Please forgive my ignomorance and lack of edumucation here, but why do certain function generators require that you use an impedance load across the output?
According to the manual, this HP 3312A requires that, "the output signal MUST be terminated into a 50 ohm load." They then recommend a 50 Ohm feedthrough terminal.
I can't help but wonder why, if it is so important, was it not just installed in the factory across the output terminals?
This is probably right in front of my eyes but please do enlighten me.
Look up transmission line theory.
The classic 50 ohm signal generator looks like an ideal AC voltage source followed by a 50ohm resistor (in practice, it's normally a less than ideal voltage source followed by a less than 50 ohm resistor).
That feeds into a 50 ohm transmission line (a piece of coax, usually) which carries the signal to a load of some sort. In order to eliminate reflections within the transmission line, the impedance of the source, the transmission line, and the load must all match.
If you put the load resistor on the output terminals of the signal generator, that would satisfy the generator, but then you wouldn't be able to attach a 50 ohm coax transmission line to carry the signal somewhere else. Furthermore, all of the signal generator's available output power would be consumed by that load resistor.
You may want to use the signal generator to actually send power into some kind of load. Signal generators aren't only for driving oscilloscopes. If you're driving a load of impedance higher or lower than 50 ohms, you can put resistance in series or parallel to make the total impedance at the end of the transmission line look like 50 ohms. Or you can use a transformer to match impedance, so that the end of the transmission line "sees" 50 ohms.
With most signal generators, terminating the transmission line with the perfect 50 ohm load is only important if you need precise voltages out of it, and if you want to avoid reflections in the transmission line. The generator itself will not be harmed by driving an open circuit.