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onegirl

  • 2 years ago

Determine all horizontal, slant, and vertical asymptotes. For each vertical asymptote, determine whether f(x) -> infinity sign of f(x) -> negative infinity sign on either side of the asymptote.

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  1. onegirl
    • 2 years ago
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    f(x) = x^2/4 - x^2

  2. satellite73
    • 2 years ago
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    is it \[\frac{x^2}{4-x^2}\]?

  3. onegirl
    • 2 years ago
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    yes

  4. satellite73
    • 2 years ago
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    ok so for the vertical asymptotes, set the denominator equal to zero and solve for \(x\)

  5. onegirl
    • 2 years ago
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    ok

  6. satellite73
    • 2 years ago
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    you get \[4-x^2=0\] or \[(2-x)(2+x)=0\] and the solutions are \(x=-2\) or \(x=2\) those are your vertical asymptotes (there are two of them)

  7. onegirl
    • 2 years ago
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    ok

  8. satellite73
    • 2 years ago
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    for the horizontal asymptote, note that the numerator and denominator have the same degree (both are degree 2) so it is the ratio of the leading coefficients

  9. onegirl
    • 2 years ago
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    okay

  10. satellite73
    • 2 years ago
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    the leading coefficient of \(x^2\) is 1 and the leading coefficient of \(4-x^2\) is \(-1\) therefore the horizontal asymptote is \(y=-1\)

  11. satellite73
    • 2 years ago
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    that is in the case where the degrees are the same. there is no slant asymptote for there to be a slant asymptote , the degree of the numerator would have to be one more than the degree of the denominator

  12. onegirl
    • 2 years ago
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    okay thx

  13. satellite73
    • 2 years ago
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    yw

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