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Math Help - Write the equation of a quadratic function given only the vertex

  1. #1
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    Write the equation of a quadratic function given only the vertex

    The function has a minimum value of 6 at x = 4.

    I know that part of the equation will be y = a(x - 4) + 6, though I have no idea how to get a. The answer is " y = 3(x - 4)^2 +6; answers may vary". The only thing I can think of would be to create a table of values to input an x and y value. Though I am unsure of how to do that. Any suggestions?
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    Quote Originally Posted by shadow6 View Post
    The function has a minimum value of 6 at x = 4.

    I know that part of the equation will be y = a(x - 4) + 6, though I have no idea how to get a. The answer is " y = 3(x - 4)^2 +6; answers may vary". The only thing I can think of would be to create a table of values to input an x and y value. Though I am unsure of how to do that. Any suggestions?
    What you'd have to do, is expand that function, to get;

    y = (ax - 4a)(x - 4) + 6

    y = ax^2 - 8ax + 16a + 6

    Then equate to y = 6 and x = 4.
    6 = 16a - 32a + 16a + 6

    Hence, 0 = 0

    Damn.
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  3. #3
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    Quote Originally Posted by shadow6 View Post
    The function has a minimum value of 6 at x = 4.

    I know that part of the equation will be y = a(x - 4){\color{red}^2} + 6, though I have no idea how to get a. The answer is " y = 3(x - 4)^2 +6; answers may vary". The only thing I can think of would be to create a table of values to input an x and y value. Though I am unsure of how to do that. Any suggestions?
    Note the correction in red.

    Without further information the value of a cannot be got. The only thing that can be said is that a > 0. So either you have not posted all of the question or the question is incomplete.
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    Just read the answers to the succeeding question. Basically, a can be any real number; hence why answers may vary.
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    Quote Originally Posted by shadow6 View Post
    Just read the answers to the succeeding question. Basically, a can be any real number; hence why answers may vary.
    No, the parabola has a minimum therefore a > 0, which is not the same as any real number.
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    Quote Originally Posted by shadow6 View Post
    The function has a minimum value of 6 at x = 4.

    I know that part of the equation will be y = a(x - 4) + 6, though I have no idea how to get a. The answer is " y = 3(x - 4)^2 +6; answers may vary". The only thing I can think of would be to create a table of values to input an x and y value. Though I am unsure of how to do that. Any suggestions?
    Here's a video explanation:
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