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anonymous

  • 5 years ago

What are the vertical asymptotes, horizontal asymptotes, slant asymptotes, and the holes of this equation: (x^2-x+1)/(x-3)

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  1. anonymous
    • 5 years ago
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    you can't factor x^2-x+1, so there will be no removable discontinuities (so-called "holes"), so x=3 will be the vertical asymptote. take the limit as x goes to infinity, and you get y=1 as the horizontal asymptote. I'm not so sure about the slant one.

  2. anonymous
    • 5 years ago
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    wait, are you sure that the horizontal asymptote is 1 because i thought that the horizontal asymptote was y=none.

  3. anonymous
    • 5 years ago
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    yeah, nvm. y goes to infinity.

  4. anonymous
    • 5 years ago
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    to find a horizontal asymptote, do polynomial long division. y=x+2 should be the slant asymptote.

  5. anonymous
    • 5 years ago
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    ok thank you! Do you think you could find the holes and slant asymptotes for this equation:(x^2-9x+20)/(2x^2-8x)

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