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Math Help - Graphing & Identifying Quadratic Functions

  1. #1
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    Graphing & Identifying Quadratic Functions

    I have to graph a quadratic function and find the vertex (simple), axis of symmetry (simple), and possible x/y intercepts (simple). I'm just having trouble converting this quadratic function into standard form (not 100% sure if that's what I need to do).

    Here's the question:


    To my understanding, I have to make the function look like this:
    (2x^2-6x+9)-9+7

    And then I would factor what's inside of the parenthesis, but I don't think it's possible with this problem... Or is it? Please help me on this one. Almost done with this assignment. Been driving me nuts for a couple of days.
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  2. #2
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    Quote Originally Posted by BeSweeet View Post
    I have to graph a quadratic function and find the vertex (simple), axis of symmetry (simple), and possible x/y intercepts (simple). I'm just having trouble converting this quadratic function into standard form (not 100% sure if that's what I need to do).

    Here's the question:


    To my understanding, I have to make the function look like this:
    (2x^2-6x+9)-9+7

    And then I would factor what's inside of the parenthesis, but I don't think it's possible with this problem... Or is it? Please help me on this one. Almost done with this assignment. Been driving me nuts for a couple of days.
    you have to complete the square ...

    2x^2-6x+7

    2(x^2 - 3x) + 7

    2\left(x^2 - 3x + \frac{9}{4}\right) + 7 - \frac{9}{2}<br />

    2\left(x - \frac{3}{2}\right)^2 + \frac{5}{2}
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  3. #3
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    Ah, ok. I forgot that you didn't have to factor out something in the whole function.

    Now I don't know how to graph the thing.

    All these other problems are starting to piss me off...
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  4. #4
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    Quote Originally Posted by skeeter View Post
    you have to complete the square ...

    2x^2-6x+7

    2(x^2 - 3x) + 7

    2\left(x^2 - 3x + \frac{9}{4}\right) + 7 - \frac{9}{2}<br />

    2\left(x - \frac{3}{2}\right)^2 + \frac{5}{2}
    Where did -\frac{9}{2} come from??
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by BeSweeet View Post
    Where did -\frac{9}{2} come from??
    2x^2-6x+7= 0

    2x^2 -6x = -7

    x^2 -3x = -\frac{7}{2}

    (x^2 - \frac{3x}{2} +\frac{9}{4}) = -\frac{7}{2} + \frac{9}{4}

    (x-\frac{3}{2}) + \frac{5}{4}
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  6. #6
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    So what would the axis os symmetry be? And I can't seem to graph it without getting it wrong...

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  7. #7
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    Quote Originally Posted by 11rdc11 View Post
    Skeeter made a mistake it suppose to be -\frac{9}{4}
    no skeeter didn't ...

    \textcolor{red}{2}\left(x^2 - 3x + \textcolor{red}{\frac{9}{4}}\right) + 7 \textcolor{red}{- \frac{9}{2}}

    2\cdot \frac{9}{4} = \frac{9}{2} ...

    \frac{9}{2} was added to the quadratic, not \frac{9}{4} ... -\frac{9}{2} is needed to balance the expression.
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  8. #8
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by skeeter View Post
    no skeeter didn't ...

    \textcolor{red}{2}\left(x^2 - 3x + \textcolor{red}{\frac{9}{4}}\right) + 7 \textcolor{red}{- \frac{9}{2}}

    2\cdot \frac{9}{4} = \frac{9}{2} ...

    \frac{9}{2} was added to the quadratic, not \frac{9}{4} ... -\frac{9}{2} is needed to balance the expression.
    My apologizes, skeeter is correct
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