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jmays14

  • 2 years ago

Horizontal asymptote of the function f(x)= 6x^3+17x^2-528/2x^3-14x+321

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  1. jmays14
    • 2 years ago
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    \[f(x) = 6x^3+17x^2-528/2x^3-14x+321\]

  2. jmays14
    • 2 years ago
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    can someone help me please

  3. Callisto
    • 2 years ago
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    Horizontal asymptotes \[=f(x) = \lim_{x \rightarrow \infty} \frac{6x^3+17x^2-528}{2x^3-14x+321}=...\]

  4. jmays14
    • 2 years ago
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    yes thats the problem

  5. Callisto
    • 2 years ago
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    Divide both the numerator and denominator by x^3

  6. Callisto
    • 2 years ago
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    And \[\lim_{x \rightarrow \infty} \frac{1}{x} =0\]

  7. jmays14
    • 2 years ago
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    i got 3

  8. Callisto
    • 2 years ago
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    Isn't it correct?!

  9. jmays14
    • 2 years ago
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    what is the next step , i believe 3 is right

  10. Callisto
    • 2 years ago
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    It is 3..

  11. Callisto
    • 2 years ago
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    \[\lim_{x \rightarrow \infty} \frac{6x^3+17x^2-528}{2x^3-14x+321}\]\[=\lim_{x \rightarrow \infty} \frac{\frac{6x^3+17x^2-528}{x^3}}{\frac{2x^3-14x+321}{x^3}}\]\[=\lim_{x \rightarrow \infty} \frac{6+\frac{17}{x}-\frac{528}{x^3}}{2-\frac{14}{x^2}+\frac{321}{x^3}}\]Then take the limit

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