Example 10 (1998 National High School Mathematics League Question) Let the function $f(x)=a x^{2}+8 x+3(a<0)$. For a given negative number $a$, there is a largest positive number $L(a)$, such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, L(a)]$. For what value of $a$ is $L(a)$ the largest? Find this largest $L(a)$ and prove your conclusion.

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The function is given by $f(x)=a x^{2}+8 x+3$ where $a<0$. For a given negative number $a$, there is a largest positive number $L(a)$ such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, L(a)]$. We need to determine the value of $a$ for which $L(a)$ is the largest and find this largest $L(a)$, along with a proof of the conclusion.
For a given negative number $a$, there is a function $L(a)>0$ related to $a$, such that $|f(x)| \leqslant 5$ holds for the entire interval $[0, L(a)]$.
Below, we use the idea of combining numbers and shapes to construct a new problem-solving context.
$$
f(x)=a x^{2}+8 x+3=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a},
$$

Since $a<0, f(x)_{\max }=3-\frac{16}{a}>0, x \in \mathbf{R}$.
Therefore, the vertex of the parabola $y=f(x)$ is in the first quadrant.
Also, $\Delta=64-12 a>0$, so the equation $a x^{2}+8 x+3=0$ has one positive root and one negative root. This means the parabola intersects the $x$-axis at two points on either side of the $y$-axis.
（1）When $3-\frac{16}{a}>5$, i.e., $-80$, so $L(a)-\frac{4}{a}$, then $L(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$. Solving this equation, we get
$$
\begin{aligned}
L(a) & =\frac{-8-\sqrt{64-32 a}}{2 a} \\
& =\frac{4}{\sqrt{4-2 a}-2} \leqslant \frac{4}{\sqrt{20}-2}=\frac{\sqrt{5}+1}{2} .
\end{aligned}
$$

The equality holds if and only if $a=-8$.
Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, it follows that when and only when $a=-8$, $L(a)_{\text {max }}=\frac{\sqrt{5}+1}{2}$.