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anonymous

  • one year ago

Help with calculus!

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  1. anonymous
    • one year ago
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  2. freckles
    • one year ago
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    just take the terms with highest degree from top and bottom

  3. freckles
    • one year ago
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    like for 2x+1 the term with highest degree is 2x for x^2-2x+1 the term with highest degree is?

  4. anonymous
    • one year ago
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    x^2

  5. freckles
    • one year ago
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    right so your model for the first one is 2x/x^2 which can be reduced

  6. anonymous
    • one year ago
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    Oh so when they ask for power function end behavior model it's asking for the highest power?

  7. anonymous
    • one year ago
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    And it would simplify to \[\frac{ 2 }{ x }\] right?

  8. freckles
    • one year ago
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    yes and I guess you could name it... g(x)=2/x and you can find the horizontal asymptote from the model

  9. freckles
    • one year ago
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    \[\lim_{x \rightarrow \pm \infty } \frac{2}{x}=?\]

  10. freckles
    • one year ago
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    here are more examples: http://chargermath.wikispaces.com/file/view/2-2+End+Behavior+Models.pdf

  11. anonymous
    • one year ago
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    Wouldn't the horizontal asymptote be 0 when graphed?

  12. freckles
    • one year ago
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    yep the horizontal asymptote would be y=0

  13. freckles
    • one year ago
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    for the first one

  14. anonymous
    • one year ago
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    In number 1 from the link you sent, how did they get g(x)=4x^2? Did they just use the 4x^2 from the f(x)?

  15. anonymous
    • one year ago
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    Okay. So for number 2 from my examples g(x)=2 and the horizontal asymptote equals.....2?

  16. freckles
    • one year ago
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    \[f(x)=\frac{4x^2-5x+6}{1} \\ g(x)=\frac{4x^2}{1}=4x^2\]

  17. anonymous
    • one year ago
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    Oh okay that makes sense for the link problem 1.

  18. freckles
    • one year ago
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    yes g(x)=2 would be the model

  19. freckles
    • one year ago
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    and to find the horizontal aymsptote just evaluate: \[y=\lim_{x \rightarrow \pm \infty}2\]

  20. anonymous
    • one year ago
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    So it isn't 2?

  21. freckles
    • one year ago
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    yep

  22. freckles
    • one year ago
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    well y=2

  23. freckles
    • one year ago
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    horizontal equations come in the form y=a number

  24. anonymous
    • one year ago
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    So the actual answer would be y=2?

  25. freckles
    • one year ago
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    just as the first one you asked about was y=0

  26. freckles
    • one year ago
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    by the way the one in the link doesn't have a horizontal asymptote

  27. freckles
    • one year ago
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    well the first one in that link I sent you

  28. anonymous
    • one year ago
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    How did they get 1 as the final answer for problem 1 in the link?

  29. freckles
    • one year ago
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    they were just checking the right and left conditions

  30. anonymous
    • one year ago
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    Why?

  31. freckles
    • one year ago
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    to see if it was a power function end behavior model

  32. anonymous
    • one year ago
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    So that step proves that using the power method is correct?

  33. freckles
    • one year ago
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    yes as the pdf stated we need both \[\lim_{x \rightarrow \pm \infty}\frac{f(x)}{g(x)}=1 \]

  34. freckles
    • one year ago
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    this is to see f and g have the same kind of end behavior at both ends

  35. freckles
    • one year ago
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    and if that holds then yes it is end behavior model

  36. anonymous
    • one year ago
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    Okay thanks! Can you check these 2 questions as well? Just want to make sure I'm getting this right. I'll say the answer I got and just check it.

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  37. anonymous
    • one year ago
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    35.) g(x)=x^2; H.A.: y=0

  38. freckles
    • one year ago
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    no ha

  39. anonymous
    • one year ago
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    36.) g(x)=x; H.A.: DNE or undetermined?

  40. anonymous
    • one year ago
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    Which part did I get wrong for 35?

  41. freckles
    • one year ago
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    no ha as in no horizontal aymptote

  42. anonymous
    • one year ago
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    How do you find the horizontal asymptote numerically? I don't think I'm understanding how to find it using the graphing method.

  43. freckles
    • one year ago
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    there is no polynomial that has a horizontal asymptote

  44. anonymous
    • one year ago
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    Oh okay. Was I right for the others?

  45. freckles
    • one year ago
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    |dw:1441235528085:dw| but |dw:1441235564894:dw| do you see in the first graph the curve is approaching a line for really big values of x you can also see the graph is approaching a line for really big negative values of x that line being y=0 but nothing like that is happening in the second graph and yes you are right for the second question

  46. freckles
    • one year ago
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    just say dne

  47. freckles
    • one year ago
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    I should add what kind of line I was talking about horizontal line

  48. freckles
    • one year ago
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    |dw:1441235738250:dw| see the curve getting closer to that dark horizontal line I drew that line being y=0

  49. freckles
    • one year ago
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    |dw:1441235784454:dw|

  50. freckles
    • one year ago
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    as we move out further and further from the origin in both left and right directions

  51. anonymous
    • one year ago
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    Oh okay I got it. Thanks so much!

  52. freckles
    • one year ago
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    np so the only real correction you need to make 35 and 36 do not have horizontal asymptotes

  53. freckles
    • one year ago
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    g(x)=x^2 and g(x)=x for 35 and 36 respectively look great

  54. anonymous
    • one year ago
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    Okay

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