9. (16 points) Given the quadratic function $f(x)$ satisfies $|f(0)| \leqslant 2,|f(2)| \leqslant 2,|f(-2)| \leqslant 2$. When $x \in[-2,2]$, find the maximum value of $y=|f(x)|$.


$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
f(0)=c, \\
f(2)=4 a+2 b+c, \\
f(-2)=4 a-2 b+c
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a=\frac{f(2)+f(-2)-2 f(0)}{8}, \\
b=\frac{f(2)-f(-2)}{4}, \\
c=f(0)
\end{array},\right. \\
\Rightarrow f(x)=a x^{2}+b x+c \\
=\frac{x^{2}+2 x}{8} f(2)+\frac{x^{2}-2 x}{8} f(-2)+\frac{4-x^{2}}{4} f(0) \\
\Rightarrow|f(x)| \leqslant\left|\frac{x^{2}+2 x}{8} f(2)\right|+ \\
\left|\frac{x^{2}-2 x}{8} f(-2)\right|+ \\
\left|\frac{4-x^{2}}{4} f(0)\right| \\
\leqslant\left|\frac{x^{2}+2 x}{4}\right|+\left|\frac{x^{2}-2 x}{4}\right|+\left|\frac{4-x^{2}}{2}\right| \text {. } \\
\end{array}
$$

When $x \in[-2,2]$,
$$
\begin{array}{l}
\frac{x^{2}+2 x}{4} \cdot \frac{x^{2}-2 x}{4}=\left(\frac{x}{4}\right)^{2}\left(x^{2}-4\right) \leqslant 0 . \\
\text { Hence }\left|\frac{x^{2}+2 x}{4}\right|+\left|\frac{x^{2}-2 x}{4}\right|+\left|\frac{4-x^{2}}{2}\right| \\
=\left|\frac{x^{2}+2 x}{4}-\frac{x^{2}-2 x}{4}\right|+\frac{4-x^{2}}{2} \\
=-\frac{1}{2}(|x|-1)^{2}+\frac{5}{2} \leqslant \frac{5}{2} .
\end{array}
$$

Therefore, the maximum value of $y=|f(x)|$ is $\frac{5}{2}$, which is attained when
$$
f(x)= \pm\left(\frac{1}{2} x^{2}-\varepsilon x-2\right) \text {, and } x=\varepsilon(\varepsilon \in\{-1,1\})
$$
