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5 February, 06:40

20-90x-50x^2

find the vertex

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Answers (2)
  1. 5 February, 06:57
    0
    20-90x-50x^2, written in standard form, would be - 50x^2 - 90x + 20. Let's use the method of completing the square to determine the vertex:

    Divide all terms in - 50x^2 - 90x + 20 by - 50, obtaining - 50 (x^2 + (9/5) x - 2/5)

    The coefficient of x is 9/5. Divide this by 2, obtaining 9/10, and then square this result: (9/10) ^2 = 81/100. Add this to x^2 + (9/5) x - 2/5 and then subtract it, resulting in x^2 + (9/5) x - 2/5 + 81/100 - 81/100, or

    x^2 + (9/5) x + 81/100 - 2/5 - 81/100, or

    (x + 9/10) ^2 - 12100

    Then the vertex location is (-9/10, - 12100).

    A faster approach would be to use x = - b / (2a) to determine the x-coordinate of the vertex and then to evaluate the function at this x-value to determine the y-coordinate. Here x = - (-90) / (-100) = - 9/10 (same as before).
  2. 5 February, 07:01
    0
    I think it's your answer
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