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# Graphing rational functions according to asymptotes

Sal analyzes the function f(x)=(3x^2-18x-81)/(6x^2-54) and determines its horizontal asymptotes, vertical asymptotes, and removable discontinuities. Created by Sal Khan.

## Want to join the conversation?

• What do you need to know before watching this video? I cant find any asymptotes or limits videos in algebra 2 here on KA.
(34 votes)
• Yea. I suppose this is the introduction video to anymptotes.

The introduction video to "End behavior functions" is given in "End behavior of polynomial functions" Algebra 2 section. And more details on anymptotes are given in "Limits and infinity" in Differential calculus section.
(25 votes)
• I learned that there are at most two (2) horizontal asymptotes and there can be an arbitrarily large number of vertical asymptotes for a function. But why at most 2 horizontal asymptotes?

Thank you
(9 votes)
• A horizontal asymptote is basically the end behavior of a function, and there can only be two end behaviors (as x approaches negative infinity or positive infinity); that's why there can be only two horizontal asymptotes.
(9 votes)
• How do you determine whether or not your function will cross your horizontal asymptote??
(1 vote)
• You find whether your function will ever intersect or cross the horizontal asymptote by setting the function equal to the y or f(x) value of the horizontal asymptote. If you get a valid answer, that is where the function intersects the horizontal asymptote, but if you get a nonsense answer, the function never crosses the horizontal asymptote.
For example, f(x) = (10x+7)/(5x-2) has a horizontal asymptote at f(x) = 2, thus:
``(10x+7)/(5x-2) = 210x+7 = 2(5x-2)10x+7 = 10x-47 = -4Since this is nonsense, the function never crosses the horizontal asymptote.``

Now let us look at an example that does cross the horizontal asymptote:
f(x) = (x²+2)/(x²+2x-6) has a horizontal asymptote at f(x) = 1, thus:
``(x²+2)/(x²+2x-6) = 1(x²+2)= (x²+2x-6)2 = 2x-62x = 8x = 4Therefore, this function crosses its horizontal asymptote at x=4``
(22 votes)
• Can there be more than 1 vertical asymptotes
(6 votes)
• Sure, as many as you like. y=tan(x) even has infinitely many.
(7 votes)
• Just to be clear,
1)Is a removable discontinuity when the numerator is a zero and denominator not and a zero?
2) Is a vertical asymptotes when the denominator is a zero and the numerator isn't?
3) Is a zero when both numerator and denominator are zero?
In this skill every time I found they followed this pattern so I got a few right, then I got one wrong it was completely wrong so I want to make sure I am not doing the incorrect thing; because it worked for me I think I just got lucky. Also the proper way to solve them would be great. The hints are just not working for me this time.

P.S. the link to the problems are here: https://www.khanacademy.org/mission/algebra2/task/5065212460400640
(6 votes)
• When you cancel, since "(x-a)/(x-a)" = 1 for all x, you don't change the graph at all, except that you need to note that x != a because /0 is undefined. Now when there are no more factors to cancel you can check the simplified expression for /0 to find asymptotes.
(1 vote)
• why is there no videos introducing concepts such as asymptotes and limits and sal just dive straight in the topic ? i have a really hard time following with the examples.
(4 votes)
• Why does the denominator = 0 when x=3 or -3?
(2 votes)
• The denominator is equal to 6*(x-3)*(x+3). If we substitute 3 for x we have 6*(3-3)*(3+3) = 6*0*6 = 0. If we substitute -3 for x, we have 6*((-3)-3)*((-3)+3) = 6*(-6)*0 = 0. So, the denominator will be 0 when x equal 3 or -3.
(7 votes)
• So I have the equation f(x)=7x/(10-3x)^4. Now I am trying to find the vertical asymptote of this equation but I do not know what to do with the ^4. I have made (10-3x)^4=0...but that is as far as I go. Any help with this?
(3 votes)
• (10-3x)^4=0 means you have 4 factors = 0
(10-3x)(10-3x)(10-3x)(10-3x)=0
Since them match, just use one: (10-3x)=0 and solve for "x" to find the vertical symptotoe..
If the factors were different, then you would need to solve each different one.
Hope this helps.
(2 votes)
• limits and continuity are calculus lessons, aren't they ?
(3 votes)
• I was taught to simplify first. Does it matter if you do that first or not?
(2 votes)
• As long as you keep track of what values aren't allowed simplifying should be fine.
(3 votes)

## Video transcript

Voiceover: We have F of X is equal to three X squared minus 18X minus 81, over six X squared minus 54. Now what I want to do in this video is find the equations for the horizontal and vertical asymptotes and I encourage you to pause the video right now and try to work it out on your own before I try to work through it. I'm assuming you've had a go at it. Let's think about each of them. Let's first think about the horizontal asymptote, see if there at least is one. The horizontal asymptote is really what is the line, the horizontal line that F of X approaches as the absolute value of X approaches, as the absolute value of X approaches infinity or you could say what does F of X approach as X approaches infinity and what does F of X approach as X approaches negative infinity. There's a couple of ways you could think about it. Let me just rewrite the definition of F of X right over here. It's three X squared minus 18X minus 81. All of that over six X squared minus 54. Now there's two ways you could think about it. One you could say, okay, as X as the absolute value of X becomes larger and larger and larger, the highest degree terms in the numerator and the denominator are going to dominate. What are the highest degree terms? Well the numerator you have three X squared and in the denominator you have six X squared. As X approaches, as the absolute value of X approaches infinity, these two terms are going to dominate. F of X is going to become approximately three X squared over six X squared. These other terms are going to matter less obviously minus 54 isn't going to grow at all and minus 18X is going to grow much slower than the three X squared, the highest degree terms are going to be what dominates. If we look at just those terms then you could think of simplifying it in this way. F of X is going to get closer and closer to 3/6 or 1/2. You could say that there's a horizontal asymptote at Y is equal to 1/2. Another way we could have thought about this if you don't like this whole little bit of hand wavy argument that these two terms dominate is that we can divide the numerator and the denominator by the highest degree or X raised to the highest power in the numerator and the denominator. The highest degree term is X squared in the numerator. Let's divide the numerator and the denominator or I should say the highest degree term in the numerator and the denominator is X squared. Let's divide both the numerator and denominator by that. If you multiply the numerator times one over X squared and the denominator times one over X squared. Notice we're not changing the value of the entire expression, we're just multiplying it times one if we assume X is not equal zero. We get two. In our numerator, let's see three X squared divided by X squared is going to be three minus 18 over X minus 81 over X squared and then all of that over six X squared times one over X squared, this is going to be six and then minus 54 over X squared. What's going to happen? If you want to think in terms of if you want to think of limits as something approaches infinity. If you want to say the limit as X approaches infinity here. What's going to happen? Well this, this and that are going to approach zero so you're going to approach 3/6 or 1/2. Now, if you say this X approaches negative infinity, it would be the same thing. This, this and this approach zero and once again you approach 1/2. That's the horizontal asymptote. Y is equal to 1/2. Let's think about the vertical asymptotes. Let me write that down right over here. Let me scroll over a little bit. Vertical asymptote or possibly asymptotes. Vertical maybe there is more than one. Now it might be very tempting to say, "Okay, you hit a vertical asymptote" "whenever the denominator equals to zero" "which would make this rational expression undefined" and as we'll see for this case that is not exactly right. Just making the denominator equal to zero by itself will not make a vertical asymptote. It will definitely be a place where the function is undefined but by itself it does not make a vertical asymptote. Let's just think about this denominator right over here so we can factor it out. Actually let's factor out the numerator and the denominator. We can rewrite this as F of X is equal to the numerator is clearly every term is divisible by three so let's factor out three. It's going to be three times X squared minus six X minus 27. All of that over the denominator each term is divisible by six. Six times X squared minus 9 and let's see if we can factor the numerators and denominators out further. This is going to be F of X is equal to three times let's see, two numbers, their product is negative 27, their sum is negative six. Negative nine and three seem to work. You could have X minus nine times X plus three. Just factor the numerator over the denominator. This is the difference of squares right over here. This would be X minus three times X plus three. When does the denominator equal zero? The denominator equals zero when X is equal to positive three or X is equal to negative three. Now I encourage you to pause this video for a second. Think about are both of these vertical asymptotes? Well you might realize that the numerator also equals zero when X is equal to negative three. What we can do is actually simplify this a little bit and then it becomes a little bit clear where our vertical asymptotes are. We could say that F of X, we could essentially divide the numerator and denominator by X plus three and we just have to key, if we want the function to be identical, we have to keep the [caveat] that the function itself is not defined when X is equal to negative three. That definitely did make us divide by zero. We have to remember that but that will simplify the expression. This exact same function is going to be if we divide the numerator and denominator by X plus three, it's going to be three times X minus nine over six times X minus three for X does not equal negative three. Notice, this is an identical definition to our original function and I have to put this qualifier right over here for X does not equal negative three because our original function is undefined at X equals negative three. X equals negative three is not a part of the domain of our original function. If we take X plus three out of the numerator and the denominator, we have to remember that. If we just put this right over here, this wouldn't be the same function because this without the qualifier is defined for X equals negative three but we want to have the exact same function. You'd actually have a point in discontinuity right over here and now we could think about the vertical asymptotes. Now the vertical asymptotes going to be a point that makes the denominator equals zero but not the numerator equals zero. X equals negative three made both equal zero. Our vertical asymptote, I'll do this in green just to switch or blue. Our vertical asymptote is going to be at X is equal to positive three. That's what made the denominator equal zero but not the numerator so let me write that. The vertical asymptote is X is equal to three. Using these two points of information or I guess what we just figured out. You can start to attempt to sketch the graph, this by itself is not going to be enough. You might want to also plot a few points to see what happens I guess around the asymptotes as we approach the two different asymptotes but if we were to look at a graph. Actually let's just do it for fun here just to complete the picture for ourselves. The function is going to look something like this and I'm not doing it at scales. That's one and this is 1/2 right over here. Y equals 1/2 is the horizontal asymptote. Y is equal to 1/2 and we have a vertical asymptote that X is equal to positive three. We have one, two ... I'm going to do that in blue. One, two, three, once again I didn't draw it to scale or the X and Y's aren't on the same scale but we have a vertical asymptote just like that. Just looking at this we don't know exactly what the function looks like. It could like something like this and maybe does something like that or it could do something like that or it could do something like that and that or something like that and that. Hopefully you get the idea here and to figure out what it does, you would actually want to try out some points. The other thing we want to be clear is that the function is also not defined at X is equal to negative three. Let me make X equals negative three here. One, two, three, so the function might look and once again I haven't tried out the points. It could look something like this, it could look something where we're not defined at negative three and then it goes something like this and maybe does something like that or maybe it does something like that. It's not defined at negative three and this would be an asymptote right now so we get closer and closer and it could go something like that or it goes something like that. Once again, to decide which of these it is, you would actually want to try out a few values. I encourage you to, after this video, try that out on yourself and try to figure out what the actual graph of this looks like.