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Math Help - Help: quadratic function whose graph is given

  1. #1
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    Help: quadratic function whose graph is given

    Determining the quadratic function whose graph is given.

    Graphic is a parabola going upwards.
    Vertex: -3,2
    Y-Intercept: 0,4

    I've tried using the following formula as the book explained, but I am not getting the right answer for some odd reason.

    f(x) = a(x-h)^2+k
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by dasani View Post
    Determining the quadratic function whose graph is given.

    Graphic is a parabola going upwards.
    Vertex: -3,2
    Y-Intercept: 0,4

    I've tried using the following formula as the book explained, but I am not getting the right answer for some odd reason.

    f(x) = a(x-h)^2+k
    The coordinates of the vertex are (h,k) = (-3,2), so the quadratic equation becomes:

    f(x) = a(x - (-3))^2 + 2 = a(x + 3)^2 + 2

    We can use the y-intercept to find the value of "a."
    We have that f(0) = 4:
    4 = a(0 + 3)^2 + 2 = 9a + 2
    9a = 2
    a = 2/9

    Therefore, the equation becomes:

    f(x) = 2/9(x + 3)^2 + 2
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dasani View Post
    Determining the quadratic function whose graph is given.

    Graphic is a parabola going upwards.
    Vertex: -3,2
    Y-Intercept: 0,4

    I've tried using the following formula as the book explained, but I am not getting the right answer for some odd reason.

    f(x) = a(x-h)^2+k
    when a parabola is written in the form f(x) = a(x-h)^2+k, the vertex is given by (h,k) and the y-intercept (when x = 0) is given by, f(0) = a(-h)^2 + k

    here we have (h,k) = (-3,2), so we have,

    f(x) = a(x + 3)^2 + 2

    Now we also have, for the y-intercept, f(0) = 4
    => f(0) = 4 = a(3)^2 + 2
    => 4 = 9a + 2
    => a = 2/9

    so our parabola is f(x) = (2/9)(x + 3)^2 + 2

    EDIT: Beaten again! Thanks a lot ecMathGeek
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  4. #4
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    That is what I got as well, however when putting into the form y=ax^2+bx+c I am gettng the wrong answer.

    Can anyone assist?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dasani View Post
    That is what I got as well, however when putting into the form y=ax^2+bx+c I am gettng the wrong answer.

    Can anyone assist?
    f(x) = (2/9)(x + 3)^2 + 2
    => f(x) = (2/9)(x^2 + 6x + 9) + 2
    => f(x) = (2/9)x^2 + (2/9)6x + (2/9)9 + 2
    => f(x) = (2/9)x^2 (4/3)x + 4
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  6. #6
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    Thank you so much, that's where the problem was. I wasn't simplyfing it correctly.
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