A MOSFET biased in saturation can convert a small change in gate voltage into a larger change in drain voltage.

Around a bias point, the drain-current variation is approximated by

$$i_d=g_m v_{gs},$$

where the transconductance is the local slope

$$g_m=\left.\frac{d I_D}{d V_{GS}}\right|_Q.$$

In a basic common-source stage with drain resistance $R_D$ and sufficiently large transistor output resistance,

$$v_o\approx -g_mR_Dv_{gs}.$$

The negative sign means the output voltage is inverted: increasing gate voltage increases drain current and therefore increases the voltage drop across the drain load.

The DC operating point and the small-signal gain are different parts of the design. Biasing keeps the transistor in the intended region; the local transconductance and surrounding circuit determine how small signals are amplified.