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Math Help - Quadratic "fun"ction

  1. #1
    Member M670's Avatar
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    Talking Quadratic "fun"ction

    So I got one problem perfect but this one is just not happening ....

    Consider the Quadratic function .
    Its vertex is (, ).
    Its -intercepts are .
    Note: If there is more than one answer enter them separated by commas.
    Its -intercept is .
    So I am being told -46 and my x intercepts are wrong?
    but to find k= c-\frac{b^2}{4a}  k=-21-\frac{10^2}{4} and then when I plug the function into this \frac{-b+-\sqrt b^2-4ac}{2a} I got  -5+-\sqrt46 Sorry I cant seem to get my root to cover the whole part over b^2-4ac

    My answer are vertex = (5,-46)  x=-5+\sqrt46 x=-5-\sqrt46 and y=-21
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  2. #2
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    Re: Quadratic "fun"ction

    f(x) = -x^2 + 10x - 21

    f(x) = -(x^2 - 10x + 21)

    f(x) = -(x - 3)(x - 7)

    try again ...
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  3. #3
    Member M670's Avatar
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    Re: Quadratic "fun"ction

    Quote Originally Posted by skeeter View Post
    f(x) = -x^2 + 10x - 21

    f(x) = -(x^2 - 10x + 21)

    f(x) = -(x - 3)(x - 7)

    try again ...
    But what about if I do it this way
       x^2 + bx + c \;=\; (x-h)^2 + k,\quad\text{where}\quad h = -\frac{b}{2} \quad\text{and}\quad k = c - \frac{b^2}{4}.
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  4. #4
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    Re: Quadratic "fun"ction

    And after you have done that, what does it tell you about the x-intercepts? Also notice that your given function is NOT of the form " x^2+ bx+ c"
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  5. #5
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    Re: Quadratic "fun"ction

    f(x) = -x^2 + 10x - 21

    f(x) = -(x^2 - 10x) - 21

    f(x) = -(x^2 - 10x + 25) - 21 + 25

    f(x) = -(x-5)^2 + 4
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  6. #6
    Member M670's Avatar
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    Re: Quadratic "fun"ction

    Quote Originally Posted by HallsofIvy View Post
    And after you have done that, what does it tell you about the x-intercepts?
    Oh I see what skeeter was trying to show me now, my x intercepts... But I still can't get the K value? and my other question why does the
    x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}, not give my x intercept?
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  7. #7
    Member M670's Avatar
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    Re: Quadratic "fun"ction

    Does this question have to be solved in the manner you are showing me?
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  8. #8
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    Re: Quadratic "fun"ction

    So does this apply to all my quadratic equations with a coefficient of -1 infront of my x^2 I need to complete the square ? and not solve by using the quadratic formula?
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  9. #9
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    Re: Quadratic "fun"ction

    for your vertex use (-b/2a, F( -b/2a)) those will be the coordinates for your vertex.
    after that factor your quadratic as skeeter did above and set each binomial equal to zero and solve for x, those will be your roots(or x-intercepts).
    for example: X^2-3x-4, when factored becomes (x-4)(x+1) set (x-4) =0 and solve do the same for (x+1). those would give you roots.
    and it seems you already know how to do y.
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