Example 8 Quadratic Function
$$
f(x)=a x^{2}+b x+c(a, b \in \mathbf{R} \text {, and } a \neq 0)
$$

satisfies the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and
$$
f(x) \geqslant x \text {; }
$$
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.

Find the largest $m(m>1)$, such that there exists $t \in \mathbf{R}$, for any $x \in[1, m]$, we have $f(x+t) \leqslant x$.
(2002, National High School Mathematics Competition)
Explanation: First, derive the analytical expression of $f(x)$ from the given conditions, then classify and discuss $m$ and $t$ to determine the value of $m$.

Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about the line $x=-1$. Therefore,
$$
-\frac{b}{2 a}=-1 \Rightarrow b=2 a \text {. }
$$

From condition (3), when $x=-1$, $y=0$, i.e., $a-b+c=0$.
From conditions (1) and (2), we get $f(1) \geqslant 1, f(1) \leqslant 1$.
Thus, $f(1)=1$, i.e., $a+b+c=1$.
From the above, we can solve for $a=\frac{1}{4}, b=\frac{1}{2}, c=\frac{1}{4}$.
Therefore, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}$.
Since the graph of the parabola $f(x)$ opens upwards, and the graph of $f(x+t)$ is obtained by translating the graph of $y=f(x)$ by $t$ units, to ensure that on $[1, m]$, the graph of $y=f(x+t)$ is below the graph of $y=x$, and $m$ is maximized, $1$ and $m$ should be the two roots of the equation $\frac{1}{4}(x+t+1)^{2}=x$.

Since $1$ is a root of the equation $\frac{1}{4}(x+t+1)^{2}=x$, we can solve for $t=0$ or $-4$.

Substituting $t=0$ into the original equation gives $x_{1}=x_{2}=1$ (which contradicts $m>1$);
Substituting $t=-4$ into the original equation gives $x_{1}=1, x_{2}=9$.
Therefore, $m=9$.
In summary, the maximum value of $m$ is 9.