# How do you find the amplitude, period, and shift for y=1/3 sin (2x - pi/3)?

Dec 21, 2017

See below.

#### Explanation:

If we write a trigonometric equation in the form:

$y = a \sin \left(b x + c\right) + d$

Then:

Amplitude is a

Period is $\frac{2 \pi}{b}$

Phase shift is $\frac{- c}{b}$

Vertical shift is d

From example:

Amplitude $= a = \frac{1}{3}$

Period $= \frac{2 \pi}{b} = \pi$

Phase shift $= \frac{- c}{b} = \frac{\frac{\pi}{3}}{2} = \frac{\pi}{6}$

Vertical shift $= d = 0$ no vertical shift.

Dec 21, 2017

$\frac{1}{3} , \pi , \frac{\pi}{6} , 0$

#### Explanation:

$\text{the standard form of the sine function is}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a \sin \left(b x + c\right) + d} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where amplitude "=|a|," period } = \frac{2 \pi}{b}$

$\text{phase shift "=-c/b" and vertical shift } = d$

$\text{here } a = \frac{1}{3} , b = 2 , c = - \frac{\pi}{3} , d = 0$

$\Rightarrow \text{amplitude "=|1/3|=1/3," period } = \frac{2 \pi}{2} = \pi$

$\text{phase shift } = - \frac{- \frac{\pi}{3}}{2} = \frac{\pi}{6}$

$\text{there is no vertical shift}$