Example 7 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 $[0, l(a)]$. Question: For what value of $a$ is $l(a)$ the largest? Find this largest $l(a)$ and prove your conclusion.
$(1998$, National High School Mathematics Competition)
Notice that
$$
f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}
$$

has the axis of symmetry at $x=-\frac{4}{a}$.
Therefore, when $x=-\frac{4}{a}$, $f(x)_{\text {max }}=3-\frac{16}{a}$.
Combining the graph, we can consider two cases: $3-\frac{16}{a}>5$ and $3-\frac{16}{a} \leqslant 5$, and find the maximum positive value of $l(a)$ for each case, then take the larger $l(a)$ when $a$ is a negative number.
(1) When $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a}$, and $l(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$. Hence
$$
\begin{array}{l}
l(a)=\frac{-8-\sqrt{64-32 a}}{2 a} \\
=\frac{4}{\sqrt{4-2 a}-2} \leqslant \frac{4}{\sqrt{20}-2}=\frac{\sqrt{5}+1}{2} .
\end{array}
$$

The equality holds if and only if $a=-8$.
Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, therefore, when and only when $a=-8$, $l(a)$ reaches its maximum value $\frac{\sqrt{5}+1}{2}$.