Question

Y=3sin(1/2 x+pi/6)please find amplitude period and phase shift

Answer 1

The given function is

`y=3\sin ((1)/(2)x+(\pi)/(6))`

We have to use the following form

`a\sin (bx-c)+d`

Where the amplitude is a, the period is 2pi/b, and the phase shift is c/b. In the given function a = 3, b = 1/2, and c = pi/6.

`\begin{gathered} a=3 \\ T=(2\pi)/(b)=(2\pi)/((1)/(2))=4\pi \\ \theta=(c)/(b)=((\pi)/(6))/((1)/(2))=(2\pi)/(6)=(\pi)/(3) \end{gathered}`

Therefore, the amplitude is 3, the period is 4pi, and the phase shift is pi/3.