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April 8th, 2014, 09:53 AM  #1 
Member Joined: Sep 2013 Posts: 75 Thanks: 0  completing the square
I have x^2  6x + y^2  4y =12 (equation of circle). I have to find the centre. x^26x+ 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(x3)^2  1/2 I dnt understand this step also. Help!!

April 8th, 2014, 10:27 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs 
We are given: $\displaystyle 6y=x^26x+6$ 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 $0=99$ to the right side so we don't change its value: $\displaystyle 6y=\left(x^26x+9 \right)+69$ Rewrite the new quadratic term as a square and combine the other terms $\displaystyle 6y=(x3)^23$ Now divide through by 6: $\displaystyle y=\frac{1}{6}(x3)^2\frac{1}{2}$ Last edited by MarkFL; April 8th, 2014 at 10:30 AM. 
April 8th, 2014, 10:31 AM  #3 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 132 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus 
And the minimum value is attained at x=3, y=1/2.


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