Question 630912: write the function to fit the function
MAXIMUM AT (0,4) and (10,4)
MINIMUM AT (5,-4)
Answer by jsmallt9(3759)   (Show Source):
You can put this solution on YOUR website! The vertical distance between the maximum and minimum is found by subtracting the y coordinates: 4 - (-4) = 8. Half of this, 4, is the amplitude for our graph.
Going down from the maximum (or up from the minimum) by the amount of the amplitude, 4, will tell us the vertical shift. 4 - 4 (or -4 + 4) = 0. So the vertical shift for our graph is 0.
The horizontal distance between corresponding points of two consecutive cycles will tell us the period. (The period will also be twice the horizontal distance between consecutive maximum and minimum points.) Horizontal distances, as you might guess, are found by subtracting x coordinates. Assuming that the given maximums and minimums are consecutive, the period will be 10 - 0 = 10 (or 2*(5 - 0) = 10).
The equation we are finding can be a sin or cos equation. The only difference between the two is a phase (horizontal) shift. With the right phase shift we can write either a sin or cos equation that will fit the given points.
For the base function, y = sin(x), y is at the vertical shift (which for the base equation is zero), on its way up to the maximum 1/4 of a cycle later, when x = 0. The phase shift for a sin equation will be how far from 0 the x is the first time y is at the vertical shift (on its way up to the maximum).
For the base function, y = cos(x), y is at a maximum value when x = 0. The phase shift for a cos equation will be how far from 0 the x is when the first maximum occurs.
As you can see from the above, it is usually easier to figure out a phase shift for a cos equation. All you have to figure out is where the first maximum is in relation to x = 0. So that is what we will find. Our first maximum where  , is at x = 0. So the phase shift for our cos equation will be zero, too.
There is only one more thing we need before we write our cos equation. The amplitude and the vertical and phase shifts all go directly into the equation. But the period does not. The period-related number that goes into the equation is:
 /period
For our equation this number will be:
We are now ready to write our cos equation. The frameworks I use when I teach this are:
For sin: y = C + A*sin(B(x-D))
and
For cos: y = C + A*cos(B(x-D))
You might have learned these with different letters. The letters used are not important. What is important is that the values we put into this equation go into the right places. For example, the amplitude is the coefficient of the sin or cos. It doesn't matter if it is called A or any other letter. It only matters that the amplitude goes there and nowhere else. The same with the other numbers we will be putting into the equation.
The "C" above is the vertical shift. Some teachers teach equations where the vertical shift is at the back:
For sin: y = A*sin(B(x-D)) + C
and
For cos: y = A*cos(B(x-D)) + C
Since addition is Commutative (i.e. the order doesn't matter), The equation is not significantly different with the "C" at the front or back. (I prefer the front because then there is no possible way for the "C" to be confused with part of the argument to sin or cos.)
Some teachers/books teach the forms:
For sin: y = A*sin(Bx-D) + C
and
For cos: y = A*cos(Bx-D) + C
where there are no inner parentheses in the argument of sin and cos. This are significantly different equations! The role of "D" is completely changed. In "my" frameworks, the "D" is where the phase shift goes. I can't begin to explain what the "D" represents in these lat two equations. Not only is it hard to explain what the "D" is, it is also harder to figure out the right number to put in that position in the equation. I cannot understand why this form is taught.
I am going to write the equation using "my" framework for cos:
A: Amplitude (use a negative if you want to "flip" the graph around its axis)
B: This the /period number
C: This is the vertical shift
D: This is the phase shift.
Using the cos framework (since we found the phase shift for cos) and the numbers we found earlier we get:
which simplifies to:
This is your equation.
P.S. Since the vertical and phase shifts were both zero, you end up with the very same equation no matter which framework you use. If we pretend that the vertical shift was 9 and the phase shift was 25, then we would get:
If your class puts the vertical shift at the back then you would get:
which is essentially the same equation. If your class uses the "no inner parentheses" frameworks then you could just take this and use the Distributive Property to multiply out the argument:
P.P.S. The sin cycle starts 3/4 of a cycle after a maximum. Our graph has a period of 10, 3/4 of 10 is 7.5. So we could write a sin equation using the same A, B and C and with a phase shift, D, of 7.5:
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