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deepthiakella1 Group Title

how do you find the horizontal asymptote of an equation

  • 11 months ago
  • 11 months ago

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

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

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

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

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

    • 11 months ago
  6. deepthiakella1 Group Title
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    how do you know

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

    • 11 months 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.

    • 11 months ago
  9. dumbcow Group Title
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    yes there are still local max\min just not any asymptotes

    • 11 months 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

    • 11 months ago
  11. deepthiakella1 Group Title
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    so is anyone gonna respond orrrr

    • 11 months ago
  12. deepthiakella1 Group Title
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    nvm gnight people

    • 11 months 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

    • 11 months ago
  14. deepthiakella1 Group Title
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    k thankss

    • 11 months ago
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