How do you identify the period and asympotes for y=-2tan(pitheta)y=−2tan(πθ)?

1 Answer
Somebody N.
Apr 11, 2018

See below.

Explanation:

If we express the tangent function in the following way:

y=atan(bx+c)+dy=atan(bx+c)+d

Then:

\ \ \bba \ \ \ ="the amplitude"

bb((pi)/b) \ \="the period"

bb((-c)/b)= "the phase shift"

\ \ \ \bbd \ \ \="the vertical shift"

For given function we have:

b=pi

So period is:

(pi)/pi=color(blue)(1)

The function will have vertical asymptotes everywhere it is undefined.

We know:

tan(theta) is undefined at pi/2 , pi/2+pi and so on. We can write this in a general way as follows:

pi/2+npi

Where n in an integer.

We now solve:

pitheta=pi/2+npi

theta=1/2+n

So vertical asymptotes occur everywhere theta=1/2+n

For:

n in ZZ

The graph confirms these findings:

enter image source here

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