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January 11th, 2007, 12:49 PM   #1
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Parabola

Complete the square to graph the parabola, y = -2x^2 - 8x - 4 .

y = ax^2 + bx + c ....this way?
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January 11th, 2007, 02:23 PM   #2
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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.
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January 11th, 2007, 09:39 PM   #3
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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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