For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

### Graphing Asymptotes in a Nutshell

- This lesson is all about learning how to graph rational functions. It's strongly recommended that you watch the previous lessons beforehand, because we'll be pulling from that work. Also, while we won't be going over it in the lesson, using a graphing utility (calculator or program) can be a great way to understand how rational functions work. Playing with function graphs can quickly build your intuition.
- Here is a step-by-step process to create an accurate graph for any rational function:

1. Begin by__factoring__the numerator polynomial and the denominator polynomial of the rational function you're working with.

2. Find the__domain__of the function by looking for where the denominator equals 0. Each of these "forbidden" locations will become one of two things: a vertical asymptote (if the zero does not occur in the numerator) or a hole in the graph (if the zero occurs in the numerator).

3. Now that we've found the function's domain,__simplify the function__by canceling out any factors that are in both the numerator and the denominator. [It's important that we don't do this in step #1 (factoring), otherwise we won't be able to find all the "forbidden" x-values in step #2 (domain).]

4. Once the function is simplified, we can find the__vertical asymptotes__. The vertical asymptotes occur at all the x-values that still cause the denominator (after simplifying) to become 0.

5. Find the__horizontal/slant asymptotes__by looking at the degree of the numerator, n, and the degree of the denominator, m. There are a total of four possible cases:- n < m ⇒ horizontal asymptote at y=0.
- n = m ⇒ horizontal asymptote at a height given by ratio of leading coefficients in numerator and denominator.
- n=m+1 ⇒ slant asymptote, which can be found by using polynomial division.
- n > m+1 ⇒ no horizontal or slant asymptote.

6. Find the__x- and y-intercepts__so we have some useful points that we can graph from the start. [It's possible for a rational function to be missing one type or both.]

7. Finally, using all this information, draw the graph. Place the asymptotes (drawn as dashed lines) and intercepts. You will probably need some more points, so plot more points as necessary until you see how to draw in the appropriate curves. - A useful concept for graphing is the idea of
*test intervals*. Because a rational function is continuous between vertical asymptotes, we know that the function can only change signs at x-intercepts or vertical asymptotes. This means we can put our x-intercepts and vertical asymptote locations in order and break the x-axis into intervals. In each of these intervals, we can test just one point to find if the function is + or − in the interval.

### Graphing Asymptotes in a Nutshell

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Introduction
- A Process for Graphing
- 1. Factor Numerator and Denominator
- 2. Find Domain
- 3. Simplifying the Function
- 4. Find Vertical Asymptotes
- 5. Find Horizontal/Slant Asymptotes
- 6. Find Intercepts
- 7. Draw Graph (Find Points as Necessary)
- Draw Graph Example
- Test Intervals
- Example 1
- Example 2
- Example 3
- Example 4

- Intro 0:00
- Introduction 0:05
- A Process for Graphing 1:22
- 1. Factor Numerator and Denominator 1:50
- 2. Find Domain 2:53
- 3. Simplifying the Function 3:59
- 4. Find Vertical Asymptotes 4:59
- 5. Find Horizontal/Slant Asymptotes 5:24
- 6. Find Intercepts 7:35
- 7. Draw Graph (Find Points as Necessary) 9:21
- Draw Graph Example 11:21
- Vertical Asymptote
- Horizontal Asymptote
- Other Graphing
- Test Intervals 15:08
- Example 1 17:57
- Example 2 23:01
- Example 3 29:02
- Example 4 33:37

### Math Analysis Online

### Transcription: Graphing Asymptotes in a Nutshell

*Hi--welcome back to Educator.com.*0000

*Today, we are going to talk about graphing and asymptotes in summary.*0002

*In the past two lessons, we learned how rational functions work.*0006

*We have studied and come to understand their behavior, along with vertical asymptotes and horizontal/slant asymptotes.*0009

*In this lesson, we use this knowledge to learn how to graph rational functions.*0015

*It is strongly recommended that you watch the previous lessons beforehand, because we will be pulling things from that work.*0019

*Also, while we won't be going over it in this lesson, using a graphing utility, whether it is a calculator or a program*0024

*or something even on a smartphone, can be an absolutely great way to understand how rational functions work.*0029

*Playing with function graphs can quickly build your intuition.*0035

*Just like it can build your intuition for any function, being able to play around and change how things are working*0038

*and what your denominator is and what your numerator is will build a really great, intuitive understanding*0042

*of how this stuff works much faster than trying to do it all by hand.*0047

*You will be able to get a really good grasp of how this stuff works, and be able to see it in your mind's eye*0051

*in a way that you just can't develop by trying to do ten problems in hand, because you can just do 100 of them*0055

*in a matter of a few minutes if you are just playing around on a calculator.*0061

*So, I highly recommend that you take the chance and use a graphing utility and play with something.*0065

*If you haven't already checked it out, there is an appendix to this course all about graphing utilities,*0069

*graphing calculators...all of that stuff... that can give you some idea of how to start in that sort of thing,*0073

*because they can be really useful for helping you in a course like this and in future math courses.*0078

*All right, let's get started: the majority of this lesson is going to be about a process for graphing rational functions.*0082

*By following the steps of this process, you will obtain what you need to graph.*0088

*You will get all of the information that you need to make a good graph.*0091

*This process also gives you a way to analyze rational functions in general, though.*0094

*So, it is not something that you can only use when you want to graph a function--just if you want to look at a rational function*0098

*and get a good idea of how it works, you might find this process useful.*0104

*So, you might use it, even if you don't need to graph anything; let's go!*0106

*The first step: a rational function starts in the form n(x)/d(x), where n(x) and d(x) are both polynomials--this division.*0110

*It is useful as a very first step to begin by factoring n and d.*0119

*While this won't directly tell us anything--factoring the numerator and the denominator doesn't immediately tell us anything directly--*0124

*it is very useful in the coming steps to have the numerator and denominator broken into smallest factors.*0130

*A lot of our steps are going to revolve around having these things already factored.*0135

*So, it is something to get out of the way, right from the beginning.*0138

*So, we are going to have a running example as we go through this.*0141

*If we have f(x) = (x ^{2} - 3x + 2)/(2x^{2} - 2x - 4), we could factor this,*0143

*and we would get (x - 1)(x - 2) for the numerator, divided by 2(x + 1)(x - 2) for the denominator.*0149

*So, we started by factoring it, by just breaking these into their factors.*0156

*By seeing what these things become, we are able to get a good sense for later steps.*0161

*It will help us out in the later steps; so we just re-format it with the numerator factored and the denominator factored; and that is all we do for right now.*0167

*The second step: find the domain of the function.*0174

*Remember: we can't have division by 0--it is one of the critical ideas here.*0177

*So, we want to find all of the x-values where the denominator is 0, because those are going to be forbidden.*0181

*We do this by finding the zeroes, the roots, of d(x); so find the zeroes of our denominator.*0186

*And since we just factored the denominator function, this is going to be pretty easy.*0191

*We just factored it, in our example, into (x + 1) and (x - 2), so we see at this point that we wouldn't be allowed -1 or +2,*0196

*because they would cause a 0 to pop up.*0203

*So, each of the zeroes of our denominator is not going to be allowed in the domain.*0206

*They will not be allowed in the domain if they are a zero of our denominator polynomial.*0210

*All other real numbers are in the domain, because everything else is fine for a polynomial.*0214

*The only problem is when we are accidentally dividing by 0; so we just have to clip those out.*0219

*Those are the forbidden locations that we can't take in.*0223

*Now, each of these forbidden locations will become one of two things: they will become a vertical asymptote,*0226

*if the zero does not occur in the numerator; or they will become a hole in the graph, if the zero does occur in the numerator.*0232

*The next step: simplify the function--once we have found the function's domain, we can simplify the function*0240

*by canceling out any factors that are in both n(x) and d(x).*0245

*So, we noticed that we had (x - 2) on the top and (x - 2) on the bottom; we have common linear factors.*0249

*So, we knock them both out, and we are left with (x - 1)/2(x + 1).*0254

*It is important to note that we couldn't do this in our very first step, factoring,*0260

*because if we did that, we wouldn't be able to find all of the forbidden x-values.*0264

*There is a forbidden x-value at x = 2; so that is something that we are not allowed to have from this.*0268

*So, the only way to find that is if we haven't already gotten rid of it.*0275

*If we start by canceling it, we will never realize that x is not equal to 2, because x = 2, that horizontal location of x = 2, is forbidden.*0278

*If we cancel out that factor before we notice that it is a forbidden location, we will never be able to figure that out.*0287

*So, we have to get that information before we cancel it; and that is why we do factoring, and then check the domain, then simplify.*0292

*The next step: find vertical asymptotes--once the function is simplified, we can find the vertical asymptotes.*0300

*The vertical asymptotes will occur at all the x-values*0305

*that cause the denominator, after we have simplified the denominator, to become 0--*0307

*so, all of the zeroes of our now-newly-simplified denominator.*0312

*So, if we have 2(x + 1) in our denominator, we are going to get a vertical asymptote at x = -1,*0316

*because that will cause our denominator to turn into a 0.*0321

*The next step: find the horizontal or slant asymptotes, or figure out if it has absolutely none of those.*0325

*Let n be the degree of the numerator; and m is the degree of our denominator.*0331

*Then, there are a total of four possible cases: n is less than m--the degree of the numerator is less than the degree of the denominator.*0339

*Our denominator grows faster than our numerator--that means that we will eventually be crushed down to nothing.*0350

*The whole fraction will be crushed down to nothing; and we have a horizontal asymptote, y = 0.*0356

*Another possibility: if n equals m--the degree of the numerator is equal to the degree of the denominator--*0360

*then there will be a horizontal asymptote at a height given by a ratio of the leading coefficients,*0367

*because they are both growing at the same class of speed, so we need to compare how their fronts go.*0373

*If we have effectively 5x ^{5} divided by 2x^{5}, the other stuff has some effect.*0379

*But in the long run, it won't be as important, and it will turn into 5/2, because the x ^{5}'s cancel out.*0385

*That is one way of looking at it: n = m means that we get a horizontal asymptote based on the ratio of leading coefficients.*0391

*Next, n = m + 1; that is a slant asymptote--we will have a slant asymptote, and we find that by using polynomial division.*0399

*You can also use polynomial division to find the horizontal asymptotes,*0408

*but it is pretty easy and fast to find it just by comparing the leading coefficients, so we don't worry about that as much.*0411

*And then finally, if we have n > m + 1, there is no horizontal or slant asymptote whatsoever,*0418

*because the numerator is growing so much faster than the denominator*0425

*that it is not going to be a horizontal thing that it goes to; it is not even going to be a slant that it goes to.*0429

*It is just going to blow into something even larger and more interesting.*0433

*But we are not going to worry about that in this course.*0436

*f(x) = (x - 1)/2(x + 1): we realize that this is a degree of 1; this is a degree of 1.*0438

*So now, we go and we compare our leading coefficients: we have a 1 on the top and a 2 on the bottom.*0445

*So, we get a horizontal asymptote of y = 1/2 in our running example.*0451

*The sixth step: find the intercepts--by finding the x- and y-intercepts, we have a couple points from in the graph that we just have to start with.*0456

*Now, it is possible for a rational function to be missing one type or both.*0464

*So, it is possible that these things won't be there, because the location of our y-intercept, x = 0, could be a forbidden location.*0468

*And all of the places where we cross over--it could either not cross the x-axis at all,*0477

*or the locations where it would cross the x-axis are actually disappeared holes in our graph, so they aren't technically intercepts.*0481

*So, it is possible to be missing these things.*0488

*But if we have them there, they are nice, and they are not that hard to find.*0490

*The x-intercepts occur wherever the function has an output value of 0, because that means we have a height of 0;*0494

*so, we are on the x-axis; we are an x-intercept with a y height of 0.*0499

*Thus, all of our zeroes mean that our numerator is at 0.*0505

*This is all of the zeroes of our simplified numerator: x - 1...what are all the places where that gives us a 0?*0509

*That gives us a 0 at x = 1, so at x = 1, we get a 0 out of it; so if we plug in 1, we get a zero out of that, because the numerator is now at 0.*0515

*The y-intercept is where the function's input is 0, because that will put us on the y-axis.*0528

*So, we plug in the x-value...*0533

*I'm sorry; I might have said the wrong thing there; I am not quite sure what I said.*0535

*The y-intercept occurs where the function's input, its x-value, the x that we are plugging in, is 0,*0537

*because that will put us right on the y-axis.*0542

*So, just evaluate our function with a 0 plugged into it, f(0): to find it, we plug in 0: (0 - 1)/2(0 + 1) from our simplified function.*0544

*That simplifies to -1/2, so we get (0,-1/2); so we have some points to start with when we are plotting.*0555

*The final thing: draw the graph--now that we have all of this information on our function, we are ready to graph it.*0563

*Begin by drawing in the asymptotes on the graph, and plot the intercept points.*0567

*If we need more points to graph the function (and we are probably going to need more points,*0571

*since the intercepts--really, there are only a couple of things that come out of the intercepts), evaluate the function*0575

*at a few more points, as you need, and plot those points, as well.*0579

*Plot those extra points; and then, at that point, you can start drawing in curves.*0583

*It is useful to plot at least one point between and beyond each vertical asymptote.*0587

*So, if we have vertical asymptotes, like here and here, you probably want to make sure you plot at least one thing in each of these locations.*0594

*And there is a good chance that you will need a couple more than that, to be able to really get a sense*0601

*for how the thing is going to come together as a picture--that is something to think about.*0605

*Then, draw in the graph, connecting the points with smooth curves, and pulling the graph along the asymptotes.*0610

*And of course, if you are not quite sure where it goes and how it works together,*0616

*you can just plot in even more points, and you will get a better sense of how the picture comes together.*0619

*Make sure you know which direction, positive or negative, the graph will go in either side of your vertical asymptotes.*0624

*And one last thing: don't forget that the forbidden values, our forbidden x locations, will create holes in the graph.*0629

*We are going to be having these locations that aren't really there, which we will denote with an open circle*0644

*to say, "Well, we would be going here; but it actually is missing that location,*0651

*because it is a forbidden thing, because it will cause us to divide by 0."*0655

*So, we are either going to have vertical asymptotes--which we would never get to anyway--*0658

*or we are going to have actual holes in our graph, which we denote with a hole,*0661

*of a just round circle that has nothing inside of it.*0665

*So, don't forget that you have to remember about the holes when you are actually drawing it in the graph.*0668

*Not every graph will have holes; but if yours does have forbidden x-values*0674

*that aren't just vertical asymptotes, you will need to do that, as we are about to see.*0678

*f(x) = (x ^{2} - 3x + 2)/(2x^{2} - 2x - 4); we simplify that into (x - 1)/2(x + 1).*0681

*We figured out that our domain had forbidden values at -1 and 2, because they caused our original denominator,*0690

*not just our simplified denominator, to turn into a 0; we are not allowed to divide by 0.*0696

*To figure out our vertical asymptote, we looked: when does our simplified denominator go to 0? That happens at x = -1.*0701

*To find our horizontal asymptote, we noticed that we have a degree of 1 on the top and the bottom; so 1/2 gave us y = 1/2 as a horizontal asymptote.*0710

*So, we plot our horizontal asymptote; we plot our vertical asymptote.*0719

*And we also figured out our intercepts; 0 comes out as -1/2, so we plot that point right there.*0725

*and 1 comes out as 0, so we plot that point right there; those are our intercepts.*0731

*Now, that is not quite enough information (for me, at least) to figure out how this is going to graph.*0736

*So, we decide, "Let's plot a few more points."*0740

*We try out -2 and -3, because -2 is one step to the left of our asymptote.*0743

*So, we plot in this point; we plot in -3; that comes out there; so we have -2 at 3/2 and -3 at positive 1.*0747

*And then also, we can say, "Well, we are not allowed to put in 2; but we are still curious to know where that would be, if it was there."*0755

*So, let's see where 2 would go if it wasn't a forbidden location.*0764

*Remember, it is forbidden here, because it caused us to divide by 0.*0769

*But in this one, it doesn't actually cause anything weird to happen in our simplified version, because we have canceled out the 0/0, effectively.*0772

*Since we have canceled it out, it doesn't cause anything weird to happen.*0779

*So, we can see where it would go by looking at our simplified version.*0781

*So, if we plug 2 into our simplified version, we get (2 - 1), 1, over 2(2 + 1), 2(3), 6; so we get 1/6.*0784

*So, we plug in, not just a point, but a hole, to tell us, "Look, there is where you would go;*0793

*you will run through that location, but you are not actually going to occupy that."*0800

*So, at this point, it seems like we are starting to get enough information.*0804

*One last thing we might want to figure out is which way we are going on each of these asymptotes.*0807

*We probably can get a sense of this at this point.*0812

*So, if we have plugged in, say, -1.0001, if we were just to the left side of our vertical asymptote,*0814

*then over here we would have 2 and -1.0001 plus 1; and up here, we would have -1.0001.*0822

*On the top we would have a negative...minus a negative, so it is a negative; and then divided by 2 times negative +1,*0834

*-1.0001, so just a little bit more negative; so it is going to be a negative number down there; 2 doesn't change it from being negative.*0841

*So, we have a negative over a negative, which cancels out to a positive.*0848

*It is going to come out being positive on this side; so we are going to be going up with our vertical asymptote on this side.*0852

*If we do the opposite, -0.999, like this, then if we plugged in -0.999 - 1, we see that that is going to end up being a negative.*0858

*And then, divided by 2 times -0.999 plus 1...we see -0.999 + 1...that stays positive, just barely.*0871

*It is a very small positive, but it is positive; so we have a positive on the bottom.*0880

*A negative divided by a positive--that remains being negative; so on the right side of our vertical asymptote,*0884

*over here, we are going to be going down, because we are going to be having negative values coming out of it.*0890

*Now, we know which way it should be going; and we know, on our horizontal asymptote, it is just going to stay on the same side and run with it.*0896

*We draw all of these things in; and sure enough, that is what happens when we draw in our picture; great.*0903

*Test intervals: this is a useful idea that can occasionally be used.*0909

*We didn't use it in our initial seven steps; but it is something that we kind of vaguely...I vaguely used it without explaining it,*0913

*on the last thing, where I said we knew that we were going to have to stay*0919

*under the horizontal, below the horizontal, in these various locations.*0922

*Let's see what it is: a useful concept for graphing is the idea of the test interval.*0925

*Because a rational function is continuous (remember, there are no jumps in a rational function,*0929

*because it is built out of polynomials, so there can't be any jumps between vertical asymptotes),*0934

*we know, by the intermediate value theorem, since we aren't allowed jumps,*0939

*that a function can only change signs at x-intercepts where it crosses from positive to negative.*0942

*It can only change signs when it hits 0 to be able to go from positive to negative--*0947

*or vertical asymptotes, because after a vertical asymptote, it does jump.*0951

*This means we can put our x-intercepts and our vertical asymptote locations in order, and then break the x-axis into intervals.*0955

*In each of these intervals, we can test just one point, because we know we can't jump*0961

*until we hit an x-intercept or we hit a vertical asymptote--we can't flip signs.*0966

*So, we can test just one point to figure out if it is going to be positive in that interval, or it is going to be negative in that interval.*0970

*So, in each of these intervals, we only have to test one point to figure out positive- or negative-ness in there.*0976

*If we have f(x) = (x - 1)/2(x + 1), our example that we have been working this whole time,*0980

*we have a vertical asymptote at x = -1, and we have an x-intercept at x = +1.*0985

*So, that means we can go from negative infinity, right here, up until -1, our first interval location stopping.*0990

*And then, from -1 up until 1 is our next thing--from -1 here to positive 1 here; and then 1 out into infinity, because we don't have any others.*0998

*There are three intervals, and the function is going to maintain its sign within that interval.*1006

*So, let's test any point: notice, -2 is inside of that interval, so let's try out -2.*1012

*We plug in -2, and we get -3 divided by -2, which becomes +3/2, so it is positive; everywhere between negative infinity and -1, it is positive.*1016

*From -1 to 1...let's look at 0: 0 is definitely in that interval.*1029

*We plug that in; we get -1 divided by 2, so we get -1/2; that means it is going to be negative everywhere inside of that interval.*1032

*And 1 to infinity--let's try out 3; 3 is in there--we plug in 3; we get 2/2(4), so we get 1/4, because we had 2/8, so 1/4.*1040

*But it is a positive, more importantly, so we know we are going to be positive everywhere from 1 out until positive infinity.*1052

*If you go back just a little bit to the previous graph, where we saw the thing actually get graphed in,*1058

*and pause it there, you will actually be able to see that it ends up being always positive between negative infinity and -1,*1063

*always negative from -1 to positive 1, and always positive from 1 out to infinity.*1069

*You will be able to see that in the graph; that is the idea of test intervals.*1074

*All right, let's look at some examples: f(x) = 3/(x + 2).*1078

*Our first step: let's figure out what is forbidden in our domain.*1084

*So, our domain can't cause anything to go to 0; so when is x + 2 equal to 0?*1088

*That happens at x = -2, so our domain will not allow -2, because we do not want a denominator to allow a 0.*1094

*Next, what are our vertical asymptotes?*1102

*That is going to happen when...are we simplified?...yes, 3/(x + 2) is already simplified,*1108

*so that is going to be whenever the denominator is 0, so we have a vertical asymptote at x =...*1113

*oh, sorry, not 0, but x = -2, where the denominator becomes 0.*1118

*So, our domain location that is not allowed is our vertical asymptote here.*1123

*And finally, a horizontal asymptote--what will this go to in the long run?*1128

*In the long run, we have a numerator's degree that is 0, and our denominator's degree is 1.*1137

*So, in the long run, the denominator will grow and eventually crush our numerator to effectively nothing.*1143

*So, in the long run, our fraction goes to 0; so it has y = 0 as its horizontal asymptote.*1148

*Now, it would probably be a good idea to find some intercept locations, so we can have a better idea of what is going on here.*1155

*Let's see, at 0, where are we? At 0, we plug in 0; 3/(0 + 2) gets us 3/2.*1160

*Do we have a y-intercept?...yes, we have a y-intercept--that is what we just figured out.*1169

*Do we have any x-intercepts?...no, because the numerator is always 3.*1173

*The numerator never goes to 0, so there are not going to be any x-intercepts.*1177

*Let's draw in our graph, and then we will fill it out.*1182

*So...oops, that got kind of curvy; I will erase that really quickly.*1186

*We know that -2 (that is our vertical asymptote) is the most interesting location; so we will give a little extra space on our left side.*1191

*1, 2, 3, 4, 5, 6, 1, 2, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5; OK.*1201

*We know about the -2...and I hope you don't mind; I am just not going to mark it down.*1219

*I think we can see pretty easily, just counting out what those are; we just know those types of marks mean 1.*1223

*So, we have a vertical asymptote at -2, and we have a horizontal asymptote at y = 0, which is just our x-axis.*1228

*Great; all right, so how can we draw this in?*1238

*We have (0,3/2); we are right here; now, we probably want some more points.*1242

*That is not quite enough information, so let's try some points; let's go just one to the left of our vertical asymptote--let's try out -3.*1248

*We plug in -3; we get 3/(-3 + 2), so -1...3/-1 gets us -3.*1258

*Let's try -4, as well: we plug in -4; 3/(-4 + 2) is -2, so that is 3/-2, or -3/2.*1266

*We plug in -5, and we will get 3/(-5 + 2), so -3, so we will get -1.*1275

*We can plot the stuff on the left side now: at -3, we are at a height of 1, 2, 3, -3.*1282

*At -4, we are at 3/2; and at -5, we are at -1.*1290

*Let's go on the other side; let's look at...we already have this point here, so let's try -1.*1296

*At -1, 3/(-1 + 2) becomes positive 1, so we are at positive 3, because 3 divided by 1 is 3.*1302

*At 0, we already figured out we are at 3/2; and at positive 1, we are going to be at 3/(1 + 2), so 3, so we will be at 1 there.*1310

*Positive 1...we are at 1; at -1, we are at 3; great; at this point, we have a pretty good idea.*1323

*We can see that we are going up here; we can see that we are going down here.*1330

*We could plug in -2.0001, and we would see that that would be 3 divided by a negative, so we are going negative.*1333

*And -1.99999 would be 3 divided by a positive, so we are going to go positive.*1342

*But we can also just see, from enough points that we have at this point--see where the curve is going.*1347

*So, we draw this in; it curves; as it approaches the asymptote, it curves more and more and more and more and more.*1351

*The other way, it is going to curve and approach its horizontal asymptote; and it approaches it, but it never quite touches it; and it goes out that way.*1359

*A similar thing is going on over here; it curves; it approaches the vertical asymptote; it won't quite touch it.*1368

*And here, it curves; it approaches the horizontal asymptote in the long run, but it doesn't quite touch it, either; great.*1373

*And there we are--the first one done.*1380

*The next one: Graph f(x) = (x + 1)/(x ^{3} + x^{2} + x + 1).*1382

*The first thing we want to do is factor this; the top is still just going to be (x + 1); what does the bottom become?*1388

*Well, we know that there is going to be an (x + 1) factor in there; we can figure that out by trying plugging in -1.*1394

*We see that it would work; but we can also eventually notice that we will eventually factor it into (x + 1)(x ^{2} + 1).*1402

*So, what is not allowed by our domain? Our domain does not allow (x + 1);*1411

*so x is not allowed to be -1, because -1 + 1 would give us a 0.*1419

*The domain is not allowed to be -1; does x ^{2} + 1 provide anything?*1424

*No, x ^{2} + 1 is an irreducible quadratic; there are no roots there, because x^{2} + 1 = 0 is always going to remain positive.*1428

*You can't solve x ^{2} + 1 if you are using real numbers, so we don't have to worry about a hole appearing there.*1435

*So, our only issue is that = -1; so we disallow -1: -1 is a forbidden location.*1440

*Now, at this point, we can simplify: we see (x + 1) and (x + 1); we cancel those out; and so, we get 1/(x ^{2} + 1).*1446

*All right, so at this point, we can figure out what is our vertical asymptote; do we have any vertical asymptotes?*1455

*We don't have any vertical asymptotes, because in our simplified form,*1461

*1/(x ^{2} + 1), there is no x we can plug in to get 0 to show up in the bottom.*1464

*x ^{2} + 1 has no solutions; it never intersects the x-axis, if we think about it.*1469

*So, a vertical asymptote...we have no vertical asymptote here.*1474

*What about a horizontal asymptote? If we plug in for a horizontal asymptote, we see--look, the denominator has a higher degree than the numerator.*1478

*So, the numerator is going to eventually get crushed by that large denominator, so we have it eventually going to y = 0.*1491

*Let's look for some intercepts: are there any intercepts?*1499

*Can we plug in anything to get an intercept on the top?*1505

*Well, in our simplified form, 1/(x ^{2} + 1), there are no intercepts that we can get*1509

*that will be x-axis intercepts, because there is nothing we can plug in to get a 0 to show up in our numerator.*1514

*But we will be able to plug in 0 for our x, so we can see what the y-axis intercept is.*1520

*We plug that in, and we get 1/(0 + 1), so we get just (0,1).*1526

*All right, at this point, we can probably draw in our graph.*1532

*We don't have a vertical asymptote; we know that -1 is interesting, and we have an intercept; so let's just make it even down the middle.*1535

*I am not quite sure how much will end up showing up on either side, so let's just draw something to begin with.*1545

*1, 2, 3, 1, 2, 3, 1, 2, 1; great; once again, every tick mark just means a distance of one--we won't worry about writing in all of those numbers.*1554

*So, at this point, we probably want a few more points.*1574

*(0,1)...well, let's see; our horizontal asymptote is just our x-axis.*1578

*The vertical asymptote...we have no vertical asymptote.*1585

*We can plot in our intercept at (0,1), so here is a point; and now, let's start looking at some other points.*1588

*If we plug in positive 1, 1/(1 ^{2} + 1) gets us 1/2; OK, (1,1/2); here we are.*1594

*We plug in positive 2: 1 over 2 ^{2}, 4, plus 1...1/5, one-fifth.*1605

*Plug in 3; with 2, we are at 1/5; we are getting pretty low here; plug in 3: 1 over 3 ^{2}, 9, plus 1, so 1/10; we get one-tenth.*1612

*Now, we are getting really low, so we get crushed down pretty quickly there.*1626

*What about on the left side? Well, look: it is x ^{2}, and we don't have any other things there.*1630

*So, it is going to do the exact same thing on the other side, because -1 is going to square to effectively what it would have been if it had been positive 1.*1635

*-2 will go to positive 3, and -3 to positive 3; so -3 will also be 1/10; -2 will also be 1/10; and -1...wait a second!*1641

*Domain...x is not allowed to be -1; so this is actually a hole, so -1 is going to be a location where we can see where it would have gone;*1653

*but we are going to have to denote it with a hole, because it is not actually allowed to show up there.*1664

*It is allowed to be very close on either side, but at -1 precisely, it is technically forbidden.*1668

*-1...when we plug that in, we would also get 1/2; but these are both going to be a hole.*1673

*So, we can plot that point as if it was there; but we will have to plot it with a hole,*1678

*because there, it is not actually there; it is at that point where it disappears briefly.*1682

*Momentarily, the function breaks, and it just sort of isn't there.*1688

*Now, let's plug in the other -2...not 1/10; sorry about that; it should have been 1/5; I wasn't paying attention...there, and then 1/10 here.*1692

*Great; at this point, we see that we are going to see some sort of curve, like this,*1704

*where, over time, it gets closer and closer, but then it flattens out; so we will never quite touch that horizontal asymptote.*1710

*But we will get very, very close...this way, as well...it gets there; it blips out of existence,*1717

*very, very briefly, for that single location of -1; it blips out of existence, and then it pops right back into existence.*1723

*And then, once again, it goes back to getting very, very close to that horizontal asymptote, but never quite touching it.*1730

*So, that is rough; my drawing is not quite perfect--sorry; but that is a pretty good sense of what that would look like.*1735

*All right, the next one: f(x) = (x ^{2} - 1)/(x^{2} + 1).*1743

*Let's factor the top into its factors: we would get (x + 1)(x - 1); and it is divided by (x ^{2} + 1); great.*1748

*So, our domain...is there anything forbidden in the domain?*1760

*No, everything is allowed, because with x ^{2} + 1,*1764

*we can never get any zeroes to show up there; so all of the real numbers are allowed in here.*1767

*So, there is nothing to cancel out; we don't have to worry about canceling out any factors,*1774

*because we have (x + 1)(x - 1); (x ^{2} + 1) is irreducible,*1777

*so it can't break into any linear factors to cancel with those linear factors up top.*1780

*So, at this point, let's see: are there any vertical asymptotes?*1784

*Vertical asymptotes would be where the denominator is equal to 0; there are no vertical asymptotes,*1788

*because x ^{2} + 1...once again, there is nothing we can plug in to make it turn into a 0.*1794

*What about horizontal asymptotes? There are two...no, sorry, there can only be one horizontal asymptote; my apologies.*1798

*For the horizontal asymptote, we compare the degrees: squared here, squared here; both a quadratic on the top and a quadratic on the bottom.*1807

*That means that we are going to see a horizontal asymptote that isn't just the x-axis.*1814

*So, now we compare, and we see--what are the leading coefficients?*1818

*We have a 1 here and a 1 here, so we are going to see a horizontal asymptote of 1/1.*1821

*And since 1/1 just simplifies to 1, we have a horizontal asymptote of y = 1.*1827

*A height of 1 is what it will eventually, slowly move towards.*1833

*Let's look at what our intercepts are: can we get intercepts for our y-intercept?*1838

*Yes, we have no problem plugging in a 0; nothing is forbidden, so we plug in 0.*1847

*We get 1(-1)...it is probably easier to actually figure it out from this equation up here, so it is -1 divided by +1, or just -1.*1851

*We have two intercepts when the numerator becomes 0; so at -1 it becomes 0; at positive 1, it becomes 0.*1860

*And we will probably want a few more points; but let's draw this down so we get a sense of what is going on first.*1868

*So, once again, we don't have any vertical asymptotes; we have nothing special that we really want to look for.*1873

*So, let's just center our graph nicely; 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 1, 2; great.*1878

*So, we can plot in our intercepts; we have intercepts...well, let's first plot our horizontal asymptote, y = 1.*1901

*We have a dashed line at the height of y = 1; OK.*1910

*There are no vertical asymptotes; we don't have to worry about that.*1918

*Intercepts--where are our intercepts? 0 is at -1; -1 is at 0; positive 1 is also at 0.*1920

*We probably want a few more points; let's see what happens at positive 2, negative 2, and things like that.*1929

*Let's try those out: at positive 2...we plug that in, and we get 2 ^{2} - 1,*1933

*so that is going to be 4 - 1, or 3, divided by 2 ^{2} + 1; that is going to be 5; so 3/5.*1938

*Great; let's try 3--plug that in; we get 9 - 1 is 8, divided by 9 + 1 is 10; so 8/10 simplifies to 4/5.*1946

*And what about if we tried -2 and -3? Well, notice: we have x ^{2} up here.*1956

*We have x ^{2} down here; so if we plug in a negative, it is going to behave just the same as if we had plugged in a positive.*1960

*So, we see that -2 will also be at 3/5; -3 will also be at 4/5; great.*1966

*Let's plot those points: -2 is at 3/5, so 3/5 of the way up...positive 3 (I should have said positive 2, as well, back there)...*1976

*positive 3 is at 4/5; -2, now, is at 3/5; -3 is at 4/5; and we will see a curve like this,*1985

*where it gets really, really close, but it will never quite touch that horizontal asymptote.*2000

*It goes through our points--a nice, smooth curve; it is really close to the horizontal asymptote, but manages to never quite touch it.*2007

*It is symmetric left and right; great.*2014

*On to our final example: this one is going to be a little bit complicated.*2016

*All right, (x ^{3} + 1)/(x^{2} - 4); and notice, there is also this negative sign--we don't want to forget about that negative sign.*2020

*So, -x ^{3} + 1, divided by x^{2} - 4...the first thing: we want to break this into factors, because that is normally our first step.*2027

*How can we factor the top? Well, x ^{3} + 1...how can we break that down?*2037

*Well, we notice that if we plug in -1, that is going to turn to 0; so we know that x + 1 has to be a factor.*2041

*x + 1...now, there is going to be some amount of x ^{2}, some amount of x, and some amount of constant.*2048

*Constant...some x...some x ^{2}...we have x^{3}, so it must be that there is only one x^{2}.*2056

*We have 1 at the end, so it must be a constant of 1 at the end, as well.*2062

*x times x ^{2} is the only x^{3} that we will get, so we want only one x^{3}.*2066

*1 times 1 is the constant that we will get, so we want to make sure that they are both 1.*2070

*So now, we need something in the middle that will cause the x ^{2}'s and the x to cancel out.*2075

*So, if we have x ^{2} here, then we need to have it be -1 here,*2079

*so that when x times x is here, it will come out as -x ^{2}, because we have positive 1x^{2}.*2086

*So, those two cancel out; so we see that x ^{3} + 1 is the same thing as (x + 1)(x^{2} - x + 1).*2091

*That is the same thing on our top; how can we factor our bottom for our rational function?*2104

*That is going to break into (x - 2)(x + 2).*2109

*Remember, we had a negative sign out front in the beginning, so we have a negative here still.*2113

*So, f(x) = -(x + 1)(x ^{2} - x + 1)/(x - 2)(x + 2).*2118

*Great; that is helpful; but there is something else that we have to do.*2126

*If we want to get to what the horizontal asymptote or slant asymptote is...let's figure out which one it is.*2129

*Cubed here; squared here...oh, so our numerator is one degree higher than our denominator; that means we have a slant asymptote.*2134

*How do we figure out a slant asymptote?--through polynomial division.*2141

*So, we want to do polynomial division on this: x ^{2} - 4 divides into...now, here is where things get a little bit tricky.*2145

*We have this negative here once again, so we can't divide into x ^{3} + 1, because then we would have to remember to deal with that negative.*2153

*So, we can make things a little bit easier on ourselves, and we can distribute that negative.*2159

*And we will get -x ^{3} - 1--that is the same thing as what was initially there--divided by x^{2} - 4.*2163

*So now, we can be safe by doing that instead.*2170

*x ^{2} - 4....-x^{3}...how many x^{2}'s do we have? 0 x^{2}'s. How many x's? 0 x's. Minus 1...*2172

*How many times does x ^{2} go into -x^{3}?*2183

*It is going to go in -x; -x times x ^{2} is -x^{3}; -x times -4 becomes positive 4x.*2184

*We subtract this whole thing, distribute the subtraction...-x ^{3} + x^{3} becomes 0; 0x - 4x becomes -4x.*2193

*We bring things down; we get 0x ^{2} - 1; so we have a remainder of -4x - 1; and what came out was -x.*2204

*So, what we have is that f(x) is also equal to (can be written in the form of) -x plus its remainder of -4x - 1, divided by x ^{2} - 4.*2220

*Great; so now we have been able to figure out what the slant asymptote is; the slant asymptote is this -x right here.*2235

*Now, we might want to check this, because it is easy to make a mistake with polynomial division, if...*2241

*well, it is just easy to make a mistake with polynomial division.*2245

*So, let's check it; let's make sure that this f is the same as the f that we started with.*2248

*So, we can put this over a common denominator, -x times x ^{2} - 4, over x^{2} - 4, plus -4x - 1, over x^{2} - 4.*2252

*We distribute up here; we have -x ^{3} - 4x; let's combine our two, since they are over a common denominator:*2265

*x ^{2} - 4 + -4x - 1...the...that was my mistake; -x times -4 becomes positive 4x, so it does cancel out.*2273

*Positive 4x and -4x cancel out; and we get -x ^{3} - 1, over x^{2} - 4,*2289

*which, as we already talked about, is the same thing as -x ^{3} + 1.*2297

*We can distribute that negative, and we get -x ^{3} - 1; so it checks out--that is good; great.*2301

*Let's see both of our two ways of looking at this.*2307

*There is the factored form that we figured, and then there is also the polynomial division form.*2309

*So, the polynomial division form is necessary, because it gives us our slant asymptote.*2316

*And the factored form is necessary, because it tells us our vertical asymptotes, and also helps us figure out some other things.*2320

*And then also, we just initially started with (just so we can still have it on our paper) (x ^{2} + 1), -(x^{2} + 1), over (x^{2} - 4).*2327

*That might be helpful, since it is not that many numbers; it might be helpful for when we actually have to calculate some extra points.*2338

*OK, what is allowed in our domain? Our domain forbids when x + 2 or x - 2 becomes 0; so that is going to happen at -2 and +2.*2344

*x is not allowed to be -2; it is not allowed to be positive 2.*2358

*Where are our vertical asymptotes? Notice that there are no common factors between the top and the bottom.*2362

*We have (x + 1) and (x ^{2} - x + 1) on the top, and (x + 2) and (x - 2); none of these things have anything in common.*2369

*They are not the same factor exactly, so we can't cancel anything out.*2375

*So, we have vertical asymptotes at where we are forbidden for our domain, -2 and positive 2.*2379

*What about horizontal asymptotes? We figured out that it wasn't a horizontal asymptote.*2384

*We figured out that it is a slant asymptote, because the degree was one higher on the top; so we will write it as a slant asymptote.*2389

*But the idea, in either case, is the same thing: what happens in the long term to this function?*2397

*That is going to be y = -x; the part that, in the long term...this part in the right, that I just circled--it goes to 0.*2401

*With very large x's, that thing will eventually get crushed down to 0; so we are left with just this thing in the box, the -x.*2411

*And that is why it is our slant asymptote.*2417

*And we might want to know the intercepts, just because they are not too hard to graph.*2420

*So, we will have intercepts: we plug in 0: -(0 + 1)/(0 - 4), so -1/-4 becomes positive 1/4.*2423

*Let's plug in some other ones; -1 and 0...we can figure that one out, because if we plug in a -1 here, the whole top goes to 0.*2439

*So, that is an x-axis intercept.*2449

*So, let's draw this thing in, because it is going to be hard to work with.*2452

*OK, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8.*2466

*OK, I am going to mark in a location of 5...3, 4, 5...just so we have a little bit of...1, 2, 3, 4, 5...reference that we can easily find our way around.*2493

*1, 2, 3, 4, 1, 2, 3, 4, 5, -5, 1, 2, 3, 4, 5, -5; great.*2502

*We have a little bit of reference there, on our thing.*2510

*Now, let's draw in our vertical asymptotes at -2 and positive 2.*2513

*And we have a slant asymptote at y = -x; so what does that look like?*2530

*It goes at a 45-degree angle; it has a slope of -1, so for every step to the right it takes, it takes a step down.*2536

*It cuts through...nice and...oops, that was not quite as nice and even as I want it to be.*2543

*OK, we see a slant asymptote like that, cutting through the whole thing.*2552

*All right, we can plot our intercepts: 0 is at 1/4, just above, and -1 is at 0.*2557

*Now, at this point, we say, "Oh, we need a lot more information."*2568

*We need to plot a lot of points, so I have brought a bunch of points; I figured them out beforehand with a calculator.*2571

*We don't have enough room to calculate it all by hand, and honestly, it would be kind of boring.*2577

*But occasionally, you will have to run through this yourself; so just be aware that, when you need more points, you just work through them.*2580

*We have -(x ^{3} + 1)/(x^{2} - 4); we can just plot in points and figure them out.*2586

*So, let's look at what happens as we get close to this vertical asymptote of -2.*2592

*We plug in -1.5, and we find out that we are at -1.35; so we can plot that point, -1.5, at -1.35.*2597

*OK, and I am just going to put down a whole bunch of points, and then we will plot them all at once.*2612

*1; we are at 0.66; 6 is repeating, so it is just 2/3.*2617

*What about -3? We are at 5.2; at -4, we are also at 5.2; that is interesting.*2625

*At -5, we are at 5.9; at positive 3, we are at -5.6; at positive 4, we are at -5.42.*2633

*At positive 5, we are at -6; OK, that should be about enough for us to figure things out.*2646

*So, at positive 1, we are at 0.66666; we are at 2/3, more accurately.*2653

*We also might want to know where we are at 1.5; I didn't do that one--oops.*2662

*But at 1.5...we could figure that out...1.5 cubed, plus 1, 3/2, cubed...that is a little difficult.*2670

*We know that is going to end up going up here, though, because we are thinking about a number that is just a little under 2,*2677

*and just a little bit under 2 is going to cause this bottom part to be negative;*2684

*and it is going to cause the top part to remain positive; so a negative is in front (remember: we can't forget about this negative in front);*2689

*negative in front; positive on top; negative on the bottom; this cancels out, and we get positive.*2695

*So, we are going up this way on this side; and when we are at -1.999, just to the right*2699

*of our vertical asymptote on the left, we know that we are going to be going down,*2706

*because we see that, at -1.5, we are going down already.*2709

*If we consider -1.99999, x ^{2} - 4 is once again going to be negative, because it is smaller.*2712

*x ^{3} + 1...-1.99999...that is going to make a number that is negative on the top, because negative cubed is larger than positive 1.*2718

*-1 point anything, cubed, is going to be larger than positive 1, so that will be a negative.*2731

*So, we have a negative on the top, a negative on the bottom, and a negative in front; it comes out to one negative left, so we are down here.*2734

*What about a little bit to the left, if we were at 2.0001?*2740

*Then, the x ^{2} - 4 will end up being a positive, because it will be just large enough to beat out that -4.*2744

*x ^{3} + 1 is still going to be negative, but we have that negative in front, so it cancels, so we will be going up on this side.*2751

*And over here at positive 2, a little bit over 2, x ^{2} - 4...if we were at 2.0001, that squared, minus 4,*2758

*is going to be larger than the -4; so it will be positive; x ^{3} + 1...it will remain positive.*2767

*We have that negative in front, so it will be going down here.*2771

*Now, we know the directions of all of our asymptotes; let's plot in the rest of our points, and we will be ready to draw this in.*2774

*1, 2, 3...5.2...5, and just a hair up; -4 is going to also be at 5.2.*2780

*My graph is a little bit high on the y = x; my graph is not quite perfect--sorry.*2797

*-5 will be at 5.9; and if we go this way, we are going to go up here.*2802

*One thing to notice is that there is something odd going on between -3 and -4.*2808

*If we were to calculate another point (we might want to try -3.5 or -3.2 to get a sense of where it is lower),*2816

*we would eventually notice that it actually dips down to its absolute lowest minimum around here.*2822

*We could figure that out precisely, if we had a calculator and a lot of time.*2827

*3...we plug in 3; we are at -5.6, so a little below...there we are.*2831

*At 4, we are at -5.42; once again, there is that interesting thing where it will curve back up.*2838

*It turns out that the minimum is somewhere between those two.*2842

*At positive 5, we are at -6; once again, my graph of the green slant asymptote isn't quite perfect.*2847

*Probably, really, my red axes aren't quite exactly the same scale.*2858

*That is the problem with drawing it all by hand, without having a ruler.*2862

*But at this point, we are finally ready to graph this thing.*2866

*We are going to get on this side of the slant asymptote.*2868

*In the middle part, where we don't have to worry about the slant asymptote, because it is too close to these verticals, we are going to be like this.*2873

*When we have the slant asymptotes, we will be pulled off this way; and then, we get pulled along the slant asymptote here.*2884

*We will get closer and closer over time.*2893

*It is pulled along the vertical asymptote, and then it dips, and then it gets pulled along the slant asymptote.*2895

*It wouldn't curve away there; that is just my imperfections; there we go.*2902

*And then, it gets closer and closer to that slant asymptote the whole time.*2908

*It is pretty tough to draw something this complex; but this is absolutely as complex as you are going to end up seeing*2911

*in a class or have any homework or tests that have to do with.*2917

*So, at the worst case, you just end up having to take a lot of points and plot a bunch of points, and you can figure out how this thing behaves.*2920

*You can figure out where the asymptotes are; you can figure the slant asymptote, horizontal asymptote, vertical asymptote...*2926

*all of that business...your domain...you can figure out all of these things.*2931

*And when it comes time to graph, you just have to punch out a lot of numbers, so you can actually see what the picture looks like.*2935

*But all in all, it is not that hard, if you just make enough numbers.*2940

*All right, I hope everything there made sense; and we will see you at Educator.com later--goodbye!*2943

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