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How to get the horizontal asymptote.

  1. Sep 18, 2004 #1
    f(x)=(9x^2-36)/(x^2-9)

    I forgot how to get the hotizontal asymptote. Is it f(x)=0?
    When I do that, there is no way to solve it

    Also, how do I find the intervals for when f is increasing?

    I'm not asking for the answer, but the equation, I totally forgot

    thanks
     
  2. jcsd
  3. Sep 18, 2004 #2

    arildno

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    Horizontal (right-hand side) asymptote: [tex]\lim_{x\to\infty}f(x)[/tex]
    Horizontal (left-hand side) asymptote: [tex]\lim_{x\to-\infty}f(x)[/tex]

    What does the derivative of a function tell you about the function's behaviour?
     
  4. Sep 18, 2004 #3
    I'm even more confused. I know that getting the vertical asymptote is setting the denominaor equal to zero and solving for it but what is it for the horizontal?
     
  5. Sep 18, 2004 #4

    Pyrrhus

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    are you familiar with limits?
     
  6. Sep 18, 2004 #5

    arildno

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    What value L do f(x) approach when x goes to infinity?
    The line g(x)=L (parallell to the x-axis!) is called the horizontal asymptote to f(x)
     
  7. Sep 18, 2004 #6
    So how would I find the equation for each horizontal asymptote of: f(x)=(9x^2-36)/(x^2-9)
     
  8. Sep 18, 2004 #7

    arildno

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    Think this way:
    Let x be a huge positive number.
    Then, surely, 9x^2 must be a lot larger than 36.
    Similarly, x^2 must be a lot larger than 9.

    Hence, you don't make a big mistake by setting:
    [tex]9x^{2}-36\approx9x^{2}[/tex]
    (the relative error is tiny)
    Similarly:
    [tex]x^{2}-9\approx{x^{2}}[/tex]
    Hence:
    [tex]f(x)=\frac{9x^{2}-36}{x^{2}-9}\approx\frac{9x^{2}}{x^{2}}=9[/tex]
    (for huge positive x's)
    Hence, the horizontal right-hand asymptote is L(x)=9
     
  9. Sep 18, 2004 #8
    I understand your process of thinking however, is there a specific formula that can calculate that number?

    I know that to obtain the vertical asymptote, it's by setting the denominator to zero.

    e.g. f(x)=(9x^2-36)/(x^2-9)
    x^2-9=0
    x=+3,-3

    so the equations for the vertical asymptote is x=3 and x=-3

    how do I do this with the horizontal asymptote?
     
  10. Sep 18, 2004 #9

    arildno

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    No, there is in general no foolproof method in determining limit values (in your case, to get the horizontal asymptotes)
    There exist a rigourous method which in principle tells you whether a chosen number is a limit or not (i.e, if you've made a right (or wrong!) guess)

    The reasoning I gave you, however, is sufficient to determine the horizontal assymptotes you'll meet.
    So here's a method, if you like:
    1. Think of x as a huge number.
    Practical meaning:
    If x appears in a sum (or difference) with a constant, discard that constant.
    Further, retain only the highest "power" of x in a sum where x appears in multiple terms.
    For example: [tex]3x^{2}-7x+14\approx3x^{2}[/tex] when x is huge?
    Why?
    [tex]\frac{3x^{2}}{7x}=\frac{3}{7}x[/tex] which is huge since x is huge.
    That is the magnitude (always positive!) of the term "3x^2" is much bigger than the magnitude of "-7x".

    2. These simplifications should be enough to find the asymptote.
     
  11. Sep 18, 2004 #10
    I understand the idea now. My second question addressed finding the intervals for when f is increasing.

    f(x)=(9x^2-36)/(x^2-9)

    how do I find the intervals for when f is increasing?
     
  12. Sep 18, 2004 #11

    arildno

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    I quote myself:
    In particular, how is information of whether a function is increasing or decreasing given by the values of the derivative?
     
  13. Sep 18, 2004 #12

    Tide

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    Analyze the behavior of the derivative of the function. If it's positive then the function is increasing.
     
  14. Sep 18, 2004 #13
    So find the derivative of: f(x)=(9x^2-36)/(x^2-9)

    then find the intervals of when it is increasing and that will tell me when the function is increasing. Correct?
     
  15. Sep 18, 2004 #14

    Tide

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    Yes, the function is increasing when f '(x) > 0.
     
  16. Sep 18, 2004 #15

    arildno

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    NO!!
    You find the intervals where f'(x) is greater than zero, not the intervals where f'(x) is increasing..
     
  17. Sep 18, 2004 #16
    f`(x)=(90x)/(x^2-9)^2

    this is positive from [0,3) and (3,infinity)

    is this correct?
     
  18. Sep 18, 2004 #17

    arildno

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    Yes; assuming your expression for f'(x) is correct.
     
  19. Sep 18, 2004 #18
    I actually cheated and used a graphing calculator to solve for when (90x)/(x^2-9)^2 is positive. How would I solve it without a graphing calculator?
     
  20. Sep 18, 2004 #19

    arildno

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    Well, the denominator is always non-negative (why?)
    Hence, only the sign of the numerator is of importance (why?)
     
  21. Sep 18, 2004 #20
    okay, I got a good understanding of it all. Thanks for the help & advice.
     
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