# How do you graph, find the zeros, intercepts, domain and range of f(x)=1/3abs(2x-1)?

Feb 20, 2018

domain $\left\{x , x \in \left(- \infty , + \infty\right)\right\}$
range $\textcolor{w h i t e}{\text{d}} \left\{y , y \in \left[0 , + \infty\right)\right\}$

See explanation.

#### Explanation:

This is of graph type $\vee$.

The part $| 2 x - 1 |$ is always positive so $\frac{1}{3} | 2 x - 1 |$ is also always
positive. Thus $y \ge 0$

$\textcolor{b l u e}{\text{Determin the vertex}}$

Set $y = 0 = \frac{1}{3} | 2 x - 1 |$

Multiply both sides by $\frac{3}{1}$

$y = 0 = | 2 x - 1 |$

$y = 0 = | 0 |$

So $2 x - 1 = 0 \implies x = \frac{1}{2}$

So the the vertex of $\vee$ is at $\left(x , y\right) \to \left(\frac{1}{2} , 0\right)$

Thus the vertex is on the x-axis. There is no plot below the x-axis.
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$\textcolor{b l u e}{\text{Determine the y-intercept}}$

Set $x = 0$ giving:

$y = \frac{1}{3} | 0 - 1 |$

$y = \frac{1}{3} \times \left(+ 1\right) = \frac{1}{3}$

${y}_{\text{intercept}} \to \left(x , y\right) = \left(0 , \frac{1}{3}\right)$
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Plot two graphs but cit them off so that there is no line below the x-axis

Line 1 $y = \frac{1}{3} \left(2 x - 1\right)$ where $x \le 0$
Line 2 $y = \frac{1}{3} \left(2 x - 1\right)$ where $x \ge 0$ 