Answer
$f(-4)=8$
$f(-1)=-1$
$f(-\frac{3}{2})=\frac{-3}{4}$
$f(0)=0$
$f(25)=-1$
Work Step by Step
Because $x=-4, x=\frac{-3}{2}, x=-1$ are in between $(-\infty,-1]$ of $f(x)=x^{2}+2x$
so $f(-4)=(-4)^{2}+2(-4)=8$
$f(-1)=(-1)^{2}+2(-1)=-1$
$f(\frac{-3}{2})=(-\frac{3}{2})^{2}+2(-\frac{3}{2})=\frac{-3}{4}$
x=0 is in between $(-1,1]$ of $f(x)=x$
So $f(0)=0$
x=25 is in between $(1,+\infty)$ of $f(x)=-1$
$f(25)=-1$
