# Finding horizontal and vertical asymptotes | Rational expressions | Algebra II | Khan Academy

We have f (x) equal to (3x ^ 2 – 18x – 81) on (6x ^ 2 – 54). In this video I want to find the equations of the horizontal ones and vertical asymptotes and I encourage you to stop the video and try to solve this on your own, before I try to solve it. I accept that he tried. Let’s think about each of them. Let’s first think about the horizontal asymptote, let’s see if there is at least one. The horizontal asymptote is that straight, the horizontal line that f (x) approximates, when the absolute value of x approaching infinity. Or you can tell what approximates f (x), when x approaches infinity, and what approximates f (x), when x approaches minus infinity. There are two ways to think about this. Let me rewrite the expression for f (x) right here. This is 3x ^ – 18x – 81, all this on (6x ^ 2 – 54). There are two ways to think about this. First, you can say that when the absolute value of x gets bigger and bigger and bigger, the members of the greatest degree in the numerator and the denominator will dominate. Who are the members of the highest degree? In the numerator you have 3x ^ 2, and in the denominator you have 6x ^ 2. When the absolute value of x approaching plus infinity, these two members will dominate. f (x) will be approx 3x ^ 2 on 6x ^ 2. These articles will be less important – obviously -54 won’t increase at all, and -18x will grow much more slowly from 3x ^ 2, the members of the highest degree will dominate. If we look at these articles, we can think of simplification this way. f (x) will get closer and closer up to 3/6 or 1/2. You can say that there is a horizontal asymptote at y, equal to 1/2. Another way to think about it is if you don’t like it claim that these two members dominate, is that we can separate the numerator and the denominator of the highest degree or x raised to the highest degree in the numerator and denominator. The member of the highest degree is x ^ 2 in the numerator. Let’s separate the numerator and denominator – or I must say, the member of the highest degree in the numerator and denominator is x ^ 2 – let’s also divide the numerator, and the denominator of this. If you multiply the numerator by 1 / x ^ 2 and the denominator of 1 / x ^ 2. Note that we do not change the value of the whole expression, we just multiply it by 1, assuming that x is not equal to 0. We get 2. In the numerator, see, 3x ^ 2 divided by x ^ 2, will be equal to 3 minus 18 / x minus 81 / x ^ 2, and then all this on 6x ^ 2 by 1 / x ^ 2 – this will be 6, and then minus 54 / x ^ 2. What will happen? If you want to think about, if you want to think about boundaries, when something approaches infinity … If you want to say the boundary when x tends to infinity … What will happen? This, this and this will tend to 0, so you will tend to 3/6 or 1/2. If you say it’s x tends to minus infinity, it will be the same thing. This, this and this tend to 0 and, again, the expression tends to 1/2. This is the horizontal asymptote. y = 1/2. Let’s think about vertical asymptotes. Let me write this down here. Let me scroll down a bit. Vertical asymptote or perhaps asymptotes. There may be more than one. It may be tempting to say: “You come to a vertical asymptote, when the denominator is equal to 0, which will make this rational expression indefinite. ” And as we shall see, in this case, this is not entirely true. Just make the denominator equal to 0, by itself will not make a vertical asymptote. It will definitely be a place where the function is not defined, but does not in itself create a vertical asymptote. Let’s think about this denominator here, we can decompose it. Let me decompose the numerator and denominator. We can convert this as f (x) is equal to – in the numerator obviously each member is divided into 3, so let’s put 3 in parentheses. That will be it 3 by (x ^ 2 – 6x – 27). All this on the denominator – in it each member is divided into 6. 6 (x ^ 2 – 9) and see if we can decompose further numerator and denominator. This will be f (x) equal to 3 by, let’s see – 2 numbers, their product is -27, their sum is -6. -9 and 3 seem to work. You can have (x – 9) by (x + 3) – I just spread the numerator on the denominator. This here is the difference of squares. This will be (x – 3) (x + 3). When is the denominator equal to 0? The denominator is equal to 0, when x is equal to 3 or when x is equal to -3. I advise you to stop the video for a moment. Consider whether both of these are vertical asymptotes? You may realize that the numerator is also equal to 0 when x is equal to -3. We can simplify this a little bit, and then it becomes clear where are our vertical asymptotes. We can say that f (x) – we can divide the numerator and the denominator of (x + 3) and you just have to … if we want the function to be identical, we need to make the function itself not be determined when x is equal to -3. This definitely made us divide by 0. We have to remember this, but this will simplify the expression. This same function will be, if we divide the numerator and denominator by (x + 3), it will be 3 (x – 9) on 6 (x – 3) for x other than -3. Note that this is an identical setting of our original function and should put this limited here – for x other than -3, because our original function is not defined for x, equal to -3. x = -3 is not part of the definition set for our original function. If we reduce (x + 3) by the numerator and denominator, we must remember this. If we just put this here, this will not be the same function, since it is without the constraint * is * defined for x = -3, but we want to have exactly the same function. You will have a breakpoint here and now we can think of the vertical asymptotes. Vertical asymptotes will be a point which makes the denominator equal to 0, but not the numerator. x = -3 makes both equal to 0. Our vertical asymptote – I’ll do this in green, just for a change, or blue. Our vertical asymptote will be at x = 3. This makes the denominator equal to 0, but not the numerator – let me write this down. The vertical asymptote is at x = 3. Using this data or what we just discovered … You can start by trying to sketch the graphic, that alone will not be enough. You may want to put a few points, to see what’s going on around the asymptotes, as we approach the two different asymptotes, but if we look at the graph … Let’s just do it for fun. To complete the picture. The graph of the function will look similar to this – and I don’t do it on a scale. That’s 1, and that’s 1/2 here. y = 1/2 is the horizontal asymptote. y is equal to 1/2 and we have a vertical asymptote at x = 3. We have 1, 2 … I will do this in blue. 1, 2, 3, again – I didn’t draw this in scale. The values \u200b\u200bof x and y are not on the same scale, but we have such a vertical asymptote. Just looking at this, we don’t know exactly what the graph of the function will look like. It may be so and can do something similar or something like that. Or he can do something like that. Or something like that. I hope he got the idea and to find out what he’s actually doing, you have to try some points. The other thing that needs to be clear is that the function is not defined at x = -3. Let me do x = -3 here. 1, 2, 3, ie the graph of the function may look like this and, again, I haven’t tried the points. It may look like this, where it is not determined at -3 and then to look like this and, perhaps, here it is, or does something similar. Not set at -3 and that here will be the asymptote, so we get closer and closer and can do something similar or do something like that. Again, to decide which of these is actually, you have to try several values. And I advise you then video to try to do it yourself and try to understand what it looks like the actual graphics of this function.