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Thread: Identifying the vertex of a Quadratic Function

  1. #1
    Sep 2009

    Identifying the vertex of a Quadratic Function

    I really don't mean to be a nusance, I am new here and hope to be more of a contributer of this site rather that bothering people, so I apologize. I seem to have a hard time with a specific step in taking a quadratic trinomial that cannot be factored and and converting it into a vertex formula [f(x)=(x-h)+k].

    In this case:

    f(x)= 2x + 8x + 7

    I know that the first step is to group the x terms leaving the leading coefficient out of the parenthesis such as:

    f(x)= 2(x + 4x) + 7

    Next step is to take the middle term and square half the coefficient: (4/2) and then add and subtract by that number.

    f(x)= 2(x + 4x+4-4) + 7

    Now here is where it gets fuzzy for me, I read this online and here is the further step according to what I read:

    Now, move the second -4 to the outside of the parenthesis, only now we have to consider that the outside coefficient isn't just a simple positive one. There's a 2 out there.
    f(x) = 2(x + 4x + 4) - 2(4) + 7

    Question, is that -2 that is underlined the -4 that was outside the parentesis? And if so where did that four to the right of it come from? I'm stumped as to how it got there because if the -2 was the -4 inside the parenthesis and we have the +4 inside the parenthesis, that (4) that is next to the -2 seems to have crashed the party. I just can't figure out how it got there according to there steps and there seems to be no sufficient information as to how it got there, please explain.
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  2. #2
    Sep 2009
    Alright, given: $\displaystyle f(x)=2x^2+8x+7$

    You want it in the form $\displaystyle f(x)=a(x-h)^2+k$, which can also be written as $\displaystyle y-k=a(x-h)^2$

    So, subtract over 7 and factor out a 2: $\displaystyle y-7=2(x^2+4x)$

    You understand the step of completing the square. Divide 4 by 2, square, and add: $\displaystyle 2(x^2+4x+4)$ Because we added 8 (you have to multiple $\displaystyle a$ through the quadratic) to one side, we add it to the other:

    $\displaystyle y+1=2(x+2)^2$, or
    $\displaystyle f(x)=2(x+2)^2-1$

    Vertex = $\displaystyle (h,k)$, or $\displaystyle (-2, -1)$
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