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

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Answer 1 · 2018-07-27T14:52:50

See graph and explanation.

A sec value #notin [ - 1, 1 )#

#y = sec ( 3 x - 2/3pi ) - 1 notin { -2 0 )#

As sec ( .. ) = 1/ cos ( .., )

the asymptotes are given by the zeros of #cos ( 3x - 2/3pi )#

#rArr # the asymptotes are

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

#x = 1/3 ( ( 2 k + 1 ) pi/2 + 2/3 pi ) = (( 3 ( 2 k + 1 ) + 4 ) pi/18#

#= 1/18 (6 k + 7) pi = ....- 5/18pi, 1/18pi, 7/18pi, ...#

See graph, with two asymptotes, near O.

graph{((y+1)cos(3x-2/3pi)-1)(y)(y+2)(x+5/18 pi+0.001y)(x-1/18pi+0y)=0}