# How do you graph y=-cot(4x)?

Sep 16, 2016

A short Table for $y = - \cot 4 x$ for one period #x in (0^o, 45^o):.

$\left(x , y\right) : \left(0 , - \infty\right) \left({15}^{o} , - \sqrt{3}\right) \left({22.5}^{o} , 0\right) \left({30}^{o} , \sqrt{3}\right) \left({55}^{o} , \infty\right)$,

#### Explanation:

$\tan \theta$ is periodic with period $\pi$. So,

$\tan 4 x$ is periodic with perid $\frac{\pi}{4}$

Note infinite discontinuity at $x = 0 \mathmr{and} x = \frac{\pi}{4} = {45}^{o}$.

Make a graph for one period, $x \in \left(0 , \frac{\pi}{4}\right) = \left({0}^{o} , {45}^{o}\right)$ and move it

laterally on either side, successively, on either side, to get the graph

on both left and right sides.

A short Table for $y = - \cot 4 x$ follows..

$\left(x , y\right) : \left(0 , - \infty\right) \left({15}^{o} , - \sqrt{3}\right) \left({22.5}^{o} , 0\right) \left({30}^{o} , \sqrt{3}\right) \left({55}^{o} , \infty\right)$,

Now, you can make your graph, with hand or using an

electronic graphic device.

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