How do you graph $y = 2 cot (2 x + \frac{π}{3}) + 1$ $y=2cot(2x+pi/3)+1$ ?

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Answer 1 · 2017-01-02T10:02:28

It is periodic, with period $\frac{π}{2}$ $pi/2$ . The graph is for $12_{+}$ $12_+$ periods.
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For $cot (2 x + \frac{π}{3}) + 1$ $cot(2x+pi/3)+1$ , the period is $\frac{π}{2}$ $(pi)/2$ . The asymptotes are given by

$2 x + \frac{π}{3}$ $2x+pi/3$ = a multiple of $π = k π$ $pi=kpi$ , giving $x =(kpi-pi/3)/2=((3k-1)/6)pi, k = 0, +-1, +-2, +-3, ...$

For one period, $x in (-pi/6, pi/3)$ ,

there are two terminal asymptotes

$uarr x=-pi/6 darr and uarr x = pi/3 darr$ .

graph{(y-1)tan(2x+1.047)-1=0 [-10, 10, -5.21, 5.21]}

How do you graph y=2cot(2x+pi/3)+1y=2cot(2x+π3)+1y=2cot(2x+pi/3)+1?