Question

How do you find the amplitude, period, and shift for `y=1 +sec ( (x-pi)/3)`?

Answer 1

Amplitude: none
Period: `6pi`
Phase: `-pi`

Explanation:

Given a trigonometric function of the form

`y=Asec(Bx - C) + D`

Amplitude: Since the `sec` function does not have a maximum or minimum, there is not amplitude.

Period: `(2pi)/abs(B)`

Phase (i.e., horizontal) shift: `C/B`

Vertical shift: `abs(D)`

Given `y=1+sec((x-pi)/3)`

`y = sec((x-pi)/3) + 1`

`y = sec(1/3 x - pi/3) + 1`

By the formula,

Period: `(2pi)/(1//3)=6pi`

Phase shift: `(-pi//3)/(1//3)=-pi`