January 11th, 2007, 12:49 PM  #1 
Newbie Joined: Dec 2006 Posts: 28 Thanks: 0  Parabola
Complete the square to graph the parabola, y = 2x^2  8x  4 . y = ax^2 + bx + c ....this way? 
January 11th, 2007, 02:23 PM  #2 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
That is the format that it is currently in. To complete the square, change the equation to the form y  h = a(x  k)². Then the vertex will be at (h,k). From there, moving n units left or right from the vertex causes the point to be an² units above the vertex. In this case, since a will be negative, you will move an² units down from the vertex, when you move left or right n units.

January 11th, 2007, 09:39 PM  #3 
Newbie Joined: Jan 2007 From: USA Posts: 12 Thanks: 0 
Here's the steps... y = 2x^2  8x  4 Divide by 2 to get rid of the leading coefficient: 1/2y = x^2 + 4x + 2 Take half of 4 squared; you need a 4 to complete the square so add 2: 1/2y+ 2 = x^2 + 4x + 4 You now have a perfect square trinomial so factor it: 1/2y + 2 = (x+2)^2 Solve for y by multiplying by 2; then add 4 to get y by itself y  4 = 2(x+2)^2 y = 2(x+2)^2 + 4 It is now in the form y = a(xh)^2 + k You can figure it out now from here 

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