How do you find the amplitude and period of $y=sec(1/3theta)$ ?

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Answer 1 · 2017-10-21T00:31:49

Amplitude = None, Period = $6pi$

Use the standard form $y = a (secbx - c) d$ equation, where
Amplitude = a, Period = $(2pi)/b$ , Phase shift = c/b & vertical shift = d.

Given equation is $y = sec ((1/3) theta)$

Since the graph of the function sec does not have a maximum or minimum value, there can be no value for the amplitude.
Amplitude = none

Period $= (2pi) / b = (2pi) /( 1/3) = 6pi$

Phase shift = c / b = 0 as ( c = 0)

Vertical shift = d = 0.

How do you find the amplitude and period of y=sec(1/3theta)y=sec(1/3theta)?