9. (14 points) (1) The quadratic function $f(x)=a x^{2}+b x + c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfies:
(1) For $x \in \mathbf{R}$, $4 x \leqslant f(x) \leqslant \frac{1}{2}(x+2)^{2}$ always holds;
(2) $f(-4+2 \sqrt{3})=0$. Find $f(x)$.
(2) Let $f_{1}(x)=\frac{3}{2+x}, f_{n+1}(x)=$ $f_{1}\left(f_{x}(x)\right)$, find $f_{2009}(0)$.
9. Solution: (1) Let $x=2$ to get $8 \leqslant f(2) \leqslant 8$, hence $f(2)=$ $8, 4a+2b+c=8 \Rightarrow b=\frac{8-4a-c}{2}$.

Also, $4x \leqslant f(x)$, so $ax^2 + (b-4)x + c \geqslant 0$ always holds, then $\Delta=(b-4)^2 - 4ac = \left(\frac{8-4a-c}{2}-4\right)^2 - 4ac = \left(2a - \frac{c}{2}\right)^2 \leqslant 0$, thus, $c=4a, b=4-4a$, substituting $f(-4 + 2\sqrt{3})=0$ gives $a=\frac{1}{3}$, so $b=\frac{8}{3}, c=\frac{4}{3}$,
hence $f(x)=\frac{x^2 + 8x + 4}{3}$.
(2) Construct the sequence $a_n = \frac{f_n(0) - 1}{f_n(0) + 3}$, then $\frac{a_{n+1}}{a_n} = \frac{f_{n+1}(0) - 1}{f_{n+1}(0) + 3} \times \frac{f_n(0) + 3}{f_n(0) - 1} = -\frac{1}{3}$
(because $f_{n+1}(0) = \frac{3}{2 + f_n(0)}$, so the sequence $\{a_n\}$ is a geometric sequence, and $a_n = \frac{1}{9} \times \left(-\frac{1}{3}\right)^{n-1}$, thus $a_{2009} = \frac{1}{3^{2010}}$, then by $\frac{f_{2009}(0) - 1}{f_{2009}(0) + 3} = \frac{1}{3^{2010}}$, solving gives $f_{2009}(0) = \frac{3^{2010} + 3}{3^{2010} - 1}$.