 In this video, I want to graph a rational function r of x equals 3x squared minus 3x over x squared plus x minus 12. And I want to do that without any graphing technology whatsoever. I just want to graph it using information about the intercepts of the function and its asymptotes, and of course using multiplicities on how it behaves near those asymptotes. So taking our function, the first thing I want to do is factor the numerator factor denominator. To factor the numerator, we have 3x squared minus 3x. There's a GCD, a common divisor, a graphing divisor of 3x. Factor that out, and that would leave behind x minus 1, and that gives the factorization of the top 3x times x minus 1. In the denominator, I have this quadratic trinomial x squared plus x minus 12. Since the leading coefficient is 1, I'm going to look for factors of negative 12 that have to be 1. You can take 4 and negative 3, and so the reverse soil would be x plus 4 times x minus 3. And so then analyzing this, what makes the denominator go to 0? Well, if we set the denominator equals 0, you get x plus 4, that could equal 0, right? Which would give negative 4 as a value. You could also have x minus 3 equals 0, which means x would equal 3. So these are going to be the forbidden values from the domain 3 and negative 4. Plugging those in would make the function not defined. It's not defined at those values. Now I want to mention that this function right here is in lowest terms. It's in lowest terms. Therefore, in lowest terms, if something makes the denominator go to 0, that means it gives us a vertical asymptote. So the vertical asymptotes of this function are going to be x equals 3 and x equals negative 4. Each of these factors shows up with a multiplicity of 1 in the denominator. So this tells us that since these things have odd multiplicity, we're going to cross infinity at these vertical asymptotes. Alright, let's come and look at the x-intercepts. The x-intercepts are the things that are going to make the numerator go to 0. So if you have 3x times x minus 1 equal to 0, by the zero product property, we get x equals 0 or x equals 1. So we get these x-intercepts right there. Now the same thing is true about their multiplicities, right? x shows up once, x minus 1 shows up once. So both of these x-intercepts have an odd multiplicity, which tells us we're going to cross the x-axis at that value. Now to find the y-intercept, we just have to look at r of 0. That's going to give us 0 on top times 4 times negative 3, which means our y-intercept is going to be 0 as well. So this function will pass through the origin. Okay? The last thing I want to mention is about the in-behavior of the graph. We want to see what happens as x approaches plus or minus infinity. Well, when it comes to the in-behavior, I actually prefer that we look at not the factored form but the multiplied out form, this one right here. Because in that situation, only the leading terms matter on top. So as x approaches plus or minus infinity, we see that r of x will be approximately the same thing as the ratio of its leading terms. 3x squared over x squared, which this implies just to be 3. This suggests to us that we have a horizontal asymptote at y equals 3. So y will approach 3 as x approaches plus or minus infinity. So we do have this horizontal asymptote in play. Now whatever you and so let's start graphing this information. Let's start graphing this thing right here. So I'm going to indicate for us the x-axis and the y-axis. Let me move that over a little bit, the y-axis right here. And so what information did we find? We found out that it goes through the origin, 0, 0. That was the y-intercept and an x-intercept. It also goes through x equals 1 to give myself a little bit more space. I'm going to make this be x equals 1. This is 2. This is 3. So this is negative 1. I think I'm going to keep the y scale still the same way. Let's still make it be 1 to 1 there. So we'll make this be 1, 2, 3. The horizontal asymptote was 3. So maybe I'm actually going to change my mind a little bit. Let's just squish these things together. I think I can live with that. So this will be our x-intercept, 0. And this will be our x-intercept, 1. Again, I can live with that. So what we discover also in addition to the intercepts, we had the asymptotes. There was a vertical asymptote we found at negative 4. So 1, 2, 3, negative 4 right here. It's always a good idea to label your picture here, especially when you're bad at drawing like me. We're going to get x equals negative 4. We had another vertical asymptote at x equals 3. And that's why I changed my scale. This picture would just be too big if I spread the scale too much. I just have to accept my intercepts are going to be a little bit close to each other. So we're going to get x equals 3 in this situation right here. We also have this horizontal asymptote at y equals 3. So I'm going to put that right here on the graph. So we end up with y equals 3 as this horizontal asymptote. So this is information we have. We can start going from here, but there's one other bit of information we don't know yet that I want to know. So the thing I care about next is, does my graph intersect its horizontal asymptote? Because our function can't cross x or vertical asymptotes, and that's because the vertical asymptotes live outside the domain of the function. But there's nothing that forbids a function from crossing its horizontal asymptote or bleak asymptote if it has one. And so to determine whether we cross our horizontal asymptote or not, we have to solve the equation r of x equals 3. And so if we do that, we would get something like the following right here, r of x equals 3. Let's solve this equation. I'm going to clear the denominators. So you times both sides by x squared plus x minus 12. And if you distribute that, you're going to get, so you basically kill this part off right here, and you're going to get x squared plus x minus 12. Distribute the 3. You end up with 3x squared plus 3x minus 36, which there's a 3x squared of both sides. If you subtract 3x squared, it cancels off. So we get a negative 3x is equal to 3x minus 36. I'm going to add 3x to both sides. This gives us that 0 equals 6x minus 36. If you move the 36 to both sides, if you add 36 to both sides right here, you're going to end up with 6x equals 36. So divide both sides by 6. We end up with x equals 6. So what this tells us is that our function does, in fact, cross its horizontal acetote. It happens at x equals 6. I'm going to add this to my function graph right here. So we get 1, 2, 3, 4, 5, 6. At 6, our function will cross its horizontal acetote right there. Now, it will cross our horizontal acetote, but this is the only place where we will cross our horizontal acetote. So starting at the y-intercept, what can happen? So I'm going to take my function. I'm going to go to consider what happens at the y-intercept. A couple of things could happen. Well, we know that our function, because this is a y-intercept, it's going to go up above and below. It's going to do one or the other. Now, because we have odd multiplicity, we have to cross at this moment. We'll come back to that in a moment, I should say. It's got to go up or below. Well, if you go up, it's going to have to go down through the x-intercept and cross to the other side. That's an option. Or if it went down, you'll have to come and cross the x-intercept like that. Because again, our x-intercept at 1, it was a crossing. So we either get this green picture or we got the yellow picture. Well, what happens with these? If you're in the green picture, you're going to have to go towards your vertical acetote, which means you either turn down, which would give us an x-intercept, which we don't have. Or you'd have to go up, right? And you have to cross the horizontal acetote, which we're only allowed to do that at 6. So it turns out with this green possibility, you get into a conflict. There's no way of fitting the picture together. And so actually it turns out this green picture has to be removed from consideration. So what this tells us is that our function at the y-intercept, actually it comes up and then comes back down. That's what had to happen. And now with our picture right here, we either, as we get close to the vertical acetote, we either turn up towards infinity, or we go down towards negative infinity. But we can't go up towards infinity, same problem. So the next intercept we don't have, and we need to cross into the horizontal acetote, which we don't have. So it turns out what must be the case is that we're going to continue on towards negative infinity right here on our picture. That's what's going to have to happen. Just by the process of elimination, just pretend like graphing this rational function is like playing Sudoku or some other logic puzzle. We follow the rules, and we make the steps based upon the only thing that doesn't lead to contradiction. Now as we approach this vertical acetote, because we're touching infinity, we have to come back from to the side. So our spaceship wraps around to the other side, like so, and then we have to go towards this, we have to go towards this number right here. Okay? In which case then our function, sweet, it's going to touch, it's going to cross our horizontal acetote. So it comes back on to the side and then wraps around right here, like so. And so this is what our picture is going to be doing. At some point it has to turn around and come back towards this horizontal acetote that's in play. So this gives us the right side of the picture. What about the left side? Well, with our intercept here, the intercept is either got to go down or it's got to go up. But wait a second, we already know the multiplicity says it's got to cross. So it's got to come back over here. So then we're going to be below the x-axis. We have to approach either our vertical acetote by going towards negative infinity or we got to go up towards positive infinity. Same problems as before. We can't cross the x-axis. We can't cross the acetote. So the only picture that's applicable is the one you see right here. At x equals negative four, we're going to cross infinity and come from the other side. This time though, we won't cross the horizontal acetote. We have to stay above it. And so we get this picture right here. Well, let's see how this compared to our computer generated image, which you can see. I guess it's, we can see it right there. Can we see both of these at the same time? There you go. And so we see the behavior we want. We see sort of like this bowl shape, concave down shape going on near the x-intercepts right there. There's some maximum that's between one and zero, but it's really small. Over here, we're going to cross our acetote. We're below it, but it's going to bend back towards this. My bump is sort of too pronounced, but that's okay. At some point it turns around here. And then over here, we have exactly what we expected to do. So in fact, we did draft a pretty good rational function without using any technology whatsoever, just by looking at the intercepts and the asymptotes of the function.