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