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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

From the graph below, name the type and location of all the asymptotes.
  • 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 long-run.
  • 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.
Vertical Asymptote: x=4,       Horizontal Asymptote: y=0
From the graph below, name the type and location of all the asymptotes.
  • 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 long-run.
  • 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.
Vertical Asymptotes: x=−2,  2,       Horizontal Asymptote: y=−4
From the graph below, name the type and location of all the asymptotes. [Note: a dashed line has been put in for one of the asymptotes to assist in identifying it.]
  • 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 long-run.
  • 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 slope-intercept line equation y=mx+b, we find that the slant asymptote is y=−x.
Vertical Asymptote: x=0,       Slant Asymptote: y=−x
Write out a rational function that has the following qualities. Vertical asymptote: x=7,    Horizontal asymptote: y=−8
  • 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
    f(x) = ?

    x−7
    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.
  • Putting these together, we get
    f(x) = −8x

    x−7
    .
    [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:
    −16x

    2x−14
           −8x(x3+1)

    (x−7)(x3+1)
           −8x3+5x2−18

    (x−7)(x+2)(x−15)
    (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.)]
f(x) = [(−8x)/(x−7)] [Note: There are other possible answers for this question, the above is just one of the easiest answers to write out. Look at the final step for some other possible answers if you're curious.]
Write out a rational function that has the following qualities. Vertical asymptotes: x=−4, −3, 4,    Horizontal asymptote: y=[1/2]
  • 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
    f(x) = ?

    (x+4)(x+3)(x−4)
    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 x3. 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]x3 as the numerator.
  • Putting these together, we get
    f(x) =
    1

    2
    x3

    (x+4)(x+3)(x−4)
    .
    [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:
    x3

    2(x+4)(x+3)(x−4)
           4x4

    8(x+4)(x+3)(x−4)(x+47)
          
    1

    2
    x3(x−4)

    (x+4)(x+3)(x−4)2
    (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.)]
f(x) = [([1/2]x3)/((x+4)(x+3)(x−4))]    [equivalently expanded: [([1/2]x3)/(x3+3x2−16x−48)]  ] [Note: There are other possible answers for this question, the above is just one of the easiest answers to write out. Look at the final step for other possible answers if you're curious.]
Graph the rational function below.
f(x) = 2

x−3
  • 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.
    x−3=0
    Thus, we have a "forbidden" location of x=3, so the domain is x ≠ 3.
  • 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 x-intercepts and the y-intercept 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.
Graph the rational function below.
g(x) = x+5

x2+3x−10
  • 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.
    (x+5)(x−2)=0
    Thus, we have "forbidden" locations at x=−5, 2, so the domain is x ≠ −5, 2.
  • Simplify the function.
    g(x) = x+5

    x2+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 x-intercepts and the y-intercept 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.
Graph the rational function below.
h(x) = 3x2+3x−36

x2−x−6
  • Begin by factoring the numerator and denominator [but don't simplify yet]. We can factor as follows:
    Numerator: 3x2+3x−36    =     3(x2 + x − 12)     =     3(x+4)(x−3)

    Denominator: x2 −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.
    (x+2)(x−3)=0
    Thus, we have "forbidden" locations at x=−2, 3, so the domain is x ≠ −2, 3.
  • Simplify the function.
    h(x) = 3x2+3x−36

    x2−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 x-intercepts and the y-intercept 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.
Graph the rational function below.
f(x) = −2x3−2x

x3−3x2−4x
  • Begin by factoring the numerator and denominator [but don't simplify yet]. We can factor as follows:
    Numerator: −2x3−2x    =     −2x(x2 + 1)

    Denominator: x3−3x2−4x     =     x(x2−3x−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.
    x(x+1)(x−4)=0
    Thus, we have "forbidden" locations at x=−1, 0, 4, so the domain is x ≠ −1, 0, 4.
  • Simplify the function.
    f(x) = −2x3−2x

    x3−3x2−4x
        =     −2x(x2 + 1)

    x(x+1)(x−4)
        ⇒     −2(x2+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 x-intercepts and the y-intercept 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.
Graph the rational function below.
g(x) = 6x2−12x+7

3x−3
  • 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 6x2−12x+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.
    3(x−1)=0
    Thus, we have a "forbidden" location at x=1, so the domain is x ≠ 1.
  • 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

    6x2
    −12x
    +7

  • 2x
    −2
    R:  1
    3x
    −3

    6x2
    −12x
    +7
    6x2
    −6x
    −6x
    +7
    −6x
    +6
    1
    Thus, through long division, we have shown that
    g(x) = 6x2−12x+7

    3x−3
        =     2x −2 + 1

    3x−3
    ,
    which means we have a slant asymptote of y = 2x−2.
  • Find points to plot on the graph. It can be useful to find any x-intercepts and the y-intercept 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 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 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

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 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 function--just 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 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) = (x2 - 3x + 2)/(2x2 - 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 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 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-values0305

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 5x5 divided by 2x5, 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 x5'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 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 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 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 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 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 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-values0674

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

f(x) = (x2 - 3x + 2)/(2x2 - 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 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 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 jump0961

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)/(x3 + x2 + 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)(x2 + 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 x2 + 1 provide anything?1424

No, x2 + 1 is an irreducible quadratic; there are no roots there, because x2 + 1 = 0 is always going to remain positive.1428

You can't solve x2 + 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/(x2 + 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/(x2 + 1), there is no x we can plug in to get 0 to show up in the bottom.1464

x2 + 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/(x2 + 1), there are no intercepts that we can get1509

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/(12 + 1) gets us 1/2; OK, (1,1/2); here we are.1594

We plug in positive 2: 1 over 22, 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 32, 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 x2, 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) = (x2 - 1)/(x2 + 1).1743

Let's factor the top into its factors: we would get (x + 1)(x - 1); and it is divided by (x2 + 1); great.1748

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

No, everything is allowed, because with x2 + 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); (x2 + 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 x2 + 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 22 - 1,1933

so that is going to be 4 - 1, or 3, divided by 22 + 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 x2 up here.1956

We have x2 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, (x3 + 1)/(x2 - 4); and notice, there is also this negative sign--we don't want to forget about that negative sign.2020

So, -x3 + 1, divided by x2 - 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, x3 + 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 x2, some amount of x, and some amount of constant.2048

Constant...some x...some x2...we have x3, so it must be that there is only one x2.2056

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

x times x2 is the only x3 that we will get, so we want only one x3.2066

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 x2's and the x to cancel out.2075

So, if we have x2 here, then we need to have it be -1 here,2079

so that when x times x is here, it will come out as -x2, because we have positive 1x2.2086

So, those two cancel out; so we see that x3 + 1 is the same thing as (x + 1)(x2 - 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)(x2 - 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: x2 - 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 x3 + 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 -x3 - 1--that is the same thing as what was initially there--divided by x2 - 4.2163

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

x2 - 4....-x3...how many x2's do we have? 0 x2's. How many x's? 0 x's. Minus 1...2172

How many times does x2 go into -x3?2183

It is going to go in -x; -x times x2 is -x3; -x times -4 becomes positive 4x.2184

We subtract this whole thing, distribute the subtraction...-x3 + x3 becomes 0; 0x - 4x becomes -4x.2193

We bring things down; we get 0x2 - 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 x2 - 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 x2 - 4, over x2 - 4, plus -4x - 1, over x2 - 4.2252

We distribute up here; we have -x3 - 4x; let's combine our two, since they are over a common denominator:2265

x2 - 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 -x3 - 1, over x2 - 4,2289

which, as we already talked about, is the same thing as -x3 + 1.2297

We can distribute that negative, and we get -x3 - 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) (x2 + 1), -(x2 + 1), over (x2 - 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 (x2 - 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 -(x3 + 1)/(x2 - 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 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, x2 - 4 is once again going to be negative, because it is smaller.2712

x3 + 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 x2 - 4 will end up being a positive, because it will be just large enough to beat out that -4.2744

x3 + 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, x2 - 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; x3 + 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 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 later--goodbye!2943