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Math Help - Completing the square to convert to vertex form

  1. #1
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    Completing the square to convert to vertex form

    I understand the operations involved in the process, but what I do not understand is this statement:

    "In the vertex form of the quadratic, the fact that (h, k) is the vertex makes sense if you think about it for a minute, and it's because the quantity "x h" is squared, so its value is always zero or greater; being squared, it can never be negative."

    I found this on the purplemath website.
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    Re: Completing the square to convert to vertex form

    Also, in the example they provide, the first-degree x term disappears at and they don't give any explanation for this. Here is the link

    Completing the Square: Finding the Vertex
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Completing the square to convert to vertex form

    Can you be more clear what you don't understand?
    I guess you understand (x-h)^2>0 this statement is always true.
    And I don't see where the first-degree x term disappears, may be can you give the example?
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    Re: Completing the square to convert to vertex form

    Now that I look back at the example from the purplemath page I see it more clearly, I understand it now.Yes, I understand that (x-k)^2 > 0, but how come it having to be zero or positive makes it obvious that it is the vertex?
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  5. #5
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    Re: Completing the square to convert to vertex form

    Just to stick my oar in- strictly speaking, (x- h)^2> 0 is NOT true. What is true is that (x- h)^2\ge 0. Of course, x= 0 gives the vertex itself. For all other x x- h is not 0, and then (x- h)^2> 0 so that x= 0 gives the lowest point of the parabola.
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  6. #6
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    Re: Completing the square to convert to vertex form

    I understand the first portion of what you are saying, but could you clarify this part "For all other x x- h is not 0, and then (x- h)^2> 0 so that x= 0 gives the lowest point of the parabola."

    Thank you
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