A community for students.
Here's the question you clicked on:
 0 viewing
BloomLocke367
 one year ago
How do you tell from looking at an equation whether there is a vertical or horizontal asymptote?
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

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1for \(\color{green}{\rm Vertical~ asy.}\) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u help me @Nnesha

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1okay.. so \(f(x)=\frac{x}{x1}\) 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 11=0.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1wow, okay. that was easy.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1so for \(q(x)=\large \frac{x1}{x}\) vertical: x=0 and horizontal:y=1?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1okay. this one is a little different but I'm PRETTY sure I got it. \(g(x)=\large\frac{x+2}{3x}\)...so v. tote: x=3 and h. tote: y=1? (I shortened it lol)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1that's right! x+3= 3 so x= 3

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1This one I don't know at all :O \(q(x)=1.5^x\)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1thanks for your help btw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1is there a denominator ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1there is just one right so no vertical asy

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1N>D ( highest degree) numerator is greater than denominator so horizontal asy would bE ?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1NO HORIZONTAL ASYMPTOTE.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1sorry, I had that in caps in my notes XD

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1okay. 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.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1because 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.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1yes h asy y=0 right but when you solve for x you will get negative one x^2=1

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1can you have an imaginary asymptote? or is the answer none?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you will get x= i which is imaginary i would say no vertical asy

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1dw: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

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1I have to go eat dinner, thank you!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1np :=) have a wonderful evening o^_^o
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.