Ecuation of Parabola from Point in vertex form I've been recently studying parabolas and I learnt about the vertex form and what the values meant. Recently on a small test I came across this problem.
For the vertex at point (2,-3) find the ecuation in vertex form. 
Well I simply thought It would be as 
$$y+3=(x-2)^2$$
But the grader told me I was wrong, and supposedly the answer was 
$$y+3=-\frac{1}{3}\left(x-2\right)^2$$
I have absolutely no clue where that 
$-\frac{1}{3}$ came from. The other piece of information I had was another point on the Parabola : (5,-6)
Could someone explain me where $-\frac{1}{3}$ came from I would appreciate it! 
Thanks! 
 A: The general equation should be $$4a(y+3)=(x-2)^2$$ and you have to use the point to get $a$ using the given point we have $$4a(-6+3)=(5-2)^2$$ This implie that $4a=-3$ and so
$$-3(y+3)=(x-2)^2
\\(y+3)=\frac{-1}{3}(x-2)^2$$
PRABOLA
Notice the importance of the constant $a$
A: After some confusion the way of solving for the vertex form of a parabola as this one is ...
Point A: (2,-3) Point B:(5,-6)
Vertex Form Ecuation : 
$$y = a(x – h)^2 + k$$
We'll first work the basic part of the vertex for by making
$$y+3=(x-2)^2$$
After this we realized that the parabola has a vertex on a negative y point, for this there must be a value in front of the $(x-2)^2$ as it's the one that defines it, 
We'll use our previosly found ecuation to find a as
$$(y+3)=a(x-2)^2$$
We'll plug into this ecuation our Point B that we know it's a point on the parabola, doing so we get 
$$(-6+3)=a(5-2)^2$$
After simplifying we get
$$a(-3)=a(9)$$
We solve for a 
$$\frac{-3=9a}{-3}$$
$$a=-\frac{-3}{9}$$
So we end up having
$$y+3=\frac{-3}{9}(x-2)^2$$
Or a simplified version
$$y+3=\frac{-1}{3}(x-2)^2$$
