How do you graph #y=-cot(4x)#?

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Answer 1 · 2016-09-16T01:59:05

A short Table for #y = -cot 4x# for one period #x in (0^o, 45^o):.

#(x, y): (0, -oo) (15^o, -sqrt 3) (22.5^o, 0) (30^o, sqrt 3) (55^o, oo)#,

#tan theta# is periodic with period #pi#. So,

#tan 4x# is periodic with perid #pi/4#

Note infinite discontinuity at #x = 0 and x = pi/4=45^o#.

Make a graph for one period, #x in (0, pi/4)=(0^o, 45^o)# and move it

laterally on either side, successively, on either side, to get the graph

on both left and right sides.

A short Table for #y = -cot 4x# follows..

#(x, y): (0, -oo) (15^o, -sqrt 3) (22.5^o, 0) (30^o, sqrt 3) (55^o, oo)#,

Now, you can make your graph, with hand or using an

electronic graphic device.

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