How do you find the amplitude and period of y=1/2sec(3theta)y=12sec(3θ)?

1 Answer
Nov 3, 2017

AMPLITUDE: 1/212
PERIOD: (2pi)/32π3

Explanation:

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

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


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.