 In a previous video, we were discussing the in-behavior of rational functions. We were looking for horizontal asymptotes that occurred when the function was either bottom-heavy or balanced. But when it was top-heavy, you don't have a horizontal asymptote. But I was making a statement that the function was approximately the same thing as a monomial function. We can do a little bit better than that. So we want to talk about in this video the idea of an oblique asymptote. That there might be a slanted line or an oblique line so that as x goes towards infinity or negative infinity our rational function starts to behave like a diagonal line, a slanted line of some situation. Now an oblique asymptote, it's going to be a line y equals mx plus b. It only happens when the numerator is exactly one degree larger than the denominator. And we find this oblique asymptote using polynomial division. So as a first example, let's take r of x to equal 3x to the fourth minus x squared over x cubed minus x squared plus one. We can see when we look at the top and the bottom, the leading term on top versus the bottom is one bigger than the bottom. 3x to the fourth over x cubed. I was saying before that the in-behavior of this thing will be approximately 3x to the fourth over 3x cubed. Which would simplify to be 3x. So this thing is approximately a line but we can even do better than that. So if we go through the division process, take 3x to the fourth minus x squared divided by x cubed minus x squared plus one. We put the numerator here, the denominator here. How many times does x cubed go into 3x to the fourth that goes in there 3x times like we observed earlier? But we want to do a little bit better than just saying 3x. We want to do a better estimate and so we're going to go through this division process. So take 3x and times it by my divisor. So you're going to get 3x to the fourth plus I should say minus 3x cubed and then a 3x. But we're subtracting this thing so I've distributed the negative sign to all these places. So these ones since we're subtracting them become negatives and as we're subtracting the negatives becomes a positive. So you're going to see 3x to the fourth minus 3x to the fourth they cancel. You get 0x cubed plus 3x cubed as the next 3x cubed bring down the negative x squared. We get that here and then bring down the negative 3x. This process then repeats itself. We then do it again. So then we have to ask ourselves how many times is x cubed divided into x 3x cube? That'll happen three times and then we go through the division. We go we get all this. But one thing I want to mention to you is that when you're looking for an oblique asymptote, the oblique asymptote will be the quotient. It's going to be the quotient. You don't actually need the remainder. So you have my permission when you're looking for oblique asymptotes to stop, stop, stop, stop, stop, stop, stop, stop. When you find the oblique asymptote here, when you find the quotient 3x plus 3. So when it comes to finding oblique asymptote, you only have to do this process one and a half times. The first the first step gives you the slope of the oblique asymptote and then the second half step is going to give you the constant. So this tells me that my function if I were to graph it. Oops, that's diagonal. If I was graphing this, this function will have a slope of three. It has a y intercept of three. So you get this like oblique line, something like this. And then as x goes towards infinity, it's going to or negative infinity, it's going to start behaving like this line. This is what we're expecting. This is why we want to find these oblique asymptotes. Let's do another example. In this one, we have x cubed over 4x squared. Again, they differ by one. So this is a situation where we're going to have an oblique asymptote because the top differs by the bottom just by one. And so how do we determine that? We're going to take x cubed divided by 4x squared. That's going to give us, you go through the details of that, x cubed divided by 4x squared. That gives you x over 4 or 1 4th x. Okay, erase that. So we get this term right here. So we're going to take our divisor and times it by 1 4th x. 4x squared times 1 4th x is going to give us an x cube. We're going to subtract it. And then we're going to take 1 4th x times negative 1. That's going to be a negative 1 4th x and then we're going to subtract that so it becomes positive. The x cubes cancel and we're going to get negative x plus 1 4th x. That combines to give you this. And then you bring down the two, you repeat the process. We ask ourselves how many times does 4x squared divided to negative 3 4th x? Oh, that's too small. You get a plus zero right here. And so like I said, you don't need to finish this thing off. You just need the quotient. The oblique acetote here will be y equals 1 4th x. The remainder does not matter. Okay, let's do another example. Let's take this rational function. We have x squared minus 4x minus 5 over x minus 3 looking just at the leading terms. They differ by a power and so x squared over x. And so we know there's going to be an oblique acetote in this on the graph of this function. Let's compute what that oblique acetote will be. We'll take x squared minus 4x minus 5 divided by x minus 3. x goes into x squared x many times. If you take x times x minus 3, you'll get x squared minus 3x. Subtract that. You're going to get negative x squared plus 3x. The x-squares cancel negative 4x plus 3x is a negative x. You then rinse and repeat. x goes into negative x negative 1 times for which you can go through the details to find the remainder of negative 8 if you want to. But we don't need that. Again, we don't need anything past this point. We now have our oblique acetote y equals x minus 1. Now, because the denominator was x minus 3, you actually could have found this using synthetic division if you wanted to. Which can be a little bit faster. Bring down the 1. 1 times 3 is 3. Minus 4 is negative 1. You could keep on going, but honestly, I don't care because this gave us the information we need right here. The oblique acetote was y equals x minus 1. We don't need the remainder to find an oblique acetote. We just need the quotient. So the oblique acetote would be y equals x minus 1. Okay. In this example, let's see. If you take the rational function r of x equals x cubed minus 2x squared plus 3 over x minus 3. In this situation, the top and bottom differ by 2 degrees. So there's no horizontal acetote, but there's also no oblique acetote. So admittedly, if you're asking for oblique acetotes, the answer would be none. There doesn't exist one. So the oblique acetote, it does not exist. Is this still a top-heavy function? And so like before, I was like, oh, the in behavior is going to look like x cubed over x, which is x squared. So this function will be approximately x squared. It'll have the same in behavior. If we want to do a little bit better than just x squared, we can actually find an acetote. It would be a parabolic acetote. Turns out your function, as you go off towards infinity, is going to look like a parabola. And so as you go off towards infinity, your function will start to behave like a parabola. If we want to find out what that parabola is, we still do long division here. So we take x cubed divided by x that gives us x squared. If we times x minus 3 by negative x squared, we get negative x cubed plus 3x squared. The x cubed will cancel. Negative 2x squared plus 3x squared gives us an x squared. We then repeat the process. Take x squared divided by x. That will give you an x times x minus 3 by a negative x. You're going to get negative x squared plus 3x. The negative x squared's cancel. We get a 3x. Bring down the 3, of course. Then you're going to ask yourself, well, x divides into 3x three times. There is some more to be done here, but we don't need it because this gives us the asymptote. The quotient, we don't need the remainder. And so our function will be approximately the same thing as the parabola. x squared plus x plus 3 as x goes towards infinity. Now again, x cubed plus x plus 3 has the same in behavior as x squared. So we often can get away with just having just the monomial right here. But if we want to even better approximation of the in behavior function, we can look for this parabolic parabola, a parabolic asymptote, excuse me. And we can do this for all, all top heavy functions. A top heavy function will be approximately the same thing as a polynomial for which if we just want the in behavior, we can ascertain that from the quotient of leading terms. We can get that from a monomial. Because again, the in behavior of a polynomial comes from its leading term. That's sufficient. But if we want the true asymptotic behavior of a rational graph, we should do long division and find out that asymptote, which will be a polynomial. It coincides with the quotient of that polynomial division. And I should also mention on this one, we could have also done synthetic division since the denominator was x minus 3.