Question #1a9c9

1 Answer
Jul 29, 2017

The period is #pi/3#, and the asymptotes are at #pi/3, -pi/3, pi, -pi#.

Explanation:

The standard form of a cotangent function is #f(x)=a cot (bx-c) +d#.

The period of a tangent or cotangent function is equal to #pi/(absb)#. (On the other hand, the period of a sine, cosine, cosecant, or secant function is equal to #(2pi)/(absb)#.)

In this case, #b=3#.

#pi/(absb) = pi/(abs3) = pi/3#

Additionally, the function #cot x# has asymptotes at #x=pin, n in Z#. However, since #f(x) = 2 cot 3x# is horizontally shrunk by a factor of #1/3#, the asymptotes are at #x=pi/3n, n in Z#. So, the asymptotes would be at #pi/3, -pi/3, pi, -pi#.