ALGEBRA

Determine whether the quadratic function has a minimum or maximum value.Then find the coordinates of the minimum or maximum point.
f(x)=2x^2-4x

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  1. if you know Calculus, then
    f'(x) = 4x - 4
    = 0 for a max/min
    4(x-1) = 0
    x = 1
    then f(1) = 2-4 = -2
    so (1,-2) is a min (since the parabola opens upwards)

    if you don't know Calculus, use "completing the square" to find the vertex

    f(x) = 2[x^2 - 2x + 1 - 1 ]
    = 2[(x-1)^2 - 1]
    = 2(x-1)^2 - 2
    so the vertex is (1,-2) as above

    or

    you could use the formula:
    the x of the vertex is -b/(2a)
    = -(-4)/2(2)
    = -1
    etc

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    posted by Reiny
  2. y = 2 (x^2-2x)
    dy/dx = 2 (2x-2) = extreme at x = 1
    d^2y/dx^2 = 4 so that extreme is a minimum
    at x = 1
    f(1) = -2

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    posted by Damon

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