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How do i solve the pair of simultaneous equations
y=2x^2-3x+1 ??

how do i find minimum value of 2x^2-3x+1 and the value of x for which the minimum occurs???

  • MATH -

    equate the two right sides, since they are both equal to y
    2x^2 - 3x + 1 = 4 - 2x
    2x^2 - x - 3 = 0
    (2x-3)(x+1) = 0
    2x-3=0 or x+1=0
    x = 3/2 or x = -1

    sub those values into y = 4-2x to get the corresponding y values of the points of intersection.

    for the second part, are you taking Calculus.
    If so, you should surely know how to do this. Take the first derivative, set it equal to zero and solve for x
    Sub that back into the function to get the y.
    If you are not taking Calculus you should know how to complete the square to find the vertex of the matching parabola.

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