And that's the point minus five comma minus four.

That point right there, the point 0, minus 2.

Let's try the point minus 1.

Negative 1, minus the second x point, minus 2, so 3 minus 1 is 2.

That's our starting y point, over our finishing x point minus our starting x point.

I'll take this point 2 minus negative 4 over 4 minus negative 2.

We're thinking of the highest data point minus the lowest data point.

If I have two point six four minus zero point zero four eight six.

So it's going to be y minus let's use this point right here y minus 25 is equal to the slope negative 13 over 5 times x minus this point, 5.

And 0 minus 2 is this point right there.

That's minus 6 24, which is equal to minus 1 4, which could be this point.

So what's d1? d1 is the distance between this point and this point minus f0.

Or, we can view it as the y value of our end point minus the y value of our starting point over the x value of our end point minus the x value of our starting point.

So this point right here is the point a comma 0, and this point right here is the point minus a comma 0.

So two point six four minus point zero four eight six is equal to two point five nine one four.

But this right here is the point minus 2, 1.

What's seven point seven seven repeating, minus point seven seven, going on and on, forever repeating?

It's going to be 6 minus your starting point y value.

Then we could say our change in y is our finishing y point minus our starting y point.

Which point shows the location of 5 minus 2i on the plane?

That means that the point 0, minus 3 is on the line.

And minus 3, you're going to get a point something like that.

You say, y minus the y point, which is just x0 squared, that's that right there, is equal to the slope of the normal line minus 1 over 2 x0 times x minus the x point that we're at.

They say that we have 0's at the point minus 1, 0 and the point 2 comma 0.

So, if this point right here is the point, and we already showed that, this is the point the center of the ellipse is the point 1, minus 2.

My function g takes any point in R2, any coordinate, this is the point 3, minus 5, whatever it is.

The first problem I have here says five point seven three minus point zero eight two one equals who knows?

If that confuses you, all I'm saying is, it's going to be equal to our finishing y point is negative 6 minus our starting y point, which is 6, over our finishing x point, which is 5, minus our starting x point, which is negative 1.

And if you did minus 2, you're going to get a point that

It would be the point x is minus 2 and y is 1.

So this is going to be the point f,0 and this is going to be the point minus f,0.

And of course this point right here this will be a, so this would be the point minus a comma 0.

But point slope form says that, look, if I know a particular point, and if I know the slope of the line, then putting that line in point slope form would be y minus y1 is equal to m times x minus x1.

Well, that is going to be negative 6, that's our end point, minus 3.

So let's write it in point slope form. y minus this y value, so y minus negative 3, is equal to the slope 3 5 times x minus this x value, this x coordinate, x minus 0.

So we do the end point, which is 3 minus the starting point, which is negative 1, which is equal to 4.

That's the exactly the same thing as five hundred twenty six dollars and ninety cents, minus minus adding a negative is the same thing as subtracting a positive minus fifteen point oh eight.

Is the center of this hyperbola is at the point x is equal to minus one.

So the points a,0, and the point minus a,0, are both on this hyperbola.

And at 2 we're at negative 4, so the vertex is at the point 2, minus four.

So that's z plus 2 times z minus 2, hopefully you can recognize that at this point.

And they tell us we have a y intercept, y intercept, at the point 0, minus 2.

In point slope form, it takes the form y minus y1 and I'll tell you what y1 is in a second is equal to m times x minus x1, where the coordinate x1, y1 is a point on the line.

Plus, minus, plus, minus, plus, minus, plus, minus, plus.

Well, the whole point of multiplying by x minus 1 is so that that cancels out with that.