Algebra Pre-Algebra and Basic Algebra Math Forum

 April 8th, 2014, 10:53 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, 11:27 AM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 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 11:30 AM. April 8th, 2014, 11:31 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 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 Tags completing, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Trainpancake Algebra 3 October 2nd, 2013 03:11 PM link107 Algebra 2 September 8th, 2012 03:42 PM pixelpaige Algebra 2 September 8th, 2011 06:39 PM -DQ- Algebra 3 September 10th, 2009 06:51 PM SteveThePirate Algebra 4 May 17th, 2009 01:44 AM

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