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

Jul 15, 2015

The amplitude is $| - 2 | = 2$, the phase shift is $0$, and the period is $\frac{2 \pi}{\frac{4}{3}} = \frac{3 \pi}{2} \setminus \approx 4.7124$

#### Explanation:

Assume that $B > 0$. For a function of the form $y = A \sin \left(B x + C\right) + D = A \sin \left(B \left(x + \frac{C}{B}\right)\right) + D$, the amplitude (half the vertical distance between the high point and low point) is $| A |$,
the period (horizontal peak-to-peak distance) is $\frac{2 \pi}{B}$,

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

Similar statements hold for functions of the form $y = A \cos \left(B x + C\right) + D = A \cos \left(B \left(x + \frac{C}{B}\right)\right) + D$, but the horizontal shift is with respect to a cosine wave.