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Math Help - Question on Quadratic Application

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    Question on Quadratic Application

    A parabola whose equation is y=x2-2x+k has a turning point with coordinates (1,-5). Find the value of K
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    Quote Originally Posted by hawk123 View Post
    A parabola whose equation is y=x2-2x+k has a turning point with coordinates (1,-5). Find the value of K
    y(1) = -5

    -5 = (1)^2 - 2(1) + k
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    y
    = x^2 - 2x + k
    = x^2 - 2x + (-2/2)^2 - (-2/2)^2 + k
    = x^2 - 2x + 1 - 1 + k
    = (x^2 - 2x + 1) + (k - 1)
    = (x - 1)^2 + (k - 1)
    = (x - H)^2 + (K)
    where (H,K) is the turning point = (1, -5)
    hence -5 = k - 1
    k = -4
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