Given that x+2y=4,determine the minimum value of x^2+xy-y^2.

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  1. x = 4-2y

    V = value =(4-2y)^2 + (4-2y)y - y^2
    V = 16 - 16 y + 4y^2 + 4 y -2 y^2 -y^2
    V = y^2 - 12 y + 16
    y^2 - 12 y = V-16

    y^2 - 12 y + 6^2 = V - 16 + 36

    (Y-6)^2 = V+20
    vertex at v = -20

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