Question

Question #51bdf

Answer 1

`y = 3cos((pi)/8 x+ (5pi)/8)+ 1`

Explanation:

First of all, we must notice that there are infinite solutions for two reasons:

1. We can always add a phase of `2 pi` in order to get the same equation basically
2. We can multiply the phase by some integer so that it hits a certain number of minima between these two points. For an example of that, here we have `cos(x)` and `cos(3x)` which both have a max at 0 and a min at `pi`.
   graph{y=cos(x) [0, 6.3, -2, 2]}
   graph{y=cos(3x) [0, 6.3, -2, 2]}

So let's just choose to make some choice and just agree that if it works, we stick with it.

As you calculated, we have an equation that is something like

`y = A cos(f x + phi) + B = 3 cos(f x + phi) + 1`

where `f` is the frequency and `phi` is that phase.

Now, we could try and plug in the numbers or we could be fancy. I'm going to do the fancy way, but that's just because I'm lazy.

So for a minimum, we can say that `f x + phi = pi` because that's a minimum of cosine. For the maximum, we can say that `f x + phi = 2pi`.

Therefore, we have a 2x2 series of equations:
`f * 3 + phi = pi`
`f * 11 + phi = 2pi`

If we subtract the two equations, we get
`8f = pi -> f = pi/8`

Plugging that in,
`(3pi) / 8 + phi = pi -> phi = (5pi)/8`

Therefore, one possible solution is
graph{y = 3cos((pi)/8 x+ 5pi/8)+ 1 [-2, 15, -4, 5]}