Example 8. The parabola $y=x^{2}-2 b x+1$ and the line $y=-\frac{1}{2} x$ $+\frac{1}{2} m$, for any real value of $b$, always have intersection points, then the range of real number $m$ is $\qquad$
(1987, Shenyang City Junior High School Mathematics Invitational Competition)
Since the parabola and the line always have intersection points, the system of equations
$$
\left\{\begin{array}{l}
y=x^{2}-2 b x+1 \\
y=-\frac{1}{2} x+\frac{1}{2} m
\end{array}\right.
$$

implies that there are real solutions. Eliminating \( y \) and simplifying, we get
$$
2 x^{2}+(1-4 b) x+2-m=0 .
$$

For this equation to always have real roots, it must be true that
$$
\Delta=(1-4 b)^{2}-8(2-m) \geqslant 0 \text {. }
$$

Clearly, when \( 2-m \leqslant 0 \) i.e., \( m \geqslant 2 \), \(\Delta \geqslant 0\).
That is, when \( m \geqslant 2 \), the parabola and the line always have intersection points.