anonymous
  • anonymous
how do you find the horizontal asymptote of an equation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
what's the equation?
anonymous
  • anonymous
Take the limit as \(x\) goes to infinity.
anonymous
  • anonymous
yes ^ and divide by the highest power of x

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

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