How do you graph $y = 6 csc (3 x + \frac{2 π}{3}) - 2$ $y=6csc(3x+(2pi)/3)-2$ ?

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Answer 1 · 2017-02-19T13:48:15

See Socratic graph and explanation.

y is of period $\frac{2 π}{3}$ $(2pi)/3$ .

As $| csc (θ) | \geq 1, | y + 2 | \geq 6$ $|csc(theta)|>=1, |y+2|>=6$ , giving $y \geq 4 and \leq - - 8$ $y >=4 and <=--8$

The graph has vertical asymptotes, when $csc (3 x + 2 \frac{π}{3}) = \infty$ $csc(3x+2pi/3)=oo$ , giving

$x= 1/3(kpi-2pi/3), k =0, +-1, +-2, +-3, ...$

For that matter, half period $1/3pi$ is the difference between two

consecutive asymptote x values.

The graph for one period #x in(-2/9pi, 4/9pi) is also included.

graph{(y+2)sin(3x+2pi/3)-6=0 [-40, 40, -21, 19]}

graph{(y+2)sin(3x+2pi/3)-6=0 [-.7, 1.4, -14.2, 12]}

Not-to-scale graph to reveal spacing $pi/3$ = 1.05, for the three

asymptotes at $x = -.7, .35 and 1.4$

How do you graph y=6csc(3x+(2pi)/3)-2y=6csc(3x+2π3)−2y=6csc(3x+(2pi)/3)-2?