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

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

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Answer 1 · 2016-09-16T01:59:05

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

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

Explanation:

$tan θ$ $tan theta$ is periodic with period $π$ $pi$ . So,

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

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

Make a graph for one period, $x \in (0, \frac{π}{4}) = (0^{o}, 45^{o})$ $x in (0, pi/4)=(0^o, 45^o)$ 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 $y = - cot 4 x$ $y = -cot 4x$ follows..

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

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

electronic graphic device.

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

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

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Explanation:

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