How do you graph and list the amplitude, period, phase shift for y=1/2tantheta?

1 Answer
Feb 25, 2018

The amplitude is 1/2, the period is pi/1=pi, and the phase shift is 0.

Explanation:

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]}