# How do you graph y = abs(3x+2) ?

Jun 20, 2015

Basically, it's a V-shape with slope $\pm 3$ and vertex $\left(- \frac{2}{3} , 0\right)$

#### Explanation:

The vertex is where $\left\mid 3 x + 2 \right\mid = 0$ at $\left(- \frac{2}{3} , 0\right)$.

When $x < - \frac{2}{3}$ the inner expression $3 x + 2 < 0$ so

$\left\mid 3 x + 2 \right\mid = - \left(3 x - 2\right)$

and the equation becomes $y = - 3 x + 2$

This is a straight line of slope $- 3$.

When $x > \frac{2}{3}$ the inner expression $3 x + 2 > 0$ so

$\left\mid 3 x + 2 \right\mid = 3 x - 2$

and the equation becomes $y = 3 x - 2$

This is a straight line of slope $3$

So the original equation represents a V shape with slope $\pm 3$ and vertex $\left(- \frac{2}{3} , 0\right)$

graph{abs(3x+2) [-10, 10, -5, 5]}