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Math Help - End-Behavior Asymptote

  1. #1
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    End-Behavior Asymptote

    Hi, guys. This is my first thread on here, so bare with me if I make some mistakes.
    I'm doing these end behavior problems and need some help...

    1. Find the end-behavior asymptote(s) in the graph of f(x)=sqrt(4x^2-9)/(x+5).
    I can't figure out a way to divide the square root.

    2. Find the end-behavior asymptote(s) in the graph of f(x)=(4x-15x^2)/(x^2-121).
    This is how I solved for it:
    (4x-15x^2)/(x^2-121)=-15+(4x-1815)/(x^2-121)
    y=-15
    I'm not sure if I did it right. I just followed example problems.

    3. Same for f(x)=(x^2-5)/(x-3).
    My answer this time was y=4+((x+4)/(x-3)), so if I did the last problem correctly, the asymptote should be y=4.

    4. Find the polynomial end-behavior asymptote in the graph of f(x)=(2x^3+6x-1)/(x+2).
    My answer was y=2x^2-4x+14

    If you could explain the process, that would help a lot. Thanks!
    Last edited by iyppxstahh; February 1st 2010 at 02:34 AM.
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  2. #2
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    Quote Originally Posted by iyppxstahh View Post
    Hi, guys. This is my first thread on here, so bare with me if I make some mistakes.
    I'm doing these end behavior problems and need some help...

    1. Find the end-behavior asymptote(s) in the graph of f(x)=sqrt(4x^2-9)/(x+5).
    I can't figure out a way to divide the square root.
    \frac{\sqrt{4x^2- 9}}{x+5}= \frac{\sqrt{4x^2- 9}}{\sqrt{(x+5)^2}}= \sqrt{\frac{4x^2- 9}{x^2+ 10x+ 25}}.
    Now divide 4x^2- 9 by x^2+ 10x+ 25.

    2. Find the end-behavior asymptote(s) in the graph of f(x)=(4x-15x^2)/(x^2-121).
    This is how I solved for it:
    (4x-15x^2)/(x^2-121)=-15+(4x-1815)/(x^2-121)
    y=-15
    I'm not sure if I did it right. I just followed example problems.
    Yes, since the remaining fraction has lower degree in the numerator than in the denominator, it will go to 0 and so f(x) will go to -15.

    3. Same for f(x)=(x^2-5)/(x-3).
    My answer this time was y=4+((x+4)/(x-3)), so if I did the last problem correctly, the asymptote should be y= 4.
    No, you did not divide correctly. Since the numerator is of degree 2 and the denominator of degree 1, you would expect a quotient of degree 2-1= 1. x-3 divides into x^2- 5 with quotient x+ 3 and remainder 4: y= x+ 3 + 4/(x-3). The (slant) asymptote is y= x+ 3.

    4. Find the polynomial end-behavior asymptote in the graph of f(x)=(2x^3+6x-1)/(x+2).
    My answer was y=2x^2-4x+14
    You appear to have divided incorrectly again. Could you show exactly how you got that?

    If you could explain the process, that would help a lot. Thanks!
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  3. #3
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    No, you did not divide correctly. Since the numerator is of degree 2 and the denominator of degree 1, you would expect a quotient of degree 2-1= 1. x-3 divides into x^2- 5 with quotient x+ 3 and remainder 4: y= x+ 3 + 4/(x-3). The (slant) asymptote is y= x+ 3.
    Oh, thank you. I accidentally switched the quotient and the remainder.

    You appear to have divided incorrectly again. Could you show exactly how you got that?
    Well, the answer was was y=2x^2-4x+14 R-29, and I took out the remainder...I haven't figured out how to post equations yet, but I've checked and re-checked the problem, and it is divided correctly. Isn't it?

    Thank you, so much!
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