Example 3 Given two quadratic functions $y_{1}$ and $y_{2}$, when $x$ $=\alpha(\alpha>0)$, $y_{1}$ reaches its maximum value of 5, and $y_{2}=25$; also, the minimum value of $y_{2}$ is $-2, y_{1}+y_{2}=x^{2}+16 x+$ 13. Find the value of $\alpha$ and the analytical expressions of the quadratic functions $y_{1}$ and $y_{2}$.
Explanation: Since when $x=\alpha(\alpha>0)$, $y_{1}$ reaches its maximum value of 5, the vertex coordinates of $y_{1}$ are $(\alpha, 5)$. Therefore, the quadratic function $y_{1}$ can be set as $y_{1}=m(x-\alpha)^{2}+5$ $(m<0)$. Thus,
$$
y_{2}=x^{2}+16 x+13-m(x-\alpha)^{2}-5 \text {. }
$$

When $x=\alpha$, $y_{2}=25$, that is,
$$
\alpha^{2}+16 \alpha+8=25 \text {. }
$$

Solving this, we get $\alpha_{1}=1, \alpha_{2}=-17$ (discard $\alpha_{2}$).
Hence, $y_{2}=x^{2}+16 x+13-m(x-1)^{2}-5$
$$
=(1-m) x^{2}+(16+2 m) x+(8-m) \text {. }
$$

Since the minimum value of $y_{2}$ is -2, we have
$$
\frac{4(1-m)(8-m)-(16+2 m)^{2}}{4(1-m)}=-2 \text {. }
$$

Solving this, we get $m=-2$.
Therefore, the analytical expressions of the quadratic functions are
$$
y_{1}=-2 x^{2}+4 x+3, y_{2}=3 x^{2}+12 x+10 \text {. }
$$