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

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

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

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

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

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

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

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

wio Group TitleBest ResponseYou've already chosen the best response.1
A 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.
 one year ago

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

deepthiakella1 Group TitleBest ResponseYou've already chosen the best response.0
ohhhh 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
 one year ago

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

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

dumbcow Group TitleBest ResponseYou've already chosen the best response.1
sorry...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
 one year ago

deepthiakella1 Group TitleBest ResponseYou've already chosen the best response.0
k thankss
 one year ago
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