# Find the maximum or minimum value of the quadratic function by completing the square.

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 :'(

• Please show us how you tried it and we can better help you by telling you where it went wrong. – naslundx Feb 26 '14 at 17:56
• you came up falsely with 7/9 but the correct term was 7/6. How did you find 7/9? – Jimmy R. Feb 26 '14 at 18:07
• @Stefanos I thought I have to divide $\frac{7}{3}$ by $3$, it was wrong. :P – Kiara Feb 26 '14 at 18:10

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.