# How do you find the amplitude, period, asymptotes (from 0-2pi), shifts, and the min/max of the following trig functions Y= 2tan ( theta - pi/4)?

Nov 9, 2017

See below.

#### Explanation:

The tan function has no max/min value and consequently no amplitude. This is because:

as $\theta \to \frac{\pi}{2} , \tan \left(\theta\right) \to \infty$ and is undefined

The shift is $\frac{\pi}{4}$, this can be seen from $2 \tan \left(\theta - \frac{\pi}{4}\right)$ subtracting $\frac{\pi}{4}$ shifts the axis to the left, or the graph to the right. This can therefore be viewed either way.

Vertical asymptotes will occur where the function is undefined. This can be calculated using:

$\theta - \frac{\pi}{4} = \frac{\pi}{2}$ and $\theta - \frac{\pi}{4} = \frac{3 \pi}{2}$ ( this is for the given interval )

$\theta - \frac{\pi}{4} = \frac{\pi}{2} \implies \theta = \textcolor{b l u e}{\frac{3 \pi}{4}}$

$\theta - \frac{\pi}{4} = \frac{3 \pi}{2} \implies \theta = \textcolor{b l u e}{\frac{7 \pi}{4}}$

Period.

This can be found from two values where the tangent is zero. i.e.

$\theta - \frac{\pi}{4} = 0 \implies \theta = \frac{\pi}{4} \mathmr{and} \theta = \frac{5 \pi}{4}$

$\frac{5 \pi}{4} - \frac{\pi}{4} = \pi$

So:

Amplitude Doesn’t exist due to being undefined.

Period $\textcolor{w h i t e}{\cdot} \pi$

max/min Doesn’t exist due to being undefined.

shift $\textcolor{w h i t e}{\cdot} \frac{\pi}{4}$

Asymptotes At $\frac{3 \pi}{4} \mathmr{and} \frac{7 \pi}{4}$

Graph: 