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Math Help - How did this happen?

  1. #1
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    How did this happen?

    I am maximizing and minimizing the optimum. Once I had this quadratic function, f(x)=-[x^2-4x+4-4-2] the book says I factor it into vertex form. Giving me two equations in vertex form, f(x)=-(x-2)^2-6 and f(x)=-(x-2)^2+6. Any help understanding the factoring of that quadratic function?
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  2. #2
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    Re: How did this happen?

    Quote Originally Posted by dLogic View Post
    I am maximizing and minimizing the optimum. Once I had this quadratic function, f(x)=-[x^2-4x+4-4-2] the book says I factor it into vertex form. Giving me two equations in vertex form, f(x)=-(x-2)^2-6 and f(x)=-(x-2)^2+6. Any help understanding the factoring of that quadratic function?
    Only the second equation is correct. The first is wrong. You had f(x) = -(x^2-4x+4-4-2) = -(x^2-4x-2). The first equation you wrote:

    \begin{align*}-(x-2)^2-6 & = -(x^2-4x+4)-6 \\ & = -x^2+4x-10 \\ & = -(x^2-4x+10) \\ & \neq -(x^2-4x-2)\end{align*}.

    So, I don't understand why the book gives you both equations. Only the second equation is correct. Anyway, to show you how the factoring works:

    \begin{align*}f(x) & = -[x^2-4x+4-4-2] \\ & = -[(x^2-4x+4)-4-2] \\ & = -[(x^2-4x+4)-6]\end{align*}

    Now, factor x^2-4x+4 into (x-2)(x-2) = (x-2)^2. Then you have f(x) = -[(x-2)^2-6] = -(x-2)^2+6.

    The process the book used is called completing the square.
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  3. #3
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    Re: How did this happen?

    It is an optimization problem. I don't know if that helps to understand why they wrote the equation that doesn't equal the quadratic. I imagine the optimization idea is why we take 1/2 of the coefficient of and multiply it (-4) to get 4. Then +4-4 to the original equation. Then factor two vertexes? hmm...
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  4. #4
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    Re: How did this happen?

    Quote Originally Posted by dLogic View Post
    It is an optimization problem. I don't know if that helps to understand why they wrote the equation that doesn't equal the quadratic. I imagine the optimization idea is why we take 1/2 of the coefficient of and multiply it (-4) to get 4. Then +4-4 to the original equation. Then factor two vertexes? hmm...
    It is because any parabola can be put into the following form: f(x) = p(x-a)^2 + b. If p>0 then b is the minimum value of f(x). If p<0 then b is the maximum value of f(x). There is only one formula. I don't know what you mean factor "two vertexes". There is only one "vertex" of a parabola.

    Given a function f(x) = a_2x^2+a_1x+a_0, a_2 \neq 0, let's put it into the form f(x) = p(x-a)^2 + b.

    First, factor out a_2 from each term: f(x) = a_2\left(x^2 + \dfrac{a_1}{a_2}x + \dfrac{a_0}{a_2}\right)
    Now, we want to complete the square. So, we take half of the coefficient of x and square it. That is \left(\dfrac{a_1}{2a_2}\right)^2 = \dfrac{a_1^2}{4a_2^2}. We add and subtract it inside the parentheses:

    f(x) = a_2\left(x^2 + \dfrac{a_1}{a_2}x + \dfrac{a_1^2}{4a_2^2} - \dfrac{a_1^2}{4a_2^2} + \dfrac{a_0}{a_2}\right)

    Now, we use the associative rule to gather together the terms we want:

    f(x) = a_2\left(\left[x^2 + \dfrac{a_1}{a_2}x + \dfrac{a_1^2}{4a_2^2}\right] - \dfrac{a_1^2}{4a_2^2} + \dfrac{a_0}{a_2}\right)

    We can now factor the expression \left[x^2 + \dfrac{a_1}{a_2}x + \dfrac{a_1^2}{4a_2^2}\right] = \left(x + \dfrac{a_1}{2a_2}\right)^2.

    So, now we have:

    f(x) =a_2\left(\left[x + \dfrac{a_1}{2a_2}\right]^2 - \dfrac{a_1^2}{4a_2^2} + \dfrac{a_0}{a_2}\right)

    Multiplying out the a_2, we get:

    f(x) = a_2\left(x + \dfrac{a_1}{2a_2}\right)^2 - \dfrac{a_1^2}{4a_2} + a_0

    So, we have p = a_2, a = -\dfrac{a_1}{2a_2}, and b = a_0 - \dfrac{a_1^2}{4a_2}.

    Now, the quadratic is in the form the book wants.
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    Re: How did this happen?

    Thank you for your help. I appreciate it.
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