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Function: expressing functions in vertex form.

  1. May 21, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-5-21_13-7-11.png

    2. Relevant equations

    upload_2016-5-21_13-13-36.png
    upload_2016-5-21_13-14-29.png

    3. The attempt at a solution
    a)
    f(x) = -5x2 + 20x + 2

    y = -5x2 + 20x + 2

    Factor -5 from the first two terms:

    y = -5x2 + 20x + 2

    = -5 (x2 – 4x) +2

    Complete the square in the bracket:

    (1/2 b)2 = [1/2 (-4)]2 = (-2)2 = 4

    Group the perfect square trinomial:

    = -5 (x2 – 4x + 4 – 4) +2

    Remove -4 from the brackets, it must be multiplied by -5:

    = -5 [(x2 – 4x + 4) – 4] +2

    = - 5 (x2 – 4x + 4) + 20 + 2

    Factor the trinomial. Add remaining constants:

    = -5 (x-2)2 + 22

    b) The vertex is (2, 22)

    c) Since the parabola is concave down, the maximum value is y = 22

    d)f(x) = -5x2 + 20x + 2

    y = -5x2 + 20x + 2

    Let y = 0

    0 = -5x2 + 20x + 2

    Substitute a = -5, b = 20, c = 2 into the quadratic formula:

    upload_2016-5-21_13-15-39.png


    The x-intercepts are x= -0.09 and x = 4.09

    Is this correct?

    Thanks.
     

    Attached Files:

  2. jcsd
  3. May 21, 2016 #2
    Correct. You really should check by graphing instead of relying on others.
     
  4. May 21, 2016 #3
    You could also verify the x-intercept values by subbing them into the original equation. The whole thing should be equivalent to 0.
     
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