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]}