Question

find the vertex of the parabola (x+3)^2=4(y+2)

Answer 1

Im sure the first part is just foiling out the x+3
which is x^2+6x+9

Answer 2

don't "foil"

Answer 3

bahaha ok, well, what should I do then?

Answer 4

I just have to say your picture made me laugh so hard. lolol

Answer 5

from the form you are in you see the first coordinate of the vertex is \(-3\) because that will make \((x+3)^2=0\)
replace \(x\) by \(-3\) and see what you get for \(y\)

Answer 6

me too, well played @satellite73. I need some laughs at this point. I'm getting burnt out on this stuff haha

Answer 7

\[(x+3)^2=4(y+2)\]
\[0=4(y+2)\iff y=-2\]

Answer 8

it was displayed for you in the form you were given
\[(x-h)^2=a(y-k)\] vertex is \((h,k)\)

Answer 9

what became of the 4?

Answer 10

oh, wait. nevermind. I guess it's just not needed.

Answer 11

and as far as taking out the opposite as in the parenthesis, thats just to make it equal zero?