# How do you graph  y=1/2+2sin(3x-pi/2)?

Oct 4, 2016

The amplitude is $2$, the period is $\frac{2 \pi}{3}$, the phase shift is $\frac{\pi}{6}$ right, and the vertical shift is $\frac{1}{2}$.

#### Explanation:

First, rewrite the problem as $y = 2 \sin \left(3 x - \frac{\pi}{2}\right) + \frac{1}{2}$

This is of the form $y = A \sin \left(B x - C\right) + D$ where

$A =$ the amplitude

$\frac{2 \pi}{B} =$ the period

$\frac{C}{B} =$ the phase shift

$D =$ the vertical shift

In this example, the amplitude $A = 2$. Amplitude is the vertical distance from the "midline" to the max or min. It is not the the distance from max to min.

Period $= \frac{2 \pi}{B} = \frac{2 \pi}{3}$. One complete cycle of the sin graph will be $\frac{2 \pi}{3}$ horizontal units wide.

Phase shift $= \frac{C}{B} = \frac{\frac{\pi}{2}}{3} = \frac{\pi}{6}$ The graph will be shifted $\frac{\pi}{6}$ units to the right.

Vertical shift $D = \frac{1}{2}$ units up.

Let's look at each transformation of the graph. First $y = \sin x$ Next, change the amplitude to $2$ Now change the period from $2 \pi$ to $\frac{2 \pi}{3}$ Next, add a phase shift of $\frac{\pi}{6}$ to the right. The red graph has the phase shift. The blue is without the phase shift. Lastly, add a vertical shift of $\frac{1}{2}$ units up. 