How do you graph $y = \frac{1}{2} + 2 sin (3 x - \frac{π}{2})$ $y=1/2+2sin(3x-pi/2)$ ?

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How do you graph $y = \frac{1}{2} + 2 sin (3 x - \frac{π}{2})$ $y=1/2+2sin(3x-pi/2)$ ?

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Answer 1 · 2016-10-04T21:43:38

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

Explanation:

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

This is of the form $y = A sin (B x - C) + D$ $y=Asin(Bx-C)+D$ where

$A =$ $A=$ the amplitude

$\frac{2 π}{B} =$ $(2pi)/B=$ the period

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

$D =$ $D=$ the vertical shift

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

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

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

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

Let's look at each transformation of the graph. First $y = sin x$ $y=sinx$

enter image source here

Next, change the amplitude to $2$ $2$
enter image source here

Now change the period from $2 π$ $2pi$ to $\frac{2 π}{3}$ $(2pi)/3$

enter image source here

Next, add a phase shift of $\frac{π}{6}$ $pi/6$ to the right. The red graph has the phase shift. The blue is without the phase shift.
enter image source here

Lastly, add a vertical shift of $\frac{1}{2}$ $1/2$ units up.
enter image source here

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How do you graph $y = \frac{1}{2} + 2 sin (3 x - \frac{π}{2})$ $y=1/2+2sin(3x-pi/2)$ ?

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Explanation:

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