# How do you graph f(x)=2abs(x-1)-3?

Jul 23, 2015

Plot the point where $\left(x - 1\right) = 0$ and one point for each of $\left(x - 1\right) > 0$ and $\left(x - 1\right) < 0$; connect the initial point through each of the other 2 points as 2 line segments.

#### Explanation:

Given $f \left(x\right) = 2 \left\mid x - 1 \right\mid - 3$

When $x = 1$
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(1\right) = - 3$
$\left(1 , - 3\right)$ is a point common to both line segments ($x \le 1$ and $x \ge 1$)

Picking $x = 3$ as an arbitrary point with $x > 1$
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(3\right) = 2 \left\mid 3 - 1 \right\mid - 3 = 1$
$\left(3 , 1\right)$ is a point on the line segment for $x > 1$

Picking $x = 0$ as an arbitrary point with $x < 1$
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(0\right) = 2 \left\mid 0 - 1 \right\mid - 3 = - 1$
$\left(0 , - 1\right)$ is a point on the line segment for $x < 1$