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anonymous

  • one year ago

I am working on Analyzing Graphs of Quadratic Functions, the problem I am stuck on is f(x)=x^2/2+4x+4. I think the area I am messing up on is the squaring. Someone please putting this in the most simplest form. Algebra is my worst enemy.

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  1. anonymous
    • one year ago
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    what exactly are you supposed to do?

  2. anonymous
    • one year ago
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    put it in the form f(x)=a(x-h)^2+k. find the vertex and graph it.

  3. anonymous
    • one year ago
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    is it \[y=\frac{x^2}{2}+4x+4\]?

  4. anonymous
    • one year ago
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    that's the original function. but I don't know how to work it out to because an order pair

  5. anonymous
    • one year ago
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    if you want to write in vertex form first factor out the \(\frac{1}{2}\) from the first two terms and write \[y=\frac{1}{2}(x^2+2x)+4\]

  6. anonymous
    • one year ago
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    easiest way to find the vertex is to use \(-\frac{b}{2a}\) for the first coordinate

  7. anonymous
    • one year ago
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    in your example it is \[-\frac{4}{2\times \frac{1}{2}}=-4\]

  8. anonymous
    • one year ago
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    i wrote something stupid \[y=\frac{1}{2}(x^2+2x)+4\] if you factor out the \(\frac{1}{2}\) you actually get \[y=\frac{1}{2}(x^2+\color{red}8x)+4\]

  9. anonymous
    • one year ago
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    thanks I got it now.

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spraguer (Moderator)
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