Graphs of Modulus Functions

Sketch y = |f(x)| for straight lines, quadratics, cubics and trig curves by reflecting the part below the x-axis, and label every intercept.

In the exam for IGCSE Additional Maths 0606 · 6 real past-paper questions below

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What the modulus does

A taxi charges you for distance, not direction. Drive 5 km north or 5 km south, and you pay for 5 km. The modulus does the same to a number: it keeps the size and drops the sign.

Rule

$∣ x ∣$ is read “mod x”. It is the size of x, so it is never negative: $∣3∣ = 3$ , $∣ - 3∣ = 3$ and $∣0∣ = 0$ .

Formula

∣ x ∣ = {x, - x, x \geq 0 x < 0

$x$ — any number. If it is already positive or zero, it stays. If it is negative, the minus sign in $- x$ makes it positive: $∣ - 4∣ = - (- 4) = 4$ .

Now do this to a whole function. For each x, work out f(x) first. Then take its size.

Worked example

1
Let $f (x) = 2 x - 6$ . Find $∣ f (x) ∣$ at $x = 0$ , $3$ and $5$ .
2
$f (0) = - 6, ∣ f (0) ∣ = 6$ — The line is below the x-axis here. The modulus turns −6 into 6.
3
$f (3) = 0, ∣ f (3) ∣ = 0$ — On the x-axis, nothing changes.
4
$f (5) = 4, ∣ f (5) ∣ = 4$ — Above the x-axis, nothing changes either.

Rule

To draw $y = ∣ f (x) ∣$ : keep every part of $y = f (x)$ that is on or above the x-axis. Reflect every part below the x-axis up, using the x-axis as a mirror.

Key idea

The two straight bars for modulus were introduced by the German mathematician Karl Weierstrass in 1841. Mathematicians still use his notation today.

Key idea

The graph of $y = ∣ f (x) ∣$ never goes below the x-axis. Every point on it has $y \geq 0$ .

Modulus of a straight line

Start with the simplest case, a straight line. The part of the line below the x-axis flips up, so the graph becomes a V.

Steps

1
1. Find where the line crosses the x-axis: solve $a x + b = 0$ . This is the corner of the V.
2. Find the y-intercept: put $x = 0$ , then take the modulus. It is $∣ b ∣$ .
3. Draw the line above the axis. Reflect the part below the axis. Label both intercepts on the sketch.

Worked example

1
Sketch $y = ∣2 x - 5∣$ .
2
$2 x - 5 = 0 \Rightarrow x = \frac{5}{2}$ — The corner of the V sits on the x-axis at this point.
3
$x = 0 : y = ∣ - 5∣ = 5$ — The y-intercept. The line was at −5, so the modulus makes it 5.
4
$x = 4 : y = ∣3∣ = 3$ — One more point on the right arm, to check the direction.
5
$(\frac{5}{2}, 0), (0, 5)$ — Draw the V through these points and write both coordinates on the sketch.

The right arm has gradient 2. The left arm is its mirror image, so it has gradient −2. The two arms are equally steep.

Common mistake

Drawing a rounded bottom. The modulus of a straight line has a sharp corner on the x-axis, not a curve.

Common mistake

Labelling the y-intercept as $(0, - 5)$ . The modulus makes every y-value positive, so it is $(0, 5)$ .

Key idea

For $y = ∣ a x + b ∣$ : the corner is at the root of the line, and the y-intercept is $∣ b ∣$ .

Modulus of a quadratic

When the $x^{2}$ term is positive, the curve is a U and only the part between the roots is below the x-axis. That part flips up, so the bottom of the U becomes a hump.

Worked example

1
Sketch $y = ∣ x^{2} - 2 x - 8∣$ and label the intercepts.
2
$x^{2} - 2 x - 8 = (x - 4) (x + 2)$ — Factorise to find the roots.
3
$x = - 2, x = 4$ — The graph meets the x-axis here. These points stay where they are.
4
$x = 0 : y = ∣ - 8∣ = 8$ — The y-intercept. The curve was at −8, so after the flip it is at 8.
5
$x^{2} - 2 x - 8 = (x - 1)^{2} - 9$ — Complete the square: the lowest point of the U is (1, −9).
6
$(1, - 9) \to (1, 9)$ — The modulus reflects the lowest point up. It becomes the top of the hump.

Rule

At each root, the graph has a sharp point on the x-axis, called a cusp. Mark schemes look for these sharp points and for the labelled intercepts.

The sketch also tells you how many solutions $∣ f (x) ∣ = k$ has. Draw the flat line $y = k$ and count where it meets the graph. For $y = ∣ x^{2} - 2 x - 8∣$ :

$k = 0$ : 2 solutions, at the two cusps.
$0 < k < 9$ : 4 solutions. The line cuts both outer arms and both sides of the hump.
$k = 9$ : 3 solutions. The line just touches the top of the hump.
$k > 9$ : 2 solutions, on the outer arms only.

Common mistake

Drawing smooth, round bottoms at the roots. The graph changes direction suddenly there, so draw sharp points.

Common mistake

Leaving the y-intercept at −8. It was below the axis, so it flips to 8.

Key idea

For $y = ∣ quadratic ∣$ : the roots become cusps and the lowest point of the U becomes a hump. If the $x^{2}$ term is negative, the two outer arms are below the axis and they flip up instead.

Modulus of a cubic

Cubics are given as a product of three brackets. Sketch the cubic first. Then reflect every part below the x-axis.

Steps

1
1. Set each bracket to 0 to find the three roots.
2. Put $x = 0$ to find the y-intercept.
3. Find the sign of the $x^{3}$ term. Positive: the curve starts bottom left and ends top right. Negative: it starts top left and ends bottom right.
4. Reflect the parts below the x-axis. Label all four intercepts.

Worked example

1
Sketch $y = ∣ (x + 2) (x - 1) (x - 3) ∣$ .
2
$x = - 2, 1, 3$ — The three roots. They become cusps on the x-axis.
3
$x = 0 : (2) (- 1) (- 3) = 6$ — The y-intercept is 6. It is already positive, so it stays.
4
$x \times x \times x = x^{3}$ — The x terms of the brackets multiply to plus x cubed, so the cubic starts bottom left.
5
$f (x) < 0 for x < - 2 and 1 < x < 3$ — Here the cubic is below the x-axis, so these two parts are reflected.
6
$(- 2, 0), (1, 0), (3, 0), (0, 6)$ — Label all four intercepts. The left arm now comes down from the top left to the cusp at −2, and between 1 and 3 there is a new hump.

Common mistake

Stopping the reflected left arm at the axis. It keeps going up to the top left of the sketch, because $∣ f (x) ∣$ grows large as x goes far left.

Common mistake

Missing a minus inside a bracket. In $(1 - x)$ the x term is $- x$ , which changes the sign of $x^{3}$ . Count the minus signs before you decide where the curve starts.

Key idea

Cubic first, modulus second. Both ends of $y = ∣ cubic ∣$ point upwards.

Modulus of a trig graph

Trig curves work the same way. Here f(x) will look like $a sin b x + c$ , $a cos b x + c$ or $a tan b x + c$ .

First find the highest and lowest values of the curve before the modulus. For $a sin b x + c$ , sin goes from −1 to 1, so the curve goes from $c - a$ up to $c + a$ .

Worked example

1
Sketch $y = ∣2 sin x + 1∣$ for $0^{\circ} \leq x \leq 36 0^{\circ}$ .
2
$highest = 1 + 2 = 3, lowest = 1 - 2 = - 1$ — The lowest value is negative, so part of the curve is below the x-axis.
3
$x = 0 : y = 1$ — The y-intercept. It is above the axis, so it stays.
4
$sin x = - \frac{1}{2} \Rightarrow x = 21 0^{\circ}, 33 0^{\circ}$ — Where the curve crosses the x-axis. These become cusps.
5
$(27 0^{\circ}, - 1) \to (27 0^{\circ}, 1)$ — The lowest point flips up. Between the cusps there is now a small hump of height 1.
6
$(9 0^{\circ}, 3)$ — The big peak was above the axis, so it is unchanged.

Rule

If the lowest value $c - a$ is 0 or more, the curve never goes below the axis and the modulus changes nothing. For example, $y = ∣2 sin x + 5∣$ is the same graph as $y = 2 sin x + 5$ .

For $y = ∣ tan x ∣$ , the parts below the axis flip up as usual. The vertical asymptotes stay exactly where they were.

Common mistake

Reflecting the whole curve, or shifting it up. Only the parts below the x-axis move, and they are reflected.

Key idea

Find the top and bottom of the curve first. If the bottom is negative, that part flips up into a new hump.

Which sketch? Read the question

Every modulus sketch uses one idea. The question tells you which curve to draw first.

You see	You do
Sketch $y = ∣ a x + b ∣$	V shape. Corner at the root, y-intercept $∣ b ∣$
Sketch $y = ∣ quadratic ∣$	Roots become cusps, the lowest point flips into a hump
Sketch $y = ∣ (x - p) (x - q) (x - r) ∣$	Cubic shape from the sign of $x^{3}$ , then reflect
Sketch $y = ∣ a sin b x + c ∣$	Find $c + a$ and $c - a$ , then reflect what is below
“State the intercepts”	Write the coordinates of every axis crossing on the sketch
How many solutions of $∣ f (x) ∣ = k$ ?	Draw $y = k$ and count the crossings

Common mistake

Forgetting to label the intercepts. A correct shape with no values written on it loses marks.

Key idea

Draw $y = f (x)$ lightly first. Then keep the top, flip the bottom, and press hard on the final graph.

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Real past-paper questions on graphs of modulus functions

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