# How do you graph y = -abs(x + 4)?

Jul 5, 2015

This is an upside-down V shape with arms having slope $\pm 1$ and vertex at $\left(- 4 , 0\right)$

#### Explanation:

$\left\mid x + 4 \right\mid = x + 4$ when $x + 4 \ge 0$, that is when $x \ge - 4$
So $y = - \left\mid x + 4 \right\mid = - x - 4$ has slope $- 1$ when $x \ge - 4$

$\left\mid x + 4 \right\mid = - \left(x + 4\right)$ when $x + 4 < 0$, that is when $x < - 4$
So $y = - \left\mid x + 4 \right\mid = x + 4$ has slope $1$ when $x < - 4$

The vertex is at the point where $\left(x + 4\right) = 0$, giving $x = - 4$ and $y = 0$. So the vertex is at $\left(- 4 , 0\right)$

The intersection with the $y$ axis will be where $x = 0$. Substituting $x = 0$ into the equation, we get $y = - \left\mid 4 \right\mid = - 4$. So the intersection is at $\left(0 , - 4\right)$

graph{-abs(x+4) [-13.38, 6.62, -5.84, 4.16]}