Question

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

Answer 1

It is periodic, with period `pi/2`. The graph is for `12_+` periods.
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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]}