# How VDB JFET circuit reverse-biases the gate-to-source voltage?

I already know how Gate-bias, Self-bias, and Current-source bias works in a JFET configuration. However, I am still stuck in voltage-divider-bias JFET circuit. (circuit below)

In a JFET circuit, the gate-source voltage must be reverse-biased. However, in a VDB JFET circuit, the gate voltage is always positive.

Posted below is also the transconductance curve, it shows that the Q point in this VDB circuit will have a negative voltage at either Q1 or Q2, how is that suppose to happen if VDB produce positive voltage at the gate?

With 2 volts applied to the gate node, current will start to flow into the JFET drain and raise the source voltage thus cutting-off the channel. Equilibrium will be found with a source voltage (with respect to ground) that is more positive than the gate voltage with respect to ground i.e. $$\V_{GS}\$$ is negative a couple of volts for instance.