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

1 Answer
Apr 21, 2017

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]}