Using period to find the equation of a function

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Homework Statement


The period of f is 5. If f(1)=4, f(2)=5, and f(4)=-2, the value of f(7) is:

Homework Equations


The standard format for trig functions is A[tex]\sin[/tex]B(x-C)+D
A is the amplitude, [tex]\frac{2\pi}{B}[/tex] is the period (for sin or cos functions), C is the phase, and D is the vertical displacement.

The Attempt at a Solution


Finding the period of a function when you have the equation is easy, but doing the opposite seems to be impossible. The only way I'm going to figure out the value of f(7) is if I figure out the equation first. Presumably the function is a trig function. If it's a sin function, then the equation must be A[tex]\sin\frac{2\pi}{5}[/tex](x-C)+D, in order for the period to be 5.

Beyond that, I haven't been able to figure out what I should do, because there doesn't seem to be any real pattern to the y values, and my textbook only shows how to find the period from the equation, not how to find values when given vague information. Could someone point me in the right direction?
 
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Well, it seems you've been given the period which is 5. What does a period of 5 suggest about f's cycle?
 
Think about the definition of a period of a function. Then look at an example of e.g. f(x) = sinx. The period is 2pi. So at x = 0 and x = 2pi there is something in common. What is it?
 
Ok, I may be completely off the mark here, but here's what I'm thinking: the period of a function is the shortest distance you have to travel along the x-axis for the function to begin another cycle (according to my lesson book). The cycle of a sine curve starts at zero and ends at 2pi, so the value of x must always be somewhere within that cycle. The cycle of the function f is 5, so it starts at zero and ends at 5. f(5) is the end of the cycle, so f(7) should be the same as f(2), because you just keep traveling the same distance, the same curve.

In other words, f(7)=5, because f(2)=5.

That is right, isn't it?

Thanks for the help.
 
That's right. Good job!