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Graphing rational functions 3

Sal graphs y=(x^2)/(x^2-16). Created by Sal Khan and CK-12 Foundation.

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  • blobby green style avatar for user http://facebookid.khanacademy.org/1197982380
    You made it perfectly clear. What about oblique (I hope I used good term, english is not my native language) asymptotes?
    (104 votes)
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    • blobby green style avatar for user GWOJNAR
      What made the video example have a horizontal asymptote was the fact that the numerator (x) & denominator (x^2 -16) polynomials both had the same degree (their greatest exponents; in the example video, this was degree 2). More generally, the numerator & denominator have different degrees. Here are the possibilities:



      Case 1: Numerator degree = Denominator degree. Effect: Horizontal Asymptote, as in the video. The horizontal asymptote line is at the y-value that equals the ratio of the numerator & denominator coefficients (multiplying numbers) of their highest power terms; example: y = (12x^5 + 7x^2 - 8x + 9)/(4x^5 - 13x^4 + 55) has the horizontal asymptote being the line y = 3 because 12/4=3.



      Case 2: Numerator degree < Denominator degree. Effect: Horizontal Asymptote is the x-axis. Reason: For huge |x| values, the bigger denominator power creates such a huger value than what the numerator makes, so when we divide the numerator by this great value, the value of the division is very close to 0. Example: (3x^2 - 5)/(x^4 +7x - 8) produces, approximately, when x=1000, the division 3,000,000/1,000,000,000,000 = 0.000003, which is indeed quite close to 0. If You don't like approximating (but You should!-- it's a powerful technique), the precise computation is 2,999,995/1,000,000,006,992.



      Case 3: Numerator degree > Denominator degree. This case includes your question (oblique asymptote [yes, "oblique" is the correct term, meaning 'slanted']), and it also includes other more complicated cases.

      METHOD: Use long-division of polynomials to divide the numerator by the denominator, obtaining a Quotient polynomial Q(x) and a Remainder polynomial R(x). When You do this, the Remainder polynomial always ends up having a degree that is less than the degree of the original denominator polynomial. In elementary school, when we divide, say, 17/3 we obtain a quotient of 5 with a remainder of 2; all this information is contained in the equations 17 = 5*3 + 2 or (equivalently) 17/5 = 3 + 2/3. In this numerical example, notice that the Remainder 2 turns into the Remainder FRACTION 2/3 when we write the second equivalent equation. Similar things happen with long-division of polynomials: Starting with y = n(x)/d(x) where n(x) is whatever the numerator polynomial is, and where d(x) is whatever the denominator polynomial is, when we do long-division we obtain the two equivalent results n(x) = d(x)*Q(x)+R(x) and n(x)/d(x) = Q(x)+R(x)/d(x). This is just like the 17/5=3+2/3 example above!



      Let's get concrete with the graph of an example of this nature. Starting with y = (4x^2+8x-60)/(x-6), the long-division [see the video that's a few steps after the one here, for how to do long-division] yields the results Q(x)=4x+32 and R(x)=132. Therefore the original equation is equivalent to

      y = (4x+32) + 132/(x-6). Now we are ready to learn about what's happening with the graph: Observe that when x is huge in magnitude (i.e., getting closer to either + or - infinity), then the Remainder Fraction 132/(x-6) gets ever closer to 0 in value. Example, when x=-1,000,000 we have 132/-1,000,006 which is pretty tiny. Because of this, we conclude that when x is huge (either + or -), then our graph is really close to the graph of y = (4x+32)+0. The straight line y=4x+32 is what is called an OBLIGUE ASYMPTOTE for the rational function y = (4x^2+8x-60)/(x-6). Recapping what this means: The graph of y = (4x^2+8x-60)/(x-6) gets really close to the slant line y = (4x+32) as we move further & further to the right (x approaching + infinity) or further & further to the left (x approaching - infinity).



      GENERAL RESULT: After conducting long-division we convert any rational function y = n(x)/d(x) into the equivalent form y = Q(x) + R(x)/d(x). As x becomes huge (either + or -), the value of the Remainder Fraction R(x)/d(x) always approaches the value 0, because the process of long-division always yields a Remainder whose degree is less than the degree of the denominator d(x). THEREFORE, the graph of the quotient, y = Q(x), always gives an asymptote for the original rational function. This asymptote is properly called the Main Asymptote or Quotient Asymptote. Every rational function has a Main Asymptote. [It's possible for a rational function to have NO vertical asymptotes. Example y = 2x^3/(x^2+1).]



      Extra #1: In the example above we obtained the Remainder polynomial to be R(x)=132, a constant function that didn't depend upon the value of x. If we had started out with the degree of the denominator d(x) to be 2 or more, then it can happen that the Remainder polynomial R(x) is a function that has x in it. Without doing a complete example, if d(x) was degree 2, then it would be possible to obtain R(x)=2x-44, which is equivalent to R(x)=2(x-22). Notice that for the special x-value of x=22, we have R(22)=0, so for this special value the general fact that n(x)/d(x) = Q(x) + R(x)/d(x) would become n(x)/d(x) = Q(x) + 0 WHEN x=22. This means that the graph we want for y = n(x)/d(x) and the graph of the Main Asymptote y = Q(x) have the very same y-value WHEN x=22, so our desired graph of y = n(x)/d(x) actually intersects the (dotted) graph of the Main Asymptote y=Q(x). The example here had R(x)=2(x-22), causing our function to cross through the graph of the Main Asymptote when x=22. IF instead we had had a situation where d(x) was degree 3, and where it so happened that R(x)=3(x-50)^2, then (because of the square on the (x-50) here) our graph of y = n(x)/d(x) would just KISS the graph of the Main Asymptote y=Q(x)WHEN x=50; this is just like how a parabola, say y = 4(x-7)^2, just kisses the x-axis when x=7. The key issue when deciding if a root, Z, of R(x) [i.e., an x-value for which R(Z)=0] is a Crossing Point -or- a Kissing Point of intersection between the actual graph and the graph of the Main (Quotient) Asymptote is whether the root value, Z, had an Even or Odd power on (x - Z) in the factored form of R(x). Above we had an example with R(x)=2(x-22)^1 and the other example with R(x)=3(x-50)^2. The term for the exponent on a root's factor is the MULTIPLICITY of the root. So if R(x) has an ODD multiplicity root, the graph of the desired function n(x)/d(x) CROSSES the graph of the Main Asymptote y=Q(x) at that particular root. If R(x) has an EVEN multiplicity root, the graph of the desired function n(x)/d(x) KISSES the graph of the Main Asymptote y=Q(x) at that particular root. Note that this distinction between crossing behaviour & kissing behaviour is just like the issue for roots of simple polynomial graphs.



      Extra #2: Recall the general idea that y = n(x)/d(x) = Q(x) + R(x)/d(x), and that because long-division yields that R(x) always has lower degree than the degree of d(x), the value of the Remainder Fraction R(x)/d(x) always approaches 0 when the value of x approaches + or - infinity. The nice consequence of this is that our desired graph approaches the graph of the Main Asymptote y = Q(x)when the value of x approaches + or - infinity. We'd like to know a little more!: we'd like to know if the true graph of y = n(x)/d(x) = Q(x) + R(x)/d(x) ends up being ABOVE or BELOW the graph of the asymptote y = Q(x). This question is easy to answer-- we just need to determine if the Remainder Fraction R(x)/d(x) [which we know shrinks to 0 as x gets huge] is positive or negative for huge values of x. This is quite easy! Suppose, for example, that we have the Remainder Fraction R(x)/d(x)= (-3x^2+5x+4)/(x^3+4x^2-7x+8). When x is huge in magnitude, the highest power terms in the numerator & denominator dominate the value of the fraction: R(x)/d(x) is approximately (-3x^2)/(x^3) which reduces to -3/x. As x grows to be huge positive this fraction shrinks to 0 as expected, but we see now that it will be a tiny NEGATIVE value. THEREFORE, we'll have n(x)/d(x)=Q(x)+(tiny negative) which causes our desired graph of n(x)/d(x) to be a bit less than the Quotient asymptote; our graph approaches the Quotient Asymptote from below it, WHEN x is approaching (+)infinity. On the other end of the picture, when x approaches (-)infinity, note that the approximated Remainder Fraction, -3/x, is now a tiny POSITIVE value. THEREFORE, we'll have n(x)/d(x)=Q(x)+(tiny positive) which causes our desired graph of n(x)/d(x) to be a bit greater than the Quotient asymptote; our graph approaches the Quotient Asymptote from above it, WHEN x is approaching (+)infinity.

      LOOKING BACK: The analysis of a rational function via long-division to consider
      y = n(x)/d(x) = Q(x) + R(x)/d(x) actually applies perfectly to all 3 Cases above, including the simplest cases of horizontal asymptotes. For the above Case 1 where the numerator n(x) and denominator d(x) both have the same degree (say this degree is N), we saw that there was a horizontal asymptote at the value y=A/B where n(x)=Ax^N + other junk, and where d(x)=Bx^N + other junk. Observe that the long-division yields this same conclusion, because Q(x)=A/B. For the above Case 2 with the numerator n(x) having a lesser degree (say L) than the degree (say M) of the denominator d(x), causing the x-axis to be the horizontal asymptote, we have the following long-division analysis. The long-division of n(x)/d(x) yields a Quotient Q(x)=0 because we're trying to divide a bigger degree polynomial into a lesser degree polynomial. Once again, we have our Main Asymptote, now given by the equation y = 0 (which is indeed the x-axis). So, overall, it's not really remembering 3 different cases; all 3 cases can be thought of through the long-division mentality.
      (231 votes)
  • piceratops tree style avatar for user Chandana Deeksha
    what is an ASYMPTOTE??
    (6 votes)
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  • blobby green style avatar for user Christine Mack
    Why was 4.01 picked when findind y? Could 7 be picked or another number?
    (12 votes)
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  • blobby green style avatar for user abhishek.singh.047
    can asymptotes of a hyperbola be considered as tangents?
    (13 votes)
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    • leaf grey style avatar for user CasualJames
      Also, by definition, tangents touch a shape (hyperbola, circle, etc.) in exactly one spot. A function will never touch an asymptote. Now, towards the extremes of the function, will an asymptote appear to be tangent to an equation? Sure. But it's not, and won't behave the same way.
      (6 votes)
  • leaf grey style avatar for user Aravind Margam
    how would you write rational functions from a graph or sata like

    a zero at 2
    vertical asymptote at x=-1
    horizontal asymptote at y=1
    ???
    (5 votes)
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  • blobby green style avatar for user Darrah
    Would (3x)^2 be 3x^2 or 9x^2 because you have to distribute it?
    (5 votes)
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    • leafers tree style avatar for user Ali Greene
      the parentheses around the 3x part is very important. (3x)^2 is different than 3x^2. The first one means you are squaring the whole thing, whereas the second means you are just squaring x and then multiplying by 3.
      (3x)^2 = (3x)(3x) = 9x^2
      (9 votes)
  • leafers seedling style avatar for user Lauren
    Can someone explain why if the numerator > denominator there is no horizontal asymptotes?
    (5 votes)
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    • piceratops tree style avatar for user Theresa Johnson
      This has to do with the nature of horizontal asymptotes. They tell you about the end-behavior of functions (i.e. the limit as x-> infinity)

      When the degree of the numerator is larger than the degree of the denominator, that means that the value of the numerator is going to increase much more quickly than the value of the demoninator.

      This, in turn, means that the value of the function will increase without bounds as x approachs infinity. So there is no number that the ends will get closer and closer to and thus no horizontal asymptote.
      (7 votes)
  • piceratops ultimate style avatar for user Jav
    Are there diagonal (for lack of a better term) asymptotes (i.e., non-horizontal or vertical)?
    (3 votes)
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  • blobby green style avatar for user Isabella Bowyer


    Why does Y never equal 1? That is, why is the horizontal Asymptote a 1?
    Why can X never equal something greater than -4 but less than 4--that is, something in between the two vertical Asymptotes?
    (5 votes)
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    • blobby green style avatar for user akruti
      If you think about it, the function x^2/(x^2-16) can never equal 1, because you know (x^2)/(x^2) = 1 (since anything over itself equals 1). So if the denominator is (x^2)-16, the denominator will always be a little smaller than the numerator, therefore the graph will never actually reach y=1.
      I'm not quite sure about what your second question means. But in any case, X can equal any number except 4 and -4 (for this particular function). Sal's graph shows this as well.
      Hope this helps! :)
      (2 votes)
  • purple pi purple style avatar for user Shifat Taushif
    This question has been lingering in my head ever since I learned the concept of asymptoptes, but it is quite a vague/unclear one.

    What happens when the line reaches a point where if it attempts to move 1 unit closer to the asymptote, it will equal the asymptote?

    Let's say that we have the function y=f(x) and it is a downward curve approaching the horizontal asymptote y=3 from the above. When the function reaches 2.999(gazillions of 9's)99, it is so close to y=3 that if the value of f(x) decreases by 0.000(gazillions of 0's)01 (moving 1 unit to the right on the x-axis), it will equal 3.
    I know that at such a point, it cannot equal 3, so what happens there?
    (2 votes)
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    • piceratops ultimate style avatar for user Just Keith
      It is not an actual requirement that the function does not ever equal the asymptote at any value as x approaches infinity. That does tend to be the case, but it is not a requirement.

      An asymptote is what the function approximates when x (or y, depending on what kind of asymptote) approaches infinity.

      So, there is no rule saying that that the function and the asymptote cannot meet for some values of x. Once x gets large, it is usually the case that the function and the asymptote won't ever be exactly the same, but there do exist functions in which that is the case (but only for some values of x, never all values of x).

      However, the case that you described won't actually happen. Even when you have 2.999..... on and on for countless trillions of 9's, you still won't have it =3. You never run out of "one more 9" that you can add.

      Though, of course, for all practical purposes, once the function gets so close to the asymptote that there is no real-world, useful distinction between the two, we can just say they have become essentially equal.
      (8 votes)

Video transcript

In this video, we're going to see if we can graph a rational function. A rational function is just a function that has an expression on the numerator and the denominator. It has a polynomial in the numerator-- Let's see, we have x squared over-- and another polynomial in the denominator --x squared minus 16. We could obviously graph this by just trying out a bunch of points and then connecting the dots. That's what a calculator would do for us, a graphing calculator. But what we want to do is, before we try out some points to kind of fill in the gaps, I want to understand the basic structure of this graph first. And to understand that, I want to see what happens as x gets really big, so x gets really big, or x gets really, really small, as x goes in the negative direction. Or another way we could think about it, I want to understand what happens as the magnitude of x, or the absolute value of x, becomes really big as it approaches really, really bigness, or as it approaches infinity. So when the size of x approaches infinity. Which essentially is saying, as x goes really far in the positive direction, or x goes really far in the negative direction, what is going to happen to the value of this function? So let's get out a calculator. Won't use the graphing part of it just yet, but let's just try out some values. What happens when x is equal to 10? It's going to be the same thing as when x is equal to negative 10, because when you put a negative 10 here, you square it, you get 100, just like 10. Same here, negative 10, you square it, you get the same thing as a positive 10. So whether you go in the super high positive direction or the super low negative direction, as you approach positive or negative infinity, you're going to approach the same thing because you're squaring the values. But let's try out some values. If I get 10 squared divided by 10 squared minus 16, I get 1.19. Now what happens if x gets a little bit bigger? This is x is equal to 10. What happens when x is equal to 100? We have 100 squared divided by 100 squared minus 16. I'm getting even closer to 1. When x was 10, around here, we're getting y is 1.19. When x is 100, 100 squared over 100 squared minus 16, y is 1.0016. Just for fun, let's try 1,000. So it's 1,000 squared divided by 1,000 squared minus 16. And we're even closer to 1. So as the size of x gets larger and larger and larger, our y gets closer and closer to 1. And that would also be true if this was a negative 10, because negative 10 squared over negative 10 squared minus 16 is going to be the exact same thing. Because the negative, when you square it, is going to be a positive. It's going to be the same thing as 10 squared, same thing over here. So whether x gets really big or x gets really small, we're going to be approaching y is equal to 1. You could try it with a million if you want, and you're going to get a number even closer to 1. So as the size of x approaches infinity, the absolute value of x, or the distance from the origin, approaches infinity, y is approaching 1. Or another way to think about it is, the graph of this function is going to approach the line y is equal to 1. So let me graph the line y is equal to 1. So I'll do it in a dotted line because this isn't the graph of our function, but this is a line that our function is approaching. So that is the graph of y is equal to 1. Now, this idea of a function, or the graph of a function, approaching a line but never quite touching it. So this is going to get closer and closer and closer to this line, y equal to one, in that direction, but never quite getting close enough to it. It'll approach 0, its distance from this y equals 1, but it'll never quite get there. This line that the graph is approaching is called an asymptote. And it'll be even more clear once I actually graph the function. We're going to work up there. And since it's a horizontal line, we call this a horizontal asymptote. This is what our graph approaches as we go in the positive direction, or really far in the negative direction. Let's think about some of the other interesting things about this, about this function right here. Well one thing that might pop out at you is this is a difference of squares. This is x squared minus 4 squared. So we can rewrite this as x squared over x plus 4 times x minus 4. So what's going to happen here, as x approaches either positive 4, or x approaches negative 4? Well, first of all, try those values out. If x is equal to 4, what is going to happen? This expression right here, this term right here, is going to be equal to 0. And we're going to be dividing by 0. We cannot do that. Similarly, if x is equal to negative 4, we'd be dividing by 0. This expression, right here, is going to be equal to 0. We can't do that. We could say that this function is undefined at x is equal to plus or minus 4. It can't equal those values because we'd be dividing by 0 in either one of those circumstances. Now, what happens as we approach those values? What happens as x approaches negative 4? Let's just do that one for fun. What happens as x approaches negative 4? Let's say we're approaching it from the negative direction. So let's try it out in our calculator. So let's say we want to go from the negative direction. So let's start with negative 4.1. So if we have negative 4.1 squared divided by negative 4.1 squared minus 16, what do we get? We get 20.75. So we get this number, whatever. Let's see if we get even closer to negative 4. So let me just get that entry there. So let's get a little bit closer to negative 4. So instead of negative 4.1, let's do negative 4.01. So let me insert a negative 4.01. And then over here, this is negative 4.01, and see what it is. Now we went to 200, so we're getting to larger and larger values. Let's try negative 4.001. Let's try that out. Whoops, that's not what I wanted to do. I wanted to do that. So let's try. No that's not what I want to do. Let's see. So we want to go to, instead of 4.01, I want to do 4.001, and over there, negative 4.001. And what do we get? We get 2,000. So as we get closer and closer to negative 4 from the negative direction, we're approaching larger and larger, super larger numbers. And you can try it, if it's 4.0000001, it's going to get to smaller and smaller numbers-- or sorry, larger and larger numbers here. If you do 4.001, it's probably going to be 20,000. And then if you add another 0 here. So as we get closer and closer, it's getting to larger and larger numbers. So we could say, as x approaches negative 4, we could say y is approaching infinity. It's getting to a larger and larger and larger value. But we can't ever quite get to x is equal to 4. It's undefined there. That will make the denominator here equal to 0. So what we want to do here is, we can never quite equal x equal negative 4. So let me see, x is equal to 1, 2, 3, 4. We can never quite get to x is equal to negative 4. Let me draw x is equal to negative 4 as a dotted line, right there. That is x is equal to negative 4. We can never quite get there, but as we approached it from the negative side, as we had 4.1, then 4.01, we went to larger and larger values. And we also know that as we went on the left-hand side, as we go to larger and larger x values, that y will get closer and closer to 1. So you have a general sense of what this part of the graph will look like. This part of the graph is going to look something like that. As x gets to super negative numbers, it gets closer and closer to 1, as x gets closer and closer to negative 4 from the negative direction, it's going to go closer and closer to infinity. You're going to get closer and closer to a very-- It's going to get larger and larger, I guess, is an easy way to say it. Now, just like x equal negative 4, x equals 4 will also be a point where the graph is undefined. So let me graph that here. 1, 2, 3, 4. Right here. Right over here. x is equal to 4. And, once again, what happens as we approach x equals 4, let's say from the positive direction? So as x approaches 4 from the positive direction, what's going to happen? So this is like trying out x is equal to 4.01, or x is equal to 4.001, or x is equal to 4.0001. So we're just getting closer and closer and closer to x is equal to 4. Well, these values are the exact same values that we just tried on our calculator, except they are the negative version of them, right? And we already saw that, just the way that this function is set up, the negative numbers, they get squared, so whether you take the negative or the positive x values, it's going to be the same thing. This graph is symmetric. When x is equal to negative 5, it's the same thing as x is equal 5. When x is equal to negative 10, it's the same thing as x equals 10. So the same thing is going to happen. You could try it out with your calculator, if you like. If you try out these values, you're going to see, as we get closer and closer to 4, we're going to approach larger and larger numbers. These same numbers over here. So the graph over here, as we get closer and closer to 4, we're going to approach larger and larger numbers. And then here, as x gets larger and larger and larger, we saw over here, we had these horizontal asymptotes, y gets closer and closer to 1. So just like we called this a horizontal asymptote, these values-- or you can even view these vertical lines: x is equal to negative 4 and x is equal to 4 --we call these vertical asymptotes. These are lines, asymptotes, once again, they are lines that the graph approaches, but never quite touches. So that's what's going on here. And then we can think about what's happening to the graph inside of here. So you could think of it in a couple of ways. You could say, well, what happens as x approaches 4 from the negative direction? So let's try that out, from the negative direction. So what happens if you do 3.9 squared divided by 3.9 squared minus 16? You get negative 19.25. Now what happens if we do 3.99? So let me put another 9 here. So we're going to get closer and closer to 4, and we're going to do it from the left-hand side as we approach 4. So insert another 9 here. So even more negative. So let's just do one more. So we're going to be even more negative. So let me make it 3.999. Get even closer to 4. You're getting even more negative. And this is also going to be true if we did negative 3.9, or negative 3.99, or negative 3.999, because when we square it, the negatives and the positives are the same thing. You square negative 1, you get a positive 1. So as we approach 4 from-- you go 3.9, 3.99, we get closer and closer to 4 --we're getting more and more negative numbers. We approach negative infinity. So as we approach-- let me just graph it here. As we approach from this direction, we're going to get smaller-- want to not touch our asymptote --it's going to look something like that. As we approach it from the left-hand side, we're getting smaller and smaller numbers. And that's also going to be true as we approach negative 4 from the right-hand side, right? As we get negative 3.9, 3.99, 3.999, we're going to drop down. It's going to look something like that. And now that we have a general sense of what the graph is, now is a good time where we could maybe plot a few points here. And the easiest one is, what happens when x is equal to 0? You have 0 squared over 0 squared minus 16. So the point when x is equal to 0, we're going to have 0 over, well, negative 16, which is just 0. So the point 0, 0 is on this curve. And then we could try some other points if you like, but the general shape here is going to look something like this. You could plot more points if you really want to nail down exactly what the curve is doing in between, but here is the general structure. And we tried out a lot of values with the calculator. And I did that because I really wanted to show you why it's dropping down like this. And if you think about it, it makes complete sense. As you get closer and closer, let's say you get closer and closer to 4. Either way, as you get closer and closer to 4, this is going to become a smaller and smaller and smaller number, because this is the difference between x and 4. So this is becoming a smaller and smaller and smaller number. Then, when you take 1 over that, right? You can essentially view this as x squared over x plus 4, or times 1 over x minus 4. If this is becoming smaller and smaller, this whole thing, 1 over a super small number, is a super large number. So as you can imagine, you're going to get larger and larger, and depending on whether you are approaching from the positive or negative, so whether this is a super small negative number or super small positive number, that's going to flip the sign. But either way, the magnitude-- So we're getting to a very large magnitude in the negative direction because the difference between x and 4 on this side is negative, right? 3.9 minus 4 is 0.1. Take the inverse of that, it's 10. So we're getting negative numbers here. You take the inverse, you're going to get super large negative numbers. So I really want to give you that intuition. But the general way of being able to graph these type of things, your first thing you want to do is identify the horizontal asymptotes. What happens as we get very-- the magnitude of our x is very large, so super positive values or super negative values. You could try it out on a calculator, if you like. You literally, if you try out the value of a million or a billion, it will kind of give you the answer. But the way you could also think about it is, as x gets really large, you could view that this thing, these terms right here grow so much faster-- I mean this is just a constant term. This term doesn't matter anymore. If this is a million and a million, who cares about the 16? So as x gets really large, you could say that y is approximately x squared over x squared. These two terms dominate. You don't need to worry about the 16 anymore. And of course, this is equal to 1, which is exactly what we got when we plugged in really large numbers. So, in a problem like this, where you have the same coefficient, or where you have the same degree on the numerator and the denominator, you look at the coefficient of those terms. So in this case, the coefficient is 1 and 1. So our horizontal asymptote is going to be 1 divided by 1, or y is equal to 1. If this was 2x squared over x squared minus 16, our horizontal asymptote would be y is equal to 2. We would approach that line, up there. If it was a negative 2, our horizontal asymptote would be y is equal to negative 2. So that's how you identify the horizontal asymptotes where you have the same degree in the numerator and the denominator. If the denominator has a larger degree, then the denominator is going to get larger much faster than the numerator, and your asymptote is going to be 0. I'll show an example of that in the future. And obviously, if your numerator has a higher degree than your denominator, it's going to grow way faster than your denominator, and you won't have any asymptote. You'll just keep growing, or keep going in the negative direction. And that's actually the case with all of the polynomials we've seen. You can do them all as being over 1. In which case, there was no horizontal asymptote. Now the vertical asymptotes you identify by essentially just factoring the denominator and figuring out where does it equal 0. Those are the points where the function is not defined. And I'll show you in the future, there are some special cases where they won't be vertical asymptotes, and I guess that special case is, for example, if you had-- Well, I won't show you the special case right now. I'll show you that in a future video. But in general, if you factor the bottom terms and they don't cancel out with anything on the numerator, then you're going to be dealing with a vertical asymptote. If I had another x minus 4 up here, if my numerator was x squared times x minus 4, and then these canceled out, and my expression simplified to this, the equation would still be undefined at x is equal to 4, because you would give the 0 in the denominator. But since that x minus 4 cancels out with the x minus 4 in the numerator, it would not have been a vertical asymptote. We'll look at that in the future. But this equation wasn't that. So the general rule of thumb for identifying the vertical asymptotes, factor the denominator, figure out where the denominator equals 0, and if those terms don't cancel out with any terms of the numerator, then those are vertical asymptotes. And then to figure out the behavior, I guess, within the asymptotes, you can plot some points. You can try out some points. You can actually substitute values for x and figure out what y is. Now just to validate that we hopefully got the right answer, let's actually graph our rational function. So let me turn it on. Let me graph it. And we say y is equal to x squared divided by x squared minus 16. And let's see what we get. Nope, I just want to graph it. My range is off. Let me do my range. Let me see, x minimum value I want for x, let's say it's negative 10. My maximum value I want for x is 10. x scale is 1. y minimum value, I want negative 10. y maximum value, I want 10. Then y scale, I want 1. Now let me graph it. There we go. Look at that. Just like what we drew. We have an asymptote, as x gets really large, or x gets really, really small, that asymptote is y is equal to 1. We have our vertical asymptote. It graphed it because it tried to connect the dots, but it essentially graphed our asymptotes for us, but that wouldn't actually be part of the graph. But as we approach 4 from 0, I guess we can say, we go super negative. As we approach negative 4 from 0, we get super negative. Because in either of those situations, as you approach 4 from this side, this term is going to be negative. As you approach negative 4 from thid side, this term, right here, is going to be-- Well, this term right here is going to be positive, but then this term right here is going to be negative. Negative times a positive, you could play with it as you like. But we approach negative infinity in either case. And then as x approaches infinity, this thing asymptotes away. So hopefully you found that fun.