a. To find the x-axis intercept, we need to
find the load line from circuit. If we find the Thevenin equivalent of the
voltage source together with the two resistors, we can easily find the relation
between v_{x} and i_{x}.

R_{Th} = R//3R = 0.75R (if we short the source).

V_{Th} = [R/(3R+R)].V_{A} = 0.25V_{A} (voltage
divider).

Now, we can apply Ohms law for the new resistor: 0.75R.

Get: 0.25V_{A} – v_{x} = 0.75Ri_{x}. Therefore,
the load line equation is i_{x} = -(4/3R)v_{x} +V_{A}/3R.

Therefore, i_{x} = -[4/(3´20)]v_{x} + 9/(3´20) i.e. i_{x} = -(1/15)v_{x} +
3/20. (see figure)

x-axis interception, so i_{x} = 0, i.e. -(1/15)v_{x} +
3/20 = 0 therefore, v_{x} = 2.25

b. The slope of the load line is -1/15 mA/V
as found above in the equation.

c. To find the bias voltage, we just need
to read the v_{x} component of the meeting point of the load line and
the current voltage characteristics of the device X. We can clearly see that
the v_{x} component of this point is at 5 units. But each unit
represents 0.3V, then the bias voltage is 5´0.3 = 1.5V.