Can You Graph and Determine the Domain of f(x,y)=√(x²-1)?

[gokugreene]

I am trying to find and graph the level curve f(x,y)=\sqrt{x^2-1} that passes throught the point (0,1), as well as its domain and range.

I am not sure if my reasoning is right, so let me know if I got the wrong idea.

For the graph I have x = 1 which is independent of y and is just a vertical line. Is this correct?

Would the domain be (-\infty, -1]\cup[1,\infty) or [1,\infty) ? Because \sqrt{x^2-1} = \sqrt{x-1}\sqrt{x+1}
I'm confused.

Range: [0,\infty)

any help would be greatly appreciated

Thanks

Update: this a function of two variables

 

[G01]

That sounds correct to me. I just graphed it and it also looks correct, Unless I am missing something.

 

[mezarashi]

If we have y = \sqrt{x^2 - 1}, then x cannot exist on the real domain for any value: \mid x \mid \leq 1.

 

[gokugreene]

I am looking for the domain, range, and graph of this level curvef(x,y)=\sqrt{x^2-1}

I have the range, but the domain and graph I am unsure of.
For the graph, I graphed on my paper x=1, you plug in one, z = 0 and x = 1 and y can be anything.
Domain: I am unsure of but yes \mid x \mid \leq 1
But if I plug say x = -1 into \sqrt{x-1}\sqrt{x+1} I am going to get \sqrt{-2}\sqrt{2}