How do you write the equation of a parabola given the vertex and y-intercept?

the standard parabola in vertiex form is y = a(x-x0)^2 + y0, (x0,y0) is (2,0), from (0,4) you can get the a value, a = (y-y0)/x0^2

Use vertex form, y = a(x-h)^2 +k where h is the x coordinate of the vertex and k is the y coordinate of the vertex.  Right now, you know the vertex (2, 0) and a point that just happens to be the y-intercept (0, 4).  Plug these values into the equation.  So, the h value is 2, the k value is 0 and we will use the x and y from the y-intercept for the x and y values in the equation.  The only value we don't have is a, and that is what we need to solve for.  Now the equation looks like this

4 = a(0-2)^2 +0

Simplify to

4 = a(-2)^2
4 = a(4)

Divide both sides by 4 and get a=1.  Use that in your answer.  Now we have

y = 1(x-2)^2 + 0

or just

y = (x-2)^2

Let's start with the answer.

The vertex is the lowest point when ax^2 opens upward. Another way to say that is that a>0.

y = (x - 2)^2 is where this happens.
y = a(x - 2)^2 + 0 is the final answer. which shortens down to y = a(x - 2)^2

The y intercept tells you that a = 1. Here's how.
y = a(x - 2) ^2 The y intercept occurs when x = 0
y = a(0  - 2)^2
4 = a(-2)^2
4 = a*4
a = 1.

Now suppose that you had 2,0 and 0,8 instead.

y = a(x - 2)^2 + 0 just as before, but the 8 has to be considered.
8 = a(x - 2)^2
8 = a (0 - 2)^2
8 = a*4
a = 2

answer y = 2(x - 2)^2

One more example. Suppose the vertex is (2 ,-4) and the y intercept is 0,6 Now what?

y = a(x - 2)^2 - 4 Notice what happened. When 2 is put in for x, the result is - 4. That's the y value of  - 4 which is as small as it gets.

Now when x = 0, we find out what a is

y = a(0 - 4)^2 - 4
6 = a(16) - 4 Add 4 to both sides.
10 = a(16)
a = 10/16 = 5/8

It isn't pretty, but that's the method. Always do this question in the same order

b,0 means a(x -b)^2

0,c means x = 0 and you solve for a.