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anonymous

  • one year ago

The change in water level of a lake is modeled by a polynomial function, W(x). Describe how to find the x-intercepts of W(x) and how to construct a rough graph of W(x) so that the Parks Department can predict when there will be no change in the water level. You may create a sample polynomial of degree 3 or higher to use in your explanations.

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  1. anonymous
    • one year ago
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    @Michele_Laino Ok

  2. Michele_Laino
    • one year ago
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    a general formula for a polynomial of third degree can be this: \[\Large W\left( x \right) = A{x^3} + B{x^2} + Cx + D\] where A, B, C, and D are real coefficients

  3. anonymous
    • one year ago
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    Ok. Know what do I do?

  4. Michele_Laino
    • one year ago
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    in order to determine those coefficients, we need to know some data about the water level of your lake. Do you have a table which collects the water level of the lake in function of time?

  5. anonymous
    • one year ago
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    No. That was the whole question. That's why I was confused :(

  6. Michele_Laino
    • one year ago
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    if we draw a graph in which time x is along the horizontal axis, and the water level W(x) is along the vertical axis, then the x-intercept, is the month at which W(x)=0

  7. Michele_Laino
    • one year ago
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    So a way to establish the x-intercept, is to record at which month, the water level W(x)=0

  8. anonymous
    • one year ago
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    Is that the answer?

  9. Michele_Laino
    • one year ago
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    the complete answer should report the values of each coefficient A, B, C, and D, and the followed procedure to get those values. Nevertheless, without any data about the water level of the lake, we are not able to evaluate those coefficients

  10. anonymous
    • one year ago
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    what about the last part? Should we create an imaginary problem with a degree higher than 3?

  11. Michele_Laino
    • one year ago
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    Using a polynomial whose degree is greater than 3, involves many real coefficients whixh have to be determined. For example, if we conjecture a polynomial of degree 4, then we can write: \[\Large W\left( x \right) = {A_1}{x^4} + {A_2}{x^3} + {A_3}{x^2} + {A_4}x + {A_5}\] As you can see, now we have to determine, by using our experimental observations, 5 real coefficients, namely: \[\Large {A_1},\;{A_2},\;{A_3},\;{A_4},\;{A_5}\]

  12. Michele_Laino
    • one year ago
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    which*

  13. anonymous
    • one year ago
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    ahhh I understand! Ok so thats the final answer, correct?

  14. Michele_Laino
    • one year ago
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    yes! You can write this: "We can establish the values of each coefficients of our sample polynomial function for W(x), by experimental observations about the water level W(x) as function of time or as function of the months of the year"

  15. anonymous
    • one year ago
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    Thank-you:)

  16. Michele_Laino
    • one year ago
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    :)

  17. Michele_Laino
    • one year ago
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    ok!

  18. Michele_Laino
    • one year ago
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    I think that better is if we make a drawing of that function: \[\Large T\left( x \right) = {\left( {x - 4} \right)^3} + 6\] That function is represented by a cubic parabola

  19. anonymous
    • one year ago
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    I agree

  20. Michele_Laino
    • one year ago
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    here is the corresponding graph:

  21. Michele_Laino
    • one year ago
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    as we can see the turning point is at x=4. At x=4 the temperature is: \[\Large T\left( 4 \right) = {\left( {4 - 4} \right)^3} + 6 = 0 + 6 = 6\]

  22. Michele_Laino
    • one year ago
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    yes! I think so!

  23. Michele_Laino
    • one year ago
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    The requested experimental procedure, which can be used in order to find that turning point, can be this: "We substitute many values for the x variable, when a change in x, produces little change in T(x), then we are close to that turning point"

  24. anonymous
    • one year ago
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    I honestly wanna say thank-you for all of your help::)

  25. Michele_Laino
    • one year ago
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    :)

  26. anonymous
    • one year ago
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    Sorry, I have one last question!

  27. Michele_Laino
    • one year ago
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    I'm pondering...

  28. Michele_Laino
    • one year ago
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    as stated in the previous exercise, a turning point can be like a vertex of a parabola, or a quadratic function

  29. Michele_Laino
    • one year ago
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    so, Tucker and Karly are both correct, if they refer to a graph like this: |dw:1436024150337:dw|

  30. Michele_Laino
    • one year ago
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    |dw:1436024243604:dw|

  31. Michele_Laino
    • one year ago
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    |dw:1436024282359:dw|

  32. anonymous
    • one year ago
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    OHHH So how could we write that into an equation for the both of them?

  33. Michele_Laino
    • one year ago
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    |dw:1436024389509:dw|

  34. Michele_Laino
    • one year ago
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    yes!

  35. anonymous
    • one year ago
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    i understand how to do it now but I just don't know how to put it into words, if that makes sense.

  36. Michele_Laino
    • one year ago
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    a possible sentence is like this: "Tucker and Karly are saying the same thing, so they can both be correct, if they refer to a graph like this, for example:" |dw:1436024700576:dw|

  37. anonymous
    • one year ago
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    Ohh :) Thank-you once again!!!!

  38. Michele_Laino
    • one year ago
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    :)

  39. anonymous
    • one year ago
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    Are you up for one more?

  40. Michele_Laino
    • one year ago
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    ok!

  41. Michele_Laino
    • one year ago
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    That graph is a polynomial of fourth degree: \[\large P\left( x \right) = {x^4} - {x^3} - 11{x^2} + 9x + 18\]

  42. anonymous
    • one year ago
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    ok

  43. Michele_Laino
    • one year ago
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    In order to establish the point x such that P(x)=0, we have to factorize those function P(x). Do you know how to factorize that function?

  44. anonymous
    • one year ago
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    yeas. Would it be x3-2x2-9x+18?

  45. Michele_Laino
    • one year ago
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    better is: \[{x^4} - {x^3} - 11{x^2} + 9x + 18 = \left( {x + 1} \right)\left( {{x^3} - 2{x^2} - 9x + 18} \right)\]

  46. anonymous
    • one year ago
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    Wow. I can't believe I got that right! Usually I get those equations wrong. R u sure lol? I mmay have gotten that wrong

  47. Michele_Laino
    • one year ago
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    now, we can factorize the polynomial: x3-2x2-9x+18, so the complete factorization of P(x) is: \[\large {x^4} - {x^3} - 11{x^2} + 9x + 18 = \left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 3} \right)\left( {x - 2} \right)\]

  48. anonymous
    • one year ago
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    i see, I see. Now what do we do?

  49. Michele_Laino
    • one year ago
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    now the points x, such that P(x)=0, are given by the subsequent conditions: x+1=0 --->x=-1 x+3=0 ---> x=-3 x-3=0 --->x=3 x-2=0 ---> x=2

  50. Michele_Laino
    • one year ago
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    now, since x is a profit, then it has to be a positive number, so our acceptable solutions are: x=2, and x=3

  51. anonymous
    • one year ago
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    Is that it?

  52. Michele_Laino
    • one year ago
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    yes! it is the second part, we have to answer to the first part

  53. anonymous
    • one year ago
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    wouldn't we factor x3-2x2-9x+18?

  54. Michele_Laino
    • one year ago
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    yes! I have factored that polynomial

  55. Michele_Laino
    • one year ago
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    \[{x^3} - 2{x^2} - 9x + 18 = \left( {x - 2} \right)\left( {x + 3} \right)\left( {x - 3} \right)\]

  56. anonymous
    • one year ago
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    so would it be (x-2)(x+3)(x-3) as an answer?

  57. anonymous
    • one year ago
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    oh okay now what do we do?

  58. Michele_Laino
    • one year ago
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    as I wrote before, the complete factorization is: \[\left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 3} \right)\left( {x - 2} \right)\]

  59. anonymous
    • one year ago
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    ok, I understand

  60. anonymous
    • one year ago
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    now we have to answer the 2nd part yes?

  61. Michele_Laino
    • one year ago
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    we have to say the graph type which represents our original polynomial: \[{x^4} - {x^3} - 11{x^2} + 9x + 18\]

  62. anonymous
    • one year ago
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    ok.

  63. anonymous
    • one year ago
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    So what would the graph look like?

  64. Michele_Laino
    • one year ago
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    here is my reasoning:

  65. Michele_Laino
    • one year ago
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    let's consider the point x=2 for exmple

  66. anonymous
    • one year ago
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    ok

  67. Michele_Laino
    • one year ago
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    if we substitute x=2.1, into the equation of P(x), we get a negative quantity, since, we have: x-2--->2.1-2 = 0.1 >0 x+1---> 2.1 +1 =3.1 >0 x-3---> 2.1-3=-0.9 <0 x+3---> 2.1+3=5.1 >0 then the product: \[\left( {x + 1} \right)\left( {x + 3} \right)\left( {x - 3} \right)\left( {x - 2} \right)\] is negative

  68. anonymous
    • one year ago
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    ohhh I see. How would we graph w/o technology?

  69. Michele_Laino
    • one year ago
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    I think that the subsequent reasoning is better: since our graph has to pass at subsequent points: \[\left( { - 3,0} \right),\;\left( { - 1,0} \right),\;\left( {2,0} \right),\;\left( {3,0} \right)\]

  70. Michele_Laino
    • one year ago
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    necessarily it is a s follows: |dw:1436026592697:dw|

  71. anonymous
    • one year ago
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    :) Thank-you so much you're a lifesaver!!!!

  72. anonymous
    • one year ago
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    how long will u be on here? I may need help later

  73. Michele_Laino
    • one year ago
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    or like this: |dw:1436026652043:dw|

  74. Michele_Laino
    • one year ago
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    I will stay here in Open Study, for at least 1 hour

  75. anonymous
    • one year ago
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    ok:) Thanks I will email you again soon

  76. Michele_Laino
    • one year ago
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    :)

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