master-determing-the-vertex-focus-and-directrix-of-a-parabola-by-completing-the-square

Subscribe! http://www.freemathvideos.com Want more math video lessons? Visit my website to view all of my math videos organized by course, chapter and section. The purpose of posting my free video tutorials is to not only help students but allow teachers the resources to flip their classrooms and allow more time for teaching within the classroom. Please feel free to share my resources with those in need or let me know if you need any additional help with your math studies. I am a true educator and here to help you out. Welcome, ladies and gentlemen. So what I'd like to do is show you how to identify the vertex, the directrix, as well as the focus, given a conic section in standard form.

So you can see that we have three equations, and they don't look like our general conic section form that we've been given. And what we've previously talked about is the vertex of a parabola we used as h and k. And in our conic section form, we have our h and our k.

And to find the focus and the directrix from the vertex, we need to define the value of p. Because the distance from the vertex to the focus is p. The opposite of p is the distance from the vertex to the directrix. And we have p in these equations.

Well, in our standard form, you can see that we don't have an h or k or do we have a p. It's not even in the same format. So we're going to have to rewrite them in standard form into our conic section form.

But to think about how we're going to do that, we've got to look at that conic section form and kind of see what kind of makes that conic section form special. And thing that makes it special is this squared term. You can see that this is a binomial squared, and we get binomial squared from having perfect square trinomials.

And right here, we don't have any perfect square trinomials. You can see we don't even have trinomials, right? But what we're going to do is we're going to create a perfect square trinomial, so therefore then we can factor down to a binomial squared. And that process of creating a perfect square trinomial is also called completing the square. So that's exactly what we're going to do, so therefore then we can identify the h and k, as well as the value p.

So the first thing you want to do here is that I kind of have xs on one side, ys on the other side. I'm going to want to isolate the values that I'm going to use for completing the square on just one side. So I'm going to keep the xs over here, and I'm just going to get the y and my value 9 to the other side. So I'm going to add 4y to both sides, and I'm going to subtract a 9 on both sides. Therefore, I now obtain x squared minus 2x equals 4y minus 9.

OK, so now it certainly looks a little bit like the conic section form, but I don't have that binomial squared. So I've got to find the value c that creates this perfect squared trinomial, and there's a formula for that that works every single time. And that formula is b divided by 2 squared.

So what I'm going to do is I'm going to take my value b, which in this case is negative 2. If you remember value b, y equals ax squared plus bx plus c. B is the coefficient of your linear term.

So in this case, my linear term is negative 2. So we take negative 2 divided by 2 and square it. Negative 2 divided by 2 is negative 1. Negative 1 squared is 1.

Now what I'm going to do is I'm going to add that value to my other two values of my quadratic to create a perfect square trinomial. But remember my properties of equality. Whatever I do on one side of an equation, I have to make sure I do to the other side of the equation. So I'm going to add a 1 to both sides. OK?

Now what I have done is I have created a perfect square trinomial. again, perfect square trinomials can factor down to binomial squares. What two numbers multiply to give you 1 that add to give you negative 2? Well, that's negative 1 and negative 1, meaning we could write this as a product of two factors of x minus 1 times x minus 1, which is the same thing as the binomial squared x minus 1 squared equals 4y minus 8.

Now here comes another little issue that we come up with. To find the value of our h and our k, we need to have 4p being multiplied by y minus k. Well, here this 4 is not being multiplied by the y and the k. This four is only being multiplied by the y.

So to make sure it's multiplied by the y and the k, I need to factor out that 4. So when I factor out the 4, I'm left with y minus 2. OK?

Now I know the values of my h and my k, which I'll go ahead and label. So remember h is always with x. y Is always with k. So therefore my vertex in this case is going to be 1, 2.

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