Answer
$
y=\frac{3}{16} \cdot(x-6)^2+5
$
Work Step by Step
Since the vertex is $(6,5)$, we use the form $y=a(x-h)^2+k$, with $h=6$ and $k=5$. We solve for $a$, substituting in the second point, $(10,8)$.
$$
\begin{aligned}
y & =a(x-6)^2+5 \\
8 & =a(10-6)^2+5 \\
3 & =16 a \\
a& =\frac{3}{16}
\end{aligned}
$$ Hence $$
y=\frac{3}{16} \cdot(x-6)^2+5
$$
