P3 1.3 - Modulus graphs and equations | Edexcel A-Level International Mathematics

Interpret modulus as distance, sketch graphs involving absolute values and solve equations and inequalities from their intersections. Distinguish changing a function's output from changing its input.

The absolute value function

How far is $-4$ from zero? Its position is negative, but its distance is $4$ . Modulus records that nonnegative distance and produces a graph with a corner.

Modulus or absolute value

For a real number $u$ , its modulus is

|u|=\begin{cases}u,&u\ge0,\\-u,&u<0.\end{cases}

Thus $|4|=4$ , $|-4|=4$ and $|0|=0$ . Modulus is never negative.

For $|2x-4|$ , ask when the whole inside expression changes sign. It is zero at $x=2$ , not at $x=0$ . To the left, replace it by its negative; to the right, keep it.

Worked example 1: locating the corner

Question. Give a piecewise rule and the axis intercepts of $y=|2x-4|$ , and describe its shape.

Plan. Solve $2x-4=0$ for the corner, then use the sign on each side.

y=\begin{cases}4-2x,&x<2,\\2x-4,&x\ge2.\end{cases}

The gradients are $-2$ on the left and $2$ on the right. At $x=0$ , $y=4$ .

Answer. A V-shaped graph with vertex and $x$ -intercept $(2,0)$ , and $y$ -intercept $(0,4)$ .

Check. At $x=1$ , both $|2x-4|$ and the left rule give $2$ . At $x=3$ , both the modulus and right rule give $2$ .

Change only the addition outside the modulus, raising the graph by $1$ . The sign change inside still occurs at the same input.

Worked example 2: moving the vertex above the axis

Question. Give a piecewise rule and the axis intercepts of $y=|2x-4|+1$ , and describe its shape.

Plan. Add $1$ to both branches from Example 1.

y=\begin{cases}5-2x,&x<2,\\2x-3,&x\ge2.\end{cases}

Answer. A V-shaped graph with vertex $(2,1)$ , $y$ -intercept $(0,5)$ , and no $x$ -intercept because $y\ge1$ .

Check. At $x=2$ , both branch rules give $1$ . At $x=0$ , the original expression gives $|-4|+1=5$ .

Both branches meet; the change of rule does not create a gap.

[DIAGRAM: asset_name: A modulus vertex and a vertical shift; asset_slug: modulus-vertices; caption: Adding one outside the modulus raises every point, including the vertex.; description: On labelled axes the V-shaped graphs y equals absolute value of 2x minus 4 and y equals that absolute value plus 1. Vertices (2,0) and (2,1) and y-intercepts (0,4) and (0,5) are labelled.]

Watch out

$|2x-4|$ is not $2x+4$ for every $x$ . At $x=3$ , the first is $2$ , while the second is $10$ . Negate the entire inside expression only where it is negative.

Find the zero inside the modulus before choosing the branches.

Graphs of modulus of f and f of modulus

Putting modulus around the output changes heights; putting it around the input changes which side of the original graph is used.

In $y=|f(x)|$ , each input stays fixed, and negative outputs become positive. Keep the graph on or above the $x$ -axis; reflect every part below that axis upwards. In $y=f(|x|)$ , the function receives only nonnegative inputs. Keep the original right-hand half, including $x=0$ , and reflect that half in the $y$ -axis to fill the left. Discard the original left-hand half.

Two different modulus graphs

For a real-valued $f$ defined at the required inputs:

$|f(x)|$ reflects negative outputs in the $x$ -axis.
$f(|x|)$ copies the nonnegative-input half across the $y$ -axis.

The second graph is symmetric about the $y$ -axis, but its heights need not be positive.

Use the same simple original graph to see the effect of moving the modulus signs.

Worked example 3: modulus of the output

Question. Given $f(x)=x-1$ for all real $x$ , sketch $y=|f(x)|$ , stating its vertex and intercepts.

Plan. Reflect the portion of $y=x-1$ below the $x$ -axis upwards.

|f(x)|=|x-1|=\begin{cases}1-x,&x<1,\\x-1,&x\ge1.\end{cases}

Answer. A V with vertex and $x$ -intercept $(1,0)$ , and $y$ -intercept $(0,1)$ .

Check. Original point $(0,-1)$ becomes $(0,1)$ ; original point $(2,1)$ stays fixed. Both agree with the rule.

Change only the placement of the modulus signs, from the output to the input. The negative-input side is now rebuilt from the right-hand side.

Worked example 4: modulus of the input

Question. For the same $f$ , sketch $y=f(|x|)$ , stating its vertex and intercepts.

Plan. Keep $y=x-1$ for $x\ge0$ , then reflect that half in the $y$ -axis.

f(|x|)=|x|-1=\begin{cases}-x-1,&x<0,\\x-1,&x\ge0.\end{cases}

Answer. A V with vertex and $y$ -intercept $(0,-1)$ , and $x$ -intercepts $(-1,0)$ and $(1,0)$ .

Check. At $x=-2$ , the function receives input $2$ , giving $f(2)=1$ . The piecewise rule gives $-(-2)-1=1$ .

The two resulting Vs have different vertices and different ranges.

[DIAGRAM: asset_name: Output modulus compared with input modulus; asset_slug: two-modulus-graphs; caption: For f(x) equals x minus one, moving the modulus signs changes the vertex.; description: Two panels show the original dashed line y equals x minus 1. Left: y equals absolute value of f(x), vertex (1,0), y-intercept (0,1). Right: y equals f of absolute value x, vertex (0,negative 1), roots negative 1 and 1. The retained and reflected portions are distinguished.]
For f(x) equals x minus one, moving the modulus signs changes the vertex.

Watch out

$f(|x|)$ can be negative: here $f(|0|)=-1$ . Only $|f(x)|$ is guaranteed to be nonnegative. If a domain is restricted, also check that the required original inputs exist.

Describe what happens to the original graph before naming the new vertex.

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P3 1.3 - Modulus graphs and equations

The absolute value function

Graphs of modulus of f and f of modulus

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