completing the square to graph the parabola y=-2x^2-8x-4 I got stuck but here is what I have 20+y=2(x-4)^2

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completing the square to graph the parabola y=-2x^2-8x-4 I got stuck but here is what I have 20+y=2(x-4)^2

Mathematics
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y+4 = -2x^2 -8x ; divide everything by -2 y-2 = x^2 +4x ; complete the square.. (4/2)^2..+4 to each side y-2+4 = x^2 +4x +4
That's perfect. This is telling you that it's a translation in y down 20 units, and to the right in x 4 units.
y+2 = (x+2)^2 1(y+2) = (x+2)^2

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forgot to divde y by -2 :)
let me show you part of how my teacher showed me how to work it
(y/-2) +2 = (x+2)^2 ; factor the -1/2 from the y (-1/2)(y-4) = (x+2)^2
Actually though, your answer isn't right Marie.
It's in the right form though
the vertex is at (-2,4) and it opens downward....
y+4=-2x^2-8x take the -8/2= -4 then square it which is 16 add 16 to both sides y+20=-2x^2-8x+16 now I take the right side and put it in paranthesis be fore it was added togather 20+y=-2(x-4)^2
the usual look to this is: y = -2(x+2)^2 +4
you cant complete the square when the first term constant is not = 1
divide everything by -2 first
\[y = -2x^2 -8x -4\] \[y = -2(x^2 + 4x + 2) \] \[y = -2(x^2 + 4x + 2 + (2 - 2))\] ^ Completing the square \[y = -2(x^2 + 4x + 4 -2)\] \[y = -2(x+2)^2 +4\] \[y -4 =-2(x+2)^2\]
y+10=(x-4)^2
Whoops, sorry. Didn't mean to jump the gun there
its ok, thats what I got too :)
You have to factor out the -2 first, then work inside to complete the square.
is the point that i'm tryin to find to graph the parabola (4, -10)?
no, its (-2,4) as the vertex
ok thanks for your help
I showed the work, so you can see where you went wrong.
you have to factor out or divide out that -2 that is sitting in front of the x^2 before you can complete the square...
Just be sure to factor off any coefficients from your \(x^2\) term
Yeah.

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