Example 3. For the cubic function $y=f(x)$, when $x=1$, $y$ has a local minimum of -3; when $x=-3$, $y$ has a local maximum of 29. Find the expression for $f(x)$ and the center of symmetry of the graph of $y=f(x)$.
Solve $\because$ when $x=1$, $y$ has a minimum value of -3, 
$\therefore$ we can set $y=a(x-1)^{2}(x-m)-3 .(a \neq 0)$
$\because$ when $\mathbf{x}=-3$, $\mathrm{y}$ has a maximum value of 29, 
$\therefore$ we can set $y=a(x+3)^{2}(x-n)+29$.
Thus, $a(x-1)^{2}(x-m)-3$
$$
=a(x+3)^{2}(x-n)+29 \text { is an identity. }
$$

Comparing the coefficients of $x^{2}$, $x$, and the constant term, we get the system of equations
$$
\begin{array}{l}
(-a(m+2)=a(6-n), \quad(a \neq 0) \\
\left\{\begin{array}{l}
a(2 m+1)=a(9-6 n), \\
-a m-3=-9 a n+29 .
\end{array}\right. \\
\therefore \quad\left\{\begin{array}{l}
a=1, \\
m=-5, \\
n=3 .
\end{array}\right. \\
\end{array}
$$

Therefore, $y=f(x)=(x-1)^{2}(x+5)-3$
$$
\begin{array}{c}
=x^{3}+3 x^{2}-9 x+2 . \\
\because f^{\prime \prime}(x)=6 x+6, \quad \therefore f^{\prime \prime}(-1)=0 .
\end{array}
$$

Thus, $y_{x=-1}=f(-1)=-1+3+9+2=13$.
Hence, the center of symmetry of the graph of $y=f(x)=x^{3}+3 x^{2}-9 x+2$ is $(-1,13)$.