283,719 views
3 votes
3 votes
Pleaseeeee hellppppdosbshsb I NEED TO KNOW!!!

Pleaseeeee hellppppdosbshsb I NEED TO KNOW!!!-example-1
User Sean Turner
by
2.3k points

1 Answer

23 votes
23 votes

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

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

=====================================================

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

---------------

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.

---------------

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

---------------

We don't have to worry about the phase shift, so C = 0.

---------------

The midline is the average of the highest and lowest y values.

D = (5+(-1))/2 = 4/2 = 2

This means D = 2.

---------------

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.

---------------

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.

User BeeGee
by
2.6k points