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Calculus Question

  1. Oct 9, 2006 #1
    I can't figure this question out, anyone have any ideas?

    "Find a formula for a rational function whose horizontal asymptote is y = -8/3 and vertical asymptotes at x = -2 and x = 4 and whose ONLY x-intercept is at x = -5."
  2. jcsd
  3. Oct 9, 2006 #2
    What is a horizontal asymptote? It is defined as a line [tex] y = L [/tex] such that:

    [tex] \lim_{x\rightarrow \infty} f(x) = L [/tex] or
    [tex] \lim_{x\rightarrow -\infty} f(x) = L [/tex].

    A vertical asymptote is defined as a line [tex] x=a [/tex] such that [tex] \lim_{x\rightarrow a} f(x) = \infty [/tex]. (or [tex] -\infty [/tex]).

    So one function is [tex] f(x) = \frac{-8x^{2}+200}{3(x-4)(x+2)} [/tex]
    Last edited: Oct 9, 2006
  4. Oct 9, 2006 #3
    that isn't right because it never passes through x = -5
  5. Oct 9, 2006 #4
    yes it does. graph it, and look at the value at [tex] x = -5 [/tex].
  6. Oct 9, 2006 #5
    oh yeah, you're right, but it also has an x value of 5 and the only x-intercept should be -5
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