Let's compare our equation to the following:
y=Atan(bθ+c)+d
From this equation, we can deduce the following:
A is the amplitude.
The period can be calculated from the formula πb.
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=12tan(θ), we see that A=12,b=1,c=0,d=0
Therefore, the amplitude is 12, the period is π1=π, and the phase shift is 0.
Since we have no phase shift, we have vertical asymptotes at x=nπ2 where n is any integer.
tan(0)=0, so we have the point (0,0) on our graph.
As we approach x=π2, 12tan(θ) begins to increase faster and faster, running nearly parallel to x=π2 but never touching it.
As we approach x=−π2, 12tan(θ) begins to decrease faster and faster, running nearly parallel to x=−π2 but never touching it.
Due to our amplitude of 12, the graph becomes slightly vertically compressed compared to the graph of y=tan(θ) (but the asymptotes remain unchanged).
We have just graphed a full period of y=12tan(θ). This pattern repeats on intervals of π, with the x-intercepts of the graph being at (πn,0) where n is any integer.
For comparison, the graphs of y=tan(θ) and y=12tan(θ) (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]}