Connecting equivalent forms of rational functions

0:00- [Instructor] What we're gonna do in this video is try to figure out as much as we can about this rational function. And in particular, appreciate how different forms of writing this rational function can help us figure out the different things that we want to know about it.

0:14So the first thing I wanna think about is horizontal asymptotes. And we covered this in a lot of detail in other videos, but this is just the behavior. What is this function approach if X gets very large or if X gets very negative? And very large in the positive direction, or I guess you could say with very large magnitude in the negative direction.

0:36So horizontal asymptotes, maybe I'll just put it as a question mark. Now this form where we just have the numerator and the denominator both expanded out in really the standard form of polynomials, you have a polynomial on the numerator, a polynomial in the denominator.

0:51This is really useful for horizontal asymptotes because as X gets really, has a very large magnitude, I should say, then what really matters is the highest degree term in both the numerator and the denominator. So what you can do is if you just look at the one way to think about it as X large, as X large, Y is going to be approximately equal to X to the third over X.

1:18Now this right over here, or over X squared, I should say. Over the largest degree in both because then these terms are going to dominate, X to the third is gonna dominate in the numerator. X to the squared is gonna dominate in the denominator. Now this is going to be approximately equal to X, which is unbounded, as X gets large and if it's approximately equal to X, then this is also going to be unbounded.

1:38It's going to get very, very large. So this does not have any horizontal asymptotes. If this was an X to the third in the denominator, then the horizontal asymptote would be one. If this was a 2X to the third, then it would be 1/2. And we go into a lot of depth in that in other videos.

1:54But the important thing to appreciate is the form that I have it right over here where the numerator is a standard form polynomial, denominator is a standard form polynomial. Very easy to figure out the horizontal asymptotes, or at least this is the best format to figure out the horizontal asymptote.

2:08Now the next thing that this form is good for is if you just wanna figure out the Y intercepts. Y intercepts is just where we intersect the Y axis. And then you just set X equal zero. So if you set X equal zero, what you get, maybe I'll do it over here on the right or I'll do it over here.

2:25When X equals zero, Y is going to be equal to, all of this stuff is going to go away and you're just going and all of this stuff is going to go away. So you're just gonna have negative 30 over negative four. So let me write that. So that's negative 30 over negative four, which is the same thing as positive 30 over four, which is the same thing as 7.5.

2:48So we have a Y intercept at X equals zero, Y is equal to 7.5. Once again, this form is very good for that. Now if we start to think about things like X intercepts and vertical asymptotes and holes, that's where it's useful to have both the numerator and the denominators actually to be factored.

3:08Now the denominator here is pretty straightforward to factor it. This is a difference of squares. So it's X plus two times X minus two. This numerator being a third degree polynomial is a little bit harder to factor. So how do we approach that? Well, one place to start, there's a lot of techniques here, but one place to start is see if any of these are factors 'cause that will help us identify holes.

3:33So let me see if X equals two is a place where that would make the numerator zero, if X equals two is a zero. So to do that I just have to say, does X equal two make the numerator zero? So when X equals two, we get eight, two to the third plus four times X squared.

3:52That would be four times four plus 16, minus, if I put a two there, minus 22 minus 30, what is this equal? This is 24 minus 22, which would be two minus 30, which is equal to negative 28. So that doesn't help. But what about if this was X equals two? Let me try X equals negative two.

4:15So X equals negative two. That would be negative eight, and then this would still be plus 16. Now this would be plus 22, plus 22 minus 30. And this is, let's see, eight plus 22 is 30 minus 30. This is indeed equals zero. So X minus two, X equals negative two is a zero for the top, which means that X plus two is a factor.

4:44Now what are we left with? Well now we can just do a little bit of algebraic long division, divide X plus two into this numerator right over here. So let me do that, maybe in another color just for variety. So if I divide X plus two into X to the third plus 4X squared minus 11X minus 30.

5:06X goes into X to the third, X squared times. So I'll just write it over here, X squared. And we go into detail on this in other videos. So review that if algebraic long division looks a little unfamiliar, X squared times X is X to the third, X squared times two is 2X squared.

5:26And then we're going to want to subtract this. And so we get these cancel, 4X squared minus 2X squared is going to be 2X squared. And then we can bring down this minus 11X minus 30. So X goes into 2X squared, 2X times. So plus 2X, 2X times all of this is 2X squared plus 4X.

5:52But we're going to subtract it and then we have, and let me scroll down a little bit so we can continue to see what's going on here. So this is going to be, let's see, these characters are gonna cancel out, negative 11 minus 4X is negative 15X minus 30, X goes into negative 15X, negative 15 times, negative 15 times X.

6:18We have negative 15X, negative 15 times two is negative 30. And we luckily have a remainder of zero. So let me bring back, go back to where we were. And so there we have, let me make sure we can see everything. So now, we can rewrite this expression as Y is equal to, in the numerator, we have X plus two times all of this business, X squared plus 2X minus 15, all of that over X plus two times X minus two.

6:54So we immediately see a hole. So this would be a place where the function is not defined, but it isn't a vertical asymptote because essentially you can cancel this out of both the numerator and the denominator. So we can really rewrite this function as Y is equal to.

7:09So we can say we have a hole at X equals negative two. And so we can rewrite the function. And actually let me factor this numerator right over here. Let's see, what two numbers when I multiply 'em, I get to negative 15. When I add 'em, I get to two. Let's see, positive five, negative three I think works.

7:27So X plus five times X minus three. That's me just factoring this part over here. That works out. Yep, that works out, all of that over X minus two. So when you factor it, it's good for finding holes. And now this form, right? Oh, we have to say, we have to make sure that X cannot be equal to, X cannot be equal to negative two.

7:51That's to preserve the fact that information that we have a hole there. So now we can see a few things. What would make the denominator zero without making the numerator zero? Well that's X equals two. So this is where we're going to have a vertical asymptote, vertical asymptote.

8:09And then where are our X intercepts going to be? Well, that's where our numerator is going to be equal to zero. So we're going to have X intercepts, and this is getting messy. My apologies, maybe do it in another color, I'll do it in purple. My X intercepts are going to be at the point are going to be at the point X equals negative five comma zero.

8:33And at the point x equals three comma zero. That's the values that make the numerator zero without making the denominator zero. If it makes both zero, then we are talking about the holes. So the big thing to appreciate, this standard form polynomial, numerator, denominators, good for horizontal asymptotes, Y intercepts.

8:51And then if you factor it as we did, then you can start to identify holes, vertical asymptotes and the X intercepts.