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

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 $\csc 2 x = \frac{1}{\sin \left(2 x\right)}$, $y$ is not defined when $\sin 2 x = 0$. So the domain is all $x$ except $n \setminus \frac{\pi}{2} , n \in Z$.

Similary the range is what $y$ values is possible for the function to output. Notice how $- 1 \le \sin 2 x \le 1$, therefore $\frac{1}{\sin 2 x}$ can never reach values between $- 1 , 1$ such as $\frac{1}{2}$ or $- \frac{1}{2}$.
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 $x = n \setminus \frac{\pi}{2} , n \in Z$

Horisontal asymptotes does not exists as ${\lim}_{x \to \pm \infty} y \left(x\right)$ 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 $\csc 2 x$.