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anonymous

  • one year ago

For the given quadratic equation convert into vertex form, find the vertex, and find the value for x = 6. y= -2x^2 + 2x +2?

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  1. anonymous
    • one year ago
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    @IrishBoy123

  2. anonymous
    • one year ago
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    @Michele_Laino

  3. Michele_Laino
    • one year ago
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    if we make this substitution: x=6 into your quadratic equation, we get: \[\Large \begin{gathered} y = - 2 \times {6^2} + 2 \times 6 + 2 = \hfill \\ \hfill \\ = - 2 \times 36 + 12 + 2 = ...? \hfill \\ \end{gathered} \]

  4. anonymous
    • one year ago
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    -58

  5. Michele_Laino
    • one year ago
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    that's right!

  6. anonymous
    • one year ago
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    So that's the value of x, correct?

  7. Michele_Laino
    • one year ago
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    that's the value of y, when x=6

  8. anonymous
    • one year ago
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    So how do we find the value of x and find the vertex?

  9. Michele_Laino
    • one year ago
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    our task was to find the value of y, when x=6

  10. Michele_Laino
    • one year ago
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    the general formula of a parabola is: \[\Large y = a{x^2} + bx + c\]

  11. anonymous
    • one year ago
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    So there is nothing else to the problem it's done that simple lol

  12. Michele_Laino
    • one year ago
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    we have to find the coordinates of the vertex, and the vertex form of the equation of your parabola

  13. Michele_Laino
    • one year ago
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    so, as I wrote before, the general equation of a parabola, whose axis is parallel to y axis, is: \[\Large y = a{x^2} + bx + c\] now, please by comparison with the equationof your parabola, what are the coefficients a, b, and c?

  14. anonymous
    • one year ago
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    6 and -58?

  15. Michele_Laino
    • one year ago
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    hint: |dw:1435870805196:dw|

  16. anonymous
    • one year ago
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    ok so its 2,2,and 2

  17. Michele_Laino
    • one year ago
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    better is a=-2

  18. anonymous
    • one year ago
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    A=-2, b=2,c=2

  19. Michele_Laino
    • one year ago
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    that's right!

  20. Michele_Laino
    • one year ago
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    now the x-coordinate of the vertex is: \[\Large x = - \frac{b}{{2a}} = ...?\]

  21. Michele_Laino
    • one year ago
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    please substitute your coefficients into taht formula

  22. Michele_Laino
    • one year ago
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    that*

  23. anonymous
    • one year ago
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    X=\[-\frac{ 2b }{ 2a}\]

  24. Michele_Laino
    • one year ago
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    hint: \[\Large x = - \frac{b}{{2a}} = - \frac{2}{{2 \times \left( { - 2} \right)}} = ...?\]

  25. anonymous
    • one year ago
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    I really dont know

  26. Michele_Laino
    • one year ago
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    why?

  27. Michele_Laino
    • one year ago
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    \[\Large x = - \frac{b}{{2a}} = - \frac{2}{{2 \times \left( { - 2} \right)}} = - \frac{2}{{ - 4}} = ...?\]

  28. anonymous
    • one year ago
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    I Oh ok i thought it was like a trick next step lol

  29. anonymous
    • one year ago
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    \[\frac{ 1 }{ 2}\]

  30. anonymous
    • one year ago
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    negative

  31. Michele_Laino
    • one year ago
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    that's right!

  32. Michele_Laino
    • one year ago
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    no, it is 1/2

  33. anonymous
    • one year ago
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    oh ok

  34. Michele_Laino
    • one year ago
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    next, the y-coordinate of the vertex is given by the subsequent formula: \[\Large y = - \frac{{{b^2} - 4ac}}{{4a}} = ...?\] as before, please substitute your coefficients into that formula

  35. anonymous
    • one year ago
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    \[y=\frac{ 2^{2}-4ac }{ 4a}\]

  36. Michele_Laino
    • one year ago
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    hint: \[\Large y = - \frac{{{b^2} - 4ac}}{{4a}} = - \frac{{{2^2} - \left\{ {4 \times \left( { - 2} \right) \times 2} \right\}}}{{4 \times \left( { - 2} \right)}} = ...?\]

  37. anonymous
    • one year ago
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    \[\frac{ 20 }{ -8}\]

  38. Michele_Laino
    • one year ago
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    I think: 20/8

  39. Michele_Laino
    • one year ago
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    am I right?

  40. anonymous
    • one year ago
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    yes

  41. anonymous
    • one year ago
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    so its \[2\frac{ 1 }{ 2 }\]

  42. Michele_Laino
    • one year ago
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    ok! so the vertex V of our parabola is this point: \[\Large V = \left( {\frac{1}{2},\frac{5}{2}} \right)\] since: \[\Large \frac{{20}}{8} = \frac{5}{2}\]

  43. anonymous
    • one year ago
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    So my answers are -58 for x inputting into the y and my vertex is (1/2) (5/2)? correct

  44. Michele_Laino
    • one year ago
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    yes! correct!

  45. anonymous
    • one year ago
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    yay we did it lol

  46. Michele_Laino
    • one year ago
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    finally, we have to write the vertex form of the equation of your parabola

  47. anonymous
    • one year ago
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    Thank you very much for your patience

  48. Michele_Laino
    • one year ago
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    :)

  49. anonymous
    • one year ago
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    ok im ready lol

  50. Michele_Laino
    • one year ago
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    first step: we can factor out -2 at the right side of your equation, so we get: \[\Large y = - 2\left( {{x^2} - x - 1} \right)\]

  51. anonymous
    • one year ago
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    so do we input abc or leave it like that

  52. Michele_Laino
    • one year ago
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    we have to continue with the second step

  53. Michele_Laino
    • one year ago
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    here is the second step: \[\Large {x^2} - x - 1 = {x^2} - x + \frac{1}{4} - \frac{1}{4} - 1\]

  54. Michele_Laino
    • one year ago
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    I added and subtracted 1/4

  55. anonymous
    • one year ago
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    so wouldnt it just x^2-x-1

  56. Michele_Laino
    • one year ago
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    yes! I cahnge the form of x^2-x-1 using the rules of algebra

  57. Michele_Laino
    • one year ago
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    change*

  58. Michele_Laino
    • one year ago
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    nevertheless the quantity is the same, I change only its algebraic shape

  59. anonymous
    • one year ago
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    \[x ^{2}-x-1\]

  60. Michele_Laino
    • one year ago
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    yes that quantity will be the same, I change its form only

  61. Michele_Laino
    • one year ago
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    next I can simplify that expression above as below: \[\Large \begin{gathered} {x^2} - x - 1 = {x^2} - x + \frac{1}{4} - \frac{1}{4} - 1 = \hfill \\ \hfill \\ = {x^2} - x + \frac{1}{4} - \frac{5}{4} \hfill \\ \end{gathered} \]

  62. Michele_Laino
    • one year ago
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    now, the quantity: \[\large {x^2} - x + \frac{1}{4}\] is a perfect square, namely it is the square of which binomial?

  63. anonymous
    • one year ago
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    i have no clue

  64. Michele_Laino
    • one year ago
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    hint: please compute this: \[\Large {\left( {x - \frac{1}{2}} \right)^2} = ...?\]

  65. anonymous
    • one year ago
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    |dw:1435879978188:dw|

  66. Michele_Laino
    • one year ago
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    perfect!

  67. Michele_Laino
    • one year ago
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    so we can write this: \[\large \begin{gathered} {x^2} - x - 1 = {x^2} - x + \frac{1}{4} - \frac{1}{4} - 1 = \hfill \\ \hfill \\ = {x^2} - x + \frac{1}{4} - \frac{5}{4} = {\left( {x - \frac{1}{2}} \right)^2} - \frac{5}{4} \hfill \\ \end{gathered} \]

  68. Michele_Laino
    • one year ago
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    namely: \[\large {x^2} - x - 1 = {\left( {x - \frac{1}{2}} \right)^2} - \frac{5}{4}\]

  69. Michele_Laino
    • one year ago
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    then I substitute that new expression, into the subsequent formula: \[\Large y = - 2\left( {{x^2} - x - 1} \right)\]

  70. Michele_Laino
    • one year ago
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    and I get: \[\Large y = - 2\left\{ {{{\left( {x - \frac{1}{2}} \right)}^2} - \frac{5}{4}} \right\}\]

  71. anonymous
    • one year ago
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    \[y=-2\left( x-\frac{ 1 }{ 2 } \right)^{2}+\frac{ 5 }{ 2 }\]

  72. Michele_Laino
    • one year ago
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    congratulations!! :)

  73. anonymous
    • one year ago
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    Yay lol

  74. anonymous
    • one year ago
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    So the vertex is 1/2 and 5/2, that's are equation above and -58 is the input?

  75. Michele_Laino
    • one year ago
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    yes!

  76. anonymous
    • one year ago
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    yay thank you so much!

  77. Michele_Laino
    • one year ago
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    :)

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