1.)PPF
2.)FFP
hope it helps
This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

where
is a compact way to indicate both solutions
and
, while
are the coefficients of the quadratic equation, i.e. we consider the polynomial
.
So, in your case, we have 
Plug those values into the formula to get

So, the two solutions are


From the problem, the vertex = (0, 0) and the focus = (0, 3)
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm