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anonymous
 3 years ago
how do you find the horizontal asymptote of an equation
anonymous
 3 years ago
how do you find the horizontal asymptote of an equation

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Take the limit as \(x\) goes to infinity.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0horizontal asymptote does not exist....f(x) > neg infinity as x>infinity

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes there are still local max\min just not any asymptotes

anonymous
 3 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry...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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