SOLUTION: Hi, I'm having trouble figuring out a parabola's equation. The Y intercept is (0,6), the vertex is (2,8). These are the only points given. Can you please explain to me how to work

Question 752747: Hi, I'm having trouble figuring out a parabola's equation. The Y intercept is (0,6), the vertex is (2,8). These are the only points given. Can you please explain to me how to work out the equation/what it is?
Thanks a lot,
Nathan
Found 2 solutions by rothauserc, josmiceli:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
Find the equation of the parabola with Y intercept (0,6) and the vertex is (2,8). These are the only points given.
The vertex form of a quadratic function is
y=a(x−h)^2 + k
where (h,k) is the vertex. Use your given vertex, and solve for a using the fact that you know the y-intercept.
Note that the vertex and any other point on the curve would be sufficient.
y = a(x-2)^2 + 8
6 = a(0-2)^2 + 8
6 = 4a + 8
4a = -2
a = -1/2
y = (-1/2)(x-2)^2 + 8

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The general form is

where is the y-intercept,
(0,6),so now you have

The x-coordinate of the vertex,
(2,8), is , so

So far, you have:

Now plug in the point ( 2,8 )

and

------------

answer
--------------------------
Is the vertex at (2,8) ?

and

OK
Here's a plot of the equation:

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