How do you graph #y=5sec(x/2)+2#?

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Answer 1 · 2018-03-11T08:15:15

As below

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

Given : #y = 5 sec(x/2) + 2#

#a = 5, b = 1/2, c = 0, d = 2#

Amplitude = none.

Period = # (2pi) / |b| = (2pi) / (1/2) = 4pi#

Phase shift = -c / b = 0#

Vertical shift = d = 2#

graph{5 sec(x/2) + 2 [-10.38, 9.62, -7.8, 2.2]}