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Math Help - Graph the quadratic equation...

  1. #1
    casper
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    Question Graph the quadratic equation...

    I really need some help here. Please be specific as this is new to me.

    Graph the quadratic equation after completing the given table of values.
    y = x^2 + 2x – 3.
    Include the axis of symmetry and show specific points, vertex, and intercepts.
    x, y
    –3, ?
    -2, ?
    –1, ?
    0, ?
    1, ?




    I don't expect anyone to try to tell me how to graph the points. I can do that. I need step by step with how to get the points, axis of symmetry and vertex. My text doesn't really help me understand and this is the only problem I can't figure out of my assignment.
    Greatly appreciated!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by casper View Post
    I really need some help here. Please be specific as this is new to me.

    Graph the quadratic equation after completing the given table of values.
    y = x^2 + 2x 3.
    Include the axis of symmetry and show specific points, vertex, and intercepts.
    x, y
    3, ?
    -2, ?
    1, ?
    0, ?
    1, ?




    I don't expect anyone to try to tell me how to graph the points. I can do that. I need step by step with how to get the points, axis of symmetry and vertex. My text doesn't really help me understand and this is the only problem I can't figure out of my assignment.
    Greatly appreciated!!
    you get the points by plugging in the given x-values into the formula given

    for instance, when x = -3, y = (-3)^2 + 2(-3) - 3 = 9 - 6 - 3 = 0

    this is an upward-opening parabola, the axis of symmetry is the vertical line passing through the vertex

    the x-value for the vertex is given by: x = \frac {-b}{2a}, where b is the coefficient of x and a is the coefficient of x^2

    to find the y-coordinate for the vertex, you again would plug in the x-value into the formula
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Santiago, Chile
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    Quote Originally Posted by casper View Post
    y = x^2 + 2x 3.
    To get quick the vertex

    y=x^2+2x-3=(x+1)^2-4

    So when x=-1,\,y=-4\,\therefore\,\mathcal V=(-1,-4)\,\blacksquare
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