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deepthiakella1
Group Title
how do you find the horizontal asymptote of an equation
 9 months ago
 9 months ago
deepthiakella1 Group Title
how do you find the horizontal asymptote of an equation
 9 months ago
 9 months ago

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

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

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

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

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

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

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

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

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

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

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