Solution:
Write $x^{2}-2x = x^{2}-2x+1-1 = (x-1)^{2}-1$. Since $(x-1)^{2} \geq 0$, it is clear that the minimum is $-1$.

Alternate method: The graph of $y = x^{2}-2x$ is a parabola that opens up. Therefore, the minimum occurs at its vertex, which is at $\frac{-b}{2a} = \frac{-(-2)}{2} = 1$. But $1^{2}-2 \cdot 1 = -1$, so the minimum is $-1$.