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Anachronism

  • 4 years ago

Identify the Horizontal Asymptotes of g(x) = (3x+2)/((x^2)+4)^(1/2))

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  1. Anachronism
    • 4 years ago
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    |dw:1321422281732:dw|

  2. tdabboud
    • 4 years ago
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    The answer is at 3 and -3

  3. Anachronism
    • 4 years ago
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    how would I go about finding those answers? :/ my teacher hasn't taught us how to find it with radicals in the denominator, so I'm lost.

  4. tdabboud
    • 4 years ago
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    well you look at the ratio of the exponents. The numerator is 3x and the denomintor will just be x. it is x^(2*1/2) = x so the ratio is just 3x/x. As x goes to infinity you get a horizontal asymptote of 3 As x goes to negative infinity you get a horizontal asymptote of -3

  5. Anachronism
    • 4 years ago
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    Thank you so much!

  6. tdabboud
    • 4 years ago
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    understand? I know it probably sounds very confusing

  7. Anachronism
    • 4 years ago
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    I kind of get it, but would it just be 3, then because if I substitute a large negative number for x it is a negative over a negative, and I would still get a positive 3

  8. Asanka
    • 4 years ago
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    if you know how to differentiate it , then differentiate it w.r.t x and equate the answer to 0

  9. tdabboud
    • 4 years ago
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    i dont think she is at that type of math yet, that will be too complex at this point

  10. Asanka
    • 4 years ago
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    yeah

  11. Asanka
    • 4 years ago
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    sorry guys i am new to this so i do not know your standards

  12. tdabboud
    • 4 years ago
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    No the answer will be -3 as you approach -infinity because you need to take into account the ratio 3/1

  13. tdabboud
    • 4 years ago
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    the top will outdo the bottom in this case, a very large negative number times 3 will approach -3 faster then the denomnator will reduce it to 3. Understand? it is confusing but take a second to think about it

  14. Anachronism
    • 4 years ago
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    It's alright, thank you for trying to explain it to me Asanka. I think I just visualized it, and it seems to make sense if the numerator is going three times as fast. Thanks a lot for all your help :)

  15. tdabboud
    • 4 years ago
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    exactly you got it! anytime

  16. tsegu_k
    • 4 years ago
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    I am attaching the solution hoping it will help

  17. tdabboud
    • 4 years ago
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    Nice work! very neat.. but you can just look at the ratio of like exponents to make it a little easier, but either way nice job

  18. tsegu_k
    • 4 years ago
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    Thank you very much

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