A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

BloomLocke367

  • one year ago

How do you tell from looking at an equation whether there is a vertical or horizontal asymptote?

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

    for \(\color{green}{\rm Vertical~ asy.}\) set the denominator equal to zero and then solve for the variable. for\(\color{green}{\rm Horizontal ~asy.}\) focus on highest degrees ~if the highest degree of the numerator is greater than the denominator then `No horizontal asy.` \[\color{reD}{\rm N}>\color{blue}{\rm D}\] example \[\large\rm \frac{ 7x^\color{ReD}{3} +1}{ 4x^\color{blue}{2}+3 }\] ~if the highest degree of the denominator is greater than the highest degree of the numerator then `y=0` would be horizontal asy. \[\rm \color{reD}{N}<\color{blue}{\rm D}\] example:\[\large\rm \frac{ 7x^\color{red}{2}+1 }{ 4x^\color{blue}{3}+3 }\] ~if both degrees are the same then divide the leading coefficient of the numerator by the leading coefficient of the denominator \[\rm \color{red}{N}=\color{blue}{D}\] \[\large\rm \frac{ 8x^\color{reD}{3}+1 }{ 4x^\color{blue}{3}+3 }\] \[\rm \frac{ 8x^3 }{ 4x^3 } =2\] horizontal asy. =2

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

    XD

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

    can u help me @Nnesha

  4. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    okay.. so \(f(x)=\frac{x}{x-1}\) the vertical asymptote is x=1 and the horizontal is y=1, right? because the highest degrees are the same and are both 1. The leading coefficients are also both 1 and 1/1=1 for the vertical asymptote 1-1=0.

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

    that's right !!

  6. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    wow, okay. that was easy.

  7. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so for \(q(x)=\large \frac{x-1}{x}\) vertical: x=0 and horizontal:y=1?

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

    you got it!

  9. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    okay. this one is a little different but I'm PRETTY sure I got it. \(g(x)=\large\frac{x+2}{3-x}\)...so v. tote: x=3 and h. tote: y=-1? (I shortened it lol)

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

    that's right! -x+3= -3 so x= 3

  11. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    This one I don't know at all :O \(q(x)=1.5^x\)

  12. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thanks for your help btw

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

    is there a denominator ?

  14. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    no? just 1. lol

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

    there is just one right so no vertical asy

  16. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    okay

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

    N>D ( highest degree) numerator is greater than denominator so horizontal asy would bE ?

  18. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    NO HORIZONTAL ASYMPTOTE.

  19. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry, I had that in caps in my notes XD

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

    yes right

  21. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    okay. now for this one, \(p(x)=\large\frac{4}{x^2+1}\) the horizontal asymptote is y=0 because it's "heavy on the bottom" (there is a higher degree on the bottom) but I'm not sure about the vertical asymptote.

  22. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    because it would be \(x^2=-1\) which would be i for the answer which is an imaginary number. I don't think you can have x=i, but I may be wrong.

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

    yes h- asy y=0 right but when you solve for x you will get negative one x^2=-1

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

    yes you're right!

  25. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can you have an imaginary asymptote? or is the answer none?

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

    you will get x= i which is imaginary i would say no vertical asy

  27. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    okay thank you

  28. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think I'm good. I did three more but it wouldn't be fair for me to have ALL of my homework checked XD thank you.

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

    |dw:1441231766886:dw| graph (from purplemath.com) in this graph red lines represent vertical asy and we can't see imaginary stuff so i would say not vertical asy

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

    aw np :=)

  31. BloomLocke367
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I have to go eat dinner, thank you!

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

    np :=) have a wonderful evening o^_^o

  33. 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.