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3 December, 19:57

How do you write the vertex form equation of the parabola #y=2x^2 + 11x-6#?

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  1. 3 December, 20:19
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    The easist way to find the vertex is to complete the square:

    y=2x^2 + 11x-6 = > y=2[ x^2 + (11/2) x + (11/2) ^2 - (11/2) ^2 ] - 6

    This can be rewritten as

    y = 2[ (x+11/2) ^2 ] - (121/4) - 24/4

    y = 2 (x+11/2) ^2 - 145/4

    this is the equation in vertex form. The vertex is at (-11/2, - 145/4).
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