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

Oct 15, 2016

Period is $2 \pi$. Phase is $\frac{\pi}{4}$ and vertical shift id 3 units of distance. As y varies continuously in $\left(- \infty , \infty\right)$, within a period, the question of finding amplitude does not arise. -

#### Explanation:

The period of $\tan \theta$ is $P = \pi$.

The period of $\tan \left(k \theta - \phi\right) + b$ is $\frac{\pi}{k}$'

Here, $y = \tan \left(\theta - \frac{\pi}{4}\right) + 3$. So,

the period $P = \frac{\pi}{\frac{1}{2}} = 2 \pi$.

Phase shift $\phi = \frac{\pi}{4}$.

Vertical shift is 3.

In brief, this could be obtained

from the graph of $y = \tan \left(\frac{x}{2}\right)$

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

in the y+ direction through 3 units.

Within one period, say $\left(- \pi , \pi\right)$, 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 $\infty$, because reaching

the crest (zenith) at $\infty$ or the lowest point (nadir) at $- \infty$ is unreal.