The performance will change the stage gain will goes up!! The important quiescent (DC) conditions will not. Suppose you need 6V at a current of 1 mA flowing into the collector or drain of a transistor. If you start of with a Vcc of 12V, then the quiescent load is (12 - 6)/1 K = 6 K ohms. So if the current due to the signal is 2 micro A P-P, then the output voltage is 6 K X 2 micro A = 12 mV P-P. If instead you use some sort of active device (or boot strapping) so the dynamic impedance is 100K ohms, then the output voltage is 2 micro A X 100K = 200 mV P-P. A worthwhile gain. These techniques normally lead to a narrowing of the useable bandwidth, as the active load becomes "less active" as the frequency rises.
Frank