 Well, welcome back to episode 13 of Math 1050 College Algebra. I'm Dennis Allison, and I teach here at UVSC in the Mathematics Department. Episode 13 is actually the first half of two episodes that are related, both dealing with rational functions. So you'll see this discussion continued in episode 14. Rational functions are functions that are ratios of two polynomials. So for example, if we go to the objectives for the course, we'll be graphing a function like f of x equals 1 over x. That's a ratio of two polynomials. We can think of one as being a constant polynomial over a linear polynomial x. And we'll also be graphing g of x equals 1 over x squared. We'll also talk about what do we mean by asymptotes, vertical and horizontal asymptotes. Then we'll be making transformations of our two original rational functions. And then we'll look at some applications of these functions. Now, when I say a rational function, what I mean is a function of this form f of x equals, and then on the top in the numerator, I put a polynomial, and in the denominator, I put a polynomial. So we'll call them p of x and q of x. As an example, we could have f of x equals x plus 1 over x squared plus 2x minus 3, for example. That would be one example of a little bit more complicated rational function. Now, the two rational functions that we want to consider today, the two that we had listed on the screen, were f of x equals x squared, excuse me, f of x equals 1 over x. And the other one is g of x equals 1 over x squared. Now, these two functions are going to be added to our list of fundamental functions that we will be translating up and down, left and right. We'll be stretching, we'll be inverting, and so forth, like we did some other functions. What were the other fundamental functions that we've considered so far in this course? What was one of them? Y equals x? Y equals x, okay, or we'll say f of x equals x. So let me just list those right below here. There was f of x equals x, that was a linear function. What was another one? F of x equals x squared. F of x equals x squared, that was a parabola, yeah. And remember, there were three target points for each one of those functions. What was another one we've had, Matt? F of x equals the absolute value of x. The absolute value function, and there were three target points for that one that we considered early on, Ginny. And f of x equals the square root of x. F of x equals the square root of x, yeah, square root of x. Now, what was different about the target points for the square root of x? No negatives. Yeah, there were no negative values allowed, so there were only two target points. There were still four others, though. Anybody remember another one that we've had? F of x equals x cubed? F of x equals x cubed, exactly. That was a function that we said at one time was called a higher order parabola, or a higher parabola, because it does sort of come down like a parabola, a little bit steeper, a little bit flatter as you approach the origin. But then it hooks down into the third quadrant. And there were three others. Anybody remember any others? F of x equals the cube root of x? Yeah, the cube root, exactly. So just like we had the square root, there was the cube root function, which had three target points. It looks very much like the cubic function, except it comes in sideways and goes across rather than coming down in the first quadrant into the third quadrant. There was also a constant function, f of x equals a constant. That one's so basic, it's almost hard to think of it as being much to graph. What do constant functions generally look like if you graph them? A straight line. A straight line, in fact, what kind of straight line? Horizontal. Horizontal straight line, exactly. And then there's the greatest integer function, that's the most mysterious one of all. And the way we wrote that was to put square brackets on the x like that. Although in your textbook, you may see it written with a square bracket and an extra bar in there. And the reason I write it both ways is because on my word processor, when I type up problems, I don't have that double bar. So I usually just use the square bracket when I represent it. What makes this function different from all the others is this one has discontinuities. You have jumps, you have these horizontal branches, these little steps that keep ascending as you go from left to right. So that makes it a little different because not all the pieces are connected. So it's not continuous as they say. Well, to add to this list of eight functions, I want to add these two other functions, one over x and one over x squared. And let's see what these look like because they're very useful in various applications of college algebra. And we'll look at some of those applications by the way before this episode is over. Well, I've never graphed f of x equals one over x before in this course. So as we would with any function that we haven't graphed before, I would make a table of values and plot points. But of course that's sort of a primitive and a rather tedious way of graphing a function. So what we want to do is look for some target points that we could graph. And then we want to keep the shape in mind so we can sketch it more quickly. Let's see, let's put our table over here on the side. I'll put x and then I'll put f of x or y on the right. Suppose I were to choose one, two, three, four for x. What would be f at one for this function? Would be one, one over one, one over one is one. What would I get if I substituted in two? One half. One half, exactly. And for three it would be one over three, so one third, and for four it's one over four. So you notice that as the x gets bigger, the function value gets smaller because it's going to be one over that. What if I were to substitute in a negative one, negative two, negative three, say negative four? What would be f of negative one? Negative one. Negative one because it's one over negative one, which is negative one. And for negative two it's one over negative two or negative one half. And for negative three I think we'll get negative one third. And for negative four we'll get negative one fourth. Let's see, what integer have I sort of carefully avoided here? Zero. Zero, yeah. Dennis, what about zero? What happens if you substitute in zero right here? It's undefined. You can't divide by zero, so there's no answer. So what that means is zero is not in the domain of this function. This function has domain, all real numbers except zero. So I'm really not allowed to choose x because it's not in the domain. Well let's plot the values that we have so far, we have quite a few of them there. So I'll just make my graph out here in this open space. And here's the x-axis and this is the y-axis. So one of the ordered pairs I'll plot will be the point one, one. And then I'll also locate the point two, one half. Two, one half will be right about there. Three, one third, well it's kind of hard to show that that's one third, but we'll guess it's about there, and four, one fourth. So it looks like these points as I connect them are approaching the x-axis, and the further out I go the bigger the x, the smaller the function value, so this approaches the x-axis. But it will never actually touch the x-axis because one over x will never be zero. My function value will never be zero, but I can get as close to zero as I wish. For example, if I substituted in a thousand, one over a thousand, I'd be so close to the x-axis you probably couldn't tell the difference when I graph it. It's not quite touching the x-axis. Over on the other side we have negative one, negative one. That comes from this ordered pair right here. We have negative two, negative one half. We have negative three, negative a third, negative four, negative one fourth. And you see the same things happening over here, but I'm approaching the x-axis from underneath. Now what we don't have is what happens in between here. So let me go back and add some other values to my table, but I'm going to choose numbers close to zero, but not exactly zero. Suppose, for example, we choose x to be one half or one third or one fourth. So one half right about here, one third right about there, one fourth right about there. If I substitute in one half for x, we get one over a half, and what does that reduce to? That's two, yeah, if we invert and multiply. And one over one third is three, and one over one fourth is four. So at one half I go up to two, at one third I go up to three, and at one fourth I go up to four. So you see we have just the opposite effect as my x's get close to zero. Instead of my function approaching the x-axis, they actually turn away from the x-axis, and they start approaching but never quite touch the y-axis. Now of course, if it ever actually crossed the y-axis, I'd have a function value at zero, and we know that zero is not the domain. So I'm approaching the y-axis and coming up from the right hand side, and if I were to substitute in minus a half, minus one third, minus one fourth, so that I choose values over on this side, minus a half, minus a third, minus a fourth, I would get negative two, negative three, and negative four. Plotting those points at negative a half, negative two, at negative one third, negative three, and at negative one fourth, negative four. So my graph curls down on this side. Now these two portions of the graph are called the branches. I have the first quadrant branch and the third quadrant branch of the function one over x. And this function has something in common with the greatest integer function that we saw earlier in the semester, and that is that it has some disconnected pieces. And at the moment that my graph jumps from the lower branch to the upper branch, I'd call that a discontinuity because it isn't connected there. Just like in the greatest integer function, I had lots of discontinuities where we jumped to each level of the graph. Now we have names for the x-axis and the y-axis in the sense that the curve's approaching the axis, but it never quite touches it. The x-axis here for this graph is a horizontal asymptote, is the term we use. It's a horizontal asymptote. And what I mean by an asymptote is it is a line that the curve approaches as I go out to infinity, but it never quite touches. Can we go to the second graphic in our list there, the one titled asymptotes? There we go. It says, when a graph approaches a horizontal line at infinity, either plus or minus infinity, we call the line a horizontal asymptote. And when it approaches a vertical line, at a vertical line like x equals a, then that line is called a vertical asymptote. Well, back on this graph, x equals zero, which is the y-axis, x equals zero is a vertical asymptote. Now, the word asymptote is a rather unusual term, but it comes from Greek, and A-S-Y-M-P-T-O-T-E. So I have a horizontal and a vertical asymptote. Now, as we will find out in the next episode, it's actually possible for a graph not this graph, but some later rational functions to cross the horizontal asymptote. For example, this one may go across the x-axis and then come back to it and approach it from underneath, because it's only at infinity that we considered an asymptote, that the curve approaches the asymptote at infinity. Okay, now, next time I graph one over x, I don't want to go to all this trouble, excuse me. So what I want to do is decide on some target points so that I can graph this more quickly. What would you choose for target points in this function? One-one. One-one, right here, yeah, there's the point one-one, and what else? Negative one, negative one. Negative one, negative one. Now, if we just keep those two points in mind and locate those points and remember the shape, then we can graph one over x quickly in the future. So here's how I would graph it the next time it comes up. If I want to graph the function f of x equals one over x, I would locate those two target points, one-one, and negative one, negative one. But not zero, zero, because zero is not in the domain of the function. And then I would remember that the function turns up and approaches the y-axis, and it turns out and approaches the x-axis as a horizontal asymptote. On the other side, it turns and approaches the x-axis from underneath, and it approaches the y-axis going down from the left-hand side. Now this isn't a perfectly accurate sketch of the function, but it certainly is a lot quicker. And we're never going to get a perfect sketch of the function, but we can try to save time, so this is how we'll graph it in the future. The other rational function that we consider a fundamental rational function is f of x, I think we called it g of x a moment ago. g of x equals one over x squared. Once again, you notice that the domain of this function is all real numbers except zero, all real numbers except zero. Now to graph this one, if I make a table of values this one time, I'll substitute in one, two, three, four. And what would the function value at one be? It'll be one. How about it two? One fourth. One fourth, yeah, it's getting smaller faster. At three, it's one over nine, and four, it's one over 16. One over 16, okay, and let's do negative one, negative two, negative three, negative four. Let's see, when I square these numbers, I'm going to get positive answers. So I get the same results, one, one fourth, one ninth, and one 16th. So this time, when I graph them, we have the point one one. I bet we'll want to pick that as a target point. At two, we go up one fourth. What did we go up for the previous function at two? One half. The y-coordinate was one half, now it's one fourth. So this one's getting closer faster. At three, it's one ninth. I can hardly tell the difference between that point and the point on the x-axis. And at four, it's one 16th. So this curve approaches a horizontal asymptote at the x-axis. But although it's difficult to show it as a different approach, it approaches more quickly. On the other side, at negative one, the function value was one. That came from right here, negative one one. And at negative two, we get one fourth. And at negative three, we get one ninth. And at negative four, we get one 16th. So I get the same sort of thing happening over here. Now on the other hand, there's a little bit of a difference in what's happening as I approach the y-axis. For one thing, it looks like both pieces are going up, both branches, rather than one going up and one going down. I'll just make a table to represent some other values that are more specific there. There's x and there's g of x. What if I were to choose either plus or minus a half? Either plus a half or minus a half. I think I'm going to get the same result either way. It'll be one over one fourth, because we're going to square the x, and that's four. What if I choose plus or minus one third? We get nine. And plus or minus one fourth, we get 16. Now look what happens. At one half, I go up to four. At one third, I go up to nine. That's going to take me way off the graph. And at one fourth, I go up to 16, way off the graph. So if I draw this graph, it's going to turn up very rapidly. Let's see. In fact, I need to turn it even faster than that. It's going to turn more rapidly, and it's going to go right through this point, which was one half and four. And now on the other side, at negative one half, four, it's going to turn and go up there. So I think if we were to compare one over x and one over x squared, we would say that one over x squared approaches the horizontal asymptote even faster than one over x did. But it approaches the vertical asymptote even more slowly than one over x did. One over x, if you remember at one half, we were at two. Now we're at four. So one over x is actually closer to the y-axis than one over x squared is. But once again, I have a horizontal asymptote that's the x-axis. And I have a vertical asymptote that's the y-axis. OK. Just curious, can you guess what a graph of this next function would look like? I'll call it capital F of x. One over x cubed. What do you think we would graph that, Stephen? It would look a lot like one over x. Only it would be a sense steeper and quicker. Exactly, yeah. Same target points as one over x. One one and negative one, negative one. But as you approach the x-axis, it approaches it even faster, even faster than one over x squared did. And out here, it's difficult to tell that it's actually above the x-axis and not on the x-axis. But it goes up even faster. And it does approach the y-axis, but it approaches the y-axis more slowly than one over x or one over x squared did. For example, at one half, you'd go up to eight to get onto the graph. And on the other side, my graph approaches the x-axis from underneath even more quickly than one over x or one over x squared did. And it would descend and approach the y-axis even more slowly. But it still approaches it. So in other words, as this power gets bigger, these two branches have more of an L-shape to it and not so much of a curve going through the point, one one. OK, and if I were going to graph one over x to the fourth, I think I'd put two target points up above the x-axis, and both branches would turn up very quickly. And they would lay out almost on the horizontal x-axis even more dramatically than what we saw before. You know, when I was talking about the graph of g of x equals one over x squared, I didn't show you how I would graph it quickly in the future using target points. So let me just mention that. For one over x squared, I would choose the target points one one and negative one one. And I would draw my graph coming down like this through this point, and I would have it approach the x-axis relatively quickly. It's kind of hard to show that accurately. And this one comes down and approaches the horizontal asymptote over here rather quickly. And that's a rough sketch of one over x squared. OK, let's go to the next graphic on transformations of rational functions. And let's consider these examples. It says sketch the following graphs by locating target points. Do not make a table of values. That is not to plot random points, but just use the target points. Let's first try graphing f of x equals three minus one over x. And then let's try graphing g of s equals two over s plus three. Now these are both variations of the function one over x. So that first one is f of x equals three minus one over x. Now what I see are two transformations being made on my fundamental graph one over x. If I turn it around, I could say minus one over x plus three. Now what are the transformations I'll need to make in that fundamental graph? Jenny, what do you think? What should we do? It's going to be shifted up to three on the y-axis. We're going to shift it up three. What else? It has a slope of negative one. Well, let's see. We don't want to use the word slope for these curves. We use slopes for straight lines. It's going to flip it over, exactly. We're going to have to flip the graph over. So when we see three minus one over x, if that looks confusing, just put it in the form where the three is added on the right-hand side. And that may seem more reasonable as a vertical translation. So when I go to graph this function, here's how I'll do it. We'll get our scaling set up here. And we want to raise the graph three units up. Now, you remember the x-axis was a horizontal asymptote? I'm going to take the horizontal asymptote with me and put it in as a dotted line right here, right through three. Now, that's not actually part of the graph. It's merely an aid in helping me sketch the graph. And then I want to flip my graph over. You notice from my new origin, I would go over one and up one to my new target point. But when I flip it over, I'll go over one and down one. And I'll have a target point right here. That's actually at the point one, two, because I went over one and down one for my new origin. And the other target point, which would normally have been in the third quadrant when I flip it over, is in the second quadrant. If I go back one, I'll go up one. So with those two points alone, I'll draw the graph. It approaches the horizontal asymptote over here. And it approaches the vertical asymptote like so. And on the other side, it will approach the vertical asymptote like that. And it'll approach the horizontal asymptote over here. And of course, we're doing this strictly so we can get a rough sketch of the graph. So we're doing this for speed rather than for accuracy. How could I figure out where is that x intercept right there? Is that y equal to zero? Is that y equal to zero? Yeah, let's just do that down here. This is the y coordinate. I'll set that equal to zero, 3 minus 1 over x. And that says that 1 over x is equal to 3. Now if I just invert both sides, the reciprocal of 1 over x should equal the reciprocal of 3. x is 1 third. So we should be, if this is totally accurate, we should be crossing the x axis at 1 third. And that actually looks reasonable. It looks like that could be 1 third. Where's the y intercept for this function? It never does. Actually, it never does. You see, because it never does cross the y axis. But even if you hadn't noticed that, to find the y intercept, you'd let x be zero. And you see, we can't let x be zero, because our function is undefined. So there is no y intercept for this function. And we can see that in a number of ways. We can see that by substituting of zero, we get no answer. And we can see that from our original graph, we had a vertical asymptote at x equals zero. And therefore, we won't be crossing that vertical asymptote. OK, the other function that was on that list, on that graphic, was g of s equals 2 over s plus 3. Now, let's see. What have I done that makes it look different? Well, for one thing, I've called the function g rather than f. And I've chosen a variable s rather than x. But that's only cosmetic. So when I draw my axes, I'll have to label my horizontal axis, s. Now, what are the changes that will be made in the fundamental graph 1 over s? Well, let's see. There's a 2 in the numerator. And you know I could bring that 2 out in front and call this 1 over s plus 3. What does a 2 in front, a multiplier by 2, what does that do to the graph? It stretches it? It's going to stretch it. Yeah, it's going to stretch it vertically and horizontally away from the x-axis. So this is a stretch. And what is the plus 3, s plus 3? What's that going to do to the graph, Matt? Move it three units to the left. To the left. Yeah, that's always a tricky one. You see a plus 3, you want to move it in the positive direction, but it moves it in a negative direction. If I might just remind you over here on the side, if you remember when we were graphing the parabola, x plus 3 squared, that parabola was shifted over three units to negative 3. And then that parabola went up like that. The reason we explained that it had its vertex at negative 3 is because this square is 0 whenever the quantity inside is 0, and the quantity inside is 0 when x is negative 3. So at negative 3, that's when I get function value 0. Now in this case, I want to shift this graph to the left 3. So here's negative 3 right here. And that means my vertical asymptote is going to be moved over three units. And I'll put my vertical asymptote here. Or if you would like a different explanation, when is this function undefined? This function is undefined if I substitute in negative 3 for s. So that's where my vertical asymptote should be. Not at plus 3, but at minus 3. OK, now I locate target points. This is basically going to be the graph of 1 over s. So I'll go over 1 and up 1, but there's a stretch. So when I go over 1, I go up 2. So I go up to 2 units for that target point. And if I go back 1, I go down 2. And I get a target point right there. Now this function has a graph then that looks like this. And on the other side, let's take out that parabola. Because we could use the space now. In this case, my parabola curls down this way. And it approaches the horizontal asymptote over here. And now when you draw this, you may draw this where it curves out a little slower. Or you may draw it where it curls out a little faster. Any of those sketches would be fine with me. Because like we said many times, this is just to get a general shape of the graph. Is there an x-intercept for this function? No. No, there's not. Is there a y-intercept? Yes. Yes, there is. Sam, how would you find the y-intercept for that function? You put a 0 in. OK. So to find the y-intercept, I guess we'll call that the y-axis whenever you label it, what you do is you let s be 0. And if I substitute in a 0, we do get an answer. We get 2 over 0 plus 3, which is 2 third. So this number right here, or this point, should be the point 0 and 2 thirds. And that looks just about right. It's about 2 thirds. OK. Let me make up a couple more functions like this, variations of fundamental functions. And I want you to tell me how I should go about graphing them. What if I were to graph, let's say, f of x equals minus 1 over x minus 2 squared. I want to sketch that function. Now, you know, if you stop and think about it before we started this lesson today, if you had been asked to graph this, probably the only thing you could think of would be to do would be to make a table of values. And we know that that's slow and tedious. But now we can graph this rather quickly. What's the fundamental graph that I'm going to be drawing here? 1 over x squared. Yeah. OK, so over here we're thinking. So we're thinking of, oh yes, we're thinking of f of x equals 1 over x squared. But if I could look exactly like that, what are the changes that we'll make on this? Flip it over. Right. It's going to be flipped over because of the negative 1. Because you can think of that as a negative 1 coefficient. And move it to the right, too. And move it to the right, too. OK, so if you think of those two changes, let's see if we can do that all at once in our fundamental graph. You see, this actually gives us power to draw a wide variety of functions that we before would have thought would have been very tedious to graph. I don't know if you find this terribly exciting. This may seem tedious, too. But it's certainly not as tedious as making a table of values. OK, so we're going to move it over two units to the right. So that means my vertical asymptote's going to move over to the right. So that's going to be the line x equals 2 right there. And the target points. Let's see, who can help me with the target points? What should I do? Go to the right one, and then what? Down one. And let's see, Jenny, why did you say go down one? Because on the normal graph, 1 over x squared, you would want up one. But since you're turning this upside down. Right. We flipped this over, so now we go down one. And when you go to the left one, what do you do? We go down one again. Because see, this is 1 over x squared. And both of its target points normally would have gone over 1 and up 1. But we flipped it over. So they both go down. And therefore, this graph looks like this. OK, so that didn't take us very long to graph that one at all. And that's the graph of f of x equals negative 1 over x minus 2 squared. Now in applications of graphing techniques to business, to physics, to engineering, to other courses, some of the key values are to know where's the vertical asymptote? OK, that's at x equals 2. Where's the horizontal asymptote? That's at y equals 0. The x-axis, y equals 0 is the horizontal asymptote. Where are the intercepts? Well, this one has no x-intercept. But it does have a y-intercept. Can anyone name the y-intercept, by the way? Will it be 1 negative 1 half? No, close. Let's see. We'd have to substitute into 0. That would be negative 1 over negative 2 squared. So it'd be what? Negative 1 fourth. So the y-intercept is at 0 negative 1 fourth, because we have to square that, of course. So we found the y-intercept right there, where it crosses the y-axis. And these are the values that are the most significant values, the horizontal asymptote, the vertical asymptote, the y-intercept, the x-intercepts, if there are any x-intercepts, weren't in this case. The target points only because they help us to draw the graph. And you might say, well, now, Dennis, why is the y-intercept, let's say, a significant point in a graph? Why is it more significant than, say, what's the value over here at 5? Well, you see, if this were a business application, and if my horizontal axis represented time, then this value would be the value at t equals 0 at the moment that we're beginning the problem, you might say, where the problem is turned on. Let's say we have a company and we're starting up, we're starting production of a new item, and this might represent the startup costs. So initially, before we make any items, this is how much we spend to start up. And as we begin to make more items, it begins to cost us more. And it looks like in this case, the function never turns around, even after you go after past the asymptote to the other side, we still have negative expenses, but we're coming back towards 0. So if this represented the profits, let's say, of a company, this company never has a positive profit, but that would be the startup cost right there. OK. Did have negative infinity for a project. Yeah, it looks like it'd go to infinity, so that we really wouldn't have much of an application for a business that wouldn't go to infinity. But in other words, there are physical interpretations to these applied problems to what various points might be. Let's take another function. How about this one? g of x equals 1 over 2 times x minus 3 squared minus 1. Now, you see what makes it different among other things is I put a 2 in the bottom, not on the top. Can anyone explain how we're going to justify that this is a fundamental graph with several transformations made on it? Just put it out of brackets in a half. Put a 1 half out in front. Yeah, put a 1 half out in front, and this is 1 over x minus 3 squared, and then put the minus 1 outside. Excuse me. Now, what is the 1 half going to do to the graph? Compress it. It's going to compress it, yeah. So here, we have a compression. Here, we have a shift to the right 3, and what does the minus 1 do? Move it down 1 on the y-axis. Yeah, it's going to move it down 1. Let's see if we can accommodate all those changes. So we'll get this set up to graph. We said we're going to move it over 3 to the right, so I'm going to take my vertical asymptote, move it over 3 to the right, and put it right here. This is the line x equals 3. And we said the horizontal asymptote's going to be moved down 1. In fact, the entire graph's going to be moved down 1. So my horizontal asymptote is the x-axis. I move it down 1, and I'll put it in a dotted line here. Now, as I was saying earlier, these asymptotes are not an official part of the graph. They're merely an aid in helping me sketch the graph. So when I draw the graph, I could just as easily wipe out the asymptotes, take the asymptotes out, and the graph would still remain. So this is my new origin, and this was originally 1 over x squared before we started fooling around with it. So if I go over 1, I would normally go up 1, but now I go up a half, because it's been compressed. And if I go to the left 1, I go up a half, because it's been compressed. And so now when I draw my graph, I have it come down, go through this point, and then it approaches very quickly the horizontal asymptote. And on the other side, I have it come down and approach very quickly the horizontal asymptote. As a matter of fact, over here, you probably wouldn't be able to tell the difference between the graph and the asymptote if we could really zoom in on it. So this is our graph. How many x-intercepts do we have? Two. Looks like we have two. How could I figure out what they are? Is that y equal to 0? Is that y equal to 0? OK, so if I set y equal to 0, this is to get the x-intercepts. I'll say 0 equals 1 over 2 times x minus 3 squared minus 1. That says that 1 is equal to 1 over 2 times x minus 3 squared. Then if I multiply both sides by 2 times x minus 3 squared, I have 2 times x minus 3 squared equals 1. So if I divide by 2, x minus 3 squared is 1 half. Let's just make a little room here so we can continue. Now if I take a square root, x minus 3 will be, will equal, plus or minus the square root of a half. OK, now here's where we, remember in the very first episode we talked about rules for simplifying radicals. This is not simplified in the official sense because I have a fraction under the radical. So what would be a better way to write that? Square root of 2 over 2. Well, yeah, actually let me fill in a step. The square root of 1 over the square root of 2. And then if you multiply by the square root of 2 over the square root of 2, we'll have Ginny's answer. So x minus 3 equals plus or minus the square root of 2 over 2. And therefore, x is equal to 3 plus or minus the square root of 2 over 2. So can anyone tell me what this x-intercept is right there, what that number is? 3 plus root 2 over 2. 3 plus the square root of 2 over 2. And if we're going to make it a point, I'll put a 0 on the end. And so you see this number is slightly more than 3. That makes sense because 3 is where the asymptote is. And the x-intercept on this side is 3 minus the square root of 2 over 2, 0. So we do have two x-intercepts and we're able to calculate both of them. And to get the y-intercept, we let x be 0. I think we can do that in this space. If I let x equal 0, what would be the g value? That'd be 1 over 2 times negative 3 squared minus 1. Can anyone simplify that for us? I'm running out of room here, so I need you to work that out for me in your head. How much will that be? Isn't it 1 over 18? 1 over 18 minus 18 over 18 minus 17 over 18. Yeah, minus 17 over 18. That's almost negative 1. And you see, that is the y-intercept right there. At this point would be 0, negative 17, 18ths. Not a pretty number at all, but that's the value. Hey, we didn't say rational numbers, rational functions were pretty, but we can sketch them nonetheless. OK, now let's go to the next graphic. And we'll see here graphs of two rational functions. And the question that I ask you is, what is the rule? What is the function rule for each one of these rational functions? In part A, there's no horizontal asymptote. But if it were drawn in, what do you think the horizontal asymptote would be? 1. It'd be y equals 1. And there's no vertical asymptote shown, but what do you think, if it were drawn in, where would it be? Negative 2. x equals negative 2. Yeah, so if you imagine that those horizontal and vertical asymptotes are drawn, what rule does this function have? Let's say we call this function f, f of x equals. Now, we want to move it two units to the left. So how would the rule begin? x plus 2, 1 over x. 1 over x plus 2. Now is that 1 over x plus 2 squared or just squared? Squared. Yeah, because you notice on both sides of the asymptote, we're going down. It looks like that graph has been flipped over. So what else do we need to do to this? Put a negative on it. And let's go back to that graphic screen so that everyone at home can see what we're doing. What's one more change I need to make in this? Plus 1. Plus 1. Looks like we need to go up 1, plus 1. So if you come back to the green screen, I think this is the function, the function rule, for the first function that was graphed there. Negative 1 over x plus 2 squared plus 1. You know, if you look on the website, you'll see these very same graphs, and you'll see answers to these questions that I'm asking you on the website. So if you're not copying this down quite fast enough, I think you'll get what you need from the website. OK, let's go back to that graphic and look at the second function graph there. Now, in the second graph, you notice there are two target points plotted. It looks like they're at 1, 1 half, and negative 1, negative 1 half. And the question that we're being asked there is what is the rule for that function? Well, would you say this is a variation of 1 over x or 1 over x squared in Part B? 1 over x. 1 over x. What modification have we made to the graph? Compressed it. We've compressed it. So how should I change the equation G of x equals 1 over x? G of x equals 1. Yeah, I would have to guess here. Looks like that's about 1 half. So I would say if we come to the green screen, this is going to be 1 half times 1 over x. There are no horizontal or vertical transformations being made. So I think what we saw there was the graph of 1 over 2x. That's all there is to it. So we have this rule for function A for the graph in A, and we have this rule for the graph in B. OK, now you know what? When you come to rational functions, the author can do a number of things to try to hide what the transformations are. Let me give you an example. In fact, let's go to the next graphic, and I think we'll see an example. Improper rational functions. Now, a rational function whose numerator has a degree as large or larger than that of the denominator is an improper rational function. So I ask you this question. How can we graph the improper rational function f of x equals x plus 1, x minus 1, over x minus 2? And where would you begin to graph this? Well, let's come to the green screen and take a look at that. We have f of x equals x minus 1 over x minus 2. And if I told you that this is actually a mere transformation of 1 over x, or 1 over x squared, you may not at first think it is, because we have two polynomials, both of them are first degree. But this is called an improper rational expression, because I have the same degree on top as on bottom. So what I'm going to do is to divide the bottom denominator expression into the numerator. Now, I suppose I could use synthetic division to do this, but this is so short, I don't think it's going to take much time anyway I approach it. If I divide x minus 2 into x minus 1, x goes into x one time. And when I multiply that out, I get x minus 2, and I need to subtract that off, just using ordinary long division. So when I subtract x minus x, I get 0. Minus 1, subtract negative 2, what will I get there? We get a 1 there. That's my remainder. I'll add that on as a fraction added on at the end. So this is the same thing as 1 plus 1 over x minus 2. Well, you see, we may not have recognized the graph in this form, but I think we do recognize the graph in this form. And they're one and the same. So if I graph this improper rational expression, it's the same thing as graphing this proper rational expression plus 1. So if I want to graph function f, I'll just graph the expression over here, because this looks simpler for me to graph. It's the very same thing as what we had there. So what transformations would I make in the graph of 1 over x? Move to the right, too. Move to the right, too. So I'm going to put my vertical asymptote there. This is at x equals 2. That's an asymptote. So I better put some spaces in there. And you said move it up 1. So if I move it up 1, this would be y equals 1. And here, of course, is x equals 2. There are no stretches or compressions. There's no reflection. So if I go over and up 1, I get a point there. If I go back and down 1, I get a point right here on the x-axis. And here is my graph. And we have just graphed here f of x equals x minus 1 over x minus 2. Now you see, when you look at this rule, it's not so obvious that the graph is merely a transformation of 1 over x. It's not until you carry out the long division that you realize that this function can be written in that form that we realize how we're going to graph it. So the author can try to hide a function like this by expressing it in this form. Now you know what? When I say an author can hide it, of course, that holds for the instructor as well. The instructor can hide the function by writing it in this form. So let me give you an example of a function that you might see, say, for example, on a test or on a quiz or homework. I'm going to call it g of x. And suppose the function is x times x plus 2 over x squared plus 2x plus 1. Now, at first glance, you might think, my gosh, this is a very complex problem to graph. But I think if I use long division, it's not so bad at all. I'm going to divide the bottom into the top. But I'm going to multiply out the numerator first before I do it. x squared plus 2x over x squared plus 2x plus 1. So I'll divide the denominator into the numerator. How many times will x squared go in x squared? One time. I'll put a 1 over there. So I have x squared plus 2x plus 1. And I have to subtract that off. The x squareds cancel. The 2x's cancel. What do I get here? Negative 1. And I can no longer divide the divisor into negative 1. So I get this remainder. What that tells me as I'm kind of running out of space here is that this is going to be 1 plus negative 1 over x squared plus 2x plus 1. Now, those of you at home and those of you right here in the studio may say, well, gee, Dennis, I'm not sure how to graph this either. Well, I don't think we've quite finished with it. How can I factor this denominator? x plus 1 squared. That's x plus 1 squared. So I'm going to make several changes here. I'm going to write minus 1 over x plus 1 squared. And then the one that's in front, I'm going to put in the back. Now, you see, g of x can be represented this way. g of x can be represented that way. g of x can be represented as this constant plus a proper fraction. And finally, g of x can be represented this way. Now, this is the one that tells me how to draw the graph. This is basically the graph of g of x equals 1 over x squared with a few modifications made. I'm going to be moving it up one. I'm going to be moving it to the left one. And I'm going to be inverting it. So here we go. We are now going to graph this function that seems so complicated originally. Let's get our scale set up. Here's the x-axis. So we're going to move our graph to the left one. So I'll put a vertical asymptote at negative 1. That's the line x equals negative 1. We're going to move the graph up one. So the horizontal asymptote comes up one. So here's my new origin. I have to invert my graph. So when I go over one, I'll go down one. And from the new origin, I'll go to the left one and go down one. And my graph looks like this. I'm not being too careful here how I graph it. It's just the general shape that we want. Some of you at home may say, gee, Dennis, you say your graph isn't very good. But when we draw them, ours don't even look quite that good. What I'm going to be looking for are the general features of the graph. So it doesn't have to be. This isn't an art course. And I'm obviously not an artist. But I'm looking for the general idea. What are the 2x intercepts here? 0 and negative 2. And look what happens. If I were to set this equal to 0, that would mean the numerator would be 0. And if the numerator were 0, the solutions would be 0 and negative 2. So I do have those 2x intercepts. What is the y-intercept? Well, the y-intercept, where it crosses the y-axis, is at 0. And I think if you substitute in 0 for x there and there and there and there, you get 0 over 1. So the y-intercept would be 0. And so this is the graph of the function g of x equals x times x plus 2 over x squared plus 2x plus 1. Now, we didn't do that in, say, 30 seconds. It took longer than that. But prior to this discussion in this episode, I think we would have had a lot of difficulty in drawing a graph of that at all. OK, well now you see that we have two new fundamental functions, f of x equals 1 over x and g of x equals 1 over x squared. So if you add that to our list, I guess now we have 10 fundamental functions. And we can graph infinitely many variations of those. You might say, well, what makes those fundamental, as opposed to, say, other functions that we haven't talked about? Well, these are the functions that come up most often in applications and other courses. Absolute value, greatest integer, linear functions, rational functions, 1 over x, 1 over x squared. So what we're trying to do is build a catalog of functions in our mind that we can sketch quickly and variations of them. Let's look at this application, and I'll show you how rational functions come up in other problems. Suppose that a drug is being administered to a patient by means of an IV, and the concentration c in the blood stream after t minutes is given by this formula. The concentration at time t is 20 minus 20 over t plus 1 squared, where c of t is measured in milligrams per liter of blood. So we have this IV, sort of a drip IV, and a drug is being administered to a patient. And let's say that this function begins at the moment t equals 0. That's when the IV is first introduced. And c of t is the amount of milligrams per liter that's in the blood at time t. Now, how could I find out what is the initial concentration of the drug in the blood stream? Let's come to the green screen, and I'll write this function right here. c of t is 20 minus 20 over t plus 1 squared. Excuse me. So the first thing I might ask is, what is the initial concentration of the drug in the blood stream? How would I answer that? Yeah, set t equal to 0, because we said we were going to begin this at the moment t equals 0. So what we do is we calculate c at the moment t is 0. So that's the value I'll substitute in. And what I get is 20 minus 20 over 0 plus 1 squared. What does that reduce to be? We get what? We get 0. We get 0 milligrams. That's right, because when you first introduce the IV, there is no concentration of drug. There is no drug in the blood stream at all. So we get 0. OK, next question. What's the approximate concentration after, let's say, 10 minutes? If t is measured in minutes, c at 10 would be 20 minus 20 over 11 squared, which is 20 minus 20 over 121. Well, let's see. This is roughly 1 sixth. So let's say it's approximately, just to speed this up, it's approximately 20 minus 1 sixth, which would be 19 and 5 sixth milligrams per liter of blood. So the concentration has gone from 0 up to 19 and 5 sixth. What would the graph of this function look like? Whoops, I shouldn't have erased my rule. 20 minus 20 over t plus 1 squared. What would the graph of this look like? Well, let's see. Shift over 1. It would be shifted up 20. It's going to be shifted up 20. Yep, so let's just say this is 20 right here. Yeah, because we have 20 added on in front, as opposed to being added on in back, it's added in front. What else? Me inverted. It's going to be inverted because of the negative. What does this do? Move it to the left one. To the left one, so let's say we go over 1 right here. This is negative 1. I'll put a vertical asymptote there. And let's see, starting from the origin, if I go over 1, I go down because it's been inverted. And what's the stretch? 20. 20, yeah, because see, there's a 20 there. So if I go over 1, I go down 20. I get a point here. And if I go back 1, I get to go down 20 right there. And my graph looks like this. But you know what? I've been graphing this for negative as well as positive values. If I throw away all the negative values, I'm left with this portion of the graph for t greater than or equal to 0. It starts off at concentration 0. That's the concentration axis. It starts off at concentration 0. And it approaches the line 20. Now, if I come out here to 10, this altitude should be 19 and 5, 6. And in the long run, what is the concentration of the drug in the bloodstream going to approach? 20. It's going to approach 20 milligrams per liter. So we've been able to draw a graph that is a visual example of this application. Well, let's see, we've introduced two new fundamental graphs. We've looked at transformations of those fundamental graphs. And we've looked at one application sort of hurriedly at the end of the tape. I'll see you next time for episode 14.