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how do you find the horizontal asymptote of an equation

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what's the equation?
Take the limit as \(x\) goes to infinity.
yes ^ and divide by the highest power of x

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Other answers:

its for calc... so its like 4x^3-x^4
horizontal asymptote does not exist....f(x) -> neg infinity as x->infinity
how do you know
so when i graph it i can still find the relative extrema ?
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.
yes there are still local max\min just not any asymptotes
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
so is anyone gonna respond orrrr
nvm gnight people
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
k thankss

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