A function generator has a 50 ohm and 600 ohm main outputs: http://www.ece.mtu.edu/labs/EElabs/EE2303/model_25.pdf What does it stand for? What is the difference? Is that about the load resistance?
Think of the function generator as a Thevenin source. If you don't know what that is, look it up since it's well worth learning. In fact, look up Norton source too. Thevenin and Norton sources are two sides of the same coin.
Briefly, you can think of your function generator as a ideal voltage source with a resistance in series. That may not be how it is actually constructed, but the point is you can't tell the difference from outside. This means that if you measure the open circuit voltage of a signal the function generator is producing on the 50 Ω output, then connect a 50 Ω load to it, the voltage will drop to half the original value.
If you connect a 1 kΩ load to the 50 Ω output, then you have a voltage divider with the top resistor 50 Ω and the bottom 1 kΩ. This means the output voltage will be (1 kΩ)/(50 Ω + 1 kΩ) = 0.95 of the open circuit voltage.
The reason 50 Ω and 600 Ω were probably chosen is because 50 Ω is a common impedance for tranmission lines, like some coax and twisted pair ethernet. 600 Ω is common in audio applications.
The way Agilent (and Rigol and probably most modern) function/arbitrary function generators work is that there is a power amplifier internally with a 50\$\Omega\$ resistor in series with the output. The amplifier produces twice the voltage that you've programmed (when you set it to low impedance 50\$\Omega\$ output).
So if you measure the output voltage of the function generator when it is set to 1.0Vp-p with an oscilloscope using a high-Z x10 probe you'll see 2.0Vp-p on the oscilloscope.
When you load the output with 50.0\$\Omega\$ you will see the voltage you programmed.
Setting the output to high-Z leaves the 50\$\Omega\$ resistor in there but just halves the voltage from the amplifier.
As the manual for this particular generator says (and this can easily confuse the unwary):
Thus the voltage delivered to the amplifier input will often be double the panel setting.
If your load resistor is some other value than 50\$\Omega\$ then the voltage will be as predicted from the voltage divider formula (between 0 and double the set voltage).
50\$\Omega\$ is a common impedance used for relatively high frequency or broadband signals since coaxial cable with a characteristic impedance of 50 ohms is of practical dimensions.
600\$\Omega\$ is commonly used in audio for telephony and the output stage would work similarly, except with 600\$\Omega\$ rather than 50\$\Omega\$.