What is the point of R2 in the following diagram:
I get that R1 controls the current to the Base, but what does R2 do?
The R2 resistor is used to bring the voltage on the base into a known state. Basically when you turn the whatever source of current you have at the other side of R1 off, the whole line would go into an unknown state. It may pick up some stray interference and that may influence the operation of the transistor or the device on the other side or it may take some time for the voltage to drop just with just the transistor base. Also note that the source of the current going through R1 may leak and that may affect the way transistor operates.
With the R2, which is in configuration called pull-down resistor, we are certain that whatever excess voltage there may be in the branch containing R1 will be safely conducted into ground.
There are two possible reasons:
In addition to the reasons that Olin has mentioned, there's one more: R2 ensures that the transistor turns off rapidly.
Let's suppose you have a source which is not a switch, but a TTL circuit like a 74LS04. TTL circuits (at least the TI SN74LS04) have a minimum output high voltage of 2.4V and a maximum output low voltage of 0.4V. And suppose that R1 is 1K, and the Vbe "on" drop is about 0.6V.
That gives you a 1.8mA (=(2.4V-0.6V) / 1K) current to turn the transistor on, but only -0.2mA to turn the transistor off. Bipolar transistors do have parasitic capacitance that needs to be charged / discharged (not quite the same behavior as MOSFETs).
Now put R2 = 1K: this pulls 0.6mA out of a Vbe=0.6V transistor, yielding a turnon current of 1.2mA and a turnoff current of -0.8mA, so the turnoff behavior will be faster.
The obvious reason for it is to serve as a pull-down resistor, to make sure the base is held low (when there is no specific signal through R1) in order to avoid spurious switching. If there's any other reason for it, it's not jumping out at me.
As well as (and partially a part of) what is said by others, the transistor produces a base-emitter leakage current. With drive to R1 open circuit and R2 ommitted the base floats and the leakage current develops a voltage across the B-E junction which can turn the transistor on. R2 provides a path for this current. As the current is small R2 may be large and the actual value used is usually far smaller than needed. As long as R2 dissipates little energy compared to the energy in R1, having R2 in the 10's to 100 of kilohm range does no harm.