Question

How do you find the amplitude and period of a function `y = (1/2) sin ((x/3)-π)`?

Answer 1

Amplitude is `1/2` and period is `6pi`

Explanation:

In `y=asin(bx+c)`,

amplitude is `a`, period is `(2pi)/b` and phase shift is `-c/b`.

In the function `y=1/2sin(x/3-pi)`

Amplitude is `1/2` and period is `(2pi)/(1/3)=6pi`