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Thread: In desperate need of Math Help!!!! +rep

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    In desperate need of Math Help!!!! +rep

    i have a few questions that are really pissing me off. if anyone could walk me through them i would greatly appreciate it.

    also i will +rep anyone who can help.




    Find the Vertex of the function (put the function in vertex for first): f(x)=8+2x-x^2

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    Ray-Chill's Avatar
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    =] Awesome! We just learned about this.
    xD
    Firstly, put your equation into the format ax^2+bx+c=0. I'm assuming you know the quadratic formula to find the x intercepts,
    x = (-b +or- SQRT(b^2-4ac))/2a

    The vertex will lie on the axis of symmetry of the parabola, which is halfway between the x-intercepts ie -b/2a. Use that to work out x, and then substitute it back into the quadratic to find the y value.

    Eg1. y=-x^2+2x+8 here a=-1, b=2 and c=8
    So -b/2a = -2/2(-1)
    = 1
    Then substitute x=1 into the quadratic
    y=-1+2+8
    =9
    Therefore the vertex is (1, 9)

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    Victoria's Avatar
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    MasterMind0wnz's Avatar
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    I have something that will help you greatly if you don't already know it. PM Me.

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    ack is it negative or positive?

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    zlolekim's Avatar
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    Its positive John

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    MasterMind0wnz's Avatar
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    Vertex: (1,9)

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    Victoria's Avatar
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    sorry for giving you the wrong answer ^_^ i just googled it seemed legit!

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    thanks guys! repping everyone who helped

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