End behavior of rational functions | Mathematics III | High School Math | Khan Academy

So we're given this function ( f(x) ) and it equals this rational expression over here. We’re asked what does ( f(x) ) approach as ( x ) approaches negative infinity? So as ( x ) becomes more and more and more and more negative, what does ( f(x) ) approach? And like always, pause the video and see if you can think about that on your own.

Well, one thing that I like to do when I'm trying to consider the behavior of a function as ( x ) gets really positive or really negative, is to rewrite it. So ( f(x) ) I'm just rewriting it once is equal to (\frac{7x^2 - 2x}{15x - 5}).

Now, an interesting technique to think about what happens to the different terms as ( x ) gets very positive or ( x ) gets very negative is to divide both the numerator and the denominator by the highest degree term of ( x ) in the denominator. The highest degree term in the denominator is the first degree term; we just have a single ( x ) there.

So let's multiply both the numerator and the denominator by (\frac{1}{x}) or another way of thinking about it is we're dividing both the numerator and the denominator by ( x ). If we're doing the same thing to the numerator and the denominator, if we're multiplying or dividing them by the same value, I should say, well then I'm just really multiplying it by one, so I'm not changing its value. But this will make it a little bit more interesting, a little bit easier for us to think about what happens when ( x ) becomes very, very, very negative.

So, ( \frac{7x^2}{x} ) or being multiplied by (\frac{1}{x}) is going to be ( 7x ). ( 2x ) times (\frac{1}{x}) or ( \frac{2x}{x} ) is just ( 2 ) and then all of that over ( \frac{15x}{x} ) or ( \frac{15x}{x}) is just going to be ( 15 ), and then you have ( \frac{5}{x} ) which is equal to ( \frac{5}{x} - \frac{5}{x} ).

Now this is equivalent for our purposes to what we started with, but it makes it a little bit easier to think about what happens when ( x ) gets very, very, very negative. Well, when ( x ) gets very, very, very, very, very negative, this is going to become a very large negative number. You subtract ( 2 ) from it, it really won’t matter much. You divide that by ( 15); well that’s not going to matter much, and this is just going to become very, very, very small.

You're taking ( 5 ) and you're dividing it by ever larger negative numbers or more and more negative numbers, so this right over here is going to go to ( 0 ). This thing over here is going to go towards infinity, and if you, or I should say, it's going to go towards negative infinity. ( 7 ) times, you know, for ( 7 ) times a negative trillion, ( 7 ) times a negative googol, ( 7 ) times a negative googol plus we’re getting more and more negative numbers.

This is going to get, this is going to approach negative infinity. It doesn’t matter that you’re subtracting ( 2 ) from that. In fact, that’ll get even more negative. It doesn’t matter if you then divide that by ( 15); you’re still approaching negative infinity. If you had negative, if you had an arbitrarily negative number you divided by ( 15), you still have an arbitrarily negative number. And so you could say that this is going to go to negative infinity.

Now another way that you could have thought about it, and this is actually how I do think about it when I’m trying to, when I see these types of problems, I say, well, which terms in the numerator and the denominator are going to dominate? What do I mean by dominate? Well, as ( x ) gets very positive or ( x ) gets very negative, another way to think about it as the magnitude of ( x ) gets large, the absolute value of ( x ) gets large, the higher degree terms are going to grow much faster than the lesser degree terms.

So we could say that for large ( x ), and when I say large, I mean high absolute value. High absolute value and if we’re going to negative infinity, that’s high absolute value. So ( f(x) ) is going to be approximately equal to the highest degree term on the top which is ( 7x^2 ) divided by the highest degree term on the bottom ( 15x ), which is going to grow. In fact, this right over here is constant so as this becomes larger and larger and larger, this is going to matter a lot, lot less.

So it's going to be approximately that which is equal to (\frac{7x}{15}). Well, even here think about what happens when ( x ) becomes very, very negative here. Well, you're just going to get larger, you're going to get more and more and more negative values for ( f(x) ). So once again ( f(x) ) itself is going to approach, is going to go to negative infinity as ( x ) goes to negative infinity.

Let's do another one of these. So here they're telling us to find the horizontal asymptote of ( q(x) ) and you could find a horizontal asymptote. You could think about it as what is the function approaching as ( x ) becomes, as ( x ) approaches infinity or as ( x ) approaches negative infinity. Just as a couple of examples here, it's not necessarily the one I'm ( q(x) ) that we're focused on, but you can imagine a function; let’s say it has a horizontal asymptote at ( y = 2 ).

So that's ( y = 2) there. Let me draw that line. So let’s see, it has a horizontal asymptote like that. Well then the graph could look something like this; it could look, it could, let me draw a couple of them that have horizontal asymptotes. So maybe it’s over here, it does some stuff but as ( x ) gets really large, it starts approaching it, the function starts approaching that ( y = 2 ) without ever quite getting there.

And it could do that on this side as well, as ( x ) becomes more and more, as ( x ) becomes more and more negative, approaches it without ever getting there. Or it could do something like this; you could have, if it has a vertical asymptote too, it could look something like this where it approaches the horizontal asymptote from below as ( x ) becomes more negative and from above, and from above as ( x ) becomes more positive or vice versa.

So this is just a sense of what a horizontal asymptote is. It's really what's the behavior? What value is this function approaching as ( x ) becomes really positive or ( x ) becomes really negative? Well, let’s just think about it. We can essentially do what we just did in that last example: what happens if we were to if we were to divide all of these terms by the highest degree term in the denominator?

Well, if we divide so ( q(x) ) is going to be equal to the highest degree term in the denominator is ( x^9 ). So we could say ( \frac{6x^5}{x^9} ) is going to be (\frac{6}{x^4}) and then minus ( \frac{2x^9}{x^9} ) all of that over ( \frac{3}{x^9}\cdot x^7 + 1).

Well, if ( x ) approaches positive or negative infinity, ( \frac{6}{\text{arbitrarily large numbers}} ) that’s going to go to ( 0 ). ( \frac{2}{\text{arbitrarily large numbers}} ), whether they are positive or negative, that’s going to go to ( 0 ). So your numerator is clearly going to go to ( 0 ).

This term of the denominator ( \frac{3}{\text{arbitrarily large numbers}} ), whether those large, whether we’re going in the positive or the negative direction it is going to approach ( 0 ). It'll approach ( 0 ) from the negative direction if we’re approaching, if our, or we could say from below if we’re dealing with very negative ( x)'s. If we’re dealing with very positive ( x)'s, then we’re going to approach ( 0 ) from above; we’re going to get smaller and smaller positive values.

So all of these things go to ( 0 ) and this right over here is going to stay at ( 1 ). So if you’re approaching ( 0 ) in your numerator and approaching ( 1 ) in your denominator, the whole thing is going to approach ( 0 ). So in the case of ( q(x) ) you have a horizontal asymptote at ( y = 0 ).

I don’t know exactly what the graph looks like but we could draw a horizontal line at ( y = 0 ) and it would approach it, it would approach it from above or below. Let’s do one more: what does ( f(x) ) approach as ( x ) approaches negative infinity?

Well, let’s divide all of these, let’s divide all of these terms by the highest degree that we see in the denominator. We see an ( x^4 ) so ( \frac{3x^4}{x^4} ) is ( 3 - \frac{7}{x^2} ) I’m just dividing by ( x^4 ) minus ( \frac{1}{x^4} ) over ( x^4 ) divided by ( x^4} ) is ( 1 - \frac{2}{x} + 3x^4 ).

This is an equivalent, this right over here is for our purposes for thinking about what’s happening on kind of end behavior as ( x ) approaches negative infinity; this will do. I’ve just divided everything by ( x^4) and so what’s going to happen as ( x ) approaches negative infinity?

This is going to approach ( 0), this is going to approach ( 0), this is going to approach ( 0), and this is going to approach ( 0). And so as all of that stuff approaches zero, what we’re left with is we’re going to approach, we’re going to approach ( \frac{3}{1} ) or we could say just ( 3 ).

Another way you could think about doing these is look at the highest degree terms ( 3x^4 ) over ( x^4 ); ignore everything else because they’re going to be overwhelmed by these higher degree terms. So you could say ( f(x) ) is approximately equal to ( \frac{3x^4}{x^4} ) for large magnitude ( x ).

And very negative is still a very large magnitude, large absolute value, and so ( \frac{3x^4}{x^4} ) ( f(x) ) is going to be approximately equal to three or it's going to approach three. So that's another way that you could think about it.