00:01
Okay, so in this problem, we're given the vertex of a quadratic, and that vertex is at 2 .0.
00:15
We're also given a point x, y, that's also on the parabola somewhere, and that point is at 4 .4.
00:27
I think the best way to go about this, because they gave us a vertex, is to go ahead and use our equation that starts out in vertex form.
00:40
So we have a times x minus h quantity squared plus k.
00:49
In this case we just have to plug in our vertex and then solve for a and then convert it to standard form.
00:57
So let's do that first.
00:58
So plugging in our vertex, we don't know our a value yet.
01:02
So we're going to have a times x.
01:05
Our h value, as you can see, is 2.
01:09
So we're going to have x minus 2 squared.
01:13
And then plus k and our k value is zero so it's just zero so our answer is just going to be you know a times x minus 2 squared so using the other point we're going to have our y value of well that the point was 4 4 so that means when the x value is 4 the y value will be 4 or the function will return 4 so we can write 4 equals a times 4 minus 2 squared, which is just a times 2 squared, which equals 4a...