For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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 stepbystep 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" xvalues in step #2 (domain).]
4. Once the function is simplified, we can find the vertical asymptotes. The vertical asymptotes occur at all the xvalues 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 yintercepts 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 xintercepts or vertical asymptotes. This means we can put our xintercepts and vertical asymptote locations in order and break the xaxis 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
 A vertical asymptote is a horizontal location that, as approached, causes the function to fly off vertically to ±∞.
 We see that as the graph nears x=4, the function flies off vertically. Thus, x=4 is a vertical asymptote.
 A horizontal asymptote is the vertical height that the function tends towards as x goes goes very far to the right or left. We can think of it as a height that the function is "pulled" towards over the longrun.
 We see from the graph that as x becomes very large (far to the right or left), the graph approaches the vertical height of y=0.
 A vertical asymptote is a horizontal location that, as approached, causes the function to fly off vertically to ±∞.
 We see that the graph flies off vertically in two locations: as the graph nears x=−2 and as it nears x=2. Thus, there are two vertical asymptotes: x=−2, 2.
 A horizontal asymptote is the vertical height that the function tends towards as x goes goes very far to the right or left. We can think of it as a height that the function is "pulled" towards over the longrun.
 We see from the graph that as x becomes very large (far to the right or left), the graph approaches the vertical height of y=−4.
 A vertical asymptote is a horizontal location that, as approached, causes the function to fly off vertically to ±∞.
 We see that as the graph nears x=0, the function flies off vertically. Thus, x=0 is a vertical asymptote.
 A horizontal asymptote is the vertical height that the function tends towards as x goes goes very far to the right or left. We can think of it as a height that the function is "pulled" towards over the longrun.
 We see from the graph that no such height exists. As the graph goes far to the right or left, it does not "pull" to a specific height. This graph continues to go up and down forever. However, we do notice that it "pulls" along a very specific line. This is a slant asymptote.
 Looking at the dashed line (indicating the slant asymptote), we can figure out an equation for it. For any interval, the amount it goes right equals the amount it goes down, so the line has a slope of m=−1. It has a vertical intercept of b=0. Putting these observations in the slopeintercept line equation y=mx+b, we find that the slant asymptote is y=−x.
 Remember, a rational function is something in the form of a fraction where the top and bottom are both polynomials. Let's figure out what each asymptote requires of the function.
 To have a vertical asymptote at x=7, the denominator needs to go to 0 at x=7. Thus, we must have a factor of (x−7) in the denominator [and we need to make sure that factor doesn't get canceled out by something in the numerator].
 To have a horizontal asymptote of y=−8, the numerator and denominator polynomials must have the same degree. Once they have the same degree, the ratio of their coefficients must be −[8/1].
 So far, we've figured out the function as
Thus, we will make the numerator have a degree of 1 as well (because the denominator already has degree 1). The easiest thing would be to just use x. However, we must also have a coefficient in front of it to get the horizontal asymptote of y=−8. Since the denominator has a coefficient of 1, we can use −8 for the numerator's coefficient. Thus, we'll have −8x as the numerator.f(x) = ? x−7  Putting these together, we get
[Note: there are other ways to write out a rational function with the given two qualities. This is just one of the easiest ways to write it. Here are some other possible rational functions that would all have the given vertical and horizontal asymptotes:f(x) = −8x x−7.
(Some of the above have additional asymptotes or "holes" in them or both, but they all have the asymptotes required by the question and so could be considered correct as well.)]−16x 2x−14−8x(x^{3}+1) (x−7)(x^{3}+1)−8x^{3}+5x^{2}−18 (x−7)(x+2)(x−15)
 Remember, a rational function is something in the form of a fraction where the top and bottom are both polynomials. Let's figure out what each asymptote requires of the function.
 To have a vertical asymptote, the denominator needs to go to 0 at the given horizontal location. Since we have three vertical asymptotes (x=−4, −3, 4), we need three factors that will each go to 0 at each respective location. Thus, we must have factors of (x+4), (x+3), and (x−4) in the denominator [and we need to make sure that those factors are not canceled out by something in the numerator].
 To have a horizontal asymptote of y=[1/2], the numerator and denominator polynomials must have the same degree. Once they have the same degree, the ratio of their coefficients must be [1/2].
 So far, we've figured out the function as
Thus, we will make the numerator have a degree of 3 as well (because the denominator already has degree 3 [if you don't see it, expand the bottom]). The easiest thing would be to just use x^{3}. However, we must also have a coefficient in front of it to get the horizontal asymptote of y=[1/2]. Since the denominator has a coefficient of 1, we can use [1/2] for the numerator's coefficient. Thus, we'll have [1/2]x^{3} as the numerator.f(x) = ? (x+4)(x+3)(x−4)  Putting these together, we get
[Note: there are other ways to write out a rational function with the given asymptotes. This is just one of the easiest ways to write it. Here are some other possible rational functions that would all have the given vertical and horizontal asymptotes:f(x) = 1 2x^{3} (x+4)(x+3)(x−4).
(Some of the above have additional asymptotes or "holes" in them or both, but they all have the asymptotes required by the question and so could be considered correct as well.)]x^{3} 2(x+4)(x+3)(x−4)4x^{4} 8(x+4)(x+3)(x−4)(x+47)1 2x^{3}(x−4) (x+4)(x+3)(x−4)^{2}

 Begin by factoring the numerator and denominator [but don't simplify yet]. In this case, it's already done, so we can move right along to the next step.
 Find the domain. The function will "break" when you divide by 0, so that occurs at the zeros of our denominator.
Thus, we have a "forbidden" location of x=3, so the domain is x ≠ 3.x−3=0  Simplify the function. Once again, quite easy because there's nothing to cancel out.
 The vertical asymptotes are wherever the denominator still has zeros (after simplifying). Since we couldn't simplify it anymore, the vertical asymptote is the same as the "forbidden" location: x=3.
 Find the horizontal asymptote. Compare the degree of the numerator to the degree of the denominator. In this case, the numerator has lower degree, so the horizontal asymptote is y=0.
 Find points to plot on the graph. It can be useful to find any xintercepts and the yintercept in addition to finding other points as needed to draw the graph.
x f(x) 0 −0.67 2 −2 2.5 −4 3.5 4 4 2 6 0.67  Set up the graph axes, draw in the asymptotes with dashed lines, plot points, and connect with curves. Remember, as the graph gets close to an asymptote, it "pulls" alongside.

 Begin by factoring the numerator and denominator [but don't simplify yet]. The numerator is already factored, and we see that we can factor the denominator into (x+5)(x−2).
 Find the domain. The function will "break" when you divide by 0, so that occurs at the zeros of our denominator.
Thus, we have "forbidden" locations at x=−5, 2, so the domain is x ≠ −5, 2.(x+5)(x−2)=0  Simplify the function.
g(x) = x+5 x^{2}+3x−10= x+5 (x+5)(x−2)⇒ 1 x−2  The vertical asymptotes are wherever the denominator still has zeros (after simplifying). Since our new, simplified denominator is just x−2, we only have a vertical asymptote at x=2. [Notice that we still have to care about the other "forbidden" value of x=−5 by making a "hole" in the graph at the end.]
 Find the horizontal asymptote. Compare the degree of the numerator to the degree of the denominator. In this case, the numerator has lower degree, so the horizontal asymptote is y=0.
 Find points to plot on the graph. It can be useful to find any xintercepts and the yintercept in addition to finding other points as needed to draw the graph. [Also, we can use the simplified version to find where the "hole" at x=−5 should go. It doesn't actually exist there, but we can graph it as if it did, then just put a "hole" in that spot.]
x g(x) −5 −0.14 / DNE 0 −0.5 1 −1 1.5 −2 2.5 2 3 1 4 0.5 5 0.67  Set up the graph axes, draw in the asymptotes with dashed lines, plot points, and connect with curves. Remember, as the graph gets close to an asymptote, it "pulls" alongside. Also, don't forget to put a "hole" at x=−5 because the rational function technically does not exist there. We do this with an empty circle.

 Begin by factoring the numerator and denominator [but don't simplify yet]. We can factor as follows:
Numerator: 3x^{2}+3x−36 = 3(x^{2} + x − 12) = 3(x+4)(x−3) Denominator: x^{2} −x −6 = (x+2)(x−3)  Find the domain. The function will "break" when you divide by 0, so that occurs at the zeros of our denominator.
Thus, we have "forbidden" locations at x=−2, 3, so the domain is x ≠ −2, 3.(x+2)(x−3)=0  Simplify the function.
h(x) = 3x^{2}+3x−36 x^{2}−x−6= 3(x+4)(x−3) (x+2)(x−3)⇒ 3(x+4) x+2  The vertical asymptotes are wherever the denominator still has zeros (after simplifying). Since our new, simplified denominator is just x+2, we only have a vertical asymptote at x=−2. [Notice that we still have to care about the other "forbidden" value of x=3 by making a "hole" in the graph at the end.]
 Find the horizontal asymptote. Compare the degree of the numerator to the degree of the denominator. In this case, the numerator and denominator have equal degrees, so we make a fraction from their leading coefficients. This gives a horizontal asymptote of y=[3/1] = 3.
 Find points to plot on the graph. It can be useful to find any xintercepts and the yintercept in addition to finding other points as needed to draw the graph. [Also, we can use the simplified version to find where the "hole" at x=3 should go. It doesn't actually exist there, but we can graph it as if it did, then just put a "hole" in that spot.]
x h(x) −6 1.5 −4 0 −3 −3 −2.5 −9 −1.5 15 −1 9 0 6 3 4.2 / DNE  Set up the graph axes, draw in the asymptotes with dashed lines, plot points, and connect with curves. Remember, as the graph gets close to an asymptote, it "pulls" alongside. Also, don't forget to put a "hole" at x=3 because the rational function technically does not exist there. We do this with an empty circle.

 Begin by factoring the numerator and denominator [but don't simplify yet]. We can factor as follows:
Numerator: −2x^{3}−2x = −2x(x^{2} + 1) Denominator: x^{3}−3x^{2}−4x = x(x^{2}−3x−4) = x(x+1)(x−4)  Find the domain. The function will "break" when you divide by 0, so that occurs at the zeros of our denominator.
Thus, we have "forbidden" locations at x=−1, 0, 4, so the domain is x ≠ −1, 0, 4.x(x+1)(x−4)=0  Simplify the function.
f(x) = −2x^{3}−2x x^{3}−3x^{2}−4x= −2x(x^{2} + 1) x(x+1)(x−4)⇒ −2(x^{2}+1) (x+1)(x−4)  The vertical asymptotes are wherever the denominator still has zeros (after simplifying). Since our new, simplified denominator is (x+1)(x−4), we have vertical asymptotes at x=−1 and x=4. [Notice that we still have to care about the other "forbidden" value of x=0 by making a "hole" in the graph at the end.]
 Find the horizontal asymptote. Compare the degree of the numerator to the degree of the denominator. In this case, the numerator and denominator have equal degrees, so we make a fraction from their leading coefficients. This gives a horizontal asymptote of y=[(−2)/1] = −2.
 Find points to plot on the graph. It can be useful to find any xintercepts and the yintercept in addition to finding other points as needed to draw the graph. [Also, we can use the simplified version to find where the "hole" at x=0 should go. It doesn't actually exist there, but we can graph it as if it did, then just put a "hole" in that spot.]
x f(x) −7 −1.52 −4 −1.42 −3 −1.43 −2 −1.67 −1.5 −2.36 −0.5 1.11 0 0.5 / DNE 1 0.67 2 1.67 3 5 3.5 11.78 4.5 −15.45 5 −8.67 8 −3.61  Set up the graph axes, draw in the asymptotes with dashed lines, plot points, and connect with curves. Remember, as the graph gets close to an asymptote, it "pulls" alongside. Also, don't forget to put a "hole" at x=0 because the rational function technically does not exist there. We do this with an empty circle.

 Normally we would begin by factoring the numerator and denominator. However, for this problem, it would be difficult to factor the numerator (give it a try if you don't believe me). It's alright though: the only reason we care about factoring the numerator is to see if it has any common factors with the denominator. We see that the denominator breaks down to 3(x−1). Furthermore, while it's difficult to factor the numerator, we can pretty easily see that (x−1) can't be a factor in 6x^{2}−12x+7 (the numerator), so we're safe from having to worry about canceling factors later on. Thus, we can just leave the numerator unfactored (which will wind up being useful later on).
 Find the domain. The function will "break" when you divide by 0, so that occurs at the zeros of our denominator.
Thus, we have a "forbidden" location at x=1, so the domain is x ≠ 1.3(x−1)=0  Because we can't cancel out any factors between the numerator and denominator, the "forbidden" location of x=1 is also a vertical asymptote.
 Notice that the function does not have a horizontal asymptote. This is because the numerator's degree (2) is higher than denominator's degree (1). Thus, no horizontal asymptote.
However, because it is only one number higher, it will have a slant asymptote. We find the slant asymptote using polynomial long division:
3x −3 ⎞
⎠6x^{2} −12x +7 
Thus, through long division, we have shown that2x −2 R: 1 3x −3 ⎞
⎠6x^{2} −12x +7 6x^{2} −6x −6x +7 −6x +6 1
which means we have a slant asymptote of y = 2x−2.g(x) = 6x^{2}−12x+7 3x−3= 2x −2 + 1 3x−3,  Find points to plot on the graph. It can be useful to find any xintercepts and the yintercept in addition to finding other points as needed to draw the graph.
x g(x) −4 −10.07 −2 −6.11 −1 −4.17 −0.5 −3.22 0 −2.33 0.5 −1.67 1.5 1.67 2 2.33 2.5 3.22 3 4.17 5 8.08  Set up the graph axes, draw in the asymptotes with dashed lines, plot points, and connect with curves. Remember, as the graph gets close to an asymptote, it "pulls" alongside.
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Graphing Asymptotes in a Nutshell
Lecture Slides are screencaptured 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
Precalculus with Limits Online Course
Transcription: Graphing Asymptotes in a Nutshell
Hiwelcome 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 program0024
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 working0038
and what your denominator is and what your numerator is will build a really great, intuitive understanding0042
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 eye0051
in a way that you just can't develop by trying to do ten problems in hand, because you can just do 100 of them0055
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 functionjust if you want to look at a rational function0098
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 polynomialsthis 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 anythingfactoring the numerator and the denominator doesn't immediately tell us anything directly0124
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 reformat 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 0it is one of the critical ideas here.0177
So, we want to find all of the xvalues 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 functiononce we have found the function's domain, we can simplify the function0240
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 xvalues.0264
There is a forbidden xvalue 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 asymptotesonce the function is simplified, we can find the vertical asymptotes.0300
The vertical asymptotes will occur at all the xvalues0305
that cause the denominator, after we have simplified the denominator, to become 00307
so, all of the zeroes of our nownewlysimplified 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 mthe degree of the numerator is less than the degree of the denominator.0339
Our denominator grows faster than our numeratorthat 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 mthe degree of the numerator is equal to the degree of the denominator0360
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 asymptotewe 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 denominator0425
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 interceptsby finding the x and yintercepts, 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 yintercept, x = 0, could be a forbidden location.0468
And all of the places where we cross overit could either not cross the xaxis at all,0477
or the locations where it would cross the xaxis 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 xintercepts 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 xaxis; we are an xintercept 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 yintercept is where the function's input is 0, because that will put us on the yaxis.0528
So, we plug in the xvalue...0533
I'm sorry; I might have said the wrong thing there; I am not quite sure what I said.0535
The yintercept occurs where the function's input, its xvalue, the x that we are plugging in, is 0,0537
because that will put us right on the yaxis.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 graphnow 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 interceptsreally, there are only a couple of things that come out of the intercepts), evaluate the function0575
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 sense0601
for how the thing is going to come together as a picturethat 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 circle0644
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 asymptoteswhich we would never get to anyway0658
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 xvalues0674
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 positivethat 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 stay0919
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 xintercepts 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 negative0947
or vertical asymptotes, because after a vertical asymptote, it does jump.0951
This means we can put our xintercepts and our vertical asymptote locations in order, and then break the xaxis into intervals.0955
In each of these intervals, we can test just one point, because we know we can't jump0961
until we hit an xintercept or we hit a vertical asymptotewe 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 negativeness 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 xintercept 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 thingfrom 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 infinitylet's try out 3; 3 is in therewe 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 asymptotewhat 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 yintercept?...yes, we have a yinterceptthat is what we just figured out.1169
Do we have any xintercepts?...no, because the numerator is always 3.1173
The numerator never goes to 0, so there are not going to be any xintercepts.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 xaxis.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 asymptotelet'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 pointsee 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 arethe 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 xaxis, 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 seelook, 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 get1509
that will be xaxis 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 yaxis 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 onewe 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 xaxis.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, onefifth.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 onetenth.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 perfectsorry; 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 xaxis.1814
So, now we compare, and we seewhat 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 yintercept?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
Interceptswhere 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 3plug 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 pointsa 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 signwe 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}  1that is the same thing as what was initially theredivided 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 outthat 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 circledit 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 xaxis 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 45degree 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 oneoops.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 right2699
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 perfectsorry.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 seeing2911
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 latergoodbye!2943
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