A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

How do I find the slant asymptote of f(x) = x + (32/x^2)

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[f(x) = x + \frac{ 32 }{ x ^{2} }\]

  2. nincompoop
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what is an asymptote?

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    A vertical asymptote (VA) is an x value where as we approach that value the function goes either toward infinity or negative infinity. The VA is the x value that would make the denominator equal 0. A Horizontal asymptote (HA) is a Y value such that when X reaches infinity the function comes very close (but not equal to) the Y value. The HA is the same value for the limit as X approaches infinity (or negative infinity). I haven't yet graphed slant asymptotes so I'm not sure yet exactly what the slant asymptote is.

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    To find the slant asymptote the numerator needs to be divided by the denominator using long division. So the original function can be re-written as: \[\frac{ x(x ^{2})+32 }{ x ^{2} }\] Using long division the slant asymptote = x

  5. Zale101
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Slant asymptote can be found if the degree in the numerator is greater than in the denominator. If you have a function that has these characteristics, then you do long division or synthetic division (whatever works best for you) to divide. After dividing, you'll get a quotient, that quotient is you slant asymptote. y= quitient is the slant asymetote

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Zale101 \[f' = 1 - \frac{ 64 }{ x^3 }\]

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The critical points are x = 4 and x = 0, but , I don't understand why. Apparently the equation can be re-written as \[\frac{ (x-4)(x^2 + 4x + 16) }{ x^3 }\] This looks like the difference of cubes formula but there is no X in the numerator of f", so I'm not sure why it can be rewritten like that. I do understand that it is equivalent to \[\frac{ 1^3 - 4^3 }{ x^3 }\] But I don't get why x = 4 is a zero

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I made a mistake in what I wrote earlier \[1 - \frac{ 64 }{ x^3 } = \frac{ (1)(x^3) - 64}{ x^3 } = \frac{ (x^3) - 4^3}{ x^3 } \] therefore the difference of cubes formula can be used

  9. Zale101
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Your slant asymptote is x, just as you said...

  10. nincompoop
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    man, that was long huh

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.