2. Given that 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 $y=x^{2}$. Then, the range of values for $m$ is $\qquad$
2. The vertex of the parabola $y=-x^{2}+2 a x+b$ is $(a$, $\left.a^{2}+b\right)$, then $\left(a, a^{2}+b\right)$ satisfies the linear equation, which gives
$$
m a-a^{2}-b-2 m+1=0,
$$

i.e., $b=m a-a^{2}-2 m+1$.
If $y=-x^{2}+2 a x+m a-a^{2}-2 m+1$ and $y=x^{2}$ have intersection points, then the discriminant of the equation $-2 x^{2}+2 a x+m a-a^{2}-2 m+1=0$ is
$$
\Delta=4 a^{2}+8 m a-8 a^{2}-16 m+8 \geqslant 0,
$$

i.e., $a^{2}-2 m a+4 m-2 \leqslant 0$.
The condition for the equation to have real solutions $a$ is
$$
\Delta^{\prime}=4 m^{2}-4(4 m-2) \equiv 0 .
$$

Solving this gives $m \leqslant 2-\sqrt{2}$ or $m \geqslant 2+\sqrt{2}$.