Answer: [tex]y = 3\sin\left(\frac{x}{3}\right)\right)+2\\\\[/tex]
This is the same as writing y = 3sin(x/3) + 2
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Explanation:
The template for a sine function is
[tex]y = A\sin\left(B\left(x-C\right)\right)+D\\\\[/tex]
where,
- A = deals with the amplitude
- B = deals with the period
- C = phase shift
- D = midline
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The highest point is at y = 5 and lowest is at y = -1. This is a gap of 6 units, which cuts in half to 3. The amplitude is 3 meaning that A = 3.
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One peak or highest point occurs when x = 3pi/2
The next peak over is at x = 15pi/2
This is a gap of (15pi/2) - (3pi/2) = 12pi/2 = 6pi units
The period is T = 6pi which leads to B = 2pi/T = 2pi/6pi = 1/3
In short, B = 1/3
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We don't have to worry about the phase shift, so C = 0.
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The midline is the average of the highest and lowest y values.
D = (5+(-1))/2 = 4/2 = 2
This means D = 2.
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In the previous sections, we found the following
This means we go from
[tex]y = A\sin\left(B\left(x-C\right)\right)+D\\\\[/tex]
to
[tex]y = 3\sin\left(\frac{1}{3}\left(x-0\right)\right)+2\\\\y = 3\sin\left(\frac{x}{3}\right)\right)+2\\\\[/tex]
which is the final answer.
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Extra info:
If you wanted, you could use a cosine function. This is because any cosine function is a phase shift of a sine function. But that greatly complicates things and sine is better suited here.
