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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what exactly are you supposed to do?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0put it in the form f(x)=a(xh)^2+k. find the vertex and graph it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it \[y=\frac{x^2}{2}+4x+4\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's the original function. but I don't know how to work it out to because an order pair

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you want to write in vertex form first factor out the \(\frac{1}{2}\) from the first two terms and write \[y=\frac{1}{2}(x^2+2x)+4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0easiest way to find the vertex is to use \(\frac{b}{2a}\) for the first coordinate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in your example it is \[\frac{4}{2\times \frac{1}{2}}=4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wrote something stupid \[y=\frac{1}{2}(x^2+2x)+4\] if you factor out the \(\frac{1}{2}\) you actually get \[y=\frac{1}{2}(x^2+\color{red}8x)+4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks I got it now.
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