How do you graph $y=tan(x+(7pi)/6)$ ?

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Answer 1 · 2016-11-23T09:48:36

graph{y-tan(x+3.665)=0 [-10, 10, -5, 5]}

$y=tan(pi+(pi/6+x))=tan(x+pi/6)$

The period of y is $pi$ . !n the graph, the graph for one period

repeats in a cycle.

A convenient choice of one period is $x in (-2/3pi, pi/3)$ . $pi=3.14$

in the graph.

As $x to pi/3, y to oo$ and as$ x to -2/3pi, y to -oo#.

$(-pi/6, 0)$ is in the middle of the graph, where the tangent

crosses the curve. It is called a point of inflexion. Here, y'' = 0.

How do you graph y=tan(x+(7pi)/6)y=tan(x+(7pi)/6)?