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deepthiakella1
 one year ago
how do you find the horizontal asymptote of an equation
deepthiakella1
 one year ago
how do you find the horizontal asymptote of an equation

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

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

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

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

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

deepthiakella1
 one year ago
Best ResponseYou've already chosen the best response.0how do you know

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

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

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

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

dumbcow
 one year 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
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