1.3.6 The modulus function (A-level only)

What you'll learn

How to interpret and calculate the modulus of a real number.
How to sketch graphs involving $∣ f (x) ∣$ and $f (∣ x ∣)$ .
How to solve equations and inequalities containing modulus signs.
How modulus can model distances and absolute differences.

What is the modulus function?

The modulus of a real number is its distance from zero on the number line. It is written using vertical bars: $∣ x ∣$ .

Because distance cannot be negative, $∣ x ∣$ is always non-negative.

For example:

$∣5∣ = 5$ because 5 is five units from zero.
$∣ - 5∣ = 5$ because negative 5 is also five units from zero.
$∣0∣ = 0$ .

Definition

The modulus function

For any real number $x$ ,

\begin{cases} x, & x\ge 0,\\ -x, & x<0. \end{cases}

The second line changes the sign of a negative input, making the output positive.

Notice that $∣ x ∣$ does not mean “remove the minus sign”. You must first determine whether the expression inside the modulus signs is positive or negative.

Example

Evaluating modulus expressions

Find the value of $∣3 - 8∣ + ∣ - 4∣$ .

Evaluate each expression inside the modulus signs: $3 - 8 = - 5$ , while the second value is already negative 4.
Take each distance from zero: $∣ - 5∣ = 5$ and $∣ - 4∣ = 4$ .
Add the resulting non-negative values to obtain $5 + 4 = 9$ .

Key Idea

Modulus means distance

You can interpret $∣ x ∣$ as the distance between $x$ and zero. More generally, $∣ x - a ∣$ is the distance between $x$ and $a$ .

The graph of the modulus function

The graph of $y = ∣ x ∣$ consists of two straight-line sections:

when $x≥0x\ge 0$ , it follows $y = x$ ;
when $x < 0$ , it follows $y = - x$ .

These sections meet at the vertex, the point where the direction of the graph changes. Here, the vertex is the origin.

Graph of y equals modulus x, showing its two linear branches and reflection in the x-axis

The graph has range $y≥0y\ge 0$ because a modulus output cannot be negative. It is also symmetric about the $y$ -axis because $∣ - x ∣ = ∣ x ∣$ .

Translations and stretches

A graph such as

y = a ∣ x - h ∣ + k

has its vertex at $(h, k)$ .

$h$ translates the graph horizontally.
$k$ translates it vertically.
$a$ produces a vertical stretch and may reflect the graph in the $x$ -axis.
If $a > 0$ , the graph opens upwards; if $a < 0$ , it opens downwards.

Example

Sketching a translated modulus graph

Sketch $y = 2∣ x - 3∣ - 4$ and state its vertex and intercepts.

Read the vertex from the form $a ∣ x - h ∣ + k$ : here $h = 3$ and $k = - 4$ , so the vertex is $(3, - 4)$ . Since the coefficient 2 is positive, the graph opens upwards.
Find the $x$ -intercepts by setting $y = 0$ :
$2∣x−3∣−4=0∣x−3∣=2x−3=2orx−3=−2x=5orx=1.\begin{aligned} 2|x-3|-4&=0\\ |x-3|&=2\\ x-3&=2 \quad \text{or} \quad x-3=-2\\ x&=5 \quad \text{or} \quad x=1. \end{aligned}$
Find the $y$ -intercept by substituting $x = 0$ :
$y = 2∣0 - 3∣ - 4 = 2 (3) - 4 = 2.$
The graph therefore passes through $(1, 0)$ , $(3, - 4)$ , $(5, 0)$ and $(0, 2)$ .

Common Mistake

Horizontal translation signs

In $∣ x - h ∣$ , the vertex has $x$ -coordinate $h$ , not $- h$ . For example, $∣ x - 3∣$ has its vertex at $x = 3$ .

Modulus transformations of general graphs

There are two different transformations you must distinguish.

The graph of $y = ∣ f (x) ∣$

The modulus is applied to the output of $f$ .

Keep every part of $y = f (x)$ that is on or above the $x$ -axis.
Reflect every part below the $x$ -axis in the $x$ -axis.

Therefore, the transformed graph has no points below the $x$ -axis.

The graph of $y = f (∣ x ∣)$

Here the modulus is applied to the input.

Keep the part of $y = f (x)$ for $x≥0x\ge 0$ .
Reflect this right-hand part in the $y$ -axis.
Discard the original part for $x < 0$ .

The resulting graph is symmetric about the $y$ -axis.

Comparison of the graph transformations y equals modulus f of x and y equals f of modulus x

Key Idea

Inside or outside?

For $∣ f (x) ∣$ , reflect negative outputs in the $x$ -axis. For $f (∣ x ∣)$ , copy the non-negative inputs across the $y$ -axis.

Example

Transforming a quadratic graph

Describe the graph of $y=|x^2-4|$ .

Begin with $y=x^2-4$ , an upward-opening parabola with roots $x = - 2$ and $x = 2$ and vertex $(0, - 4)$ .
Identify where its output is negative: $x^2-4<0$ between the roots, so the section for $- 2 < x < 2$ lies below the $x$ -axis.
Reflect only that section in the $x$ -axis. It becomes $y=4-x^2$ for $- 2 < x < 2$ , while the outer sections remain $y=x^2-4$ .
The full graph can therefore be written
$\begin{cases} x^2-4, & x\le -2 \text{ or } x\ge 2,\\ 4-x^2, & -2<x<2. \end{cases}$

Solving modulus equations

The equation $∣ A ∣ = k$ asks which values of $A$ are a distance $k$ from zero.

If $k > 0$ , there are two possibilities:

\quad\Rightarrow\quad A=k \quad\text{or}\quad A=-k.

If $k = 0$ , then $A = 0$ . If $k < 0$ , there are no real solutions because a modulus cannot be negative.

Example

Solving a modulus equation

Solve $∣2 x - 5∣ = 7$ .

Split the equation into the two possible cases:
$\quad\text{or}\quad 2x-5=-7.$
Solve the first case:
$\Rightarrow x=6.$
Solve the second case:
$\Rightarrow x=-1.$
The solutions are therefore $x = 6$ and $x = - 1$ .

Equations with modulus on both sides

For equations such as $∣ f (x) ∣ = g (x)$ , remember that the right-hand side must be non-negative. Squaring both sides can help, but you must check the resulting solutions in the original equation.

Example

Solving an equation by squaring

Solve $∣ x - 1∣ = x + 3$ .

Since the left-hand side is non-negative, require $x+3≥0x+3\ge 0$ , giving $x≥−3x\ge -3$ .
Square both sides:
$(x−1)2=(x+3)2x2−2x+1=x2+6x+9−8x=8x=−1.\begin{aligned} (x-1)^2&=(x+3)^2\\ x^2-2x+1&=x^2+6x+9\\ -8x&=8\\ x&=-1. \end{aligned}$
Check in the original equation: $∣ - 1 - 1∣ = 2$ and $- 1 + 3 = 2$ , so $x = - 1$ is valid.

Solving modulus inequalities

Distance provides the clearest interpretation.

For $k > 0$ :

\quad\Longleftrightarrow\quad a-k<x<a+k,

because $x$ is less than $k$ units from $a$ .

Similarly,

\quad\Longleftrightarrow\quad x<a-k \quad\text{or}\quad x>a+k,

because $x$ is more than $k$ units from $a$ .

The same patterns apply to $≤\le$ and $≥\ge$ , with the boundary values included.

Tip

Inside and outside

A “less than” modulus inequality usually gives values between two boundaries. A “greater than” modulus inequality usually gives values outside them.

Example

Finding an acceptable range

A manufactured component is accepted when its length $x$ mm differs from 50 mm by no more than 0.3 mm. Find the acceptable values of $x$ .

Translate “differs from 50 by no more than 0.3” into the distance inequality $∣x−50∣≤0.3|x-50|\le 0.3$ .
Rewrite this as a compound inequality:
$−0.3≤x−50≤0.3.-0.3\le x-50\le 0.3.$
Add 50 throughout:
$49.7≤x≤50.3.49.7\le x\le 50.3.$
Therefore, accepted components have lengths from 49.7 mm to 50.3 mm, including both limits.

Common Mistake

Forgetting two regions

When solving $∣ x - a ∣ > k$ , do not give only one side of the answer. Values can lie more than $k$ units to the left or to the right of $a$ .

Intersections and graphical reasoning

An equation involving modulus can also be solved by finding intersections between graphs. For example, the solutions of $∣ f (x) ∣ = g (x)$ are the $x$ -coordinates where $y = ∣ f (x) ∣$ and $y = g (x)$ meet.

This is especially useful when an exact algebraic solution is difficult or when a question asks for the number of solutions as a parameter changes.

Example

Counting intersections

Determine the number of solutions of $∣ x ∣ = 2 - x$ .

The right-hand side must be non-negative, so any solution must satisfy $x≤2x\le 2$ .
For $x≥0x\ge 0$ , use $∣ x ∣ = x$ . Then $x = 2 - x$ , so $x = 1$ .
For $x < 0$ , use $∣ x ∣ = - x$ . This gives $- x = 2 - x$ , which simplifies to $0 = 2$ and has no solution.
There is therefore exactly one solution, $x = 1$ .

Exam technique

In the exam

Decide whether the modulus acts on an input, an output, or a distance before starting any algebra.
When solving $∣ A ∣ = k$ , consider both $A = k$ and $A = - k$ , but first check that $k$ is non-negative.
Check solutions in the original equation after squaring or splitting into cases, and use a quick graph or number line to verify inequality regions.
For graph transformations, mark roots and turning points before reflecting the relevant sections.

Self review

Check yourself

What is the difference between constructing $y = ∣ f (x) ∣$ and constructing $y = f (∣ x ∣)$ ?
How would you solve $∣3 x + 1∣ = 8$ , and why should you expect up to two solutions?
What interval is represented by $∣x−4∣≤1.5|x-4|\le 1.5$ ?

What you'll learn

How to interpret and calculate the modulus of a real number.
How to sketch graphs involving $∣ f (x) ∣$ and $f (∣ x ∣)$ .
How to solve equations and inequalities containing modulus signs.
How modulus can model distances and absolute differences.

What is the modulus function?

The modulus of a real number is its distance from zero on the number line. It is written using vertical bars: $∣ x ∣$ .

Because distance cannot be negative, $∣ x ∣$ is always non-negative.

For example:

$∣5∣ = 5$ because 5 is five units from zero.
$∣ - 5∣ = 5$ because negative 5 is also five units from zero.
$∣0∣ = 0$ .

Definition

The modulus function

For any real number $x$ ,

\begin{cases} x, & x\ge 0,\\ -x, & x<0. \end{cases}

The second line changes the sign of a negative input, making the output positive.

Notice that $∣ x ∣$ does not mean “remove the minus sign”. You must first determine whether the expression inside the modulus signs is positive or negative.

Example

Evaluating modulus expressions

Find the value of $∣3 - 8∣ + ∣ - 4∣$ .

Evaluate each expression inside the modulus signs: $3 - 8 = - 5$ , while the second value is already negative 4.
Take each distance from zero: $∣ - 5∣ = 5$ and $∣ - 4∣ = 4$ .
Add the resulting non-negative values to obtain $5 + 4 = 9$ .

Key Idea

Modulus means distance

You can interpret $∣ x ∣$ as the distance between $x$ and zero. More generally, $∣ x - a ∣$ is the distance between $x$ and $a$ .