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Math Help - function

  1. #1
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    function

    1. Find the minimum value of the quadratic y = 2x2 - 8x + 10 by completing the square. Graph the resulting parabola.

    Is this correct?

    y= 2(x^2-4x)+5
    y= 2(x^2-4x+4-4)+5
    y= 2(x-2)^2+1
    Vertex being (2,1)

    If correct, how do I graph this? and which value is the minimum?
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  2. #2
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     y=2x^2-8x+10
     y=2(x^2-4x+5)
     y=2(x^2-4x+4)+(10-8)
     y=2(x^2-4x+4)+2
     y=2(x-2)^2+2

    So the vertex is (2,2)
    Where does the mini occur on a parabolic function?
    You can graph this by plotting points with (2,2) as the vertex.
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  3. #3
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    Quote Originally Posted by Linnus View Post
     y=2x^2-8x+10
     y=2(x^2-4x+5)
     y=2(x^2-4x+4)+(10-8)
     y=2(x^2-4x+4)+2
     y=2(x-2)^2+2

    So the vertex is (2,2)
    Where does the mini occur on a parabolic function?
    You can graph this by plotting points with (2,2) as the vertex.
    So the min is 2?
    why is it 4 inside the bracket and not 5? I don't understand how you did that.
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  4. #4
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    Quote Originally Posted by william View Post
    So the min is 2?
    why is it 4 inside the bracket and not 5? I don't understand how you did that.
    Oh haha sorry. It is a bit confusing.
    Here is a less confusing work

     y=2x^2-8x+10
     y=2(x^2-4x)+10
     y=2(x^2-4x+4)+(10-8)
     y=2(x^2-4x+4)+2
     y=2(x-2)^2+2
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  5. #5
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    Quote Originally Posted by Linnus View Post
    Oh haha sorry. It is a bit confusing.
    Here is a less confusing work

     y=2x^2-8x+10
     y=2(x^2-4x)+10
     y=2(x^2-4x+4)+(10-8)
     y=2(x^2-4x+4)+2
     y=2(x-2)^2+2
    Thanks.
    So the min is 2, correct?
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  6. #6
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    Quote Originally Posted by william View Post
    Thanks.
    So the min is 2, correct?
    Yes, the min is 2.
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