How do you sketch the graph of #y=2sec3x#?

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Answer 1 · 2018-03-07T07:17:25

As below

Standard form of equation is #y = a sec (bx - c) + d#

#y = 2 sec 3x#

#a = 2, b = 3, c = 0, d = 0#

Period # = (2pi) / |b| = (2pi) / 2 = pi#

Phase shift # = c / b = 0#

Phase shift 0 to the right

graph{3 sec (2x) [-10, 10, -5, 5]}