Question

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

Answer 1

Amplitude = None, Period = `6pi`

Explanation:

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.