How do you graph $y = - 2 csc (2 x) + 1$ $y=-2csc(2x)+1$ ?

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Answer 1 · 2016-11-18T02:14:53

Using the Socratic graphic facility, I have inserted a graph.

The period of y is $\frac{2 π}{2} = π$ $(2pi)/2=pi$ . It is convenient to choose one

period as #x in (-pi/2, pi/2). Note that y has infinite discontinuity at

the ends and the middle x = 0.

A Table for making the graph:

$(x, y) : (3.31, - \frac{π}{3}) (3, - \frac{π}{4}) (3.1, - \frac{π}{6}) (\infty, 0)$ $(x, y): (3.31, -pi/3) (3, -pi/4) (3.1, -pi/6) (oo, 0)$

$(- 1.3, \frac{π}{6}) (- 1, \frac{π}{4}) (- 1.3, \frac{π}{3})$ $(-1.3, pi/6) (-1, pi/4) (-1.3, pi/3)$

The part of the inserted graph that braces the y-axis is the graph for

this period.

ygraph{(y-1) sin (2x)+2=0 [-9.91, 10.09, -3.68, 6.32]}

How do you graph y=−2csc(2x)+1y=-2csc(2x)+1?