Can you tell/show me the vertex...?

Question

I'm in precalculus and we just started Introduction to Conics: Parabolas. I want to know what's the vertex of:
A. y-7= -2(x-3)^2

B. y=4(x-2)^2

C. x^2 +6x+4y-3=0

Please it will help me alot. The ones with the best explanation on how to do it will get the 10pts.

Answer 1

A. y-7= -2(x-3)²

y = -2(x-3)² + 7
Vertex (3,7)

______

B. y=4(x-2)²

y=4(x-2)² + 0
Vertex (2,0)

_____

C. x^2 +6x+4y-3=0

-4y = x² + 6x -3
-4y = [x² + 6x + 9] -9 -3
-4y = [x² + 2·3·x + 3²] -12
-4y = [x + 3]² -12
divide by (-4)
y = -1/4 [x + 3]² + 3

Vertex (-3, 3)

__________

In the equation

y = A · [x + B]² + C,

the coordinates of the vertex are (-B, C)!

Hope this helps.

Answer 2

A (3,7)
B (2,0)
C 4y =-(x^2+6x-3) = -(x+3)^2 +12
y=-1/4(x+3)^2+3 The vertex is (-3,3)

In general if you complete the square and get
y=a(x-b)^2+c the vertex is (b,c)