Example 16 (2002 National High School Competition Question) Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfy the conditions: (1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$; (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 which, if $x \in[1, m]$, then $f(x+t) \leqslant x$.
Given: $\because f(x-4)=f(2-x), \therefore$ the graph of the function is symmetric about $x=-1$,
$\therefore-\frac{b}{2 a}=-1$, i.e., $b=2 a$
From (3) $f(-1)=0, \therefore a-b+c=0$
From (1) we have $f(1) \geqslant 1$, from (2) we get $f(1) \leqslant 1$,
$\therefore f(1)=1$, i.e., $a+b+c=1$
From (1)(2)(3), we get $a=c=\frac{1}{4}, b=\frac{1}{2}$, hence $f(x)=$ $\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}$.
Assume there exists $t \in \mathbf{R}, x \in[1, m]$, such that $f(x+t) \leqslant x$.
Taking $x=1$, we have $f(1+t) \leqslant 1$. That is, $\frac{1}{4}(t+1)^{2}+$ $\frac{1}{2}(t+1)+\frac{1}{4} \leqslant 1$, yielding $-4 \leqslant t \leqslant 0$.

For a fixed $t \in[-4,0]$, taking $x=m$, we have $f(m+$ $t) \leqslant m$. That is, $\frac{1}{4}(t+m)^{2}+\frac{1}{2}(t+m)+\frac{1}{4} \leqslant m$.
Simplifying, we get $m^{2}-2(1-t) m+\left(t^{2}+2 t+1\right) \leqslant 0$.
Solving, we get $1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t}$,
$\therefore 1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t} \leqslant$ $1-(-4)+\sqrt{16}=9$.
That is, when $t=-4$, for any $x \in[1,9]$, we always have
$$
f(x-4)-x=\frac{1}{4}\left(x^{2}-10 x+9\right)=\frac{1}{4}(x-
$$
1) $(x-9) \leqslant 0$,
i.e., $f(x-4) \leqslant x$ always holds.
$\therefore$ The maximum value of $m$ is 9.