00:02
You have f of x is equal to x squared over x minus 2 squared.
00:11
You want to find the vertical and horizontal asymptotes.
00:16
So the vertical asymptote is where the function doesn't exist.
00:23
So that is x is equal to 2 because then the denominator would be 0.
00:29
And for the horizontal asymptote, you can use limits, limits to infinity.
00:35
You can also use the idea that the power, the degree of the numerator, the degree of the numerator, is equal to the degree of the nominator.
00:45
So it is the ratio of the leading coefficients, which is one.
00:50
Also, if you take the limit to infinity, you'll get one.
00:55
And then from here, find where the function is increasing and where it's decreasing.
01:06
So to do that, we need to take the first derivative.
01:17
So this f prime of x is equal to the denominator times the derivative of the numerator, so x times 2x2 squared, times the numerator, the degree, sorry, the derivative of the denominator all over the denominator squared.
02:02
This reduces to 8 times x plus 1 over x minus 2 to the fourth.
02:11
We're going to set this equal to 0 to find the critical points.
02:16
So when we set this equal to 0, the only time, oh, i skipped ahead.
02:33
That would be the second derivative.
02:37
Let's try that again.
02:39
So the first derivative, x minus 2, cancels.
02:46
So one of those, one of these, and that becomes three.
02:52
So this is negative 4x over x minus 2 cubed.
02:59
My apologies for that.
03:01
Set this equal to zero.
03:02
It's only equal to zero when x is equal to zero.
03:05
So that becomes our critical point.
03:08
So we look at zero.
03:09
But we also need to look at where you are outside of the domain.
03:14
So you need to look at two.
03:18
So to the left from negative infinity to 0, that first derivative would be negative.
03:28
From 0 to 2, that first derivative is positive, and from 2 to infinity, that first derivative is negative.
03:38
So it is increasing from 0 to 2...