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Thread: "Graphing Quadratic Functions to find Roots"

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    Default "Graphing Quadratic Functions to find Roots"

    That's what my worksheet calls this.

    Alright, so, I have a quadratic equation in standard form.

    It's this:

    Y = -X^2 + 16X - 18

    So, X is equal to Negative B over 2A.

    -16/2(-1) = -16/-2 = 8

    I've figured out that, if I'm right, X is equal to 8.

    Following that, I get this to find Y:

    -1(8)^2 + 16(8) - 18 = 46

    However, my graph (Cartesian plane, obviously) only goes to 10/-10. Did I do something wrong here? Or is it just a problem with the graph?

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    Graphs are infinite in all four directions. You've just got to convince whatever you're using to show you the interesting regions of that graph at an interesting scale.

    That said, your solution is incorrect. You're missing a rather huge chunk of the quadratic equation.

    http://www.wolframalpha.com/input/?i=+0+%3D+-X^2+%2B+16X+-+18

    http://en.wikipedia.org/wiki/Quadratic_equation
    Wow. I've been gone a very long time indeed. So much has changed.

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    i'd relabel the units by 5's or something

    here's the graph:


    Quote Originally Posted by ForgotMyName View Post
    Graphs are infinite in all four directions. You've just got to convince whatever you're using to show you the interesting regions of that graph at an interesting scale.

    That said, your solution is incorrect. You're missing a rather huge chunk of the quadratic equation.

    http://www.wolframalpha.com/input/?i=+0+%3D+-X^2+%2B+16X+-+18

    http://en.wikipedia.org/wiki/Quadratic_equation
    LOL no

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    Quote Originally Posted by ForgotMyName View Post
    Graphs are infinite in all four directions. You've just got to convince whatever you're using to show you the interesting regions of that graph at an interesting scale.

    That said, your solution is incorrect. You're missing a rather huge chunk of the quadratic equation.

    http://www.wolframalpha.com/input/?i=+0+%3D+-X^2+%2B+16X+-+18

    http://en.wikipedia.org/wiki/Quadratic_equation
    To the bold: Good luck convincing the piece of paper on my desk.

    To the italics: How so? That's what I was given on the paper. Can you explain further?

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    Quote Originally Posted by Sandstorm View Post
    To the bold: Good luck convincing the piece of paper on my desk.

    To the italics: How so? That's what I was given on the paper. Can you explain further?
    That's odd. I originally assumed you were using a graphing calculator or maybe computer software. Then one of the roots isn't even in the range that you're graphing.

    Anyway, I misspoke. Space is infinite. Graphs are a finite representation to show a particular region of space, and ideally you would pick the location/size/resolution of what you're representing to be meaningful. (And even that explanation is wrong, if you ask a math major. But I'm not a math major.)

    The roots of
    ax^2+bx+c
    are
    (-b +/- sqrt(b^2 - 4ac))/(2a)
    using the quadratic equation

    Those roots are your "x-values", and by definition of being roots, your "y-values" there had better be 0.

    You were leaving out the whole radical.

    The roots are roughly 1 and 15.
    Precisely, they are 8 +/- sqrt(46)
    as the Wolfram Alpha link says

    So at least one of them works out in your -10 to 10 window of x-values. Maybe you're not meant to care about the other one if you're meant to be doing it graphically in that window.
    Last edited by ForgotMyName; 11-06-2010 at 07:02 AM.
    Wow. I've been gone a very long time indeed. So much has changed.

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    Quote Originally Posted by ForgotMyName View Post
    That's odd. I originally assumed you were using a graphing calculator or maybe computer software. Then one of the roots isn't even in the range that you're graphing.

    Anyway, I misspoke. Space is infinite. Graphs are a finite representation to show a particular region of space, and ideally you would pick the location/size/resolution of what you're representing to be meaningful. (And even that explanation is wrong, if you ask a math major. But I'm not a math major.)

    The roots of
    ax^2+bx+c
    are
    (-b +/- sqrt(b^2 - 4ac))/(2a)
    using the quadratic equation

    Those roots are your "x-values", and by definition of being roots, your "y-values" there had better be 0.

    You were leaving out the whole radical.

    The roots are roughly 1 and 15.
    Precisely, they are 8 +/- sqrt(46)
    as the Wolfram Alpha link says

    So at least one of them works out in your -10 to 10 window of x-values. Maybe you're not meant to care about the other one if you're meant to be doing it graphically in that window.
    roots aren't the problem on his graph, it's the vertex
    if the vertex is represented as, we'll say, (f, g), then
    f = -b/21
    and g = f plugged into the original equation

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