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anonymous

  • one year ago

How do I identify the intercepts, removable continuities, and asymptotes? (2x^3-3x+1)/x^3-5x+7

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. ganeshie8
    • one year ago
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    are you allowed to graph the function ?

  3. anonymous
    • one year ago
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    nope :/ i think i'm supposed to solve it based on just the equation

  4. ganeshie8
    • one year ago
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    Finding horizontal asymptotes is easy for a rational function

  5. ganeshie8
    • one year ago
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    \[y = \dfrac{\color{Red}{2}x^3-3x+1}{\color{red}{1}x^3-5x+7}\] As \(x\) gets large, the lower degree terms don't matter much, so the value approaches \[y \approx \dfrac{\color{Red}{2}x^3}{\color{red}{1}x^3} = \dfrac{\color{Red}{2}}{\color{red}{1}}=\color{red}{2}\] so \(\color{red}{y=2}\) is an horizontal asymptote

  6. anonymous
    • one year ago
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    thank you!! is the vertical asymptote about 2.75?

  7. anonymous
    • one year ago
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    x^3-5x=-7 right? and solve for x

  8. ganeshie8
    • one year ago
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    it should be around x= -2.75 but how did u get that ?

  9. ganeshie8
    • one year ago
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    Yes

  10. anonymous
    • one year ago
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    okay thanks!! i'm super confused on finding the x intercepts though

  11. ganeshie8
    • one year ago
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    did you find y intercepts ?

  12. anonymous
    • one year ago
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    yeah i got 1/7 by plugging in 0 in place of x

  13. ganeshie8
    • one year ago
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    Good. use the same trick simply plugin y=0 to get the x intercepts

  14. ganeshie8
    • one year ago
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    \[y = \dfrac{\color{black}{2}x^3-3x+1}{\color{black}{1}x^3-5x+7}\] plugin y=0 for x intercepts \[0 = \dfrac{\color{black}{2}x^3-3x+1}{\color{black}{1}x^3-5x+7}\]

  15. ganeshie8
    • one year ago
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    which is same as \[0 =\color{black}{2}x^3-3x+1\] solve

  16. ganeshie8
    • one year ago
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    wolfram gives 3 x intercepts http://www.wolframalpha.com/input/?i=solve+2x%5E3-3x%2B1%3D0

  17. anonymous
    • one year ago
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    could i use the formula bc of the plus and minus?

  18. anonymous
    • one year ago
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    quadratic formula*

  19. ganeshie8
    • one year ago
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    nope, the equation is cubic, not quadratic. i would simply use calculator/wolfram

  20. anonymous
    • one year ago
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    i'm confused on how i would get 3 answers :/

  21. ganeshie8
    • one year ago
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    why ? it is fine to have multiple x intercepts

  22. ganeshie8
    • one year ago
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    that just means, the graph cuts the x axis at 3 different places

  23. anonymous
    • one year ago
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    i used mathway and it gave me (1,0) and (-1+-sqrt3)/2 but i'm not sure how it got that

  24. ganeshie8
    • one year ago
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    |dw:1437717579760:dw|

  25. anonymous
    • one year ago
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    oh i see it haha thank you so much :)

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