Question

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

Answer 1

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

Answer 2

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.

Answer 3

y+2 = (x+2)^2

1(y+2) = (x+2)^2

Answer 4

forgot to divde y by -2 :)

Answer 5

let me show you part of how my teacher showed me how to work it

Answer 6

(y/-2) +2 = (x+2)^2 ; factor the -1/2 from the y

(-1/2)(y-4) = (x+2)^2

Answer 7

Actually though, your answer isn't right Marie.

Answer 8

It's in the right form though

Answer 9

the vertex is at (-2,4) and it opens downward....

Answer 10

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

Answer 11

the usual look to this is:

y = -2(x+2)^2 +4

Answer 12

you cant complete the square when the first term constant is not = 1

Answer 13

divide everything by -2 first

Answer 14

\[y = -2x^2 -8x -4\]
\[y = -2(x^2 + 4x + 2) \]
\[y = -2(x^2 + 4x + 2 + (2 - 2))\]
^ Completing the square
\[y = -2(x^2 + 4x + 4 -2)\]
\[y = -2(x+2)^2 +4\]
\[y -4 =-2(x+2)^2\]

Answer 15

y+10=(x-4)^2

Answer 16

Whoops, sorry. Didn't mean to jump the gun there

Answer 17

its ok, thats what I got too :)

Answer 18

You have to factor out the -2 first, then work inside to complete the square.

Answer 19

is the point that i'm tryin to find to graph the parabola (4, -10)?

Answer 20

no, its (-2,4) as the vertex

Answer 21

ok thanks for your help

Answer 22

I showed the work, so you can see where you went wrong.

Answer 23

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...

Answer 24

Just be sure to factor off any coefficients from your \(x^2\) term

Answer 25

Yeah.