 New Zealand
Level 7 - NCEA Level 2

Solutions to 2-variable Linear Absolute Value Functions (|ax+b| only)

Lesson

An equation of the form $|ax+b|=c$|ax+b|=c, where $a,b$a,b and $c$c are constants, can have zero, one or two solutions depending on the constants values.

Example 1 The graph of the linear absolute value function, depicted in green in the diagram, has the formula $y=|\frac{1}{2}x+1|$y=|12x+1|

By letting the function value $y=c$y=c be fixed successively at $c=-1.5$c=1.5, $c=0$c=0 and $c=1$c=1, we form three equations

$|\frac{1}{2}x+1|=-1.5$|12x+1|=1.5
$|\frac{1}{2}x+1|=0$|12x+1|=0
$|\frac{1}{2}x+1|=1$|12x+1|=1

From the graph we see that the first of these has no solution, the second has exacly one solution ($x=-2$x=2) and the third has two solutions ($x=-4$x=4 and $x=0$x=0).

It must always be true that an equation $|ax+b|=c$|ax+b|=c has no solutions if $c$c is negative, exactly one solution if $c=0$c=0 and two solutions if $c$c is a positive number.

When $c=0$c=0, the solution is $x=-\frac{b}{a}$x=ba. You should check that this is the value that makes the expression equal to zero.

Example 2

Solve $\left|\frac{2}{3}x+\frac{1}{3}\right|=0$|23x+13|=0.

We need to solve either $\frac{2}{3}x+\frac{1}{3}=0$23x+13=0 or $-\left(\frac{2}{3}x+\frac{1}{3}\right)=0$(23x+13)=0. (The two equations are the same.)

Thus, $x=-\frac{1}{2}$x=12

Note that this is indeed $-\frac{b}{a}=-\frac{\frac{1}{3}}{\frac{2}{3}}$ba=1323.

Example 3

Solve $\left|\frac{2}{3}x+\frac{1}{3}\right|=12$|23x+13|=12.

To obtain both solutions, we consider the two parts of the function definition and hence write down the two equations

$\frac{2}{3}x+\frac{1}{3}=12$23x+13=12
$-\left(\frac{2}{3}x+\frac{1}{3}\right)=12$(23x+13)=12

The first of these gives $x=\frac{3}{2}\left(12-\frac{1}{3}\right)=17\frac{1}{2}$x=32(1213)=1712.

The second equation gives $x=\frac{3}{2}\left(-12-\frac{1}{3}\right)=-18\frac{1}{2}$x=32(1213)=1812.

Worked Examples

Question 1

Consider the function $f(x)=\left|4x-4\right|$f(x)=|4x4|.

1. Plot the function $f(x)$f(x).

2. Hence determine the number of solutions to the equation $\left|4x-4\right|=-2$|4x4|=2.

Question 2

Consider the function $f(x)=\left|2x+4\right|$f(x)=|2x+4|.

1. The graph of the function is given. On the same set of axes, plot the points at which $f(x)=2$f(x)=2.

2. How many solutions does the equation $\left|2x+4\right|=2$|2x+4|=2 have?

3. Using the result of part (a) or otherwise, find the solutions to the equation $\left|2x+4\right|=2$|2x+4|=2.

State your solutions on the same line, separated by a comma.

Outcomes

M7-2

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

91257

Apply graphical methods in solving problems