Finding oblique asymptotes of a rational function
About this video
The video introduces oblique (slanted) asymptotes and shows how to find one with algebraic long division. Dividing x^3 + 8x - 4 by x^2 - x + 3 gives x + 1 with remainder 6x - 7; because the remainder fraction goes to 0 for very large or very negative x, the oblique asymptote is y = x + 1, which the graph confirms.
You'll be able to
- describe an oblique asymptote and when a rational function might have one.
- use algebraic long division to rewrite a rational function as a quotient plus a remainder fraction.
- explain why the quotient gives the oblique asymptote.
Key words
- Oblique asymptote
- A slanted line that a graph approaches as x becomes very large or very negative.
- Algebraic long division
- Dividing one polynomial by another term by term to get a quotient and a remainder.
- Quotient
- The main result of a division; here x + 1.
- Remainder
- What is left over after dividing; here 6x - 7, which becomes the fraction (6x - 7)/(x^2 - x + 3).
Transcript
- [Instructor] So let's talk a little bit about oblique asymptotes. So oblique tends to refer to something that's at a slant. So you could imagine these are asymptotes that are not vertical nor are they horizontal. And just to imagine what this could look like, let's just imagine some type of a function here.
I'll draw the axes. So that's my y-axis. This is my x-axis. And let's say you have an oblique asymptote. Let's say it looks something like, say it looks something like that. And it's for a function, which I will do in this blue color. So it might do so all sorts of things for small xs.
But as x gets larger, it will get closer and closer and closer to that asymptote. And similarly on that side is x gets more and more negative, it might get closer and closer to that asymptote. So that's what we mean by an oblique asymptote. What we're going to do in this video is think about how can we find the equation for that oblique asymptote if we're just given the function.
So for example, if our function is, let me do it in this color right here. If our function is, I'll do it in white, F of x is equal to x to the third plus 8x minus 4 over x squared minus x plus 3. When we look at just what's happening in this function, the numerator is going to be growing faster than the denominator when X gets very positive or very negative.
Because the highest degree term in the numerator is a third degree term. While in the denominator it's only x squared. But the question is how is that going to, what line will that potentially approach? And the way to understand that better is we'll do a little bit of algebraic long division to simplify this and we'll be able to, we'll not simplify this, to write it differently in a way that we can think about what happens as x gets very positive or very negative.
So let's do our algebraic long division. I'm going to divide the denominator into the numerator. I'm really just rewriting the function. So let's see, we'll do, let me do it over here. So x squared minus x plus 3 algebraic long division goes into x to the third.
And actually I'm going to leave a little bit of space because I want a second. I might need some space up here for my second degree term. So I'll put plus 8x minus 4. And now let's do it. So look at the highest degree terms. X squared goes into x to the third x times.
I'll write that in the x place, X times all of this business. X times x squared is x to the third. X times negative x is -x squared. X times 3 is 3x. Plus 3x. And then we are going to want to subtract that. That goes away. You're left with negative of a negative as a positive, positive x squared 8x minus 3x is 5x.
And then you have -4. X squared goes into x squared one time. So plus 1. 1 times all of that is x squared minus x plus 3. Subtract. And then we get, that goes away. 5x minus -x is going to be 6x. And then -4 minus 3 is -7. And so that's our remainder, because x squared won't go into that more times.
And so we can rewrite F of X. We could say that F of X is equal to this x plus one plus 6x minus 7 over x squared minus x plus 3. Now why is this interesting? Well, let's think about what happens when X gets very large or very negative or very positive or very negative.
As I could write as x approaches positive or negative infinity, our function is approximately as X becomes very positive or very negative, this denominator grows much faster than that numerator. So this thing, as you get very positive or very negative, it's going to go to zero or it's, well, yes, it's going to go to zero.
It's going to become a lot less consequential than what we see over here. So F of X is going to be approximately x plus 1. So there you have it. We figured out our oblique asymptote in this case, the oblique asymptote is y is equal to x plus 1. And we can actually see that on a graph.
I graphed it ahead of time and let me bring it in right there. And so in this situation, our function is in that teal color. That is our F of x. No, let me make. Our function is in that teal color. And you could see it does all this stuff when x is relatively small in magnitude, but as it gets larger and larger and larger or more and more negative, it is approaching y equals x plus 1 which is that yellow line that y equals x plus 1 right over here.
That is that yellow line, which is the oblique asymptote.
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