10. Let the function $f(x)=a x^{2}+b x+c(a \neq 0)$ satisfy $|f(0)| \leq 2,|f(2)| \leq 2,|f(-2)| \leq 2$, find the maximum value of $y=|f(x)|$ when $x \in[-2,2]$.
From the problem, we know
$$
\left\{\begin{array} { l } 
{ c = f ( 0 ) } \\
{ 4 a + 2 b + c = f ( 2 ) } \\
{ 4 a - 2 b + c = f ( - 2 ) }
\end{array} \Rightarrow \left\{\begin{array}{l}
c=f(0) \\
a=\frac{f(2)+f(-2)-2 f(0)}{8} \\
b=\frac{f(2)-f(-2)}{4}
\end{array}\right.\right.
$$

Thus, when \( x \in [-2, 2] \),
$$
\begin{aligned}
y=|f(x)| & =\left|\frac{f(2)+f(-2)-2 f(0)}{8} x^{2}+\frac{f(2)-f(-2)}{4} x+f(0)\right| \\
& =\left|\frac{x^{2}+2 x}{8} f(2)+\frac{x^{2}-2 x}{8} f(-2)+\frac{4-x^{2}}{4} f(0)\right| \\
& \leq\left|\frac{x^{2}+2 x}{4}\right|+\left|\frac{x^{2}-2 x}{4}\right|+\frac{4-x^{2}}{2}
\end{aligned}
$$

Since \( x \in [-2, 2] \) implies \(\frac{x^{2}+2 x}{4} \cdot \frac{x^{2}-2 x}{4} \leq 0\), we have
$$
|f(x)| \leq\left|\frac{x^{2}+2 x}{4}\right|+\left|\frac{x^{2}-2 x}{4}\right|+\frac{4-x^{2}}{2}=\left|\frac{x^{2}+2 x}{4}-\frac{x^{2}-2 x}{4}\right|+\frac{4-x^{2}}{2}=-\frac{x^{2}}{2}+|x|+2
$$

It is easy to see that when \( x \in [0, 2] \),
$$
-\frac{x^{2}}{2}+|x|+2=-\frac{x^{2}}{2}+x+2 \leq \frac{5}{2}
$$

When \( x \in [-2, 0] \),
$$
-\frac{x^{2}}{2}+|x|+2=-\frac{x^{2}}{2}-x+2 \leq \frac{5}{2}
$$

Thus,
$$
\max _{|x| \leq 2}|f(x)| \leq \max _{|x| \leq 2}\left(-\frac{x^{2}}{2}+|x|+2\right) \leq \frac{5}{2}
$$

Finally, taking \( f(x)=-\frac{1}{2} x^{2}+x+2 \), we have \( |f(2)|=|f(-2)|=|f(0)|=2 \).
Therefore, this function satisfies the problem's conditions and achieves the maximum value of \(\frac{5}{2}\) on \([-2, 2]\). Hence, the maximum value of \( y=|f(x)| \) is \(\frac{5}{2}\).
(20 points)