How do you graph, identify the domain, range, and asymptotes for #y=-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 #x#'s #y# cannot take on.

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

Similary the range is what #y# values is possible for the function to output. Notice how #-1<=sin2x<=1#, therefore #1/{sin2x}# can never reach values between #-1, 1# such as #1/2# or #-1/2#.
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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