 Welcome back to our lecture series math 1050 called jobs for 5 students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misadine. In lecture 32 in our series, we're going to continue what we were doing in lecture 31. That is, I want to give us some examples of graphing rational functions without the use of any technology whatsoever. And we're going to base these pictures upon the X intercepts, the Y intercepts, vertical and horizontal asymptotes if they exist. And then the behavior as the function approaches these asymptotes and behavior as it approaches the X intercepts that type of information. So in this example, consider the rational function R of X equals X squared minus one over X. So the first thing I would want us to do is first identify the domain. This will happen by factoring. So really, we should just factor first. Let's factor the numerator X squared minus one without factors of difference of squares X minus one and X plus one. Then factor the denominator. Well, it's an X. So it's already there, right? And so we end up with just an X right there. So without any cancellation, first identify the domain. What makes the denominator go to zero? Well, our denominator, it goes to zero when X equals zero. So that is a forbidden value from our domain. X cannot equal zero. Okay, now the fact that X can equal zero suggests that we actually don't have a Y intercept, because the Y intercept is when you plug in X equals zero. But that would end up with us having negative one over zero, which is undefined, does not exist. So this function doesn't have a, it doesn't have a Y intercept. In fact, this something else is happening at X equals zero. We'll come back to that in just a second. So we don't have a Y intercept. The X intercepts. So now if you can simplify the fraction, which there's no simple terms, there's no common terms on numerator and denominator cancel out here. It's already in lowest terms. So the numerator, what makes the numerator go to zero? That's where X intercepts are going to come from. So if we have X minus one times X plus one is equal to zero by the zero product proper, that suggests that X equals negative one. Well, actually the first one would be X equals one, the next one would be negative one. So that is X minus one equals zero, that's where the one comes from. Or if we have X plus one equals zero, that means X would equal negative one. So we're going to have two X intercepts, one and negative one right there. Notice that their multiplicities, they both show up an odd number of times in the numerator. These both have odd multiplicities, which means that at these points we will cross the X axis. Okay, if you want to, you could check for symmetry. Notice that if you plug in negative X in this picture right here in this function, you're going to get negative X squared minus one in the numerator, but taking a negative to an even power actually produces the original quantity X squared minus one. So notice that the numerator is unaffected. Well, the denominator when you plug in negative X, you end up with just a negative X. And so you can actually factor out the negative sign in front of the whole thing and you end up with a negative R of X right there. This suggests that our function is an odd function. That is, it's symmetric with respect to the origin. So there actually is symmetry in this picture. We can use that to help us if we want to. It's more of just like at the end, I'm going to check to see if it's symmetric or not. When we go through this graphing process, we're going to get some redundant information and that redundant, those redundancies are not bad things because the redundancies help us double check that if we made a mistake, then the extra information would indicate, oh, there was a mistake and then we could try to correct the mistake there. In terms of vertical asymptotes, notice that we have this factor X in the denominator. If X is zero, that would mean if we divide by zero as this thing is in lowest terms. This is going to give us a vertical asymptote. It's multiplicity is also going to be one. So you have a vertical asymptote at X equals zero. It's multiplicity, like I said, is also odd. So we're going to cross infinity at this vertical asymptote. In terms of in behavior, are there any horizontal or vertical, we know there's vertical asymptote horizontal asymptote. When you look at this one, we have a degree two on top, we have a degree one. So this is a top heavy function. It's top heavy here. So we know that there's not going to be any horizontal asymptotes, but there actually isn't a bleak asymptote here. Notice the top is bigger than the bottom by one. So normally this would mean we would just go through a long division process X squared minus one divided by X. And we could find the quotient up here to be that oblique asymptote. But let me provide a shortcut. First of all, the denominator is X. It's a linear function. So you actually could use synthetic division. But in fact, it's even better since it's just a monomial. What we can do here is that this will just be X squared over X minus one over X, which then gives us X minus one over X. So the quotient, it would be an X and then the remainder would be one, we don't need the remainder, because as X goes to infinity. This is the important observation here. As X approaches plus or minus infinity, this one over X is going to go to zero, and this thing will become a prox, this thing will just converge towards X. So Y equals X is our oblique asymptote in play right here. And I already mentioned the multiplicities are all one. So what can we say about this function? So what I'm going to do is I'm actually, I'm actually kind of graph this to the side, and I'll show you a second what the computer generated image is going to be. So let's do our X and Y axes here. So let's indicate what we've discovered so far. So we found out that we have an X intercept at one. And we have an X intercept at negative one. We know we're going to cross infinity at those points. We had a vertical asymptote at X equals zero. That's the, that's the Y axis right here, right? Let's kind of just draw this right here. So there's no X, there's no Y intercept. We have this vertical asymptote. Okay. And then we also know that we have this oblique asymptote, which is going to be the line Y equals X, which you see right here. So what this means as, as X goes to the far left or the far right, it's going to get close to this line Y equals X, like so. Now some information we want to figure out about this graph right here. So one thing we haven't discovered yet is, does our function cross its oblique asymptote? This is an important question that students sometimes get the misconception that because we can't cross a vertical asymptote, that means we can't cross any asymptotes. That's not true. A rational function will never cross its vertical asymptotes because those are outside the domain of the function. The function's not defined there. The way that we've defined horizontal and oblique asymptotes, we have no, there's no, there's no, we didn't forbid that we can cross a horizontal asymptote. What we would instead consider is that we take the function X squared minus one over X. Could this ever equal to oblique asymptote X? Is this even possible? When we solve this equation, I'm going to times both sides by X. We get X squared minus one equals X squared. If you subtract X squared from both sides, you get negative one equals zero, which is a contradiction. That would suggest there's no solution. That is, there's no place where you cross the, or the oblique asymptote. So that's the observation we're going to make here. It's like, okay, when it comes to our oblique asymptote, there's going to be no crossings here. That's a useful information as well. So because of that information right here, I want you to see what we can say that with this X intercept, we have to get closer and closer and closer to this vertical asymptote. So there's an option that could happen, right? When we're at this X intercept, do we go up or do we go down? Well, if we were going down, think about that thought experiment for a moment. If we were going down at some point, we'd have to turn around and come closer and closer towards this horizontal asymptote. We'd have to do that. But if we were, if we started down when we were below this X intercept here, this would tell us that we'd have to cross the X axis again, but there's no other X intercept for that to be, for that to happen. So this second case is actually impossible. So what we instead can see here is that our function must approach its horizontal asymptote like this. We also know that we can't cross our horizontal asymptote and come back down because like we mentioned, there's no crossings whatsoever. So with the information we have from the X intercept, it has to get closer and closer to this oblique asymptote. We have to have something like that. Now at the X intercept one, there has to be a crossing, right? And so we have to cross the other side to get below the X axis, for which then we would have to continue towards our vertical asymptote. You'd have to go towards negative infinity. We couldn't go back up towards this direction because when you get close to a vertical asymptote, you either go towards infinity or negative infinity. We can't go towards positive infinity for two reasons. One, there's no X intercept here that allows us to cross. Also, we're not allowed to cross our oblique asymptote like we discovered earlier. So both of those are kind of silly things here. So what we're going to then say is that, okay, we cross at the X intercept and we have to approach towards our vertical asymptote as we're going to go towards negative infinity. But then when we get towards negative infinity, because we have odd multiplicity at our vertical asymptote, we're going to cross infinity, come from the other side, and we're going to have to come closer and closer and closer towards our X intercept right here. We get a picture like this. We would have to cross our X axis because we have odd multiplicity again, and then we'd have to converge towards this oblique asymptote. And so we get this picture right here, which if you do check, if you were to take this graph and spin it a half spin, you would get the exact same picture again. I might have to kind of extend my line a little bit so you can see that. But this actually is an odd function that is symmetric with respect to the origin here, exactly like we thought it would. And so now let's see what the computer generated image did. That was this business right here. We see we got the exact same picture and we did it without any technology whatsoever. This sort of gave us a skeleton right here. We knew the oblique asymptote. We knew the vertical asymptote. We knew we had these X intercepts. We have to cross. We have this vertical asymptote. We have to cross. And so piecing together those dots, we get this picture right here. But I really want to emphasize that we're able to do all of this without any technology whatsoever, just using the information about the asymptotes and the intercepts and how the function behaves near its intercepts and asymptotes. We are able to put together this picture.