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haleyelizabeth2017

  • one year ago

Determine the equation of the horizontal asymptotes, if any, of the function. f(x)=(2x+1)/(x+1)

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  1. haleyelizabeth2017
    • one year ago
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    \[f(x)=\frac{ 2x+1 }{ x+1 }\]

  2. Rizags
    • one year ago
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    are you allowed to use limits?

  3. haleyelizabeth2017
    • one year ago
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    I'm not sure.

  4. Rizags
    • one year ago
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    what class is this for?

  5. haleyelizabeth2017
    • one year ago
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    PreCalculus

  6. Rizags
    • one year ago
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    ok then maybe some limits are allowed

  7. Rizags
    • one year ago
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    are you familiar with limits?

  8. haleyelizabeth2017
    • one year ago
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    Nope. But...the book gives us two methods we can use. Solving for x in terms of y is one method, and divide the numerator and denominator by the highest power of x is the second.

  9. Rizags
    • one year ago
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    yes use the second

  10. Rizags
    • one year ago
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    multiply the whole thing by \[\large \frac{\frac{1}{x}}{\frac{1}{x}}\]

  11. anonymous
    • one year ago
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    the numerator and denominator are both polynomials the degrees are the same (they are both of degree 1) the horizontal asymptotes is the ratio of the leading coefficients

  12. anonymous
    • one year ago
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    in your example it is \[y=\frac{2}{1}=2\]

  13. anonymous
    • one year ago
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    it always works this way if the degrees are the same if they are not, then it is different

  14. haleyelizabeth2017
    • one year ago
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    Okay, sorry...was afk for a moment there

  15. anonymous
    • one year ago
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    if the degree of the denominator is larger, then it is \(y=0\) if the degree of the numerator is larger, then there is no horizontal asymptote

  16. haleyelizabeth2017
    • one year ago
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    Okay, good to know. Thank you

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