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Math Help - Quadratic Equation

  1. #1
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    Quadratic Equation

    Sketch the curve y = 2x^2 -4x + 1 , indicating the coordinates of the turning point and the exact values of the x-intercepts.

    I am able to deduce the coordinates of the turning point, but how do we calculate the values of the x-intercepts?

    According to my calculation, the coordinates of the turning point are (1, -1).
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  2. #2
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    Quote Originally Posted by Ilsa View Post
    Sketch the curve y = 2x^2 -4x + 1 , indicating the coordinates of the turning point and the exact values of the x-intercepts.

    I am able to deduce the coordinates of the turning point, but how do we calculate the values of the x-intercepts?

    According to my calculation, the coordinates of the turning point are (1, -1).
    The x intercepts are simply the roots of the function. So set 2x^2 - 4x + 1 = 0 and solve for x.

    As far as the turning point is concerned, if we have a quadratic function y = ax^2 + bx + c then the line of symmetry is x = -\frac{b}{2a}. The vertex is either the lowest (when a is positive) or the highest (when a is negative) point on the parabola.

    -Dan
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