Answer
$f(x)=\displaystyle \frac{3}{5}(x-4)^{2}-7$
Work Step by Step
$f(x)=\displaystyle \frac{3}{5}x^{2}$
In order to have a vertex in the point $(4,-7)$, this function
needs to be moved $4$ units to the right and $7$ units down.
$f(x)=a(x-h)^{2}+k$
Substitute $a=\displaystyle \frac{3}{5},h=4,k=-7$
$f(x)=\displaystyle \frac{3}{5}(x-4)^{2}-7$