Question

Pleaseeeee hellppppdosbshsb I NEED TO KNOW!!!

Answer 1

Answer:
`y = 3\sin\left((x)/(3)\right)\right)+2\\\\`

This is the same as writing y = 3sin(x/3) + 2

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Step-by-step explanation:

The template for a sine function is

`y = A\sin\left(B\left(x-C\right)\right)+D\\\\`

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

• A = 3
• B = 1/3
• C = 0
• D = 2

This means we go from

`y = A\sin\left(B\left(x-C\right)\right)+D\\\\`

to

`y = 3\sin\left((1)/(3)\left(x-0\right)\right)+2\\\\y = 3\sin\left((x)/(3)\right)\right)+2\\\\`

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.