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April 8th, 2014, 09:53 AM   #1
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completing the square

I have x^2 - 6x + y^2 - 4y =12 (equation of circle). I have to find the centre. x^2-6x+ 9 + y^2 -4y +4 = 12 +9 +4. (x^2 -3)^2 + (y^2 - 2) =25. Centre is C(3;2). I understand this one BUT the next one. I have 6y = x^2 - 6x + 6 i.e y = 1/6 x^2 - x + 1 its parabola. Write the minimum value of the equation. ? now I have to complete the square. y = 1/6(x^2 - 6x + 9 - 9 + 6) I dnt understand why we add and subtracted 9. 1/6(x-3)^2 - 1/2 I dnt understand this step also. Help!!
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April 8th, 2014, 10:27 AM   #2
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Math Focus: Calculus/ODEs
We are given:

$\displaystyle 6y=x^2-6x+6$

We take the coefficient of the linear term, which is -6, cut it in half which is -3 and then square it which is 9, so we want to add $0=9-9$ to the right side so we don't change its value:

$\displaystyle 6y=\left(x^2-6x+9 \right)+6-9$

Rewrite the new quadratic term as a square and combine the other terms

$\displaystyle 6y=(x-3)^2-3$

Now divide through by 6:

$\displaystyle y=\frac{1}{6}(x-3)^2-\frac{1}{2}$
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Last edited by MarkFL; April 8th, 2014 at 10:30 AM.
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April 8th, 2014, 10:31 AM   #3
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And the minimum value is attained at x=3, y=-1/2.
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