Answer
$f(x)=\displaystyle \frac{3}{5}(x-9)^{2}-6$
Work Step by Step
$f(x)=\displaystyle \frac{3}{5}x^{2}$
In order to have a vertex in the point $(9,-6)$, this function
needs to be moved $9$ units to the right and $6$ units down.
$f(x)=a(x-h)^{2}+k$
Substitute $a=\displaystyle \frac{3}{5},h=9,k=-6$
$f(x)=\displaystyle \frac{3}{5}(x-9)^{2}-6$