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Math Help - graphing a function

  1. #1
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    graphing a function

    can someone show me how i go about graphing this?

    f(x) = 15 - x - x^2
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  2. #2
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    Quote Originally Posted by el123 View Post
    can someone show me how i go about graphing this?

    f(x) = 15 - x - x^2

    Graphing Quadratic Functions: Introduction
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  3. #3
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    thanks for that link , so i've just attempted a graph looks like a squiggly line lol.



    I was just curious, i can write f(x) = 15 - x - x^2
    or y = 15 - x - x^2

    Is that ok then i just make a table with my x values and substitue them ionto the equation to find each value of y for any given x value?
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  4. #4
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    Quote Originally Posted by el123 View Post
    thanks for that link , so i've just attempted a graph looks like a squiggly line lol.



    I was just curious, i can write f(x) = 15 - x - x^2
    or y = 15 - x - x^2

    Is that ok then i just make a table with my x values and substitue them ionto the equation to find each value of y for any given x value?
    That should be fine, yes.
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  5. #5
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    Quote Originally Posted by el123 View Post
    can someone show me how i go about graphing this?

    f(x) = 15 - x - x^2
    If your graph looks like a "squiggly line," then you got your signs mixed up somewhere.

    To graph this, you could rewrite in vertex form:
    f(x) = a(x - h)^2 + k

    So,
    f(x) = 15 - x - x^2
    f(x) = -x^2 - x + 15
    f(x) = -(x^2 + x) + 15
    f(x) = -\left(x^2 + x + \frac{1}{4}\right) + 15 + \frac{1}{4}
    f(x) = -\left(x + \frac{1}{2}\right)^2 + \frac{61}{4}

    Since the leading coefficient was negative, the parabola opens downward. The vertex is given as (h, k), so it's \left(-\frac{1}{2},\;\frac{61}{4}\right). You can also find the x- and y- intercepts to get additional points.


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    Quote Originally Posted by yeongil View Post
    If your graph looks like a "squiggly line," then you got your signs mixed up somewhere.

    To graph this, you could rewrite in vertex form:
    f(x) = a(x - h)^2 + k

    So,
    f(x) = 15 - x - x^2
    f(x) = -x^2 - x + 15
    f(x) = -(x^2 + x) + 15
    f(x) = -\left(x^2 + x + \frac{1}{4}\right) + 15 + \frac{1}{4}
    f(x) = -\left(x + \frac{1}{2}\right)^2 + \frac{61}{4}

    Since the leading coefficient was negative, the parabola opens downward. The vertex is given as (h, k), so it's \left(-\frac{1}{2},\;\frac{61}{4}\right). You can also find the x- and y- intercepts to get additional points.


    01
    Hello : look here the graph Thank you
    Attached Thumbnails Attached Thumbnails graphing a function-0707.jpg  
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