Year 2 Pure · Chapter 4

Modulus and Graph Transformations

Specification: Edexcel Pure 2.7, 2.9 — Algebra and functions · AQA section B7, B9

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In this chapter

1. The modulus function
2. Modulus equations and inequalities
3. The graphs of |f(x)| and f(|x|)
4. Combining transformations
Practice

Key facts — this chapter

The modulus $|x|$ is the size of $x$ , ignoring its sign: $|x| = x$ when $x \geq 0$ , and $|x| = -x$ when $x < 0$ .
$y = |ax + b|$ is a V shape with its vertex on the $x$ -axis at $x = -\tfrac{b}{a}$ .
$|x - a| < b$ means $x$ is within $b$ of $a$ : $a - b < x < a + b$ .
To solve $|f(x)| = g(x)$ , solve $f(x) = g(x)$ and $-f(x) = g(x)$ , then check each answer, since $g(x)$ can’t be negative.
$y = |f(x)|$ : reflect the parts of $y = f(x)$ below the $x$ -axis in the $x$ -axis.
$y = f(|x|)$ : keep the part for $x \geq 0$ and reflect it in the $y$ -axis.
Combining transformations: apply them one at a time. Changes inside the bracket act on $x$ , in reverse.

The modulus takes the sign off a number, so $|-5| = |5| = 5$ . On a graph that is a fold: whatever dips below the axis is flipped back up. This chapter draws those folds, solves equations with them in, and finishes the Year 1 work on transformations by combining several at once.

1. The modulus function

Key result

|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

$|x|$ is the distance of $x$ from $0$ , and $|x - a|$ is the distance between $x$ and $a$ . The graph of $y = |ax + b|$ is the line $y = ax + b$ with its negative part reflected in the $x$ -axis: a V with its vertex at $\left(-\tfrac{b}{a},\, 0\right)$ .

Worked example 4.1 · A V-shaped graph

Sketch $y = |2x - 3|$ , giving the coordinates of the points where it meets the axes.

The vertex is where $2x - 3 = 0$ , at $\left(\tfrac{3}{2}, 0\right)$ .
When $x = 0$ , $y = |-3| = 3$ , so it meets the $y$ -axis at $(0, 3)$ .
To the right of the vertex the graph is $y = 2x - 3$ . To the left it is $y = -(2x - 3) = 3 - 2x$ . Both arms have gradient of size $2$ .

2. Modulus equations and inequalities

The equation $|f(x)| = g(x)$ says that $f(x)$ is either $g(x)$ or $-g(x)$ . So there are two equations to solve, but only answers that make $g(x) \geq 0$ can work, because a modulus is never negative.

Solving modulus equations

Solve $f(x) = g(x)$ and $-f(x) = g(x)$ .
Check each solution in the original equation, rejecting any that make $g(x)$ negative.

A sketch of both graphs shows how many solutions to expect.

Worked example 4.2 · A V meets a line twice

Solve $|2x - 3| = x + 1$ .

First case, $2x - 3 = x + 1$ : $x = 4$ .
Second case, $-(2x - 3) = x + 1$ , so $3 - 2x = x + 1$ and $x = \tfrac{2}{3}$ .
Check: at $x = 4$ both sides are $5$ , and at $x = \tfrac{2}{3}$ both sides are $\tfrac{5}{3}$ . Both solutions stand.

The V-shaped graph y = |2x − 3| with vertex at (1.5, 0), and the line y = x + 1 crossing it at x = 2/3 and x = 4; between the crossings the V is below the line — y = |2x − 3| meets y = x + 1 at x = ⅔ and x = 4. Between them the V is below the line.

Worked example 4.3 · One answer rejected

Solve $|x - 2| = 2x + 1$ .

First case, $x - 2 = 2x + 1$ : $x = -3$ . But then $2x + 1 = -5$ , and a modulus can’t equal a negative number. Reject it.
Second case, $-(x - 2) = 2x + 1$ , so $2 - x = 2x + 1$ and $x = \tfrac{1}{3}$ . Check: $\left|\tfrac{1}{3} - 2\right| = \tfrac{5}{3}$ and $2\left(\tfrac{1}{3}\right) + 1 = \tfrac{5}{3}$ .
The only solution is $x = \tfrac{1}{3}$ .

Common mistake

Always check your answers. The two cases produce candidates, not solutions: here $x = -3$ solves $x - 2 = 2x + 1$ perfectly well, but it doesn’t solve the modulus equation.

For inequalities, find where the graphs meet, then read off from the sketch where one is above the other.

Modulus inequalities

$|x - a| < b$ means $a - b < x < a + b$ : $x$ is within $b$ of $a$ .
$|x - a| > b$ means $x < a - b$ or $x > a + b$ .
For anything more complicated, sketch both sides and use the points where they meet.

Worked example 4.4 · An inequality from the graph

Solve $|2x - 3| > x + 1$ .

The graphs meet at $x = \tfrac{2}{3}$ and $x = 4$ (Example 4.2). The V is above the line outside those points, so

x < \tfrac{2}{3} \quad\text{or}\quad x > 4

3. The graphs of |f(x)| and f(|x|)

The modulus can go on the outside of a function or on the inside, and the two give different graphs.

Key result

$y = |f(x)|$ : any part of $y = f(x)$ below the $x$ -axis is reflected in the $x$ -axis. Nothing is ever below the axis.
$y = f(|x|)$ : the part of $y = f(x)$ for $x \geq 0$ is kept, and reflected in the $y$ -axis to replace the left half. The graph is symmetrical about the $y$ -axis.

Three graphs side by side: the cubic y = f(x) crossing the x-axis at −1, 1 and 3; y = |f(x)| with the parts below the axis flipped up; and y = f(|x|), the right half of the cubic mirrored in the y-axis — Left to right: y = f(x), y = |f(x)| and y = f(|x|), for f(x) = (x + 1)(x − 1)(x − 3).

Common mistake

The two are easy to swap. Ask what the modulus acts on. In $|f(x)|$ it acts on the output, the $y$ -value, so it folds the graph up over the $x$ -axis. In $f(|x|)$ it acts on the input, the $x$ -value, so it mirrors across the $y$ -axis.

Worked example 4.5 · Four solutions

Solve $|x^2 - 4x| = 3$ .

The graph of $y = |x^2 - 4x|$ is the parabola with its dip between $x = 0$ and $x = 4$ folded up, so its bump has height $4$ at $x = 2$ . The line $y = 3$ lies below that height, so it meets the graph four times.
Case $x^2 - 4x = 3$ : $x^2 - 4x - 3 = 0$ , so $x = 2 \pm \sqrt{7}$ .
Case $x^2 - 4x = -3$ : $x^2 - 4x + 3 = 0$ , so $x = 1$ or $x = 3$ .
The right-hand side is positive, so all four stand: $x = 2 - \sqrt{7}$ , $1$ , $3$ , $2 + \sqrt{7}$ .

Put the modulus outside the function to fold the graph up over the x-axis, or inside to mirror its right half across the y-axis. Then move the line y = k and count the solutions.

Function Graph

4. Combining transformations

In Year 1 you met the four basic transformations one at a time. A function such as $y = 3f(2x) + 1$ applies several at once.

Combining transformations

Outside the bracket acts on $y$ , in the order you would calculate: $y = af(x) + d$ is a vertical stretch by scale factor $a$ , then a translation up by $d$ .
Inside the bracket acts on $x$ , in reverse: $y = f(bx + c)$ is a translation left by $c$ , then a horizontal stretch by scale factor $\tfrac{1}{b}$ .
A point $(p, q)$ on $y = f(x)$ moves to $\left(\dfrac{p - c}{b},\ aq + d\right)$ on $y = af(bx + c) + d$ .

Why it works

The graph of $y = af(bx + c) + d$ passes through $(x, y)$ when the original function passes through $(bx + c,\ \tfrac{y - d}{a})$ . So a point $(p, q)$ on $y = f(x)$ reappears where $bx + c = p$ and $y = aq + d$ . Undoing “multiply by $b$ , then add $c$ ” means subtracting $c$ first, then dividing by $b$ , which is why the inside changes run in reverse.

Worked example 4.6 · Where a point moves to

The curve $y = f(x)$ has a minimum at $(4, -2)$ . Find the minimum of $y = 3f(2x) + 1$ .

Inside: $2x = 4$ , so $x = 2$ .
Outside: $y = 3(-2) + 1 = -5$ .
A vertical stretch by a positive factor keeps a minimum a minimum, so the minimum is at $(2, -5)$ .

Worked example 4.7 · A transformed V

Sketch $y = 2|x - 1| - 3$ , giving the vertex and the points where it meets the axes.

Start from $y = |x|$ : translate right by $1$ , stretch vertically by $2$ , then translate down by $3$ . The vertex moves from $(0, 0)$ to $(1, -3)$ .
$x$ -axis: $2|x - 1| = 3$ , so $|x - 1| = \tfrac{3}{2}$ and $x = -\tfrac{1}{2}$ or $x = \tfrac{5}{2}$ .
$y$ -axis: $y = 2|-1| - 3 = -1$ , so it meets it at $(0, -1)$ .

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [2 marks]

Find the coordinates of the points where $y = |3x + 6|$ meets the coordinate axes.

Show solution

The vertex is where $3x + 6 = 0$ , so it meets the $x$ -axis at $(-2, 0)$ . When $x = 0$ , $y = |6| = 6$ , so it meets the $y$ -axis at $(0, 6)$ .

2 [3 marks]

Solve:

$|x - 4| = 3$
$|2x + 1| = 5$

Show solution

(a) $x - 4 = 3$ or $x - 4 = -3$ , so $x = 7$ or $x = 1$ .

(b) $2x + 1 = 5$ or $2x + 1 = -5$ , so $x = 2$ or $x = -3$ .

3 [2 marks]

Solve $|x + 1| < 4$ .

Show solution

$|x + 1| = |x - (-1)|$ is the distance from $x$ to $-1$ , and it must be less than $4$ :

\begin{gathered} -1 - 4 < x < -1 + 4 \\ -5 < x < 3 \end{gathered}

4 [3 marks]

The point $P(3, 5)$ lies on the curve $y = f(x)$ . Find the image of $P$ on each curve.

$y = f(x + 2)$
$y = 2f(x) - 1$
$y = f(3x)$

Show solution

(a) A translation left by $2$ : $(1, 5)$ .

(b) A vertical stretch by $2$ , then down by $1$ : $(3, 2 \times 5 - 1) = (3, 9)$ .

(c) $3x = 3$ gives $x = 1$ : $(1, 5)$ .

Standard

5 [4 marks]

Solve $|3x - 2| = x + 4$ .

Show solution

First case, $3x - 2 = x + 4$ : $x = 3$ .

Second case, $-(3x - 2) = x + 4$ , so $2 - 3x = x + 4$ and $x = -\tfrac{1}{2}$ .

Check: at $x = 3$ both sides are $7$ ; at $x = -\tfrac{1}{2}$ both sides are $\tfrac{7}{2}$ . So $x = 3$ or $x = -\tfrac{1}{2}$ .

6 [4 marks]

Solve $|x + 3| = 2x$ .

Show solution

First case, $x + 3 = 2x$ : $x = 3$ . Check: $|6| = 6 = 2(3)$ . ✓

Second case, $-(x + 3) = 2x$ : $x = -1$ . But then $2x = -2$ , which is negative, so reject it.

The only solution is $x = 3$ .

7 [4 marks]

Solve $|2x - 1| > |x + 2|$ .

Show solution

Both sides are non-negative, so squaring keeps the inequality the same way round:

\begin{aligned} (2x - 1)^2 &> (x + 2)^2 \\ 4x^2 - 4x + 1 &> x^2 + 4x + 4 \\ 3x^2 - 8x - 3 &> 0 \\ (3x + 1)(x - 3) &> 0 \end{aligned}

A quadratic with positive $x^2$ coefficient is positive outside its roots, so $x < -\tfrac{1}{3}$ or $x > 3$ .

8 [4 marks]

$f(x) = x^2 - 2x - 3$ .

Sketch $y = |f(x)|$ , giving the coordinates of the points where it meets the axes and of the maximum between them. [3]
State the values of $k$ for which $|f(x)| = k$ has exactly four solutions. [1]

Show solution

(a) $f(x) = (x + 1)(x - 3)$ , so the graph meets the $x$ -axis at $(-1, 0)$ and $(3, 0)$ . Between them $f(x)$ is negative, with minimum $f(1) = -4$ ; the modulus folds this up to a maximum at $(1, 4)$ . It meets the $y$ -axis at $(0, |-3|) = (0, 3)$ .

(b) A horizontal line $y = k$ meets the graph four times when it cuts the folded bump and both outer arms: $0 < k < 4$ .

9 [3 marks]

$f(x) = x^2 - 5x + 4$ .

Write down an expression for $f(|x|)$ when $x < 0$ . [1]
Solve $f(|x|) = 0$ . [2]

Show solution

(a) For $x < 0$ , $|x| = -x$ , so $f(|x|) = (-x)^2 - 5(-x) + 4 = x^2 + 5x + 4$ .

(b) $f(|x|) = (|x| - 1)(|x| - 4) = 0$ , so $|x| = 1$ or $|x| = 4$ . That gives $x = \pm 1$ or $x = \pm 4$ .

Exam style

10 [6 marks]

$f(x) = 2|x - 1| - 3$ .

Sketch $y = f(x)$ , giving the coordinates of the vertex and of the points where the graph meets the axes. [3]
Solve $f(x) = -x$ . [3]

Show solution

(a) The vertex is at $(1, -3)$ . It meets the $x$ -axis where $|x - 1| = \tfrac{3}{2}$ , at $\left(-\tfrac{1}{2}, 0\right)$ and $\left(\tfrac{5}{2}, 0\right)$ , and the $y$ -axis at $(0, -1)$ .

(b) For $x \geq 1$ : $2(x - 1) - 3 = -x$ , so $3x = 5$ and $x = \tfrac{5}{3}$ , which is at least $1$ . ✓

For $x < 1$ : $-2(x - 1) - 3 = -x$ , so $-2x - 1 = -x$ and $x = -1$ , which is less than $1$ . ✓

So $x = \tfrac{5}{3}$ or $x = -1$ .

11 [6 marks]

The curve $y = f(x)$ has a maximum at $(2, 5)$ and meets the $y$ -axis at $(0, 1)$ . For each curve below, find the images of these two points, and say whether the image of the maximum is a maximum or a minimum.

$y = f(x - 3) + 2$ [2]
$y = -2f(x)$ [2]
$y = f(2x + 4)$ [2]

Show solution

(a) Right $3$ , up $2$ : the maximum moves to $(5, 7)$ and is still a maximum; $(0, 1)$ moves to $(3, 3)$ .

(b) Each $y$ -value is multiplied by $-2$ : $(2, -10)$ , which is now a minimum, and $(0, -2)$ .

(c) Solve $2x + 4 = 2$ : $x = -1$ , so the maximum is at $(-1, 5)$ and is still a maximum. Solve $2x + 4 = 0$ : $x = -2$ , so $(0, 1)$ moves to $(-2, 1)$ .

12 [6 marks]

Sketch $y = |x^2 - 4|$ , giving the coordinates of the points where it meets the axes. [2]
Solve $|x^2 - 4| = 3x$ . [4]

Show solution

(a) $x^2 - 4 = 0$ at $x = \pm 2$ , so the graph meets the $x$ -axis at $(-2, 0)$ and $(2, 0)$ . Between them the parabola dips to $-4$ , which folds up to a maximum at $(0, 4)$ , where it meets the $y$ -axis.

(b) Any solution needs $3x \geq 0$ , so $x \geq 0$ .

Case $x^2 - 4 = 3x$ : $x^2 - 3x - 4 = 0$ , so $(x - 4)(x + 1) = 0$ . Only $x = 4$ has $x \geq 0$ .

Case $-(x^2 - 4) = 3x$ : $x^2 + 3x - 4 = 0$ , so $(x + 4)(x - 1) = 0$ . Only $x = 1$ has $x \geq 0$ .

Check: $|16 - 4| = 12 = 3(4)$ and $|1 - 4| = 3 = 3(1)$ . So $x = 1$ or $x = 4$ .