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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^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!!
 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: 461 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: $\displaystyle y=\frac{1}{6}(x-3)^2-\frac{1}{2}$ Thanks from agentredlum 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: 131 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. Thanks from agentredlum

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