Question

How do you find the amplitude, period, and shift for `y=tan(theta/2-pi/4)+3`?

Answer 1

Period is `2pi`. Phase is `pi/4` and vertical shift id 3 units of distance. As y varies continuously in `(-oo, oo)`, within a period, the question of finding amplitude does not arise. -

Explanation:

The period of `tan theta` is `P=pi`.

The period of `tan (ktheta-phi)+b` is `pi/k`'

Here, `y = tan (theta-pi/4)+3`. So,

the period `P = pi/(1/2)=2pi`.

Phase shift `phi= pi/4`.

Vertical shift is 3.

In brief, this could be obtained

from the graph of `y = tan (x/2)`

by moving it in x+-direction through `phi` units and then shifting it

in the y+ direction through 3 units.

Within one period, say `(-pi, pi)`, continuous #y in (-oo, oo)..

So, the question of finding global/local mini/max values is

irrelevant.Indeed, this applies to the finding of the related

amplitude as well,

I welcome a graphical depiction by an interested reader.

I do not want to say that the amplitude is `oo`, because reaching

the crest (zenith) at `oo` or the lowest point (nadir) at `-oo` is unreal.