What are the maximum and minimum impact speeds for an asteroid that would strike the Earth? I gather that the impact speed depends on the radius of the orbit of the asteroid.

Is the orbit of asteroids in the same direction around the sun, or can they move in the "opposite" direction?
Although most objects orbit the Sun in the same direction — having emerged out of the same rotating gas cloud that spawned the Solar System — some asteroids and other minor planets do move in opposite, or retrograde, orbits (see this Wikipedia article for a list of such objects).

The minimum speed for an asteroid is achieved if it has more or less the same velocity around the Sun as the Earth. In this case the gravitational attraction of Earth will accelerate the object to the escape velocity of Earth, i.e.
$$
v_\mathrm{min} = v_{\mathrm{esc,}\oplus} = \sqrt{\frac{2GM_\oplus}{R_\oplus}} \simeq 11\,\mathrm{km}\,\mathrm{s}^{-1}.
$$
Here, $G$, $M_\oplus$, and $R_\oplus$ are the gravitational constant and the mass and radius of Earth, respectively.

The maximum speed is achieved at a "head-on" collision. Earth's speed around the Sun is
$$
v_{\mathrm{orb,}\oplus} = \sqrt{\frac{GM_\odot}{d}} \simeq 30\,\mathrm{km}\,\mathrm{s}^{-1},
$$
where $M_\odot$ and $d$ are the mass of the Sun and the distance from Sun to Earth (1 AU).

If the asteroid travels on the same orbit, but in the opposite direction, the impact will then be at 60 km/s. However, if the asteroid comes from far away (e.g. the Oort Cloud), it will be accelerated by the Sun and achieve a velocity equal to the escape velocity from the Sun at the location of Earth. As is seen from the two equations above, the orbital speed and the escape velocity differ by a factor of $\sqrt{2}$. That is, an object falling from infinity toward the Sun, will have a speed equal to $30\,\mathrm{km}\,\mathrm{s}^{-1}\times\sqrt{2}=42\,\mathrm{km}\,\mathrm{s}^{-1}$ when it reaches Earth.

Hence, the maximum impact velocity is
$$
v_\mathrm{max} = 30\,\mathrm{km}\,\mathrm{s}^{-1} + 42\,\mathrm{km}\,\mathrm{s}^{-1} = 72\,\mathrm{km}\,\mathrm{s}^{-1}.
$$