14. The center of symmetry of the graph of the function $f(x)=x^{3}+2 x^{2}+3 x+4$ is . $\qquad$
14. $\left(-\frac{2}{3}, \frac{70}{27}\right)$.

Let the point $(a, b)$ be the center of symmetry of the graph of the function $f(x)$.
Then $f(a+x)+f(a-x)=2 b$, that is,
$$
\begin{aligned}
2 b= & {\left[(a+x)^{3}+(a-x)^{3}\right]+} \\
& 2\left[(a+x)^{2}+(a-x)^{2}\right]+6 a+8, \\
b= & a^{3}+3 a x^{2}+2 a^{2}+2 x^{2}+3 a+4 \\
= & (2+3 a) x^{2}+a^{3}+2 a^{2}+3 a+4 .
\end{aligned}
$$

Thus, we should have
$$
\begin{array}{l}
2+3 a=0 \Rightarrow a=-\frac{2}{3}, \\
b=-\frac{8}{27}+\frac{8}{9}-2+4=\frac{70}{27} .
\end{array}
$$