How do you sketch the graph of #y = sec(3x + pi/2)#?

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Answer 1 · 2018-07-22T11:01:27

See explanation and graph.

#y =sec ( 3 x + pi/2 ) = - csc 3x notin ( - 1, 1 )#

and #3x ne# an integer multiple of #pi #

#rArr x ne # integer multiple of #pi/3#

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

The period = period of sin 3x = (2pi)/3

Rewrite #y sin 3x + 1 = 0#, by cross multiplication..

For this form, the Socratic graph is immediate.
graph{(ysin(3x) + 1)(y^2-1)(x^2-1.0966)= 0[-10 10 -5 5]}

Observe

# y notin ( - 1, 1 )# and the asymptotes #x = +- pi/3#, near O and also the period #(2pi)/3#.