Then we have the point 1, negative 1.

So this is the point 1, 1.

The point 1 comma e.

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At the point, x is equal to 1.

So it's the point x is equal to 1.

Let's try the point minus 1.

The point 1, negative 1 is also on this top graph.

So what is the derivative of this function at the point x is equal to 1, or at the point 1 comma e?

So this is the point, 1 comma e.

So 1, negative 2 is right there.

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.

It's 1 over negative 1, so it's this point right here.

When x is equal to 1, or at the point 1 comma e.

So we're saying at the point, at the point 1 comma e, so at the point 1 comma 2.71, whatever, whatever.

So this right here must be the point 1 1 3.

So it's negative 1, so the point, 2, negative 1 is right there.

Here the vertex occurs at x is equal to 1, y is equal to negative 1.

So this is the point x is equal to one.

2 point, this is e right here, the point 1 comma e.

The center is going to be at the point 1, negative 2.

Then we have the point x, we have f of 1, which is the point 1, 1, right?

So if we have the point x is negative 1, y is negative 1.

It becomes a negative 2 So it becomes 2, negative 1.

And y is 1, so that point is right there.

So that point will show up, see if I do, 1, 2.

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

The focal point is going to be 1 over 4A above this point.

And at this point, the depth is 3 2 which is 1.

And so my center is now at minus 2, 1.

So the point 1, negative 2, is on our inverse graph.

Or f of x equals 0 at the point x equals negative 1 and x equals 1 10.

Then we go to y is equal to negative 1, so the point would be right here.

So that's 1, 2, 3, so that's this point right here.

The line has a slope of negative 1 and contains the point 4 5 comma 0.

So the point 0, 1, 2, 3, 4, 5, 6.

Jump down at this point

(0, 1) right over there and I could keep doing this. when x is 1, y is 1 when x is 2, y is 3 actually let me do that in the same purple colour when x is 2, y is 3 2,3 and then this one right over here in orange was 1,1 and this is neat by itself,

So the point 0, 3 1, 2, 3 is on the line.

And when I say exactly back to that same point, this point doesn't count.

That is my y f(x) axis, and then this over here is my x axis, and then let's say this is the point x 1, this over here would be x 1, this is y 1, right up there I can do 1 but that doesn't do much relative to this function right over here, and and let me graph it.

Because this point right here, our vertex, is equidistant, it's 1 4 below our focus and it's 1 4 above our directrix.

Move the orange dot to 1.9 on the number line.

OK, let's get to the point.

So let's proceed to the next point.