How do you find the amplitude, phase shift and period of #y = -2 sin (4/3x)#?

1 Answer
Jul 15, 2015

The amplitude is #|-2|=2#, the phase shift is #0#, and the period is #(2pi)/(4/3)=(3pi)/2\approx 4.7124#

Explanation:

Assume that #B>0#. For a function of the form #y=Asin(Bx+C)+D=Asin(B(x+C/B))+D#, the amplitude (half the vertical distance between the high point and low point) is #|A|#,
the period (horizontal peak-to-peak distance) is #(2pi)/B#,

the vertical shift is #D#, up if #D>0# and down if #D<0# (this is the vertical location of the "midline" or "average value" of the function),

and the phase shift is #C/B# as a horizontal shift of a sine wave (to the left if #C/B>0# and to the right if #C/B<0#) and as a phase "angle" it's #C/(2pi)# (this last quantity represents the horizontal shift as a signed fraction of a period).

Similar statements hold for functions of the form #y=Acos(Bx+C)+D=Acos(B(x+C/B))+D#, but the horizontal shift is with respect to a cosine wave.