|: Reflect the part between x = −3 and x = 3 in the x-axis

The vertex (0, −9) becomes (0, 9) — a local maximum

--- y = x² − 9 ━ y = |x² − 9|

Key features of y = |x² − 9|:

•

x-intercepts: (−3, 0) and (3, 0) — where graph "bounces"

•

Local maximum: (0, 9) — the reflected vertex

•

The outer parts curve inward (concave up)

Graphs of y = f(|x|)

The Symmetry Rule

To sketch y = f(|x|) from y = f(x):

Take only the part of y = f(x) where x ≥ 0 (the right-hand side)

Reflect this part in the y-axis to create the left-hand side

Key Result

The graph of y = f(|x|) is always symmetric about the y-axis.

Don't Confuse

y = |f(x)|y = f(|x|)Reflect negative y-values in x-axisReflect right side in y-axisGraph is on or above x-axisGraph is symmetric about y-axis

Memory Aid

|f(x)| - modulus is OUTSIDE, affects the y-values (vertical)

f(|x|) - modulus is INSIDE, affects the x-values (horizontal)

Worked Example04

Sketch the graph of y = 2|x| − 3, stating the vertex and intercepts.

1

This is y = 2|x| − 3, which is a V-shaped graph (a transformation of y = |x|)

2

Vertex: The minimum of |x| occurs at x = 0

When x = 0: y = 2(0) − 3 = −3. So vertex is (0, −3)

3

x-intercepts: Set y = 0: 2|x| − 3 = 0

2|x| = 3, so |x| = 1.5

Therefore x = 1.5 or x = −1.5

4

y-intercept: When x = 0, y = −3 (same as vertex)

5

The graph is a V-shape, symmetric about the y-axis, with gradient ±2

━ y = 2|x| − 3

Vertex: (0, −3)x-intercepts: (−1.5, 0) and (1.5, 0)

Transformations of Modulus Functions

You can apply transformations to modulus functions just like any other function.

General Form

For the function y = a|x + b| + c:ParameterEffecta (outside)Vertical stretch by factor |a|. If a < 0, reflects in x-axis (inverted V)b (inside)Horizontal translation by −b (vertex moves to x = −b)c (outside)Vertical translation by c (vertex moves up/down)

Finding the Vertex

For y = a|x + b| + c:

Vertex is at (−b, c)

This is where |x + b| = 0, i.e., x = −b

Finding the Range

If a > 0 (V-shape): Range is y ≥ c

If a < 0 (inverted V): Range is y ≤ c

Transformation Order

To sketch y = a|x + b| + c from y = |x|:

Translate horizontally by −b (vertex to x = −b)

Stretch vertically by factor |a|, reflect if a < 0

Translate vertically by c

05Worked Example

The function f is defined by f(x) = 3|x − 1| − 2, x ∈ ℝ.

(a) Sketch the graph of y = f(x) (b) State the range of f (c) Solve f(x) = 4

(a)

Vertex: |x − 1| = 0 when x = 1. At x = 1: f(1) = 3(0) − 2 = −2

So vertex is at (1, −2)

•

y-intercept: f(0) = 3|0 − 1| − 2 = 3(1) − 2 = 1. Point: (0, 1)

•

x-intercepts: Set f(x) = 0: 3|x − 1| − 2 = 0, so |x − 1| = 2/3

x − 1 = 2/3 → x = 5/3 OR x − 1 = −2/3 → x = 1/3

y = 3|x − 1| − 2

(b)

The vertex is the minimum point at y = −2. Since a = 3 > 0, the V opens upward.

Range: f(x) ≥ −2 or in set notation: {y ∈ ℝ : y ≥ −2}

(c)

Solve 3|x − 1| − 2 = 4, so 3|x − 1| = 6, hence |x − 1| = 2

x − 1 = 2 → x = 3 OR x − 1 = −2 → x = −1

Inverted Modulus Functions

When the coefficient a < 0, the modulus function forms an inverted V-shape (Λ shape).

Key Differences

Propertya > 0 (V-shape)a < 0 (Λ-shape)Vertex is aMinimum pointMaximum pointRange of y = a|x + b| + cy ≥ cy ≤ cInverse exists?No (many-to-one)No (many-to-one)

Example

y = 6 − 2|x + 3| has a = −2 < 0, so it's an inverted V with vertex at (−3, 6) and range y ≤ 6.

06Worked Example

The function f is defined by f(x) = 6 − 2|x + 3|, x ∈ ℝ.

(a) Sketch y = f(x) (b) State the range of f (c) Explain why f⁻¹ does not exist

(a)

This is y = −2|x + 3| + 6. Since a = −2 < 0, this is an inverted V (Λ-shape).

Vertex: |x + 3| = 0 when x = −3. At x = −3: f(−3) = 6 − 2(0) = 6

So vertex is at (−3, 6) — this is the maximum point.

•

y-intercept: f(0) = 6 − 2|0 + 3| = 6 − 6 = 0. Point: (0, 0)

•

x-intercepts: Set f(x) = 0: 6 − 2|x + 3| = 0, so |x + 3| = 3

x + 3 = 3 → x = 0 OR x + 3 = −3 → x = −6

(b)

The vertex (−3, 6) is the maximum point (since a < 0).

Range: f(x) ≤ 6 or {y ∈ ℝ : y ≤ 6}

(c)

f is a many-to-one function (e.g., f(0) = f(−6) = 0).

Since it's not one-to-one, f⁻¹ does not exist.

07Worked Example

The graph shows y = h(x) with points A(−1, 0), B(−2.5, 15), C(0, 11), D(3, 0), E(6, −5).

Sketch y = |h(x)| and y = h(|x|), showing the new positions of each point.

y = |h(x)|:

E(6,−5) → E'(6, 5)

A, B, C, D unchanged

y = h(|x|):

D(3,0) → D'(−3, 0) also

E(6,−5) → E'(−6,−5) also

A, B deleted (x < 0)

Key insight: For h(|x|), the graph becomes symmetric about the y-axis. Points with negative x-values on the original disappear; points with positive x-values get reflected.

Solving Modulus Equations

Always Sketch First

Always sketch the graphs to see how many solutions exist and which branch (positive or negative) each solution lies on.

To Solve |f(x)| = k

Point A

(reflected branch)−f(x) = kPoint B

(original branch)+f(x) = k

Remember

The sketch tells you which equation to use — solve −f(x) = k for the reflected branch, and +f(x) = k for the original branch.

Worked Example08

Solve the equation |3x − 2| = 7

1

Sketch first: y = |3x − 2| is a V-shape with vertex at x = 2/3.

The line y = 7 intersects both branches.

2

Right branch (+f(x) = k): 3x − 2 = 7

3x = 9

x = 3

3

Left branch (−f(x) = k): −(3x − 2) = 7

−3x + 2 = 7

−3x = 5

x = −5/3

4

Check:

x = 3: |3(3) − 2| = |7| = 7 ✓

x = −5/3: |3(−5/3) − 2| = |−7| = 7 ✓

Solutions: x = 3 or x = −5/3

Worked Example09

Solve |2x − 3| = x + 1

1

Sketch y = |2x − 3|:

Vertex at x = 3/2 where 2x − 3 = 0, giving y = 0. Vertex: (1.5, 0)

At x = 0: y = |−3| = 3. Point: (0, 3)

V-shape opening upward with gradient ±2

2

Sketch y = x + 1:

Straight line with gradient 1, y-intercept 1

At x = 0: y = 1. At x = 2: y = 3

3

Find intersections algebraically:

Left branch (x < 1.5): −(2x − 3) = x + 1 → −2x + 3 = x + 1 → 2 = 3x → x = 2/3

Right branch (x ≥ 1.5): 2x − 3 = x + 1 → x = 4 → x = 4

4

Verify solutions:

x = 2/3: |2(2/3) − 3| = |−5/3| = 5/3, and 2/3 + 1 = 5/3 ✓

x = 4: |2(4) − 3| = |5| = 5, and 4 + 1 = 5 ✓

Solutions: x = 2/3 or x = 4

10

Worked Example

Solve the equation |2x + 1| = |x − 4|

1

Sketch first: y = |2x + 1| has vertex at x = −1/2, y = |x − 4| has vertex at x = 4.

2

Case 1: 2x + 1 = x − 4 (both expressions have the same sign)

2x − x = −4 − 1

x = −5

3

Case 2: 2x + 1 = −(x − 4) (expressions have opposite signs)

2x + 1 = −x + 4

3x = 3

x = 1

4

Check:

x = −5: |2(−5) + 1| = |−9| = 9, |−5 − 4| = |−9| = 9 ✓

x = 1: |2(1) + 1| = |3| = 3, |1 − 4| = |−3| = 3 ✓

Solutions: x = −5 or x = 1

Solving Modulus Inequalities

Key Idea

Solving modulus inequalities uses the same approach as solving modulus equations:

Step 1: Sketch the graphs

Draw both sides of the inequality on the same axes

Step 2: Find the solutions

Solve the corresponding equation to find where the graphs intersect

Step 3: Identify the regions

From the graph, read off where one curve is above/below the other

Reading from the Graph

|f(x)| < g(x) → find where the modulus graph is below the other curve

|f(x)| > g(x) → find where the modulus graph is above the other curve

11

Worked Example

The function f is defined by f(x) = 6 − 2|x + 3|, x ∈ ℝ.

(a) State the range of f (b) Solve the inequality f(x) > 4

(a)

This is y = −2|x + 3| + 6. Since a = −2 < 0, this is an inverted V (Λ-shape).

Vertex: At x = −3, f(−3) = 6 − 2(0) = 6 (maximum point)

Range: f(x) ≤ 6

(b)

Method: Draw y = 4 on the graph and find where f(x) is ABOVE this line.

3

Find intersection points A and B: Solve 6 − 2|x + 3| = 4

2|x + 3| = 2, so |x + 3| = 1

x + 3 = 1 → x = −2 (Point B)

x + 3 = −1 → x = −4 (Point A)

4

f(x) > 4 where the graph is above the line y = 4.

This is the region between points A and B.

Solution: −4 < x < −2

Key insight: For an inverted V, ">" gives a bounded interval between the intersection points.

12

Worked Example

Solve the inequality |x − 2| < 4 − |x|

1

Sketch both graphs:

y = |x − 2| is a regular V with vertex at (2, 0)

y = 4 − |x| is an inverted V with vertex at (0, 4)

2

Find intersections: Solve |x − 2| = 4 − |x|

For x < 0: (2 − x) = 4 + x → 2 = 4 + 2x → x = −1 ✓

For x ≥ 2: (x − 2) = 4 − x → 2x = 6 → x = 3 ✓

3

Read from graph: We need |x − 2| < 4 − |x|, i.e., where the blue V is below the red inverted V.

Solution: −1 < x < 3

Finding Number of Roots

Equations involving modulus functions can have 0, 1, 2, or more roots depending on where the lines intersect.

Case 1: |f(x)| = k (horizontal line)

Case 2: |f(x)| = kx + c (straight line)

Gradient and y-intercept both affect number of roots

Essential Method

Always sketch both graphs to determine how many roots an equation has. The number of intersection points = number of roots.

Key Point

The vertex is crucial — in many exam questions, the vertex of the modulus graph is a critical point for determining the range of values of k (or other parameters) that give a specific number of roots.

13

Worked Example

The function f is defined by f(x) = 2|3 − x| + 5, x ≥ 0.

Given that the equation f(x) = kx + 8 has exactly two distinct roots, find the set of possible values for k.

1

Sketch the modulus function:

Vertex at x = 3 where |3 − x| = 0, giving f(3) = 5. Vertex: (3, 5)

At domain boundary x = 0: f(0) = 2|3| + 5 = 11. Point: (0, 11)

2

Sketch the line y = kx + 8:

Passes through (0, 8) for all values of k — between vertex and domain boundary

Gradient k determines the slope

3

Find critical values of k:

Tangent to vertex: Line through (0, 8) and (3, 5) has gradient k = (5−8)/(3−0) = −1

Parallel to right arm: Right arm has gradient 2, so k = 2 gives only 1 root

4

Analyse for two roots:

k < −1: Line too steep downward → misses V-shape → 0 roots

k = −1: Line touches vertex → 1 root

−1 < k < 2: Line crosses both arms → 2 roots ✓

k ≥ 2: Line parallel or steeper than right arm → 1 root

Answer: −1 < k < 2

Practice

Question 1: Sketch the graph of y = |3 − 2x|, stating the coordinates of the vertex and any intercepts.

Step 1: Find the vertex by solving 3 − 2x = 0

2x = 3, so x = 1.5

When x = 1.5, y = |3 − 2(1.5)| = |0| = 0

Vertex: (1.5, 0)

Step 2: Find y-intercept (when x = 0)

y = |3 − 2(0)| = |3| = 3

y-intercept: (0, 3)

Step 3: The gradient of y = 3 − 2x is −2, so:

•

For x ≤ 1.5: y = 3 − 2x (gradient −2)

•

For x > 1.5: y = −(3 − 2x) = 2x − 3 (gradient +2)

--- y = 3 − 2x ━ y = |3 − 2x|

Practice

Question 2: Solve the equation |4x − 1| = |2x + 5|

Case 1: 4x − 1 = 2x + 5

4x − 2x = 5 + 1

2x = 6

x = 3

Case 2: 4x − 1 = −(2x + 5)

4x − 1 = −2x − 5

4x + 2x = −5 + 1

6x = −4

x = −2/3

Check x = 3: |4(3) − 1| = |11| = 11, |2(3) + 5| = |11| = 11 ✓

Check x = −2/3: |4(−2/3) − 1| = |−11/3| = 11/3, |2(−2/3) + 5| = |11/3| = 11/3 ✓

Solutions: x = 3 or x = −2/3

Practice

Question 3: Solve the inequality |x + 2| > 4

Method: |x + 2| > 4 means x + 2 < −4 OR x + 2 > 4

Case 1: x + 2 < −4

x < −6

Case 2: x + 2 > 4

x > 2

Geometric interpretation: |x + 2| = |x − (−2)| represents distance from −2

We need x to be more than 4 units away from −2

Solution: x < −6 or x > 2

Summary

Key Concepts

ConceptKey Points

Modulus Definition|x| = x if x ≥ 0, |x| = −x if x < 0. Always non-negative.

y = |f(x)|Reflect parts below x-axis upward. Result is always ≥ 0.

y = f(|x|)Take right side (x ≥ 0), reflect in y-axis. Result is symmetric.

Equations|f(x)| = k gives f(x) = k or f(x) = −k. Always check solutions.

Inequalities|f(x)| < k gives −k < f(x) < k. |f(x)| > k gives f(x) < −k or f(x) > k.

Exam Skills

Sketch modulus graphs accurately with vertex and intercepts labelled

Solve modulus equations by considering both cases

Convert modulus inequalities to compound inequalities

Use the geometric interpretation (distance) to check answers

Always verify solutions by substituting back