One, (20 points) If the vertex of the parabola $y=-x^{2}+2 a x+b$ moves on the line $m x-y-2 m+1=0$, and it has a common point with the parabola $y=x^{2}$, find the range of changes for $m$.

Given the parabola $y=-x^{2}+2 a x+b$ has its vertex at $\left(a, a^{2}+b\right)$, then $\left(a, a^{2}+b\right)$ satisfies the linear equation. We have
$$
\begin{array}{l}
m a-a^{2}-b-2 m+1=0, \\
b=m a-a^{2}-2 m+1 .
\end{array}
$$

If $y=-x^{2}+2 a x+m a-a^{2}-2 m+1=0$,
the discriminant is
$$
\Delta=4 a^{2}-8 m a-8 a^{2}-16 m+8 \geqslant 0,
$$

which simplifies to $a^{2}-2 m a+4 m-2 \leqslant 0$.
For this inequality to have real solutions for $a$, the condition is
$$
\Delta^{\prime}=4 m^{2}-4(4 m-2) \geqslant 0 \text {. }
$$

Solving this, we get
$$
m \leqslant 2-\sqrt{2} \text { or } m \geqslant 2+\sqrt{2} \text {. }
$$