College Algebra: Graphing Rational Functions Ex

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So let's take a look at some more examples of graphing these rational functions, first by finding the asymptotes and then taking a look at the regions and seeing where these things fit in. Let's take a look at the first thing--f(x) = x² - 1/x + 1. Okay, well, it looks like there's going to be a vertical asymptote when x = -1. We have to be careful, because we have to make sure that we factor everybody. And this thing actually can be factored a little bit more, and if you factor the top, it's the difference of two perfect squares, what we see is the product of x + 1 times x -1, and then I've got an x + 1 on the bottom. So, in fact, I can cancel these people away as long as I promise that this is not zero, which means that as long as I promise x doesn't equal -1. So, in fact, what I see here is if x doesn't equal -1, then this function is just x - 1. Well, that's just a line. It's a rational function that's just a line. So, in fact, I can just graph the line. So the y intercept is -1 and then its slope is 1. So what I do is I look at this, and so I see--so that's f(x).
However, I promised you that x cannot equal -1. So at -1 I lied. I actually colored that in. What level is that? Well, you can see -1, that would be at -2. So this value here is -2, but I'm not supposed to put a point in there. I have to remove that point, so it's like drilling a hole in there. Now, I don't have a drill with me, so instead what I'll just use is some white thing, and I'll just remove that point. So, in fact, the graph of this rational function, which looks like it should be some sort of asymptote sort of stuff, I turns out what it really is is just a line with a whole through it. Neat! So sometimes these things can just be very simply objects, but with a little hole. And the hole came drilled in because I could cancel those things away. So notice there's no asymptotes here at all; it's a line with a hole.
One last one to sort of wrap up these ideas. Let's take 9x - 1, multiply it by x + 3, divide the whole thing by 2x + 6(x - 5). There's a lot of stuff going on here. Well, first of all, one thing I noticed is that if I factor out the 2 there I can cancel the x + 3 with the x + 3 here. So, in fact, let me do that. I see that this actually equals 9x - 1 divided by 2(x - 5). Let me say again what I did. I factored out a 2 here, which gave me a 2(x + 3). But then the x + 3 cancelled with this x + 3, as long as I promise you that x doesn't equal -3, so I can't divide by zero. Well now I'm looking at this function with this promise. Well, let's find the vertical asymptotes first. So that's where the bottom equals zero. Well, the bottom equals zero when x = 5. So that's easy. X = 5 is the vertical asymptote. How about a horizontal asymptote?
Aha! So now we're doing the racehorse thing. We're seeing which thing, top or bottom, gets to infinity faster as x goes off to the horizon. Well, you can see the highest power here is an x. The highest power here is an x, so this actually one of those tie things. So then you have to look at the coefficients. The coefficient here is a 9. Is the coefficient here a 1? No, because I have to distribute that too. So it's actually 2x - 10. So the coefficient here is a 2, on top is a 9, so there's a horizontal asymptote at y = . Cool. So now we have our asymptotes, and now we can take a look at what's going on. Let's see if we can make some progress here.
So , that's about 4-1/2, so 4-1/2 I have a horizontal person--one, two, three, four, five. So I've got a horizontal person at 4-1/2. By the way, a horizontal person is a cryptic way of saying asymptote. So y = . And then at x = 5 we have a vertical one--one, two, three, four, five. So I put a vertical one right there. And that's x = 5. And now I see I've got to figure out where everything goes. Okay, well let's take a look and see if we can figure this out. What could I do here? Well, one thing I could do is just take a look in this region here. For example, what about when x = 0? When x = 0, if you put in zero for x, I see just , so that's 1/5. So at zero I'm just at 1/5, so I'm way over here. What happens if I plug in, for example, 4? If I plug in 4 here, what do I see? Well, here I see 4 times 9, which is 36. 36 - 1 is 35, so I have a 35 on top. And on the bottom if I put in a 4, I see 4 - 5, which is -1, so that's -2, and so this is , which is like -17-1/2. So at 4 I'm at -17-1/2. I'm way down here. And what happens as I put in negative values? As I put in negative values, if you look at this, this is negative and that's negative, so the whole thing is positive, so that actually tells me that I must be sort of living life like this.
And what about in this region? Am I here or am I here? Well, let's take a look at, for example, x = 6. If I put in 6 what do I see? I see 9 times 6, which is 54, minus 1 is 53, and I divide that by putting a 6 in here. I see 6 - 5 = 1, so I just have a 2 here. So that's a positive number. It's around 26 or something, so over here I'm way up high. I'm way, way up high. So if I'm way, way up high just here, I must be doing this kind of action. So my guess is I'm doing this kind of action. So that's what the thing looks like. But if you submit that right now for your quiz, you would get it wrong, because you made a covenant with me that x would never equal -3, and now at -3 I've got that point colored in. So you've got to remember that if you make a promise, you'd better stick to it, otherwise, people are going to get really mad. So, in fact, you've got to remove that one point because we cancelled... And so, in fact, you cut out that point right there and there will be a hole right there. So, in fact, the graph of this has this hyperbole-like-looking thing, but then at -3, we have a little hole because we cancel it. Be really careful when you look at this stuff. Anyway, there you've got the graph. Have some fun trying these on your own, and don't forget to watch out for potholes. They're important.
Polynomial and Rational Functions
Graphing Rational Functions
Examples with Quadratics Page [1 of 2]