 Episode 14 of Math 1050, College Algebra, here at Utah Valley State College. I'm Dennis Allison, and I teach in the math department here. This episode is a continuation of the previous episode, number 13, where we introduced rational functions. And in this episode, we'll be looking at some more complex rational functions. The objectives for today are, first of all, to look at graphs of a wide variety of new rational functions that include slant and oblique asymptotes and multiple vertical asymptotes. And then if we have time, we'll look at a graphical solution for a rational inequality. OK, well, let's begin by just reviewing our two fundamental graphs from the last episode. We had f of x equals 1 over x, and f and g of x equals 1 over x squared. If you remember, both of these graphs had two target points. If I want to graph 1 over x, at least in the future when I graph 1 over x, let's see. Let me get another marker here. This one seems to have run out. If I want to graph 1 over x, what I'll do is plot two points. And class, who remembers what the two target points are for this function? Is it negative 1 and negative 1? Negative 1, negative 1 is a target point. And 1 and 1? And positive 1, positive 1. And then we have to know the general shape of the graph. And it's in the first quadrant. So it comes down like this, turns, and goes off to the right. And then it's in the third quadrant. It comes up from the negative y-axis, and it turns and it approaches the x-axis from underneath. The x-axis here is referred to as a horizontal asymptote. This is a horizontal asymptote. And then the y-axis is a vertical asymptote. And what does it mean to say that we have an asymptote in a graph? Who can explain that? Steven? The graph approaches the asymptote, but never actually touches it or crosses it. Yeah, for example, for the vertical asymptote, we notice that the graph here approaches the y-axis, the positive y-axis on the right. It approaches the negative y-axis on the left. And the horizontal asymptote, it approaches, as x goes out toward infinity, it approaches the horizontal asymptote, but never quite touches it as it goes off to the right-hand side. And it approaches it from underneath on the left. Now, you know, one of the things we're going to see today in the graphs that we draw is that when it comes to horizontal asymptotes, we can actually have a graph that comes down, crosses the asymptote, and then it comes back and approaches it on the way out. So the horizontal asymptote is only effective in the sense that it represents what the graph approaches near the ends, but not necessarily in the middle. So we can actually cross a horizontal asymptote, and we'll see this happen in an example later today. OK, the other fundamental graph, just to make a little room for this one here, is g of x equals 1 over x squared. And if we set up the axes, we have a slight change in our target points this time. What would be the target points for 1 over x squared? Is anybody? 1 1, yeah? Negative 1 1? And negative 1 1, exactly. You notice that 1 over x squared can never be negative, because we have 1 in the numerator, so that's clearly positive, and we have an x squared in the denominator, so that's never going to be negative. So I have a positive number over a positive number, and that means I'm always going to get a positive function value, so we'll never see the graph down here in the third or fourth octants or quadrants. So when I draw my graph, it turns up and levels off like this, and in the second quadrant, it turns up and it levels off over here. Now just to work one more example of a problem like we saw in the last episode, let's take a transformation of one of these graphs, and suppose we wish to graph the function f of x equals 2 over x minus 1. What are the changes that I'll make in one of those two fundamental graphs? And while we're at it, what is the fundamental graph that I should start with? 1 over x? We should start with 1 over x, yeah? So over here on the side, what we're thinking of is the graph of f of x equals 1 over x. That was the one that's in the first and third quadrants. And what are the changes that we'll make in this graph if I want to graph this new function, f? Vertical stretch of 2. There's going to be a stretch of 2, yeah? Stretch of 2. And it's going to be moved to the right 1. To the right 1 unit, exactly. OK, so let's take this graph of 1 over x and let's make those two changes. So we'll put it right here in the middle. So that says if I move 1 unit to the right, the vertical asymptote moves 1 unit to the right. So I'll just dot in the vertical asymptote right here. And that's going to be along the line x equals 1. And then the target points would say, from the new origin, move over 1 and up 1. But we're stretching this by a multiple of 2. So I'll go over 1 and up 2. And I'll put that target point up above. And if I go to the left 1, instead of going down 1, I'll go down 2 and I'll get a point right here. That's going to be the y-intercept. And my graph then looks like 1 over x, but it's been moved over a bit. And this is the graph of the function f of x equals 2 over x minus 1. Now, we had a rather lengthy discussion about this in the last episode. But I want to remind you of these things because we want to keep these ideas in the foreground as we introduce these new functions today. You notice this function has no x-intercept? If I were to look for x-intercepts, even though it appears there aren't any, if I were to look for x-intercepts, what I would do would be to let y be 0. And you notice over here, if I let y be 0, I have 2 over x minus 1. And if I multiply both sides by x minus 1, I'll have 0 equals, and multiplying by x minus 1, 0 equals 2. And of course, that's impossible. So what that means is it's impossible for 2 over x minus 1 to be 0. And that means there are no x-intercepts. And that's exactly what our graph tells us. It does look like we have a y-intercept here at negative 2. And if I had not known that, I would have found the y-intercept algebraically by letting x be 0. And if I let x be 0, I get 2 over 0 minus 1. And that says that f at 0, let's see, 2 over negative 1 is negative 2. And yes, that's exactly what we get right here, is that y-intercept. OK, so with that review, let's move to some new functions that we'd like to graph. And let's go to the first or to the next graphic. Now, here we want to graph some more complex rational functions. And in this example, we have two parts. Let's first graph f of x equals x over x minus 2. And then we'll come back in a minute and we'll graph part b. So let's come to the green board. And part a says we want to graph the function capital f of x equals x over x minus 2. Let's see, now, is this a proper or an improper fraction from what we said last time, class? Remember how we defined what we called a proper or an improper, should we say, ration? That's an improper fraction. This is an improper rational expression. And why is that? Because they are both of the same power at the bottom. Because we have a numerator with a degree as large, and in this case, just as large, is the denominator. We have a first degree numerator and have a first degree denominator. Or to put it another way, we have a linear numerator and a linear denominator. So in this case, what I should do is divide x minus 2 into x. I'll do that over here on the side. x minus 2 divided into x. Let's see, that goes one time. So I have x minus 2. I want to subtract that off. The x is canceled. And what I subtract here is 0. Subtract negative 2 is 2. So I'll put my remainder on the end as a rational expression. So this is the same thing as writing 2 over x minus 2 plus 1. You notice I'm putting the plus 1 on the right so that it looks to be in a more standard format where I have the vertical shift on the right-hand side. So if I want to graph x over x minus 2, I'll get the same graph as if I graph 2 over x minus 2 plus 1. The x minus 2 on the bottom says I should move it to the right 2. The plus 1 on the right says I should move it up 1. And the 2 in the numerator says there should be a stretch in the graph. So let's put all that to work for us and see how we would draw this graph. Let's see. First of all, let's move the graph 2 units to the right. So I'll take my vertical asymptote, move it 2 units to the right. So here's 2. And I'll get a vertical asymptote right here at x equals 2. Then there is a vertical transformation, vertical shift. I'll move it up 1. That says the horizontal asymptote moves up 1 with it. So I'll put the horizontal asymptote right here. And that'll be at y equals 1. So this point is my new origin. And let's see, we've taken care of the vertical shift, the horizontal shift. We still have the stretch. If I go over 1, I should go up 2 units. 1, 2, 2 units right there. And if I go to the left 1, I should go down 2 units. And I'll get a point right there. Now I'm about ready to draw my graph here and over here. But while we're at it, let's go ahead and locate the x and the y intercepts. The x intercepts. Let's see, to find x intercepts. Someone remind me again, what do I do to find the x intercept? Set y equal to 0. Set y equal to 0. Now I could either set 0 equals to this expression or 0 equals to this expression. I think that one's going to be easier to work with. So I'm going to set 0 equal to x over x minus 2. So 0 equals x over x minus 2. And if I multiply both sides by x minus 2, I get, well, 0 times x minus 2 is 0. And on the right-hand side, the x minus 2 is a cancel. And I get 0 equals x. So I have an x intercept at the point 0, 0 right there. Now that will be the y intercept because that points not only on the x-axis but the y-axis. So this is both the x and the y intercept. So when I draw my graph, I have one more point there that I can use to check for accuracy. And my graph looks like this. And my graph looks like this. OK, so this is the graph of f of x equals x over x minus 2. Let me erase some of this extra writing around it so we can look at the graph more clearly. And we'll take out this division up here. Are there any questions about that graph? OK, let's go to the other problem that was on that graphic. Yeah, there we are. Now let's graph the function g of x equals x times x minus 4 over x minus 2 squared. Now, you see, this is getting to look a little complex here. But I think we can graph it just as simply as we graph the last one. So for part b, let's graph g of x equals x times x minus 4 over x minus 2 squared. Well, let's see. What can we say about this graph before we begin? First of all, it's an improper rational expression because I have a quadratic expression on top, if I multiply it out. And I have a quadratic expression on the bottom. So I have second degree over second degree. It's only a proper rational expression if the larger degree is on the bottom. So just as in the last example, I need to divide the denominator into the numerator. And to do that division, I'll need to multiply out the numerator and the denominator and express each one of those as a single polynomial. So the numerator is x squared minus 4x. And the denominator is, let's see, what is x minus 2 squared? Jeff, what's x minus 2 squared? x squared minus 4x plus 4. Minus 4x plus 4, OK? So in some instances, this is the form I'll want to look at. In some instances, this is the form I'll want to look at. And I think I'll take this latter form right now to carry out my division. I'll do my division right here below. I'll divide x squared minus 4x plus 4 into the quantity x squared minus 4x. Now you see, because the numerator has a degree as large or larger than the denominator, I can actually divide this trinomial into this binomial. And how many times will this trinomial go into the binomial? Once. One times. I'll put that in the constant column. And multiply 1 times x squared minus 4x plus 4. And I'll subtract that off. The x-squares cancel. In this case, the negative 4x is canceled. And what I get here is a negative 4. That's my remainder. So I'll put that negative 4 over my divisor. And I think maybe I can just put that in up here. I'll say this is negative 4 over x minus 2 squared plus 1. You notice when I put negative 4 over my divisor, I went ahead and factored that again to write it in that form. You might say, well, why are you factoring it again? You started off with it factored, then you've multiplied it out. Now you're factoring it again. Well, in this form, I recognize this as a transformation of the function 1 over x squared. I don't recognize it in that form, but in this form, I do. So I think we're ready to draw a graph. The fundamental graph that I'll be using here is g of x equals 1 over x squared. But there are 1, 2, 3, 4 changes in this graph. We have to shift it to the right two units. We have to shift it up 1. We have to stretch it by 4. And we have to flip it over because of the negative. So I have to make all four of those changes when I draw this graph. So let's put that over here. x-axis, y-axis. OK, so let's see. What's the first thing you would say we should do when I start to draw this graph? What's the first of those changes we should make? Shift it to the right. How far? Shift it to the right. Two. Two. Right, shift it to the right two. So here's two. And if I shift the vertical asymptote over 2, that's the line x equals 2 right there. What's the next thing we should do to this graph? Shift it up 1. Shift it up 1, OK, because of the plus 1. So when I shift the horizontal asymptote up 1, this is the line y equals 1, OK. Why don't we do the stretch and the inversion at the same time? You see, if we're going to stretch it 4, I would go over 1 and up 4. But I'm going to invert that. So when I go over 1, I'll go down 4. 1, 2, 3, 4 right here. And we're actually opposite negative 3. And if I go to the left one, I'm going to go down 4 also because we're talking about the fundamental graph 1 over x squared. And both sides go down or up together. So if I go to the left one, I go down 4. 1, 2, 3, 4 right here. I'll take out those points. OK, so it looks like our graph. Let me move this one over just a little bit right there. It looks like our graph should start down here, come up, go through those points, and level off at 1. But I think there are a few other points I could plot that would be of interest. Let's find the x-intercepts. Which one of these expressions will tell me the x-intercepts most quickly? Probably the first one. I think the first one, yeah. Because when is this going to be 0? Well, to see, this is 0 whenever the numerator is 0. And the numerator is 0 when x is 0 or when x is 4. So for my x-intercepts, I think we're going to get x equals 0 and x equals 4. Now, I'm not working out all the details. But instead, I'm talking my way through it to save a little space and to save some time. And you can do the same thing yourself. Whenever this, I guess the question I ask is when is this expression 0? And this expression is 0 when the numerator is 0. And the numerator is 0 when either this factor or this factor is 0. And so I get 0 and 4 for my intercepts. So here's 0 right here and 4 right there. This was 2. This is 4. What about the y-intercepts? Well, this is the y-intercept because this is the point on the y-axis. So there's no need to look for an additional point. So when I draw my graph, it comes up, goes through those two points and levels off. And this one comes up, goes through these two points, and it levels off over here. Let me remove some of the other writing around it and you get a better picture of what that graph looks like. Let's go to the next example then. And now we'll look at something a little different called a slant asymptote. We haven't seen a slant asymptote in any previous examples. But this is sort of an extension of the idea of graphing rational functions. In the next example, it says sketch the graph of f of x equals x squared over x minus 1 showing its slant asymptote. Well, let's write that one down here on the screen. f of x equals x squared over x minus 1. Is this a proper or an improper rational expression? Improper. It's improper. But this time, what makes it a little different from the other examples is I have a larger degree on top than on bottom. So what I'm going to do is to divide the bottom to the top, and it's going to go more than just one time. I think we'll get a variable as part of our quotient. x minus 1 divided into x squared. Let's see, x goes into x squared x times. And when I multiply x times x is x squared, x times negative 1 is negative x. And when I subtract, I get a plus x. And x minus 1 goes into x one time. 1 times x minus 1 is x minus 1, and I'll subtract that. And when I subtract, the x is canceled and I get a 1. At this point, x minus 1 will no longer divide into the 1. That's my remainder. So I'll write that as a rational expression, 1 over x minus 1. So as an alternative form for my problem, x squared over x minus 1, I could also write as 1 over x plus 1 plus the quantity x plus 1. You notice what I've done is to take the polynomial portion of the quotient, and I put it on the right instead of on the left. I think that should be x minus 1 on that numerator. Oh, thank you. It should be x minus 1. Thank you. You're exactly right. The other one. In the denominator, I'm sorry. Oh, that's right. That should be a plus. This should be a minus right here. Yeah, thank you. OK, now, how do we go about graphing this? Well, I think I'm going to want to use both of those forms to determine what the final graph is going to look like. This x minus 1 in the denominator tells me that there's a vertical asymptote at 1, because 1 is not in the domain. And so I'm going to have a vertical asymptote at 1. So we'll put that one in right here. x equals 1. And let's see, what about x-intercepts? Are there any x-intercepts for this graph? Let's see, to find x-intercepts, I would set y equal to 0. And if I set this equal to 0, then the numerator is 0. And if the numerator is 0, what would I get for x? 0. I get 0. So for the x-intercepts, I have x equals 0. So that point is right here. And that is the y-intercept once again, so I won't go about looking for the y-intercept. If I wanted to find a y-intercept, I would set x equal to 0 and solve for y. But I think I'll come up with that same point. OK, now here's something that's different about this graph from the other graphs. You notice that if I pick x equals 3, or 5, or 10, or 20, whatever, as x gets bigger, what happens to this fraction right here? The number gets smaller. Exactly, this fraction is going to get smaller. And therefore, f of x is going to look more and more like the expression over here. As this fraction plays a smaller role, that means f of x looks more and more like the graph of f of x equals x plus 1. Now, what does the graph of f of x equal x plus 1 look like? What does that look like, Jeff? It's going to go through the y-axis at y equals 1. 1, yeah. And then just go up diagonally, up 1 over 1. OK, it's going to be a straight line. It's going to cross the y-axis at 1. And if I go over 1 and up 1, I'm going to get a diagonal line. Now, this is going to be my slant asymptote right there here. And what this tells me is that as you go further and further out to the right on the x-axis, as x's get larger, the function f of x, the original function f of x, looks more and more like x plus 1. It looks more and more like that diagonal line, because the thing that makes them different is this rational expression added on here. And this becomes smaller. This approaches 0. So what this tells me is that the function is going to turn up around in the vicinity of x equals 1. It's going to turn up there, or maybe turn down. And it's going to somehow be turning out diagonally over here. Now, I don't have any x-intercepts on this side. I had one x-intercept at the origin. So if the graph were to turn down, I would have to cross the x-axis. So apparently it doesn't turn down. It has to turn up. And so that tells me my graph should look like this. It has to approach the vertical asymptote, either above or below. And it has to approach the slant asymptote. And because there was no x-intercept, I figure it can't go down, so it has to turn up. Now, of course, there's a question here of how low do you go to turn before you go back up. We don't really know with this information. If I started plotting discrete points, I could make a better estimation of that. But I'm trying to avoid plotting a lot of points. But I figure this is the general shape of the graph. OK, now, what's happening over on the other side? Well, either the graph is going to go up near the vertical asymptote, or it's going to turn down. And if it goes up, it has to come back and approach this slanted asymptote. But there's no x-intercept over there. But there's an x-intercept on this side. So I think what that tells me is my graph comes up. It touches that x-intercept. And by the way, what is the multiplicity of that root in the numerator? So it's going to look like a, shall we say, sort of like a parabola there. It's going to turn. And that's the high point. It turns, and it comes back and goes out this way. This should be a smooth curve as it goes through the origin. And this is the graph of x squared over x minus 1. Now, you see, when you put a larger degree in the numerator, what that does is takes the horizontal asymptote, and it turns it, and it makes it slanted like this. I think we should take another example like that. Let's see. Let's take a problem like this. Suppose we have g of x equals x times x plus 3 over x plus 1 over x plus 1. Now, as I make this up, here's what I'm looking for. I want to have a rational function. And I want to have a larger degree on top than on bottom. And on top, I have a second degree polynomial. When you multiply that out, it's a quadratic. And on the bottom, I have a linear polynomial. So I have an improper rational expression with a larger degree on top. This tells me I'm going to have a slant asymptote in the problem. So I need to divide the bottom into the top. And to do that, I better multiply out the numerator first. x squared plus 3x divided by x plus 1. Now, I'll do my division over here. x plus 1 divided into x squared plus 3x. x goes into x squared x times x squared plus x. I'll subtract that off. When I subtract, I get a 2x. And x goes into 2x two times. So when I multiply, I get 2x plus 2. And I subtract this off. And I get negative 2. This is my remainder. So that says I'll have to express my answer as a negative 2 over x plus 1 added on to the quotient. See, I've taken my negative sign and put it out in front in this case. OK, so this is. Let's see, let me just rewrite this down below. This can also be written as negative 2 over x plus 1 plus the quantity x plus 2. I'm taking my polynomial expression and putting it on the right because this is going to become my slant asymptote. By the way, when I did this division, some of you may be thinking, gee, shouldn't you be using synthetic division there? You certainly could. You could use synthetic division. But because this is so short, it's just a linear divided into a quadratic. I decided to show all of the details there since speed wasn't really such an issue for that division. OK, let's see what I can say about this. There's going to be a vertical asymptote at x equals negative 1 because there's going to be a shift to the left. So the vertical line, x equals negative 1, is my vertical asymptote. There's going to be a slant asymptote, which is what? What would you guess? This slant, y equals x? y equals x plus 2. Yeah, this entire thing, y equals x plus 2. What about x intercepts? x intercepts. Well, rather than looking at this form or this form, let's go to this form. Now, when is this expression going to be 0? x equals 0 or x equals negative 5? I'm sorry, that's a 3 there. It's kind of hard to read. Yeah, that's a 3. So there are two x intercepts, x equals 0 and x equals negative 3. There's a y intercept, but because I got the origin again, that seems to be coming up a lot here, that's going to be my y intercept as well. Let's take this information to see what we can make of it. I'll put the graph over here. OK, vertical asymptote at negative 1. So I'll put my vertical asymptote right here. That's x equals negative 1. Slant asymptote, y equals x plus 2. Let's see, going back to the origin, if I go up 2, the slope is 1. Go over 1 and up 1. So I get my slant asymptote right here. This is the line y equals x plus 2 right there. Those aren't target points that I plotted there. Those are just merely aids in helping me sketch the slant asymptote. There's an x intercept at the origin, and there's an x intercept at negative 3. Negative 1, negative 2. So here's a negative 3 right here. So this is negative 3, and that was negative 1. By the way, what is the multiplicity of those roots? 0 and negative 3. 1. They're both 1. So what does that mean about the graph? Will it be turning like a parabola or will it pass through? Pass through. It'll pass through. OK, now let's put that information together. On the right-hand side, I know that my graph has to approach this slant asymptote from either above or below. It's got to come to this point, and it's going to be passing through that point, and it has to approach this vertical asymptote. I think all that information means that my graph is going to look like this. I could probably make that a little bit smoother right in there. So I think that's what the graph looks like on the right. And on the left, let's see. I know I'm going to be coming through this x intercept. I have to approach this vertical asymptote. I have to approach this slant asymptote. So I think the graph is going to come down like this. Yes? Do you take into consideration that negative 2 at all? Does it just? The negative 2. Oh, right here? Yeah. Well, you see, what that says is that there's been a graph that's been inverted. And if that negative had not been there, then the graph would have been maybe up above and here, and it would have been down below over there. It would have been inverted. But it wouldn't be just a simple inversion, because if you take off that negative, this doesn't change. So it's not a matter of putting a negative there and putting a negative here. Let me see if I explain that a little bit better. If I take off that negative, that only affects this one term. It doesn't affect that term. Whereas if you put a negative in here, that affects the entire graph. So by removing that negative, it merely flips this to the other two to two. Those aren't exactly quadrants, but these two regions. And I'd have a graph up above and I'd have a graph down below in that case. Let's see now. Let's just consider some of the options here. We said that at negative 3, the graph was going to pass through, was going to pass through the x-axis. But someone may ask, well, couldn't it have passed through this way? Could it have come down that way? Well, the problem would be if it had come down this way, I'd have to come back to this asymptote. And to get back to that asymptote, I'd have to turn and cross the negative x-axis to get down here, but I don't have any more x-intercepts. So this is the only x-intercept. So the graph must be rising as it goes to that point and not falling as it goes through that point. Because if it were coming down, it would force me to have another x-intercept further out the negative x-axis. Let's take an example now of a function that has two vertical asymptotes. Oh, let's say, no, actually we have an oblique asymptote in the next example. Let's go to this one. In this example, let's sketch the graph of G of t equals t cubed plus 1 over t. Now, this is more than a slant asymptote because I have a cubic on top over a linear denominator. And this time, we're using the independent variable t rather than x, but that's not a significant change in the function. Let's see, because I have an improper fraction, improper rational expression, I want to divide the bottom into the top. And this one I think I can do by sight because I'm just simply dividing by a t. I'm going to get a t squared plus 1 over t. I divided t into t cubed and t into the 1. And if I reverse that, that's going to be 1 over t plus its, well, this is what I would have called the slant asymptote a moment ago. This time, I'm going to call this my oblique asymptote because it's not linear, but it's a quadratic function. Now, to put this information together, I would say, first of all, I'll call that the t-axis. And when I draw in my vertical asymptote, let's see, my vertical asymptote is going to be at t equals 0. So I have a vertical asymptote right here. I can't really dot it to show you because I have a solid y-axis there, but I have a vertical asymptote there. And instead of having a slanted asymptote, I'll have a parabolic asymptote, like y equals t squared. That's going to look like this. This is the parabola. I should probably make that come up a little bit steeper. So I have this parabolic asymptote and I have a vertical asymptote. Do I have any horizontal intercepts? Let's say, I guess we'd call these t-intercepts. t-intercepts. Well, to find a t-intercept, you'd let y be 0. So I'd let 0 equal t cubed plus 1 over t. Or another way to write that is to say that 0 is equal to t cubed plus 1. I've multiplied both sides by t. Or another way, I've set the numerator equal to 0. This factors into 0 equals, well, who can tell me? How do you factor t cubed plus 1? It's t plus 1. And then it'll be t squared minus t plus 1. Very good. That's exactly right. A lot of students make mistakes on that. Most common mistake is they want to say t squared minus 2t plus 1, which would make that a trinomial square. This quadratic expression has only complex roots, but this has the real root t equals negative 1. So t equals negative 1 is my only t-axis intercept. Let me erase this so that I can put that in my graph. I'm going to have a t-intercept at negative 1. There's negative 1 right there. By the way, what was the multiplicity of that root? 3. No, let's see. It was only in one factor. It was in the factor t plus 1. So it had multiplicity 1. That tells me that my graph is going to be crossing the axis right there. But I have to approach this vertical asymptote and I have to approach this parabolic asymptote. Well, the only way I think we can do that is to have the graph come down on the outside over here. It's going to pass through that point and it's going to approach this vertical asymptote. So it's approaching this parabolic asymptote on the outside. Now, what happens on the right-hand side? Well, there are no t-axis intercepts. So can the graph come up from below and approach that? No, I don't think so. So what do you think it's going to do on this side? It's going to be on the inside of that problem. Very good. It's going to be on the inside. It's going to come down like this. It's going to turn. It's going to go back up and approach the parabolic asymptote. Of course, there's always the question how low should we draw this portion of the graph? Should we come down maybe a little bit higher? Should we come down a little bit lower? Well, this is all relative to the person drawing the graph. We don't have more information at the moment to locate that point. So if you were to draw this graph, I'd want you to just show a portion of it in there. Let me ask you this. Oh, Susan. When you're finding your x-intercepts, why do you only use your numerator and not the denominator? Well, I think that's a good question. You see, back over here when we said 0 equals t-cube plus 1 over t, one shortcut for this is just take the numerator and say, when is the numerator 0? So when is t-cube plus 1, 0? Susan's asking, how can you just go to the numerator? What about the denominator? Well, see, if I set this equal to 0, I need to multiply both sides by t to get rid of this expression. t-cube plus 1 over t times t. Over here, the t's cancel. And t times 0 is 0. So you see, what we're left with is just the numerator. So if I want to find out when a fraction is 0, it's when the numerator is 0. You might say, well, shouldn't you set the denominator equal to 0? No, that's when I have my rational expression undefined. That tells me when I have a vertical asymptote, not an intercept. So the numerator gives me intercepts, so the denominator gives me asymptotes. OK. So here's a situation where we have a rational expression, and it had not a slit asymptote, but it had this oblique or curved asymptote. And we took what little information we had, and we drew a fairly accurate sketch. OK, let's go to the next graphic. In this next instance, we have multiple vertical asymptotes. And I'm sure by now you're thinking, my gosh, when is this ever going to end? It just keeps getting more and more complex. But actually, I think as you look at this next function graph, I think you'll see the functions are getting more complex, but the effort it takes to graph them is not particularly more complicated than it was before. The function we have in this case is f of x equals x minus 1 over the product x minus 2 times x plus 1. Well, here's what goes through my mind when I look at this. First of all, I say this is a proper rational expression, because I have a larger degree on the bottom. Therefore, if I were to multiply out the denominator, I'd get a quadratic. I wouldn't be able to divide it into the x minus 1. So that tells me there's no vertical shift. This is basically no extra expression added on at the end. There's no vertical shift. And so the horizontal axis, the x-axis, is my horizontal asymptote. But I do have two vertical asymptotes. Let's see. We said that because the larger degree was on the bottom, I'm not able to carry out any division here. And therefore, there's no vertical or slant asymptote produced. However, there are two places where this function is undefined. Where would they be? x equals 2 and x equals negative 1. x equals 2. OK, that's right here. x equals 2 is undefined. So I'll put a vertical asymptote there. And x equals negative 1. There's a vertical asymptote right there. So this is x equals 2, and this is x equals negative 1. Let's see. So I have two vertical asymptotes. I have a horizontal asymptote on the x-axis. Do I have any x-intercepts? Yeah, you would. We do. Yeah, what x-intercepts do we get? 1. Very good, at 1. Because you see, the way we find x-intercepts is we set this rational expression equal to 0. And the rational expression is 0 if the numerator is 0. And that's only 0 at 1. So I have an x-intercept right here. Just as important here, what is the multiplicity of that root? 1. Is 1. So that tells me the graph is going to be passing through that point, either going up or passing through that point, going down. If that had been a square on the x-minus 1, I know that it would turn, either turn back up or it would turn back down. OK, this is all important information. Now, where will it cross the y-axis? Where is its y-intercept? Susan. I have a question. Oh, OK. Does it only pass through the point if it's an odd, if it's raised to an odd power? Right, if it's even, it turns and goes back up or it turns and goes back down. Yeah, does that make sense? Because just in that vicinity, it behaves like a parabola. So it turns back down. Or in that vicinity, it behaves like a straight line and it passes on through. Listen, how do we find x-intercepts? I mean, y-intercepts. Y-intercepts, thank you. I mean, set x equal to 0. Set x equal to 0. Let's do that right below it. If I set x equal to 0, f at 0 would be negative 1 over negative 2 times plus 1. We get plus 1 half. Plus 1 half. Now, where will that be? Let's see. Plus 1 half would be right about here. Now, a person might say, well, that doesn't seem to give us much information, but actually, that reveals a lot. Because we said the graph was going to pass through 1, either going up or coming down. Well, look, we have an appointment to be to the y-axis at a half. It must be rising as you go from right to left. So it must be coming down like this. I don't think it ever levels off in there, but it looks like it comes fairly close to it. Now, a person might say, Dennis, couldn't it pass through that point and then turn down suddenly? But if it turned down, what would have to appear between negative 1 and 0? An x-intercept? Another x-intercept. So you see, it's important to know where the x-intercepts aren't, just as it's important to know where they are. If there's no x-intercept here, the graph has to turn up and approach the vertical asymptote on the rise as it approaches it from the right. OK, now, what about over on this side? We have no information at all. So what I would do in this case is I would pick a point like this one right here, 3. And I would say, if I substitute in a 3 here, here, and here, will I be plotting a point above or below the x-axis? Just to pick this as a test point. Well, if I substitute in 3 here in the numerator, I'll get a positive 2. And if I plug in a 3 here and here, I think both of those will be positive also, won't they? Which tells me this rational expression will be positive at 3. Well, if it's positive at 3, it's positive everywhere to the right of 2 because there are no x-intercepts. It can't cross the axis. So that tells me my graph has to look like that. You notice I didn't even bother computing the exact function value at 3. I just wanted to know, should I be above or below? Now, what about on the other side? Suppose we choose, let's pick negative 3 right here. If you substitute in negative 3, let's get this out of the way, I think. If you substitute in a negative 3, will the numerator be positive or negative? Negative. It'll be a negative. Actually, it's a negative 4. And if you put it in a negative 3 there and a negative 3 there, what will each of those factors be? Negative. They'll both be negative. Negative and negative. Now, if all of those are negative, will the function value at negative 3 be positive or negative? Negative. It'll be negative. That says I should go down. I don't really care how far down, because we're not trying to get that much accuracy. But I know that my graph is below, and so it has to stay below. And it levels off like that. You might say, well, why couldn't it go up above? Well, if it ever went up above, you'd have to cross the axis. And we can't cross the axis, because there are no intercepts. So it has to stay below. So if you pick one test point, like plus 3, plus 4, plus 5, doesn't really matter. And then on the other side, pick one test point, figure the sign, is all you have to do. You don't have to figure an exact numerical value, and then you can sketch the graph. So here we have a situation of two vertical asymptotes. We still have a horizontal asymptote, and our graph is divided into three pieces. The piece to the right of 2, the piece to the left of negative 1, and the portion in between negative 1 and 2. OK, let's do one more example like that. This time I'm going to make up a problem. How about g of x, no, we'll call it g of x, equals x squared minus 2x plus 1 over x squared. Now, this example has several features that are reminiscent of the previous examples, but I think it has enough differences that we can learn something from it. Now, you notice this numerator is a perfect square. This is x minus 1 squared, and the denominator is a perfect square. x squared is already factored. So this is a rather interesting function. I don't think this function can ever be negative, because we have a square in the numerator and a square in the denominator. So I think this entire graph should either be on or above the x-axis. There's nothing below the x-axis. Will there be a vertical asymptote in this graph? Yes. Yes, there will. Susan, where will the vertical asymptote be? At very good x equals 0, because that's where the function's undefined. So I'll just sort of lightly dot in a vertical asymptote there. I can divide the bottom into the top, the denominator into the numerator, because those are both second degree. And in fact, if I go back over here, I think we can just visually divide that into all three terms. x squared goes into x one time. x squared will not go into the x or the 1. So over here, I'll put minus 2x plus 1 over x squared. What does the plus 1 tell me about the graph? Will we shift it up one? There's going to be a shift up of one unit for the horizontal asymptote. Very good. Horizontal asymptote is up here at y equals 1. OK, so why don't we kind of center that a little bit better there, I think? OK, now if you look at this format, are there any x-intercepts? Is the numerator ever zero? At 1. At 1. There's an x-intercept at 1. Multiplicity 2. Multiplicity 2. Yeah, now what's significant about multiplicity 2, Steven? It's going to touch it and then turn back without ever going through it. Yeah, either it's going to come up from below and touch it and come back down, or it's going to touch it from above and go back up. Now, do you think it will turn and go back up? Or do you think it will come from underneath and go back down? We have enough information here to say. What do you think, Steven? Come from above and. OK, what makes you think it should come down from above and go back? Because it has to be approaching the asymptote and the asymptote is above the x-axis. You see, we have an appointment to approach the horizontal asymptote and it's above the x-axis. If we come up from underneath and go back down, we can never cross the axis again to get to that asymptote. So we have to come down from above, touch it, turn, go back up, sort of like a parabola there, turn back up, and then it approaches the horizontal asymptote. Now, what makes this graph a little different is, look, I've crossed the horizontal asymptote because the horizontal asymptote only tells me what's happening out at the two ends. In the middle, sometimes you can cross the horizontal asymptote or even a slant asymptote, but then you have to come back and approach it. Now, over on the other side, on the negative side, I have no x-intercepts, which makes me think that the graph should come down and just level off. And it doesn't touch the x-axis. So this is my guess as to what the graph would look like. OK, so we've graphed yet another rather complicated, rational function by using a minimal amount of information. We've done so many graphs today that I think you kind of lose perspective on what we've accomplished, but we've graphed maybe, I don't know, six, seven rational functions. All of them are very, very complex compared to anything we've been graphing earlier this semester. And yet, if we put all the clues together, it's like solving a mystery. You can put the clues together. You can draw the graph with a rough sketch and have good accuracy in that. OK, one more idea about graphs. And this one has to do with holes that can appear in graphs. Let's take, for example, this next graphic. Sketch the graph of f of x equals x squared over x. This graph has a hole in it. This is a rational function. It's an improper rational expression. f of x equals x squared over x. We have a quadratic over a linear factor, but the difference in this one is the denominator could cancel out. Let me write that one up here. f of x equals x squared over x. Now, in earlier algebra classes, they would say divide both sides by x and simplify it to be x. But of course, when I cancel off the x, I'm assuming the x is not 0. So f of x equals x squared over x is identical to f of x equals x whenever x isn't 0. And at 0, this function is undefined. So when I draw this graph, I'm going to draw f of x equals x, which is a 45 degree line. Let's try to do that a little bit better. But I have to put a hole in it at the origin. It's undefined at the origin, but it's exactly like f of x equals x everywhere else. So this is a function that has a hole in it. And how am I going to recognize such functions? Well, whenever I have a factor in the denominator that will cancel with a factor in the numerator, I should reduce that. And see if the function, see if I've eliminated that value. If I have, I should put a hole in the graph. Let's take a look at a few others. In the next graphic, we have several examples. And let's do at least a couple of these before we run out of time. Let's take capital f of x equals x minus 2 over x minus 2. Now of course, what we all want to do is cancel off the x minus 2's and say this is equal to 1. Well, it's equal to 1 if x is not 2. Because if x is 2, then I'm canceling 0's and you can't divide by 0. So what I should do is graph the function f of x equals 1 and put a hole in it at 2. OK, let's see. Here's the x-axis, the y-axis. Here's 2. I'm going to graph the function 1. But I'm going to go back and put a hole in it at 2. And this is the graph of f of x equals x minus 2 over x minus 2. Now you might say, Dennis, it seems to me like you're really splitting hairs on the kind of functions that you're graphing here. I mean, these are getting very, very technical. Well, you know, if you move on to calculus, if you take calculus after taking college algebra, you will see functions like this. And they play a very important role in a calculus course, functions that have holes in them. One more example, and I think we're going to be just about out of time. Let's take that last example. h of u equals u plus 1 over u squared minus 1. Now I'm thinking if I factor that denominator, I'll see a u plus 1 and a u minus 1 appear. And if I cancel off the u plus 1s, this is 1 over u minus 1. But I'm assuming here that u is not negative 1, because if u is negative 1, I'm canceling 0s. So on the u y axis system, I'm going to graph 1 over u minus 1, which is one of our fundamental rational functions translated one unit to the right. And there's a target point at plus, over up 1, go back and down 1 right here. And my graph looks like this. But when I finish, I have to put a hole in it at negative 1. There's a hole right there. And this is the graph of h of u equals u plus 1 over u squared minus 1. If you want to find out where that hole is precisely, put in a negative 1 in this expression and you get negative 1 half. And this is at the point negative 1, negative 1 half. I think we're just about out of time. So if you look back over the examples that we've done today, you'll see, I think, a rather rich variety of rational functions that have horizontal, oblique, slant, asymptotes, holes in the graphs. My goodness, we've come a long way in 60 minutes. In the next episode, we'll be reviewing for our second exam for this course. And so you'll want to tune in for that. We'll go back over each of the sections that we've covered and highlight some of the important things you need to remember. There's a more extensive list of review topics on our website. And I'll see you next time for episode 15.