1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Calculus Question
  1. Question: Calculus (Replies: 2)

  2. Calculus Question. (Replies: 2)

  3. Calculus question (Replies: 2)

  4. Calculus question (Replies: 3)

  5. Calculus Question (Replies: 4)

Loading...