How do you graph, identify the domain, range, and asymptotes for y=-2csc2x-1y=−2csc2x−1?

1 Answer
sf
Jun 15, 2018

too long..

Explanation:

I won't use the whole function but a similar one, you can explore urs by lowering it 1 step and the negative version.

Find the domain by checking what xx's yy cannot take on.

As csc2x=1/{sin(2x)}csc2x=1sin(2x), yy is not defined when sin2x=0sin2x=0. So the domain is all xx except n\pi/2, n in Znπ2,n∈Z.

Similary the range is what yy values is possible for the function to output. Notice how -1<=sin2x<=1−1≤sin2x≤1, therefore 1/{sin2x}1sin2x can never reach values between -1, 1−1,1 such as 1/212 or -1/2−12.
So, y in (-\infty, infty)∖(-1,1)

The asymptotes is also closely related to the domain, when y is divided by 0 vertical asymptotes appear. As earlier this is when x=n\pi/2, n in Z

Horisontal asymptotes does not exists as lim_{x->+-infty} y(x) does not exist.

How would you graph this?
We aquired a lot of information of y so far, so you could start sketching the asymptotes, and by noticing its periodicity. To continue you should plug in known values for csc2x.

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