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Math Help - Please reply my exam is tomorrow!!!!!!!!!

  1. #1
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    [B][U]Please reply my exam is tomorrow!!!!!!!!![/U][/B]

    how does one work out the maximum point by completing the square


    thnks reply quickly plz


    osmosis786
    Last edited by osmosis786; May 17th 2009 at 11:09 AM.
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  2. #2
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    Quote Originally Posted by osmosis786 View Post
    how does one work out the maximum point by completing the square


    thnks reply quikly plz


    osmosis786
    We cant understand your problem, unless you post a problem..
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  3. #3
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    Quote Originally Posted by osmosis786 View Post
    how does one work out the maximum point by completing the square


    thnks reply quikly plz


    osmosis786
    For the quadratic equation: ax^2+bx+c = 0

    Divide through by a: x^2+\frac{b}{a}x + \frac{c}{a} = 0

    Take c/a from both sides: x^2+\frac{b}{a}x = -\frac{c}{a}

    Then make it into the form (ax+b)^2 and add b^2 to both sides: (x+\frac{b}{2a})^2 = \frac{b^2}{4a^2} -\frac{c}{a} = \frac{b^2-4ac}{4a^2}

    Can you go from there?
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  4. #4
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    Exclamation how do i find the maximum point of this equation using completing the square

    how do i find the maximum point of this equation using completing the square

    7x + 15 − 2x^2

    can u please help as i know how to do minimum points but not maximum points thanks you
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  5. #5
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    Quote Originally Posted by osmosis786 View Post
    how do i find the maximum point of this equation using completing the square

    7x + 15 − 2x^2

    can u please help as i know how to do minimum points but not maximum points thanks you
    ???? I thought the method in finding the minimum point or the maximum point of a quadratic is the same -- you put the equation into the form y = a(x - h)^2 + k by completing the square. It's just that when the leading coefficient is positive the vertex is the minimum point, and when the leading coefficient is negative the vertex is the maximum point.

    y = 7x + 15 - 2x^2
    y = -2x^2 + 7x + 15
    y = -2\left(x^2 - \frac{7}{2}x\right) + 15

    Now complete the square. When you can get the quadratic into the form
    y = a(x - h)^2 + k, the vertex is (h, k).


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  6. #6
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    Quote Originally Posted by yeongil View Post
    ???? I thought the method in finding the minimum point or the maximum point of a quadratic is the same -- you complete the square. It's just that when the leading coefficient is positive the vertex is the minimum point, and when the leading coefficient is negative the vertex is the maximum point.

    y = 7x + 15 - 2x^2
    y = -2x^2 + 7x + 15
    y = -2\left(x^2 - \frac{7}{2}x\right) + 15

    Now complete the square. When you can get the quadratic into the form
    y = a(x - h)^2 + k, the vertex is (h, k).


    01
    yes but in the minimum point one must get 'x' to be the smallest possible value so it must be 0

    however i do not know how to get 'X' to be as big as possible

    maybe i am being stupid and there is a simple thing to do but i cant get my head round it
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  7. #7
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    Quote Originally Posted by osmosis786 View Post
    yes but in the minimum point one must get 'x' to be the smallest possible value so it must be 0

    however i do not know how to get 'X' to be as big as possible

    maybe i am being stupid and there is a simple thing to do but i cant get my head round it
    Sorry, I do not understand what you're saying here. Maybe because of our locations the terminology is different, but I always understood that in a parabola the minimum or maximum point is at the vertex.


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  8. #8
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    Cool

    Quote Originally Posted by yeongil View Post
    Sorry, I do not understand what you're saying here. Maybe because of our locations the terminology is different, but I always understood that in a parabola the minimum or maximum point is at the vertex.


    01
    errmmm i think i know how to do it know...

    but thanks to everyone who tried to help even though i wasnt being clear ......
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  9. #9
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    Quote Originally Posted by osmosis786 View Post
    yes but in the minimum point one must get 'x' to be the smallest possible value so it must be 0

    however i do not know how to get 'X' to be as big as possible

    maybe i am being stupid and there is a simple thing to do but i cant get my head round it
    You mean y, correct? And you do realize our minimums can be negative (i.e. -5 which is less than 0)?
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