Given the quadratic function
$$
y=a x^{2}+b x+c
$$

the graph intersects the $x$-axis at one point with coordinates $(8,0)$, and the vertex has coordinates $(6,-12)$. Find the analytical expression of the quadratic function.
Solution 1 From the problem, we have
$$
\left\{\begin{array} { l } 
{ 8 ^ { 2 } a + 8 b + c = 0 } \\
{ - \frac { b } { 2 a } = 6 , } \\
{ \frac { 4 a c - b ^ { 2 } } { 4 a } = - 1 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=3, \\
b=-36, \\
c=96 .
\end{array}\right.\right.
$$

Therefore, the analytical expression of the quadratic function is
$$
y=3 x^{2}-36 x+96 \text {. }
$$

Solution 2 Given that the axis of symmetry of the graph is $x=6$, we know that the other intersection point of the quadratic function with the $x$-axis is $(4,0)$.
Let the analytical expression of the quadratic function be
$$
y=a(x-4)(x-8) \text {. }
$$

Since the graph passes through the point $(6,-12)$, substituting gives
$$
-12=a(6-4)(6-8) \Rightarrow a=3 \text {. }
$$

Therefore, the analytical expression of the quadratic function is
$$
y=3(x-4)(x-8)=3 x^{2}-36 x+96 .
$$