Op-amps implementation
As @beccaboo mentions, this is possible with a single op-amp, and in this particular case, the numbers fall out nicely, making it relatively simple.  Where this circuit fails in practice will be its input impedance being set by the opamp itself rather than the resistor network.  Of course, the same concept would apply if we were to add a resistor divider on the V1 input and modify the rest of the resistor network...but now I'm rambling :)







simulate this circuit &ndash; Schematic created using CircuitLab

To the math, the key to finding op-amp gains is to remember that with negative feedback, there's a virtual short between the inputs.  That is, the opamp will do whatever it can to the output to make V- equal to V+.

Once that's understood, it is a matter of node analysis to find everything you need.  In the case of the drawing above, vi(t) is the sine wave connected to V+, with the DC offset provided by the DC source V2.  To solve the circuit, recall that the current through the feedback resistor (R2) must be the same as the curren through R1, as there should be negligible current flowing into the opamp itself.

Copy/paste an image from MS Word because I don't know how to write stylized math in this editor:



Now that we have a formula for now Vo relates to Vi, it is almost a matter of inspection to see that the ratio of R2/R1 should be 2 and V2 should be 1V to make the expression Vo=3*Vi-2 work.  I like simple resistances, so I chose R2 to be 2k and R1 to be 1k, but any 2:1 ratio will work, depending on other constraints in the circuit.

Cheers!