anonymous
  • anonymous
Identify the Horizontal Asymptotes of g(x) = (3x+2)/((x^2)+4)^(1/2))
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
The answer is at 3 and -3
anonymous
  • anonymous
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.

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anonymous
  • anonymous
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
anonymous
  • anonymous
Thank you so much!
anonymous
  • anonymous
understand? I know it probably sounds very confusing
anonymous
  • anonymous
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
anonymous
  • anonymous
if you know how to differentiate it , then differentiate it w.r.t x and equate the answer to 0
anonymous
  • anonymous
i dont think she is at that type of math yet, that will be too complex at this point
anonymous
  • anonymous
yeah
anonymous
  • anonymous
sorry guys i am new to this so i do not know your standards
anonymous
  • anonymous
No the answer will be -3 as you approach -infinity because you need to take into account the ratio 3/1
anonymous
  • anonymous
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
anonymous
  • anonymous
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 :)
anonymous
  • anonymous
exactly you got it! anytime
anonymous
  • anonymous
I am attaching the solution hoping it will help
anonymous
  • anonymous
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
anonymous
  • anonymous
Thank you very much

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