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anonymous

  • one year ago

f(x)=-x^2(x-1)(x+3) Determine the end behavior: find the power function that the graph of f resembles for large values of |x|. Find the x-and y-intercepts of the graph. Determine whether the graph crosses or touches the x-axis at each x-intercept.

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  1. ganeshie8
    • one year ago
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    look at the given polynomial f(x) whats the degree of f(x) ?

  2. anonymous
    • one year ago
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    im not given any additional information.. just was given the problem and asked to find that information..

  3. ganeshie8
    • one year ago
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    Ahh, actually im asking you a question to see if you know how to find the "degree"

  4. anonymous
    • one year ago
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    I'm pretty sure i understand how to find the x and y intercepts.. should i treat this as a normal polynomial function and cross multiply?

  5. anonymous
    • one year ago
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    oh!! i thought would be the exponent

  6. ganeshie8
    • one year ago
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    it is indeed a polynomial, but do not multiply yet, i think we can finish the problem with out multiplying

  7. ganeshie8
    • one year ago
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    yes, whats the degree of f(x) ?

  8. anonymous
    • one year ago
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    i mean the degree is the exponent*

  9. anonymous
    • one year ago
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    2

  10. ganeshie8
    • one year ago
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    exponent of what

  11. anonymous
    • one year ago
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    but then again I'm thinking its just 1

  12. anonymous
    • one year ago
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    x^2?

  13. ganeshie8
    • one year ago
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    f(x)=-x^2(x-1)(x+3) notice that there are 3 factors here, multiplying them out gives the exponent of leading term as : 2+1+1 = 4 so the degree of f(x) is 4

  14. ganeshie8
    • one year ago
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    for part a, the power function that the graph of f resembles for large values of |x| is \(\large -x^4\)

  15. anonymous
    • one year ago
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    so as if i was to cross multiply id get an exponent of 2 for the first value and 1 and 1 on the middle and last term?

  16. anonymous
    • one year ago
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    ok...

  17. ganeshie8
    • one year ago
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    Exactly! \(f(x)=-\color{red}{x}^2(\color{red}{x}-1)(\color{red}{x}+3)\) its easy to see that \(-\color{Red}{x}^2*\color{Red}{x}*\color{red}{x}\) gives you \(-\color{Red}{x}^{2+1+1} = -\color{red}{x}^4\)

  18. anonymous
    • one year ago
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    what does finding the degree do for me though?

  19. ganeshie8
    • one year ago
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    good question, when you plugin large values of |x|, the function behaves almost same as the power function\(\large -x^4\)

  20. anonymous
    • one year ago
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    I've been working similar problems and haven't needed that information to find the x and y intercepts

  21. ganeshie8
    • one year ago
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    that degree thing helps in answering part a of question : `Determine the end behavior: find the power function that the graph of f resembles for large values of |x|.`

  22. anonymous
    • one year ago
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    wow... ok

  23. ganeshie8
    • one year ago
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    you can see here, how closely the graph of f(x) matches with the power function \(-x^4\) https://www.desmos.com/calculator/vj2xhutn6z

  24. anonymous
    • one year ago
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    i havent been shown how to find the solution to these questions except for finding the x and y intercepts, domain/range, but not the power function.

  25. anonymous
    • one year ago
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    ok yes!! i see that now.

  26. anonymous
    • one year ago
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    ok so finding the degree will give me the power function..

  27. anonymous
    • one year ago
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    Sorry lot of questions... i think I'm just having a hard time understanding what I'm being asked to find.

  28. ganeshie8
    • one year ago
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    good, lets find the x intercepts

  29. anonymous
    • one year ago
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    ok so from what I've understood, to find the x-intercept id want to find the value of the equation by f(x)=0, make the equation equal to 0, and finding the value of x.. thats why i assumed id cross multiply..

  30. anonymous
    • one year ago
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    but now I'm not seeing how that would work.

  31. ganeshie8
    • one year ago
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    thats right! to find x intercepts, you want to solve \[\large -x^2(x-1)(x+3)=0\]

  32. anonymous
    • one year ago
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    is this problem similar to the problem f(x)={3+x If x<0 // x^2 if x> not equal to 0....

  33. ganeshie8
    • one year ago
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    use zero product property to get : \(-x^2=0\) or \(x-1= 0\) or \(x+3=0\)

  34. ganeshie8
    • one year ago
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    that gives you x intercepts : \(0, 1, -3\)

  35. anonymous
    • one year ago
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    awesome!! ok so thats what i was thinking..

  36. anonymous
    • one year ago
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    then id have to find the y intercept?

  37. ganeshie8
    • one year ago
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    yes, simply put \(x=0\) in f(x) to get the y intercepts

  38. ganeshie8
    • one year ago
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    f(x)=-x^2(x-1)(x+3) f(0) = ?

  39. anonymous
    • one year ago
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    -3

  40. anonymous
    • one year ago
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    wouldn't i substitue all x values with 0?

  41. ganeshie8
    • one year ago
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    yes, but -3 is wrong

  42. ganeshie8
    • one year ago
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    f(x)=-x^2(x-1)(x+3) f(0) = -0^2(0-1)(0+3) = ?

  43. anonymous
    • one year ago
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    +3

  44. ganeshie8
    • one year ago
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    no, try again

  45. ganeshie8
    • one year ago
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    what do you get when u multiply something by 0 ?

  46. anonymous
    • one year ago
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    0

  47. ganeshie8
    • one year ago
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    so..

  48. anonymous
    • one year ago
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    so the answer is 0

  49. anonymous
    • one year ago
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    cuz im retarded and looked at the problem as if the 0 in front of the parentheses was invisible lol...

  50. ganeshie8
    • one year ago
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    haha happens :)

  51. ganeshie8
    • one year ago
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    so the y intercept is 0

  52. ganeshie8
    • one year ago
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    how do we answer the last part `Determine whether the graph crosses or touches the x-axis at each x-intercept.`

  53. anonymous
    • one year ago
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    plot the points and read the graph and how it falls

  54. ganeshie8
    • one year ago
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    look at the graph https://www.desmos.com/calculator/bdlowvla6f

  55. ganeshie8
    • one year ago
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    notice that the graph is crossing x axis at x=-3 and x=1 but it is just touching x axis at x=0

  56. anonymous
    • one year ago
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    ok so it touches the x axis at 0,1,-3 but only crosses at 1 and -3

  57. anonymous
    • one year ago
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    ooh you answered before i could finish typing.. but that is what i saw

  58. ganeshie8
    • one year ago
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    very good!

  59. anonymous
    • one year ago
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    but then how do know the shape of the graph just by the points.. how do we know how far up it goes and where it lies on the rest of the graph

  60. ganeshie8
    • one year ago
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    the graph goes forever on both ends, it sinks down : |dw:1440390580159:dw|

  61. ganeshie8
    • one year ago
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    zoomout in that desmos link and see...

  62. anonymous
    • one year ago
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    but what about the part that is between (0,0) and (1,0)?

  63. ganeshie8
    • one year ago
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    what about it ?

  64. anonymous
    • one year ago
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    never mind i think I'm just over analyzing what i know and the "whys" to everything.. thank you for your help... i was on the right track for some of it.

  65. ganeshie8
    • one year ago
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    its all good :) here is a question for you : whats the reason for the graph of f(x) to just touch the x axis at x=0

  66. ganeshie8
    • one year ago
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    |dw:1440390931586:dw|

  67. ganeshie8
    • one year ago
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    if you see, the graph is just touching the x axis at x=0, it is NOT crossing the x axis

  68. anonymous
    • one year ago
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    Knowing, the x-and y-intercepts, the end behavior, the power function and where the graph is crossing and touching the x-axis at each x-intercept... how should all of that be applied to graphing the equation?

  69. anonymous
    • one year ago
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    I'm supposed to label all intercepts and turning points... I understand that ill plot the (0,0), (1,0), (-3,0) and label those as my x-intercepts.. but i confused about how to label the turning points and how the power function is used.

  70. dan815
    • one year ago
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    the turning points happen where the function has 0 slope

  71. dan815
    • one year ago
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    |dw:1441149899004:dw|

  72. dan815
    • one year ago
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    f(x)=-x^2(x-1)(x+3) you have to solve the equation d/dx(f(x)) =0 find d/dx(f(x)) and set to 0 and find at what x that happens

  73. anonymous
    • one year ago
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    i know where the turning points occur.. I've already found the answers to everything.. the only thing I'm unsure about is how exactly is a turning point to be labeled on a graph and if and how a power function is to be labeled

  74. dan815
    • one year ago
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    you put a point where u calculating the turning point for example(0,0) is a turning point

  75. dan815
    • one year ago
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    |dw:1441150082317:dw|

  76. anonymous
    • one year ago
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    is there a way to send you what I've written down and graphed?

  77. dan815
    • one year ago
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    take a picture and attach file on here

  78. dan815
    • one year ago
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    or click on draw and roughly draw what u have

  79. anonymous
    • one year ago
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