I'll answer the question about tangent functions, as the questions posted right before this one will show you how to directly solve this for sine.

Working with tangents is not much different from working with sines and cosines. The only difference, really, is that the tangent does not have an amplitude. The canonical form of the equation remains the same:

`f(x) = atan(b(x-phi))+c`

Here, *a *determines the vertical stretch of the graph, *b* determines the horizontal stretch, `phi` determines the phase shift, and *c* determines the vertical shift.

Because we are concerned mainly about the period and vertical shift, we will look into those more exactly. The vertical shift is exactly *c*, no equations or change to worry about. The period has a different formula:

`T = (2pi)/b`

So, let's solve for the desired *b *now:

`(2pi)/5 = (2pi)/b`

`b=5`

Given our vertical shift of 2, it is clear that *c* = 2. Because we are not given any other information, we will set *a* = 1 and `phi` = 0. This gives us **our final equation:**

`f(x) = tan(5x)+2`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.