Mathematics 9709 · Pure Mathematics 2

Graphs of y = |f(x)|

Drawing $y=|f(x)|$ is one move: sketch the line $y=f(x)$ , then fold the part below the x-axis straight up.
A modulus can't be negative, so nothing is left below the axis. The graph also shows how many answers an equation has before you do any algebra.

$y=|f(x)|$ : draw $f$ , then flip everything below the axis up above it. The point where it crosses the x-axis becomes the corner. The graph never goes below zero.

The one move: fold the negative part up

Where $f(x)$ is positive, the graph stays the same. Where it dips below the axis, flip it up.
The point where the line crosses zero becomes the sharp corner of the V.

y = ½x − 1 crosses zero at x = 2; fold the dashed (negative) part up and you get the V, corner at (2, 0).

So the corner of $y=|ax+b|$ sits where $ax+b=0$ , i.e. $x=-\tfrac{b}{a}$ . The two arms are the line and its mirror image.

Split into pieces:

$\left|\tfrac12 x-1\right|=\begin{cases}\tfrac12 x-1 & \text{if } x\ge 2\\[2pt] -\left(\tfrac12 x-1\right) & \text{if } x<2\end{cases}$

Drag it: where the V meets the line are the solutions

The V is $y=|x-a|$ ; the dashed line is $y=k$ .
The two green crossings are the answers to $|x-a|=k$ . They sit at $x=a-k$ and $x=a+k$ . The gold band between them is where $|x-a|<k$ .
Drag $k$ below zero and the crossings disappear. No answers, because a modulus can never be negative.

corner aheight k

|x − 3| = 2·solutions x = 1 and x = 5

The two answers sit one step either side of the corner. That is why every $|ax+b|=k$ gives a $\pm$ . A line below the corner gives nothing.

Solving from the picture: where two graphs cross

An equation like $|2x-1|=3$ just asks “where does the V $y=|2x-1|$ meet the line $y=3$ ?”
The line cuts both arms of the V. You get the same two answers as the algebra. The picture also shows how many answers to expect.

Worked example

How many solutions does $|2x-1|=3$ have?

The V has its corner at $x=\tfrac12$ on the axis. The line $y=3$ sits above it and cuts both arms, so there are two answers (here $x=2$ and $x=-1$ ). A line below the corner would miss the V: no answers.

Answer

\text{two solutions}

Common mistake

The corner is where

f(x)=0

, not on the y-axis (unless

b=0

). And a modulus graph never dips below the x-axis, so

|f(x)|=k

with

k<0

has no answers at all.

Now you try

Sketch $y=|x-3|+2$ , and give the coordinates of its lowest point. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$y=|x-3|$ is a V with corner at $(3,0)$ . The $+2$ lifts the whole graph up by 2. So it is the same V raised up, lowest point at:

Answer

(3,\,2)

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Graphs of y = |f(x)|

Drawing $y=|f(x)|$ is one move: sketch the line $y=f(x)$ , then fold the part below the x-axis straight up.
A modulus can't be negative, so nothing is left below the axis. The graph also shows how many answers an equation has before you do any algebra.

$y=|f(x)|$ : draw $f$ , then flip everything below the axis up above it. The point where it crosses the x-axis becomes the corner. The graph never goes below zero.

The one move: fold the negative part up

Where $f(x)$ is positive, the graph stays the same. Where it dips below the axis, flip it up.
The point where the line crosses zero becomes the sharp corner of the V.

y = ½x − 1 crosses zero at x = 2; fold the dashed (negative) part up and you get the V, corner at (2, 0).

So the corner of $y=|ax+b|$ sits where $ax+b=0$ , i.e. $x=-\tfrac{b}{a}$ . The two arms are the line and its mirror image.

Split into pieces:

$\left|\tfrac12 x-1\right|=\begin{cases}\tfrac12 x-1 & \text{if } x\ge 2\\[2pt] -\left(\tfrac12 x-1\right) & \text{if } x<2\end{cases}$

Drag it: where the V meets the line are the solutions

The V is $y=|x-a|$ ; the dashed line is $y=k$ .
The two green crossings are the answers to $|x-a|=k$ . They sit at $x=a-k$ and $x=a+k$ . The gold band between them is where $|x-a|<k$ .
Drag $k$ below zero and the crossings disappear. No answers, because a modulus can never be negative.

corner aheight k

|x − 3| = 2·solutions x = 1 and x = 5

The two answers sit one step either side of the corner. That is why every $|ax+b|=k$ gives a $\pm$ . A line below the corner gives nothing.

Solving from the picture: where two graphs cross

An equation like $|2x-1|=3$ just asks “where does the V $y=|2x-1|$ meet the line $y=3$ ?”
The line cuts both arms of the V. You get the same two answers as the algebra. The picture also shows how many answers to expect.

Worked example

How many solutions does $|2x-1|=3$ have?

Answer

\text{two solutions}

Common mistake

The corner is where

f(x)=0

, not on the y-axis (unless

b=0

). And a modulus graph never dips below the x-axis, so

|f(x)|=k

with

k<0

has no answers at all.

Now you try

Sketch $y=|x-3|+2$ , and give the coordinates of its lowest point. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$y=|x-3|$ is a V with corner at $(3,0)$ . The $+2$ lifts the whole graph up by 2. So it is the same V raised up, lowest point at:

Answer

(3,\,2)