How do you graph #y=2cot(2x+pi/3)+1#?

1 Answer
Jan 2, 2017

Answer:

It is periodic, with period #pi/2#. The graph is for #12_+# periods.
I am satisfied that this edition is bug-free.

Explanation:

For #cot(2x+pi/3)+1#, the period is #(pi)/2#. The asymptotes are given by

#2x+pi/3#= a multiple of #pi=kpi#, giving #x =(kpi-pi/3)/2=((3k-1)/6)pi, k = 0, +-1, +-2, +-3, ...#

For one period, #x in (-pi/6, pi/3)#,

there are two terminal asymptotes

#uarr x=-pi/6 darr and uarr x = pi/3 darr#.

graph{(y-1)tan(2x+1.047)-1=0 [-10, 10, -5.21, 5.21]}