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anonymous

  • one year ago

An expression is shown below: f(x) = -16x2 + 22x + 3 Part A: What are the x-intercepts of the graph of the f(x)? Show your work. (2 points) Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points) Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

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  1. anonymous
    • one year ago
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    @UsukiDoll this is the last one im having trouble with, can you help me? please

  2. Michele_Laino
    • one year ago
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    part A: we have to solve the subsequent quadratic equation: \[\Large - 16{x^2} + 22x + 3 = 0\]

  3. anonymous
    • one year ago
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    can you help me with that? again im new to this...

  4. Michele_Laino
    • one year ago
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    I rewrite that equation as below: \[\Large 16{x^2} - 22x - 3 = 0\]

  5. Michele_Laino
    • one year ago
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    Next I compare that equation with this more general equation: \[\Large a{x^2} + bx + c = 0\] so we can write: a=16, b=-22, and c=-3 am I right?

  6. anonymous
    • one year ago
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    yes

  7. Michele_Laino
    • one year ago
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    now, in order to find the requested x-intercepts, you have to apply this formula: \[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}\] what do you get?

  8. anonymous
    • one year ago
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    completely lost

  9. Michele_Laino
    • one year ago
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    you have to substitute the values of a, b, and c into that formula: \[\Large \begin{gathered} x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}} = \hfill \\ \hfill \\ = \frac{{ - \left( { - 22} \right) \pm \sqrt {{{\left( { - 22} \right)}^2} - 4 \times 16 \times \left( { - 3} \right)} }}{{2 \times 16}} = ...? \hfill \\ \end{gathered} \]

  10. anonymous
    • one year ago
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    i dont have a calculator at the moment, its hard doing it in my head

  11. Michele_Laino
    • one year ago
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    please use windows calculator

  12. Michele_Laino
    • one year ago
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    \[\Large \begin{gathered} x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}} = \hfill \\ \hfill \\ = \frac{{ - \left( { - 22} \right) \pm \sqrt {{{\left( { - 22} \right)}^2} - 4 \times 16 \times \left( { - 3} \right)} }}{{2 \times 16}} = \hfill \\ \hfill \\ = \frac{{22 \pm \sqrt {484 + 192} }}{{32}} = ...? \hfill \\ \end{gathered} \]

  13. anonymous
    • one year ago
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    mine isnt working at the moment...

  14. Michele_Laino
    • one year ago
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    next step: \[\Large \begin{gathered} x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}} = \hfill \\ \hfill \\ = \frac{{ - \left( { - 22} \right) \pm \sqrt {{{\left( { - 22} \right)}^2} - 4 \times 16 \times \left( { - 3} \right)} }}{{2 \times 16}} = \hfill \\ \hfill \\ = \frac{{22 \pm \sqrt {484 + 192} }}{{32}} = \hfill \\ \hfill \\ = \frac{{22 \pm 26}}{{32}} = \begin{array}{*{20}{c}} {\frac{{22 + 26}}{{32}} = ...?} \\ {\frac{{22 - 26}}{{32}} = ...?} \end{array} \hfill \\ \end{gathered} \]

  15. anonymous
    • one year ago
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    48/32 and -4/32

  16. anonymous
    • one year ago
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    leaves you with -12/32

  17. Michele_Laino
    • one year ago
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    yes! and we can simplify them as below: \[\Large {x_1} = \frac{{48}}{{32}} = \frac{3}{2},\quad {x_2} = - \frac{4}{{32}} = - \frac{1}{8}\]

  18. Michele_Laino
    • one year ago
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    what is -12/32?

  19. anonymous
    • one year ago
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    your right, i was overr thinking..

  20. Michele_Laino
    • one year ago
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    so we got your x-intercepts

  21. Michele_Laino
    • one year ago
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    Now, part B: the vertex of a parabola is the point at the top or the point at the bottom of a parabola

  22. Michele_Laino
    • one year ago
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    we have these subsequent cases: |dw:1439456183056:dw|

  23. Michele_Laino
    • one year ago
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    now, in our original equation: \[\Large f\left( x \right) = - 16{x^2} + 22x + 3\] is a= -16, so what is your parabola?

  24. anonymous
    • one year ago
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    lost

  25. Michele_Laino
    • one year ago
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    the coefficient of the second degree monomial od f(x) is -16, right?

  26. anonymous
    • one year ago
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    yes

  27. Michele_Laino
    • one year ago
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    in that case, being -16 < 0, your parabola is concave down, namely your parabola is the one at the left place in my drawing |dw:1439456653946:dw|

  28. anonymous
    • one year ago
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    okay...

  29. Michele_Laino
    • one year ago
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    now, the coordinate of the vertex, are given by the subsequent formulas: \[\Large {x_V} = - \frac{B}{{2A}},\quad {y_V} = - \frac{{{B^2} - 4AC}}{{4A}}\] where A= -16, B=22, and C=3

  30. Michele_Laino
    • one year ago
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    what do you get?

  31. Michele_Laino
    • one year ago
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    hint: \[\Large {x_V} = - \frac{B}{{2A}} = - \frac{{22}}{{2 \times \left( { - 16} \right)}} = ...?\]

  32. anonymous
    • one year ago
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    -22/256 =22/-32

  33. Michele_Laino
    • one year ago
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    I got this: \[\Large {x_V} = \frac{{22}}{{32}} = \frac{{11}}{{16}}\]

  34. anonymous
    • one year ago
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    thank you

  35. Michele_Laino
    • one year ago
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    now, we have to compute the y-coordinate, so we have: \[\Large \begin{gathered} {y_V} = - \frac{{{B^2} - 4AC}}{{4A}} = - \frac{{{{22}^2} - 4 \times \left( { - 16} \right) \times 3}}{{4 \times \left( { - 16} \right)}} = \hfill \\ \hfill \\ = - \frac{{484 + 192}}{{ - 64}} = ...? \hfill \\ \end{gathered} \]

  36. anonymous
    • one year ago
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    my computer is acting up and this is all confusing to me, its my last question and im stressing

  37. anonymous
    • one year ago
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    676/-64

  38. Michele_Laino
    • one year ago
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    more precisely, we get this: \[\Large {y_V} = \frac{{676}}{{64}} = \frac{{169}}{{16}}\]

  39. Michele_Laino
    • one year ago
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    so the vertex V, of your parabola is the subsequent point: \[\Large V = \left( {\frac{{11}}{{16}},\frac{{169}}{{16}}} \right)\]

  40. anonymous
    • one year ago
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    thank you

  41. Michele_Laino
    • one year ago
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    please wait, we have to solve part C

  42. anonymous
    • one year ago
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    yes, im understanding everything, just upset that my computer isnt letting me use my cal.

  43. Michele_Laino
    • one year ago
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    in order to draw your parabola, it is suffice to draw the x-intercepts and the vertex V into a xy-plane, like this: |dw:1439457756334:dw|

  44. anonymous
    • one year ago
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    yes

  45. Michele_Laino
    • one year ago
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    then you have to draw a curve like this: |dw:1439457886482:dw|

  46. anonymous
    • one year ago
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    i understand so far

  47. Michele_Laino
    • one year ago
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    :)

  48. Michele_Laino
    • one year ago
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    that's all!

  49. anonymous
    • one year ago
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    thank you!

  50. Michele_Laino
    • one year ago
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    :)

  51. anonymous
    • one year ago
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    trying to put everything together in sentences for part c... sadly i cant send in my drawling

  52. anonymous
    • one year ago
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    @Michele_Laino

  53. Michele_Laino
    • one year ago
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    we can write this: step #1 I draw the x-intercepts, and the vertex of my parabola, step#2 I connect those three points each other using a continuous line.

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