Answer
$f(x)=\displaystyle \frac{3}{5}(x+4)^{2}-2$
Work Step by Step
$f(x)=\displaystyle \frac{3}{5}x^{2}$
In order to have a vertex in the point $(-4,-2)$, this function
needs to be moved $4$ units to the left and $2$ units down.
$f(x)=a(x-h)^{2}+k$
Substitute $a=\displaystyle \frac{3}{5},h=-4,k=-2$
$f(x)=\displaystyle \frac{3}{5}(x+4)^{2}-2$
