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

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Answer 1 · 2018-08-11T13:18:00

See explanation and graph.

#y = 3 cot (x/2) - 2,

#x ne #( asymptotic zero of the denominator sin (x/2) ) # 2kpi#,

#k = 0, +-1, +-2, +-3, ...#

The period is the period of #tan ( x/2) = pi/(1/2) = 2pi#.

Vertical shift for the axis is # - #, and so, the axis is

#y = - 2#.

See graph, depicting all these aspects.
graph{((y+2)sin (x/2) - cos (x/2))(y+2+0x)(x-2pi+0.0001y)(x+2pi+0.0001y)(x+0.0001y) = 0[-20 20 -12 8] }