A community for students.
Here's the question you clicked on:
 0 viewing
deepthiakella1
 2 years ago
how do you find the horizontal asymptote of an equation
deepthiakella1
 2 years ago
how do you find the horizontal asymptote of an equation

This Question is Closed

kari.lynee
 2 years ago
Best ResponseYou've already chosen the best response.0what's the equation?

wio
 2 years ago
Best ResponseYou've already chosen the best response.1Take the limit as \(x\) goes to infinity.

kari.lynee
 2 years ago
Best ResponseYou've already chosen the best response.0yes ^ and divide by the highest power of x

deepthiakella1
 2 years ago
Best ResponseYou've already chosen the best response.0its for calc... so its like 4x^3x^4

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.1horizontal asymptote does not exist....f(x) > neg infinity as x>infinity

deepthiakella1
 2 years ago
Best ResponseYou've already chosen the best response.0so when i graph it i can still find the relative extrema ?

wio
 2 years ago
Best ResponseYou've already chosen the best response.1A horizontal asymptote is defined as a line \(y=L\) such that \[ \lim_{x\to \infty }f(x)=L \]Or\[ \lim_{x\to \infty }f(x)=L \]It's a definition. You can not argue with a definition.

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.1yes there are still local max\min just not any asymptotes

deepthiakella1
 2 years ago
Best ResponseYou've already chosen the best response.0ohhhh so can i still find the first der. and get the critical number then 2nd der. to find the point of inflection then do the number line thing to find the concavity then label the relative extrema? my only question is how do you find all of the asymptotes and where does the critical number go on the graph

deepthiakella1
 2 years ago
Best ResponseYou've already chosen the best response.0so is anyone gonna respond orrrr

deepthiakella1
 2 years ago
Best ResponseYou've already chosen the best response.0nvm gnight people

dumbcow
 2 years ago
Best ResponseYou've already chosen the best response.1sorry...i think you are confusing local extrema with asymptotes f(x) = 4x^3  x^4 f'(x) = 12x^2 4x^3 = 0 > x = 0,3 (critical points) f''(x) = 24x  12x^2 = 0 > x = 0 , 2 (inflection points) x=0 is inflection point x=3 is local max since f''(3) <0 indicating concave down
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.