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Math Help - Quadratic Graph

  1. #1
    Member
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    Quadratic Graph

    Draw the graph with the following equation, choosing appropriate values for x. For the graph, write down the equation of line of symmetry.

    y=2x^2-3x-4

    I couldn't factorise it, and when I completed the square I got x=4 or x=-1. Then I tried quadratic formula and got x=\frac{3\pm\sqrt{41}}{4}. Therefore I got the line of symmetry to be 3. However, the line of symmetry is \frac{3}{4} so I must have done something wrong =\ .
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  2. #2
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
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    I think that you completed the square wrongly.

    2x^2-3x-4
    2(x^2-\frac{3}{2}x)-4
    2(x^2-\frac{3}{2}+(-\frac{3}{4})^2-4-(-\frac{3}{4})^2
    2(x-\frac{3}{4})^2-4-\frac{9}{16}

    hence symmetry occurs at x=\frac{3}{4}

    Or if you use calculus

    \frac{dy}{dx}=4x-3
    Let \frac{dy}{dx}=0
    4x-3=0
    4x=3
    x=\frac{3}{4}
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