Let's compare our equation to the following:
y=Atan(btheta+c)+d
From this equation, we can deduce the following:
A is the amplitude.
The period can be calculated from the formula pi/b.
The phase shift is c units to the left if c>0 and c units to the right if c<0 and no shift at all if c=0.
The vertical shift is d units up if d>0, d units down if d<0, and no vertical shift if d=0.
For y=1/2tan(theta), we see that A=1/2, b=1, c=0, d=0
Therefore, the amplitude is 1/2, the period is pi/1=pi, and the phase shift is 0.
Since we have no phase shift, we have vertical asymptotes at x=(npi)/2 where n is any integer.
tan(0)=0, so we have the point (0,0) on our graph.
As we approach x=pi/2, 1/2tan(theta) begins to increase faster and faster, running nearly parallel to x=pi/2 but never touching it.
As we approach x=-pi/2, 1/2tan(theta) begins to decrease faster and faster, running nearly parallel to x=-pi/2 but never touching it.
Due to our amplitude of 1/2, the graph becomes slightly vertically compressed compared to the graph of y=tan(theta) (but the asymptotes remain unchanged).
We have just graphed a full period of y=1/2tan(theta). This pattern repeats on intervals of pi, with the x-intercepts of the graph being at (pin,0) where n is any integer.
For comparison, the graphs of y=tan(theta) and y=1/2tan(theta) (respectively) so we can see the horizontal compression and periodic behavior:
graph{y=tan(x) [-10, 10, -5, 5]}
graph{y=1/2tan(x) [-10, 10, -5, 5]}