Answer
$f(x)=-\frac{3}{4}(x+8)^2+5$
Work Step by Step
$f(x)=a(x-k)^2+h~~$
(Standard form, where $(k,h)$ is the vertex)
$f(x)=a[x-(-8)]^2+5$
$f(x)=a(x+8)^2+5$
Use the point $(-4,-7)$ to find $a$:
$-7=a(-4+8)^2+5$
$-7-5=a(4)^2$
$-12=16a$
$a=\frac{-12}{16}=-\frac{3}{4}$
$f(x)=-\frac{3}{4}(x+8)^2+5$
