15 Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}, a \neq 0)$ satisfy the following 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 any $x \in[1, m]$, we have $f(x+t) \leqslant x$.
Analyze: First, derive the analytical expression of $f(x)$ from the given conditions, then discuss $m$ and $t$, and finally determine the maximum value of $m$.

Solution: Since $f(x-4) = f(2-x)$ for $x \in \mathbf{R}$, it is known that the axis of symmetry of the quadratic function $f(x)$ is $x = -1$. From (3), we know that $f(x)$ opens upwards, i.e., $a > 0$, so we have
$$
f(x) = a(x + 1)^2 \quad (a > 0).
$$

From (1), we get $f(1) \geq 1$, and from (2), we get $f(1) \leq \left(\frac{1 + 1}{2}\right)^2 = 1$, thus $f(1) = 1$, i.e., $a(1 + 1)^2 = 1$, so
$$
a = \frac{1}{4},
$$

hence
$$
f(x) = \frac{1}{4}(x + 1)^2.
$$

Since the graph of the parabola $f(x) = \frac{1}{4}(x + 1)^2$ opens upwards, and the graph of $y = f(x + t)$ is obtained by shifting 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, then 1 and $m$ should be the two roots of the equation
$$
\frac{1}{4}(x + t + 1)^2 = x.
$$

Substituting $x = 1$ into equation (1), we get $t = 0$ or $t = -4$.
When $t = 0$, substituting into (1), we get $x_1 = x_2 = 1$ (which contradicts $m > 1$);
When $t = -4$, substituting into (1), we get $x_1 = 1, x_2 = 9$, so $m = 9$.
When $t = -4$, for any $x \in [1, 9]$, we always have
$$
\begin{aligned}
& (x - 1)(x - 9) \leq 0 \\
\Leftrightarrow & \frac{1}{4}(x - 4 + 1)^2 \leq x,
\end{aligned}
$$

i.e.,
$$
f(x - 4) \leq x.
$$

Therefore, the maximum value of $m$ is 9.