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1. Homework Statement
Problem: [itex]y=2sin\dfrac{2\pi x}{3}[/itex]
2. Homework Equations
Standard Equation: [itex] y = A sin(B(x  C)) + D [/itex]
3. The Attempt at a Solution
First, rearrange the equation: [itex]y=sin\dfrac{2\pi x}{3}+2[/itex]
Amplitude: [itex]1[/itex] (avalue)
Period: [itex]\dfrac{2\pi}{2\pi/3}=\dfrac{2\pi}{1}\cdot\dfrac{3}{2\pi}=\dfrac{6\pi}{2}=3\pi[/itex]
Phase Shift: [itex]0[/itex] (no cvalue)
Vertical Shift: [itex]2[/itex] (dvalue)
Table (I don't know how to make):
See my attachment for the table of values.
How did I pick my xvalues to calculate my yvalues?
I used count formula: [itex]\dfrac{1}{4}\cdot \dfrac{3\pi}{1}=\dfrac{3\pi}{4}[/itex]
So, I start from the phase shift and continue adding [itex]\dfrac{3\pi}{4}[/itex] : [itex]0+\dfrac{3\pi}{4}=\dfrac{3\pi}{4}[/itex]
[itex]\dfrac{3\pi}{4}+\dfrac{3\pi}{4}=\dfrac{6\pi}{4}=\dfrac{3\pi}{2}[/itex]
[itex]\dfrac{3\pi}{2}+\dfrac{3\pi}{4}=\dfrac{6\pi}{4}+\dfrac{3\pi}{4}=\dfrac{9\pi}{4}[/itex]
[itex]\dfrac{9\pi}{4}+\dfrac{3\pi}{4}=\dfrac{12\pi}{4}=3\pi[/itex]
I don't want to graph it yet because my table of values doesn't seem correct, in my book the answer is in my second attachment. They don't even use pi for their xvalues :(
Problem: [itex]y=2sin\dfrac{2\pi x}{3}[/itex]
2. Homework Equations
Standard Equation: [itex] y = A sin(B(x  C)) + D [/itex]
 A: amplitude is A
 B: period is [itex]\dfrac{2\pi}{b}[/itex]
 C: phase shift is [itex]\dfrac{C}{B}[/itex]
 D: vertical shift is D
3. The Attempt at a Solution
First, rearrange the equation: [itex]y=sin\dfrac{2\pi x}{3}+2[/itex]
Amplitude: [itex]1[/itex] (avalue)
Period: [itex]\dfrac{2\pi}{2\pi/3}=\dfrac{2\pi}{1}\cdot\dfrac{3}{2\pi}=\dfrac{6\pi}{2}=3\pi[/itex]
Phase Shift: [itex]0[/itex] (no cvalue)
Vertical Shift: [itex]2[/itex] (dvalue)
Table (I don't know how to make):
See my attachment for the table of values.
How did I pick my xvalues to calculate my yvalues?
I used count formula: [itex]\dfrac{1}{4}\cdot \dfrac{3\pi}{1}=\dfrac{3\pi}{4}[/itex]
So, I start from the phase shift and continue adding [itex]\dfrac{3\pi}{4}[/itex] : [itex]0+\dfrac{3\pi}{4}=\dfrac{3\pi}{4}[/itex]
[itex]\dfrac{3\pi}{4}+\dfrac{3\pi}{4}=\dfrac{6\pi}{4}=\dfrac{3\pi}{2}[/itex]
[itex]\dfrac{3\pi}{2}+\dfrac{3\pi}{4}=\dfrac{6\pi}{4}+\dfrac{3\pi}{4}=\dfrac{9\pi}{4}[/itex]
[itex]\dfrac{9\pi}{4}+\dfrac{3\pi}{4}=\dfrac{12\pi}{4}=3\pi[/itex]
I don't want to graph it yet because my table of values doesn't seem correct, in my book the answer is in my second attachment. They don't even use pi for their xvalues :(
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