Graph y=csc(2x)

Problem

y=csc(2*x)

Solution

1. Identify the parent function and its properties. The function y=csc(2*x) is the reciprocal of y=sin(2*x)

2. Determine the period of the function. The period of sin(b*x) or csc(b*x) is (2*π)/|b|

Period=(2*π)/2=π

3. Locate the vertical asymptotes by finding where the sine function is zero. For sin(2*x)=0 we have 2*x=n*π for any integer n

x=(n*π)/2

Asymptotes:x=0,±π/2,±π,…

4. Find the relative extrema by looking at the peaks and valleys of y=sin(2*x) The maximums of the sine curve occur at 2*x=π/2+2*n*π and minimums at 2*x=(3*π)/2+2*n*π

Relative Minimums (where *y=1*):x=π/4+n*π

Relative Maximums (where *y=−1*):x=(3*π)/4+n*π

5. Sketch the graph by drawing the vertical asymptotes, plotting the local extrema, and drawing the U-shaped curves that approach the asymptotes.

Final Answer

y=csc(2*x)

The graph has a period of π vertical asymptotes at x=(n*π)/2 and relative extrema at (π/4,1) and ((3*π)/4,−1)

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