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 January 11th, 2007, 11:49 AM #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, 01:23 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 |a|n² units down from the vertex, when you move left or right n units.
 January 11th, 2007, 08:39 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(x-h)^2 + k You can figure it out now from here

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