Standard Form of Parabola with given point and vertex?

Question

How do I find the standard form of an equation with a given vertex and a given point on the parabola? I only know how to do it if they give me a focus... thank you.
For example:
Vertex: (0,0)
Point on Parabola: (3,6)
Find standard form

Answer 1

I assume that this question assumes y = f(x). That is the parabola has "standard orientation", which means that its axis of symmetry is parallel to the y-axis. If you don't assume that, then you don't have enough information to answer.

The general form for a parabola in standard orientation is:

y = y0 + α(x - x0)².

where the location of the vertex is (x0, y0). Thus you know, in this case, that x0 = 0 and y0 = 0. So you just have to find α.

The parabola has to go thru those two points ((0,0) and (3,6)) and its extreme value (vertex) is at (0,0). Since the vertex is (0,0) f(x) has the form a α x². To solve for a use (3,6):

6 = α 3² = 9α.

Therefore α = 6/9 = 2/3.
Thus f(x) = (2/3) x².

Answer 2

It's confusing because your vertex and the other point both have the same y coordinate, which seems impossible. Maybe you meant the point is (-5, 10)? Then you'd use x - 0 = a(y - 0)^2 if the parabola has a horizontal axis of symmetry x = ay^2 and you know x = -5 when y = 10 so -5 = 100a; a = -5/100 = -1/20 so x = -1/20 y^2, or multiplying by -20, -20x = y^2 But that's a sideways parabola. How did you know it had to be a sideways parabola? I guess I'm still confused too.

Answer 3

This is not enough information. I assume your parabola is aligned to one of the axes.

1) Let's assume it is vertical.

y - 0 = a(x - 0)²
y = ax²

Plug in the point.

y = ax²
6 = a(3²) = 9a
a = 2/3

y = (2/3)x²

2) The parabola could be sideways.

x - 0 = b(y - 0)²
x = by²

Plug in the point.

x = by²
3 = b(6²)
3 = 36b
b = 3/36 = 1/12

x = (1/12)y²