 This video will talk about holes and oblique asymptotes and rational functions. So a hole occurs when you have a common factor between your numerator and your denominator. The zeros of the denominator are usually asymptotes and you can't have a zero there because it can't reach that value. So we have a problem and it is actually a hole there. So if we want to find the common zero and evaluate it, we would have f of x is going to be equal to, the factor of this would be x minus five and x plus two and on the bottom we have x minus five and you can see that there's a common factor and that would be the zero. The common zero would be x minus five equals zero or x equals five. Well, if we evaluate it, when we put the five down here at the bottom, we're going to end up with zero in the denominator which is undefined. So we end up with f of x equaling x plus two but then we have to say well x can't be five. And so let's see what that looks like in a graph. Here's our x plus two, that rational function became a line when we simplified it but you really can't see the hole here. So it says let's look at it again but let's let our window start at four and then go to six so we want to see what happens between four and six which is five. And let's look at point zero one and this might be a little hard to see but we can look at it and see what happens. You may not be able to see it right here and we'll highlight that. Right there, there's a gap in that graph. So that's where the hole is. It literally makes a hole. All right, so oblique asymptotes. We have these rational functions and we talked about finding horizontal asymptotes if the numerator was less than the denominator degree or equal. But when we have a greater, that's when we're going to have an oblique asymptote and we find that by finding the quotient when we divide. So when the degree is greater, this makes a fraction of an improper fraction like three over two. We would have put it in proper form. We would use long division and I've worked this particular problem out for you to show you what we get. 2x minus and then the result when we subtracted these two, it gave us negative 2x as a remainder and we put that over our divisor. So the remainder that negative 2x gets close to zero for very large x values. So f of x equals 2x for large values. That's what it will be. This part right here becomes our oblique asymptote. And it says note that graph can cross oblique at the zeros of the remainder. So here, if it's going to cross, it's going to cross at zero because this and the numerator would be at zero. All right, so now they're going to ask us to graph and we know all kinds of things now. X intercepts, remember those are the zeros of the numerator. So we're going to have to factor this thing. And if we take two terms and two terms, we would have x squared, x minus two, and then minus nine and x minus two to get the plus 18. And then that's going to give us x squared minus nine times x minus two. And one more says that that's going to be x minus three, x plus three for the first factor to completely factor it and x minus two. So right now we know that our x intercepts are at three, negative three, and two. So now we want to look at the y-intercept. And remember the y-intercept is just that x equals zero. So if this is zero and this is zero, anything times zero is zero and anything times zero is zero, and we have 18. And if we have 18 and this be zero, then we don't have one. In our case, it does not exist. That must be part of our asymptote of some sort for us or something because we can't divide 18 by zero. So the vertical asymptote. Remember the vertical asymptote is going to be the zeros of the denominator. And the zeros of the denominator is just going to be x equals zero. It's vertical, so it's x equal. So that means that we have here the y-axis as our vertical asymptote. And then horizontal or oblique, it's going to be one or the other. And so we are going to look at our degrees. And our degree is three and degree is two. So here we have that it's n is greater than n. So it's an oblique. So we have to do long division to be able to find out what that is. So x squared on the outside, x cubed minus 2x squared. That's a cubed minus 9x plus 18. And if you want to pause right now, you could do that division. I will give you the final answer in just a moment. We end up with x minus 2 plus that negative 9x plus 18. So our horizontal asymptote is going to be at x minus 2 is what y is equal to. And when we talk about does it cross? What we're really going to do is say negative 9x plus 18. I want to find the zero. Remember we used just the quotient factor out of negative 9. And you've got x minus 2 equals zero. So x is going to be equal to 2. So it will cross at x equals 2. So it's going to cross right here. So let's draw what we know. We have the line x minus 2. And it's right here. Up one over one, up one over one, up one over one. And I get this oblique asymptote. And you can see that it crossed at that x-intercept to graph. We want physically graph. We're just going to look at our calculator to verify everything that we just found. And remember it won't show the oblique asymptote. And if we want to come in here and see if it kind of looks like ours, we could come back in and say that this y2 is our x minus 2. Remember this isn't really part of our graph. We think of it at the dotted line. But that looks very much like the graph that we had. You can see the x-intercepts of 2, 3 and negative 3 and the vertical asymptote of the y-axis. x-intercepts. Let's do those in purple. x minus 3x plus 2. So the zeros are at positive 3 and negative 2. And they're both going to go through. They're both going to change signs. The y-intercept. Again, we're going to have this 0 down here and that we're dividing by because we have to let x equals 0. So this does not exist. I'm not going to have anything red on my graph. Vertical asymptote. Remember that's the zeros of my denominator. So these are going to be x equal 0. Again, that came from the zero of the denominator. And then horizontal asymptote or oblique asymptote. We've got a degree 2 and a degree 1. So 2 is bigger than 1. So it's oblique. So again, we have to divide. x goes into x squared minus x minus 6. This will be x. Bring down the negative x. That'll be minus 1. Bring down the negative 6. And that's our remainder plus a negative 6 over x. So our oblique asymptote is y equal x minus 1. And there is no zero up here. It's just a negative 6. So it's not going to cross. So we have our x minus 1. It looks something like this is my oblique. Again, we want physically graph. We're just going to look at our calculator to verify everything that we just found. We have our function in here. And I have this asymptote, but we're not going to look at it at first. We'll come back and look at it. So standard window again. And here's our graph. Looks a lot like what we thought we were going to have. You can see that it's definitely tending toward this y-axis. And the x intercepts at 3 and negative 2. And if we come in here and go back to this, then you can see that there's our asymptote. And now our graph looks like ours. If you think of this line as being dotted.