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anonymous

  • one year ago

Find the horizontal asymptote of f(x)= x^2+3x-2/x-2

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  1. anonymous
    • one year ago
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    I appreciate any help

  2. anonymous
    • one year ago
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    You sure that is 3x - 2 in numerator and not 3x + 2 ??

  3. anonymous
    • one year ago
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    Yes it's in the numerator

  4. anonymous
    • one year ago
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    As you can see, degree of numerator is greater than degree of denominator, you will get Slant Asymptote...

  5. anonymous
    • one year ago
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    Ok

  6. anonymous
    • one year ago
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    Do you know the Long Division Method??

  7. anonymous
    • one year ago
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    No :(

  8. anonymous
    • one year ago
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    You have to just divide numerator by denominator.. As you do simple division and then find out the quotient..

  9. anonymous
    • one year ago
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    Ok so divide x^2+3x-2 by x-2?

  10. anonymous
    • one year ago
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    yep...

  11. anonymous
    • one year ago
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    Would it be the same as using synthetic division?

  12. anonymous
    • one year ago
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    |dw:1437159263381:dw|

  13. anonymous
    • one year ago
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    yes, you can do synthetic division too..!!!

  14. anonymous
    • one year ago
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    So, \(y = x + 5\), this is the Slant Asymptote in this case..

  15. anonymous
    • one year ago
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    Where degree of numerator is greater than degree of denominator, there is no horizontal asymptote but there is slant asympote, as you can find that by \(y = quotient\), for that just use Long Division and find the quotient..

  16. anonymous
    • one year ago
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    So I got x+1 - (4/x+2) :/

  17. anonymous
    • one year ago
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    Wait so the answer is none?

  18. anonymous
    • one year ago
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    yes None is the answer for this.. :P

  19. anonymous
    • one year ago
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    There is no horizontal asymptote? What a trick question 😅

  20. anonymous
    • one year ago
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    Thanks

  21. anonymous
    • one year ago
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    I just told how to find slant asymptote equation if you do not have Horizontal asymptote..

  22. anonymous
    • one year ago
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    Oh on :) thanks

  23. anonymous
    • one year ago
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    Ok :D

  24. anonymous
    • one year ago
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    Thanks again bye~

  25. anonymous
    • one year ago
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    See, horizontal asymptote means you will get line parallel to x-axis or where y = 0, but if asymptote is not horizontal, then it must be inclined and y \(\ne\) 0 there, so how to find y there, I just told you that.. :)

  26. anonymous
    • one year ago
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    \(\dagger\) ..

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