How do you graph #y=6csc(3x+(2pi)/3)-2#?

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Answer 1 · 2017-02-19T13:48:15

See Socratic graph and explanation.

y is of period #(2pi)/3#.

As #|csc(theta)|>=1, |y+2|>=6#, giving #y >=4 and <=--8#

The graph has vertical asymptotes, when #csc(3x+2pi/3)=oo#, giving

#x= 1/3(kpi-2pi/3), k =0, +-1, +-2, +-3, ...#

For that matter, half period #1/3pi# is the difference between two

consecutive asymptote x values.

The graph for one period #x in(-2/9pi, 4/9pi) is also included.

graph{(y+2)sin(3x+2pi/3)-6=0 [-40, 40, -21, 19]}

graph{(y+2)sin(3x+2pi/3)-6=0 [-.7, 1.4, -14.2, 12]}

Not-to-scale graph to reveal spacing #pi/3# = 1.05, for the three

asymptotes at #x = -.7, .35 and 1.4#