yuki
  • yuki
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Do you mean algebraically?
yuki
  • yuki
exactly
yuki
  • yuki
I tried implicit differentiation but I couldn't work it out.

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More answers

anonymous
  • anonymous
You solve for y, then check out what happens as you move through x. You check for where the function collapses.
anonymous
  • anonymous
I assume you're plotting y vs x?
yuki
  • yuki
by 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.
yuki
  • yuki
woops, I meant -10
anonymous
  • anonymous
Well, 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.
yuki
  • yuki
I appreciate it
anonymous
  • anonymous
\[y=\frac{x \pm \sqrt{x^2-4(10-x^3)}}{2}\]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^2-4(10-x^3), is less than zero.
anonymous
  • anonymous
For x < about 2.07, the function shouldn't exist. Apart from that, but there are no vertical asymptotes...
yuki
  • yuki
it 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
  • anonymous
So did that help? Was your question answered?
anonymous
  • anonymous
Do you have plotting software?
yuki
  • yuki
I just graphed it on my graphing calculator TI-86 and it did all the job I wanted it to do.
anonymous
  • anonymous
http://www.geogebra.org/cms/
anonymous
  • anonymous
It's free ^^
yuki
  • yuki
Thanks.

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