Actually, what difference does it make if he/she (don't know the gender of the name "wiki", but I'll call it "he" from now on so i don't go nuts!) reads it here or somewhere else...and if he chooses to "cheat" and merely copy this, then that's his responsibility--karma will take care of that ;)
So...
The reason a switching transistor is biased to avoid it's active region, is because the idea is to send as much of the power, as possible, to the load that is being switched. When the transistor is, say, half on, it will be dropping half the voltage, and thus for the current being drawn, only half the power will reach the device. The other half will be dissipated across the transistor. Why do you suppose that is not such a good thing?
Now, as a thought experiment, using the diagram, below, consider what happens when the transistor is at saturation, and then when it's at cutoff. How much power is lost in the transistor and how much reaches the load (hint, in one case, neither dissipates any power). And then, think about what happens if the transistor is only partially on.
Regarding the H-bridge, look at the current paths through the transistors for each of the states in the truth table using what you learned from Dorin's post. In the case of the H-bridge, there are two switching transistors in series to drive the load. See if you can relate the current paths and their affect on the motor, to the label in the truth table (e.g. "Stop", "Forward", "Backward" and "Off").
