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    A parabla has equation y=axSQUARED + bx + c, where a, b and c are constants. Show that the axis of the parabola has equation 2ax + b = 0, and find the co ordinates of its vertex. Hence show that the vertex lies on the x axis when bsquared = 4ac

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    Differentiate to find x at the minimum point. That should give you a start

    What have you tried so far?
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    i really dnt get it :mad:
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    You differentiate your equation to find when the gradient is zero (either a maximum or minumum point). The constants a and b stay as they are as you are differentiating with respect to x.

    The vertex of your parabola will be either a maximum or minimum (depending on whether a is positive or negative), and the axis of symmetry will go vertically through the vertex
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    (Original post by samw)
    A parabla has equation y=axSQUARED + bx + c, where a, b and c are constants. Show that the axis of the parabola has equation 2ax + b = 0, and find the co ordinates of its vertex. Hence show that the vertex lies on the x axis when bsquared = 4ac

    help
    Complete the square.
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