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

Jul 8, 2015

This is an inverted V shape, with sides of slope $\pm 2$, vertex at $\left(- \frac{3}{2} , 3\right)$, intersection with the $y$ axis at $\left(0 , 0\right)$ and intersections with the $x$ axis at $\left(0 , 0\right)$ and $\left(- 3 , 0\right)$.

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

#### Explanation:

When $x \le - \frac{3}{2}$, $2 x + 3 \le 0$ so $\left\mid 2 x + 3 \right\mid = - 2 x - 3$

and $f \left(x\right) = 3 - \left\mid 2 x + 3 \right\mid = 3 - \left(- 2 x - 3\right) = 2 x + 6$

So this part of the curve is a line of slope $2$ which intersects the $x$ axis where $2 x + 6 = 0$, that is at $\left(- 3 , 0\right)$

When $2 x + 3 = 0$, $x = - \frac{3}{2}$ and $f \left(x\right) = 2 x + 6 = 3$.

So the point $\left(- \frac{3}{2} , 3\right)$ is the minimum of $\left\mid 2 x + 3 \right\mid$ and the maximum of $f \left(x\right)$, thus the vertex.

When $x \ge - \frac{3}{2}$, $2 x + 3 \ge 0$ so $\left\mid 2 x + 3 \right\mid = 2 x + 3$

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

So this part of the curve is a line of slope $- 2$ which intersects the $x$ and $y$ axis at $\left(0 , 0\right)$.