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Math Help - Help with Quadratics

  1. #1
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    Help with Quadratics

    I am having trouble understanding a problem that has been explained to me. I previously believed that I understood how to convert an equation from standard to vertex form (completing the sqaure) however, this question, and the subsequent answer has stumped me. Can anyone make sense of this for me?

    ---

    Solve for vertex:

    h(d) = -0.051(d)^2 + 2.04(d) + 1.6

    Now, here I will try to put the answer I was given, but it is rather complicated and I am not sure of my ability to explain it using a keyboard. I have no idea how to use math coding.

    h(d) = -0.051(d^2 - 2.04(d) + (2.04)^2) + 0.051(2.04)^2 + 1.6
    -------------------0.051------0.102------------0.102

    h(d) = -0.051(d-20)^2 + 22

    Therefore, the vertex is (20,22)

    ---

    Am I missing something or is this a problem that is not solved by the standard complete the square method? To put it simply I do not understand the solution at all, can anyone help my confusion?
    Last edited by Aerillious; June 19th 2008 at 07:36 PM. Reason: Trying to make equations appear neater.
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  2. #2
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    Hi, Aerillious!

    Quote Originally Posted by Aerillious View Post
    Solve for vertex:

    h(d) = -0.051(d)^2 + 2.04(d) + 1.6

    Now, here I will try to put the answer I was given, but it is rather complicated and I am not sure of my ability to explain it using a keyboard. I have no idea how to use math coding.
    First let me fix your notation here, since HTML will by default eat extra spaces like that (put your stuff between [code] and [/code] tags to fix that, or better yet, learn to use \text{\LaTeX}):

    Quote Originally Posted by Aerillious View Post
    h(d) = -0.051\left(d^2 - \frac{2.04}{0.051}\cdot d + \left(\frac{2.04}{0.102}\right)^2\right) + 0.051\left(\frac{2.04}{0.102}\right)^2 + 1.6

    h(d) = -0.051(d-20)^2 + 22

    Therefore, the vertex is (20,\;22)

    ---

    Am I missing something or is this a problem that is not solved by the standard complete the square method? To put it simply I do not understand the solution at all, can anyone help my confusion?
    This does, more or less, use the standard method of completing the square. Let me try to elucidate it some more:

    We have h(d) = -0.051d^2 + 2.04d + 1.6. First, when completing the square, you want to get rid of the coefficient of the 2nd-degree term. So they factored a -0.051 from the first two terms:

    h(d) = -0.051\left(d^2 - \frac{2.04}{0.051}d\right) + 1.6

    Now we take half of the coefficient of the linear term and square it.

    Since our coefficient is b = -\frac{2.04}{0.051}, we find \left(\frac b2\right)^2 = \left(-\frac{2.04}{2\cdot0.051}\right)^2 = \left(\frac{2.04}{0.102}\right)^2.

    We can add and subtract this quantity like so:

    h(d) = -0.051\left(d^2 - \frac{2.04}{0.051}d + \left(\frac{2.04}{0.102}\right)^2 - \left(\frac{2.04}{0.102}\right)^2\right) + 1.6

    But instead of leaving the second constant term inside the parentheses, let's move it outside, where it belongs, so that we can group it with the 1.6. To do this we must distribute the -0.051:

    h(d) = -0.051\left(d^2 - \frac{2.04}{0.051}d + \left(\frac{2.04}{0.102}\right)^2\right) - \left(-0.051\right)\left(\frac{2.04}{0.102}\right)^2 + 1.6

    \Rightarrow h(d) = -0.051\left(d^2 - \frac{2.04}{0.051}d + \left(\frac{2.04}{0.102}\right)^2\right) + 0.051\left(\frac{2.04}{0.102}\right)^2 + 1.6

    Now we factor the inner expression:

    h(d) = -0.051\left(d - \frac{2.04}{0.102}\right)^2 + 0.051\left(\frac{2.04}{0.102}\right)^2 + 1.6

    Note that \frac{2.04}{0.102} = 20, so we have

    h(d) = -0.051\left(d - 20\right)^2 + 0.051(20)^2 + 1.6

    \Rightarrow h(d) = -0.051\left(d - 20\right)^2 + 22

    Looking at the standard form for the equation of a parabola with vertex at (h,\;k) and focus at (h,\;k+p): y = \frac1{4p}(x - h)^2 + k, we see that our vertex is at (20,\;22).
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  3. #3
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    I apologize for the messy text, I will try to learn LATEX when I have the time.

    As for your answer, I can't thank you enough. I believe I was confused as a result of the answer trying to squeeze everything into one line instead of breaking it down like you did.

    Thanks again.
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