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Homework Help: Find the horizontal asymptote

  1. Jul 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the horizontal asymptote(if there is one) using the rule for determining the horizontal asymptote of a rational function for (x^2+x-12)/ (x^2 -4)


    2. Relevant equations



    3. The attempt at a solution

    the degree of the numerator and denominator are both 2.

    Y=(An)/(Bn)
    Y=1/1
    Y=1

    When I do the math, the horizontal asymptote is the line y=1.

    However when I graph this equation on a TI- 84 plus graphing calculator, if i use the trace, or table functions, the part of the graph that does not appear to cross the line y=1, does. Why is this?
     
    Last edited: Jul 8, 2010
  2. jcsd
  3. Jul 8, 2010 #2

    Mark44

    Staff: Mentor

    No, the degree of the numerator and denominator is 2. How did you get 1?
    The whole idea about a horizontal asymptote is to describe behavior of the function for very large x or very negative x. For very large (or very negative) values, the graph of the function won't cross the asymptote. For values of x that are relatively close to 0, the graph can cross the asymptote.
     
  4. Jul 8, 2010 #3
    Sorry, yes the degree is 2, I typed the wrong number. Thank you for the quick response.

    * I will correct that in the original posting, as it was a repeated typo
     
  5. Jul 8, 2010 #4

    Mark44

    Staff: Mentor

    Now this part is wrong -
    There are two things going on here: (1) the degrees of numerator and denominator, (2) the coefficients of the leading terms in the numerator and denominator.

    In a rational function, when deg(numerator) = deg(denominator), the equation of the horizontal asymptote is y = an/bn. Here, an is the coefficient of the highest degree term in the numerator, and bn is the coefficient of the highest degree term in the denominator.

    For your problem an = bn = 1.
     
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