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Find the maximum or minimum function of the quadratic function by completing the squares. State the value of $x$ at which the function is maximum or minimum.

$y=3x^2+7x+9$

I already posted similar question on this topic before, and got great solutions, but as soon as I try another sum my answer doesn't come out right.

Have a look at this: find max or min by completing the square

Maybe I have misunderstood something, can you please tell me the actual and perfect way for solving such problems?

I tried in this way:

$y=3(x^2+\frac{7}{3}x+\frac{9}{3})$

$y=3[(x+\frac{7}{9})^2+\frac{7}{9}+3]$

Is this the right way? If I proceed, $x$ = $-\frac{7}{9}$ and my book says it is not right! Help please :'(

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    $\begingroup$ Please show us how you tried it and we can better help you by telling you where it went wrong. $\endgroup$ – naslundx Feb 26 '14 at 17:56
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    $\begingroup$ you came up falsely with 7/9 but the correct term was 7/6. How did you find 7/9? $\endgroup$ – Jimmy R. Feb 26 '14 at 18:07
  • $\begingroup$ @Stefanos I thought I have to divide $\frac{7}{3}$ by $3$, it was wrong. :P $\endgroup$ – Kiara Feb 26 '14 at 18:10
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By using that $(x+a)^2=x^2+2ax+a^2$, you have that $$\begin{align*}y&=3\left(x^2+(7/3)x+3\right)=3\left(x^2+2(7/6)x+(7/6)^2-(7/6)^2+3\right)\\&=3\left(\left(x+(7/6)\right)^2+3-(7/6)^2\right)\end{align*}$$ (where $a=7/6$). So the minimum is for $x=-(7/6)$ since a square is always $\ge0$. For the named $x$ the term in the square is $0$ so it attends it's minimum value.

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