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

Dec 16, 2015

amplitude$= 1$
period$= {270}^{\circ}$
phase shift$=$none
vertical shift$=$none

#### Explanation:

The general equation for a sine graph is:

y=color(red)asin(color(orange)k(x-color(green)d)+color(blue)c

where:
$\textcolor{red}{a} = | \text{amplitude} |$
$\textcolor{\mathmr{and} a n \ge}{k} = {360}^{\circ} / \left(\text{period}\right)$
$\textcolor{g r e e n}{d} =$phase shift
$\textcolor{b l u e}{c} =$vertical shift

In this case, since the value of $a$ is $1$, the amplitude is $1$. It would not be $- 1$ because the amplitude is always an absolute value, or always positive.

The period is found by substituting the $k$ value, $\frac{4}{3}$, into the equation, period$= {360}^{\circ} / k$:

period$= {360}^{\circ} / k$
$= {360}^{\circ} / \left(\frac{4}{3}\right)$
$= {360}^{\circ} \div \frac{4}{3}$
$= {360}^{\circ} \cdot \frac{3}{4}$
$= {\textcolor{red}{\cancel{\textcolor{b l a c k}{{360}^{\circ}}}}}^{{90}^{\circ}} \cdot \frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}$
$= {270}^{\circ}$

The phase shift is $0$ in this case since there is no indication after the $x$ variable in the equation.

The vertical shift is $0$ in this case since there is no indication at the end of the equation.