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yuki

  • 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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  1. anonymous
    • 5 years ago
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    Do you mean algebraically?

  2. Yuki
    • 5 years ago
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    exactly

  3. Yuki
    • 5 years ago
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    I tried implicit differentiation but I couldn't work it out.

  4. anonymous
    • 5 years ago
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    You solve for y, then check out what happens as you move through x. You check for where the function collapses.

  5. anonymous
    • 5 years ago
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    I assume you're plotting y vs x?

  6. Yuki
    • 5 years ago
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    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.

  7. Yuki
    • 5 years ago
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    woops, I meant -10

  8. anonymous
    • 5 years ago
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    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.

  9. Yuki
    • 5 years ago
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    I appreciate it

  10. anonymous
    • 5 years ago
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    \[y=\frac{x \pm \sqrt{x^2-4(10-x^3)}}{2}\]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.

  11. anonymous
    • 5 years ago
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    For x < about 2.07, the function shouldn't exist. Apart from that, but there are no vertical asymptotes...

  12. Yuki
    • 5 years ago
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    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.

  13. anonymous
    • 5 years ago
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    So did that help? Was your question answered?

  14. anonymous
    • 5 years ago
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    Do you have plotting software?

  15. Yuki
    • 5 years ago
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    I just graphed it on my graphing calculator TI-86 and it did all the job I wanted it to do.

  16. anonymous
    • 5 years ago
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    http://www.geogebra.org/cms/

  17. anonymous
    • 5 years ago
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    It's free ^^

  18. Yuki
    • 5 years ago
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    Thanks.

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