That's not exactly the same alpha as before. We set this alpha over here to 0.5.

So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic is a 1, you can just say, all right, what two numbers add up to this coefficient right here?

Right Move to the next control

We have an x squared term.

Now we just do the same thing over.

So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b.

The first term has one 5.

And now we've cleaned up the equation a good bit.

That's because I've fit my parameters to the test set.

So they give us a function, f of x.

The maximum value for the second term

Set the maximum value for the second term

So that lets us know that we're dealing with a hyperbola.

And actually one quick clue to tell you what this is you look at the x squared and the y squared terms if there are.

How many times does x minus 2 go into 2x squared minus 8?

So let's add our x terms.

Or we could write this as x minus 7 squared is equal to 0.

And then all of that over v1 dot v1 times v1 dot v1.

Well, that's all good, but let's approximate it with a quadratic equation, with adding another x squared term.

So the first term gives me 0.8 times 0.2 times 0.2 times 0.5.

So that lets us know, since we're inside here, this tells us that this is kind of a vertical asymptote, and another way to guess it is, if you see that y squared term is positive.

So our first term we saw is 15.

looks like a squared.

Plus, and what's the derivative of a next term, of a first degree term?

Well if you factor the 2 times 5 times u times v out, all you're going to be left with in this first term is the 2 times u, so 2u here.

And the things we're going to add, those are what complete the square.

If I want to represent a, what I do is, I draw an arrow to minus 4 plus this first term, minus 1.

4 minus 1 is 3.

So let me just draw a little bit of a graph.

So in this example what's A?

Set the maximum for the second term

Set the value for the maximum term of the sequence

I mean, if I just wrote this right here, you would just cancel these two terms out and you would get a over x plus 5.

So you get x minus 2 goes into x to the third and actually, what I'm going to do is, I'm going to write I leave space for the second degree term, which is 0, the first degree term, and then minus 8 is the constant term, so minus 8 I

You just have to look at the highest degree terms.

So 25r over 5r is 5.

We're just multiplying the corresponding terms, the third term times the third term, the second term times the second term, the first term times the first term.

You want to split the first degree term into two terms.

So 2x squared.

Now let's consider the degree of all the nodes in the graph.

But just as a reminder, what you want to do is you want to split this middle term.

Also on the web, when you're filling in one of these forms like your addresses,

So this first degree so this is going to be a 0 degree or a constant term Here the degree is 2.

left with a squared plus 2ab plus b squared.

So this it does fit the pattern of a perfect square.