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

Apr 5, 2017

Domain is $\left(- \infty , \infty\right)$, range is $\left[- 2 , 2\right]$.

$y$-intercept is $2$ and $x$-intercept is $x = - 1$

#### Explanation:

For $x \le - 2$,

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

for $x \ge 0$,

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

and for $- 2 < x < 2$ $f \left(x\right) = x + 2 - \left(- x\right) = 2 x + 2$

Hence domain is $\left(- \infty , \infty\right)$, range is $\left[- 2 , 2\right]$

and for $x = 0$ i.e. $y$-intercept is $2$ and $x$-intercept being at $y = 0$, is given by $2 x + 2 = 0$ i.e. $x = - 1$

The graph appears as follows:

graph{|x+2|-|x| [-10, 10, -5, 5]}