1. If on the interval $[2,3]$, the function $f(x)=x^{2}+b x+c$ and $g(x)=x+\frac{6}{x}$ take the same minimum value at the same point, then the maximum value of the function $f(x)$ on $[2,3]$ is
$$
-1.15-4 \sqrt{6} \text {. }
$$

Notice that, $g(x)=x+\frac{6}{x} \geqslant 2 \sqrt{6}$, when $x=\sqrt{6}$ $\in[2,3]$, the equality holds, i.e., $g(x)$ reaches its minimum value $2 \sqrt{6}$ at $x=\sqrt{6}$.
Thus, by the problem statement, we have
$$
f(x)=(x-\sqrt{6})^{2}+2 \sqrt{6} \text {. }
$$

Also, $3-\sqrt{6}>\sqrt{6}-2$, so the maximum value of $f(x)$ on $[2,3]$ is $f(3)=15-4 \sqrt{6}$.