Question

How do you find the amplitude, period and phase shift for `y=4sin(1/2(theta-pi/2))+5`?

Answer 1

The amplitude is `A=4`
The period is `=4pi`
The phase shift is `=pi/2`

Explanation:

`y = A sin(Bx + C) + D`

Amplitude is `A`

Period is `(2π)/B`

Phase shift is `−C/B`

Vertical shift is `D`

Here, we have

`y=4sin(1/2(theta-pi/2))+5`

`y=4sin(1/2theta-pi/4)+5`

The amplitude is `A=4`

The period is `=(2pi)/B=(2pi/(1/2))=4pi`

The phase shift is `=(pi/4)/(1/2)=pi/2`

graph{4sin(1/2(x-pi/2))+5 [-11.71, 13.61, -1.49, 11.17]}