Question

How do you find the amplitude and period of `y=1/2sec(3theta)`?

Answer 1

AMPLITUDE: `1/2`
PERIOD: `(2pi)/3`

Explanation:

Standard Form: `y=asec(b(x-c))+d`
`a=`amplitude

So for `y=1/2sec(3θ) -> a=1/2`
Therefore, the amplitude is `1/2`.

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Period of Secant: `(2pi)/|b|`
So for `y=1/2sec(3θ) -> b=3`
`(2pi)/|b|=(2pi)/|3|=(2pi)/3`
Therefore, the period is `(2pi)/3`.