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deepthiakella1 Group Title

how do you find the horizontal asymptote of an equation

  • one year ago
  • one year ago

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  1. kari.lynee Group Title
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    what's the equation?

    • one year ago
  2. wio Group Title
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    Take the limit as \(x\) goes to infinity.

    • one year ago
  3. kari.lynee Group Title
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    yes ^ and divide by the highest power of x

    • one year ago
  4. deepthiakella1 Group Title
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    its for calc... so its like 4x^3-x^4

    • one year ago
  5. dumbcow Group Title
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    horizontal asymptote does not exist....f(x) -> neg infinity as x->infinity

    • one year ago
  6. deepthiakella1 Group Title
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    how do you know

    • one year ago
  7. deepthiakella1 Group Title
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    so when i graph it i can still find the relative extrema ?

    • one year ago
  8. wio Group Title
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    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
  9. dumbcow Group Title
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    yes there are still local max\min just not any asymptotes

    • one year ago
  10. deepthiakella1 Group Title
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    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
  11. deepthiakella1 Group Title
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    so is anyone gonna respond orrrr

    • one year ago
  12. deepthiakella1 Group Title
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    nvm gnight people

    • one year ago
  13. dumbcow Group Title
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    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
  14. deepthiakella1 Group Title
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    k thankss

    • one year ago
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