Example 3 Given that the parabola $y=-x^{2}+b x+c$ is tangent to the parabola $y=x^{2}$. Find the geometric position of the vertex of the parabola.
(17th All-Russian Mathematical Olympiad)
Explanation: Let the point of tangency of the two parabolas be $Q\left(x_{1}, y_{1}\right)$. Then their equations at the point of tangency are
$$
\begin{aligned}
\frac{y_{1}+y}{2} & =-x_{1} x+b \cdot \frac{x_{1}+x}{2}+c, \\
\frac{y_{1}+y}{2} & =x_{1} x .
\end{aligned}
$$

Since the two parabolas are tangent to each other, they have the same tangent line at the point of tangency. Therefore, the above two equations are the same, giving us
$$
\left\{\begin{array}{l}
-x_{1}+\frac{b}{2}=x_{1}, \\
\frac{b x_{1}}{2}+c=0 .
\end{array}\right.
$$

From this, we can get $x_{1}=\frac{b}{4}, c=-\frac{b x_{1}}{2}=-\frac{b^{2}}{8}$. 
Thus, the equation of the parabola $y=-x^{2}+b x+c$ can be written as $y=-x^{2}+b x-\frac{b^{2}}{8}$. Therefore, the vertex is $\left(\frac{b}{2}, \frac{b^{2}}{8}\right)$.