Answer
$f(x)=\displaystyle \frac{3}{5}(x-2)^{2}+8$
Work Step by Step
$f(x)=\displaystyle \frac{3}{5}x^{2}$
In order to have a vertex in the point $(2,8)$, this function
needs to be moved $2$ units to the right and $8$ units up.
$f(x)=a(x-h)^{2}+k$
Substitute $a=\displaystyle \frac{3}{5},h=2,k=8$
$f(x)=\displaystyle \frac{3}{5}(x-2)^{2}+8$