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anonymous
 5 years ago
x^3 + xy y^2 = 10.
is there any way to find the vertical asymptote by hand?
I can do it graphically, but it looks like a problem that is solvable by hand.
anonymous
 5 years ago
x^3 + xy y^2 = 10. is there any way to find the vertical asymptote by hand? I can do it graphically, but it looks like a problem that is solvable by hand.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you mean algebraically?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I tried implicit differentiation but I couldn't work it out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You solve for y, then check out what happens as you move through x. You check for where the function collapses.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I assume you're plotting y vs x?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0by letting \[2y^2+xy+(x^3+10) = 0 \] and solving for y using the quadratic formula gave me a curve and I was able to figure out the solution, but if there is any way to do it algebraically it would be great. I tried to differentiate the explicit function, too but that took a very long time and I doubt that it is the natural way.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, you solve for y as per quadratic and then look at the result to see if there are points where parts of the function will be undefined. Let me do the problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=\frac{x \pm \sqrt{x^24(10x^3)}}{2}\]will be a function for y for either the positive or negative square root. The only place this should fall apart is for those x's such that the radicand x^24(10x^3), is less than zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For x < about 2.07, the function shouldn't exist. Apart from that, but there are no vertical asymptotes...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it seems like the problem looks easy but the answer is not. Thanks for the help though, it feels good to have other people making same conclusions as me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So did that help? Was your question answered?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you have plotting software?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I just graphed it on my graphing calculator TI86 and it did all the job I wanted it to do.
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