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Lecture Comments (2)

1 answer

Last reply by: Professor Selhorst-Jones
Mon Sep 30, 2013 9:05 AM

Post by Mike Olyer on September 29, 2013

Why is 1 not a vertical asymptote for the function in example one? Why does canceling out allow this?

Rational Functions and Vertical Asymptotes

  • A rational function is the quotient of two polynomials:
    f(x) = N(x)

    D(x)


     
    ,
    where N(x) and D(x) are polynomials and D(x) ≠ 0.
  • Since a rational function f(x) = [N(x)/D(x)] is inherently built out of the operation of division, we must watch out for the possibility of dividing by zero. The domain of a rational function is all real numbers except the zeros of D(x).
  • As the denominator of a rational function goes to 0 (and assuming the numerator is not also 0), the fraction becomes very large. While it can't actually divide by 0, as it gets extremely close to 0, the function "blows out" to very large values. We call this location a vertical asymptote. A vertical asymptote is a vertical line x = a where as x gets close to a, |f(x)| becomes arbitrarily large. Symbolically, we show this as
    x → a     ⇒     f(x) → ∞   or   f(x) → − ∞.
  • On a graph, we show the location of a vertical asymptote with a dashed line. This aids us in drawing the graph and in understanding the graph later.
  • To find the vertical asymptotes of a rational function, we need to find the x-values where the denominator becomes 0 (the roots of the denominator function). However, not all of these zeros will give asymptotes. It's possible for the numerator to go to 0 at the same time, which will cause the function to just have a hole at that x-value, but not "blow out" to infinity.
  • We can find the vertical asymptotes of a rational function by following these steps:

      1. Begin by figuring out what x-values are not in the domain of f: these are all the zeros of D(x).
      2. Determine if N(x) and D(x) share any common factors. If so, cancel out those factors. [Alternately, this step can be done by checking that the zeros to D(x) are not also zeros to N(x).]
      3. After canceling out common factors, determine any zeros (roots) left in the denominator. These are the vertical asymptotes.
      (4.) If graphing the function, figure out which way the graph goes (+∞ or −∞) on either side of each asymptote. Do this by using test values very close to the asymptote. For example, if the asymptote is at x=2, look at f(1.99) and f(2.01). [You can also do this in your head by thinking in terms of positive vs. negative, which we discuss in the Examples.]

Rational Functions and Vertical Asymptotes

Find the domain of the function below, then find any vertical asymptote(s).
f(x) = 7

x−5
  • The domain of a function is the set of all values that the function can accept. The only thing that could "break" a function like this would be dividing by 0. That means we need to figure out where the denominator can equal 0.
  • Set the denominator equal to 0:
    x−5=0
    We see that the only place where the function "breaks" (is not defined) is at x=5. Thus, x=5 is the only x-value not allowed. Therefore any x value other than 5 is in the domain.
    Domain:    x ≠ 5
  • A vertical asymptote is some specific x-value that, as x approaches it, the function grows very large (positive or negative). This happens when the numerator is non-zero and the denominator is getting very close to 0.
  • To figure out where this occurs, begin by seeing if there are any common factors that can be canceled in the numerator and denominator:
    7

    x−5
    The above cannot be simplified any more, so there are no common factors. [This means we don't have to worry about the numerator also being 0 when the denominator goes to 0.]
  • Once any common factors have been canceled (or there aren't any, such as in this problem), find what zeros (roots) are left in the denominator:
    7

    x−5
    The above denominator still only has a zero at x=5, as we figured out before. Thus, x=5 is a vertical asymptote.
Domain:    x ≠ 5;        Vertical asymptote at x=5
Find the domain of the function below, then find any vertical asymptote(s).
g(x) = 17x

3x+9
  • The domain of a function is the set of all values that the function can accept. The only thing that could "break" a function like this would be dividing by 0. That means we need to figure out where the denominator can equal 0.
  • Set the denominator equal to 0:
    3x+9=0
    We see that the only place where the function "breaks" (is not defined) is at x=−3. Thus, x=−3 is the only x-value not allowed. Therefore any x value other than −3 is in the domain.
    Domain:    x ≠ −3
  • A vertical asymptote is some specific x-value that, as x approaches it, the function grows very large (positive or negative). This happens when the numerator is non-zero and the denominator is getting very close to 0.
  • To figure out where this occurs, begin by seeing if there are any common factors that can be canceled in the numerator and denominator:
    17x

    3x+9
    The above cannot be simplified any more, so there are no common factors. [This means we don't have to worry about the numerator also being 0 when the denominator goes to 0.]
  • Once any common factors have been canceled (or there aren't any, such as in this problem), find what zeros (roots) are left in the denominator:
    17x

    3x+9
    The above denominator still only has a zero at x=−3, as we figured out before. Thus, x=−3 is a vertical asymptote.
Domain:    x ≠ −3;        Vertical asymptote at x=−3
Find the domain of the function below, then find any vertical asymptote(s).
h(x) = 3

x2−2x−24
  • The domain of a function is the set of all values that the function can accept. The only thing that could "break" a function like this would be dividing by 0. That means we need to figure out where the denominator can equal 0.
  • Set the denominator equal to 0:
    x2−2x−24=0
    We can find the solutions to this equation by factoring.
  • (x+4)(x−6)=0, so the function "breaks" (is not defined) at x=−4 and x=6. These are the only two x-values not allowed. Therefore any x value other than them is in the domain.
    Domain:    x ≠ −4,  6
  • A vertical asymptote is some specific x-value that, as x approaches it, the function grows very large (positive or negative). This happens when the numerator is non-zero and the denominator is getting very close to 0.
  • To figure out where this occurs, begin by seeing if there are any common factors that can be canceled in the numerator and denominator:
    3

    x2−2x−24
    = 3

    (x+4)(x−6)
    We see there are no common factors, so we can continue. [This means we don't have to worry about the numerator also being 0 when the denominator goes to 0.]
  • Once any common factors have been canceled (or there aren't any, such as in this problem), find what zeros (roots) are left in the denominator:
    3

    (x+4)(x−6)
    The above denominator still has zeros at x=−4,  6, as we figured out before. Thus, x=−4 and x=6 are the vertical asymptotes.
Domain:    x ≠ −4,  6;        Vertical asymptotes at x=−4 and x=6
Find the domain of the function below, then find any vertical asymptote(s).
v(x) = 2x+14

x3+5x2−14x
  • The domain of a function is the set of all values that the function can accept. The only thing that could "break" a function like this would be dividing by 0. That means we need to figure out where the denominator can equal 0.
  • Set the denominator equal to 0:
    x3+5x2−14x=0
    We can find the solutions to this equation by factoring.
  • x(x−2)(x+7)=0, so the function "breaks" (is not defined) at x=0, x=2, and x=−7. These are all the x-values not allowed. Therefore any x value other than them is in the domain.
    Domain:    x ≠ −7,  0,   2
  • A vertical asymptote is some specific x-value that, as x approaches it, the function grows very large (positive or negative). This happens when the numerator is non-zero and the denominator is getting very close to 0.
  • To figure out where this occurs, begin by seeing if there are any common factors that can be canceled in the numerator and denominator:
    2x+14

    x3+5x2−14x
    = 2(x+7)

    x(x−2)(x+7)
    We see that there is a common factor: (x+7), so we cancel this factor out before trying to find the vertical asymptote(s).
    2

    x(x−2)
    [Note that the (x+7) factor still affects the domain, as we figured out before. It just cannot contribute to the creation of vertical asymptotes because both the top and bottom of the fraction produce 0 at the x-location of −7.]
  • Once any common factors have been canceled, find what zeros (roots) are left in the denominator:
    2

    x(x−2)
    The above denominator only has zeros at x=0 and x=2. These are the vertical asymptotes.
Domain:    x ≠ −7,  0,   2;        Vertical asymptotes at x=0 and x=2
From the graph below, identify the location of its vertical asymptote.
  • A vertical asymptote is some specific x-value that, as x approaches it, the function grows very large (positive or negative).
  • Look on the graph for a horizontal location where the graph shoots toward ±∞ vertically.
  • We see that happen at x=2, so that is the vertical asymptote.
x=2
From the graph below, identify the location of its vertical asymptotes.
  • A vertical asymptote is some specific x-value that, as x approaches it, the function grows very large (positive or negative).
  • Look on the graph for the horizontal locations where the graph shoots toward ±∞ vertically.
  • We see that happen at x=−4 and x=6, so those are the vertical asymptotes.
x=−4 and x=6
Draw the graph of the rational function below.
f(x) = 1

x+3
  • Just like we did in the previous questions, begin by identifying the function's domain and the vertical asymptotes. They'll help when we need to draw the graph.
  • Because the fraction is as simplified as possible, the "holes" in the domain are the same as the location of the vertical asymptote:
    x+3 = 0
    The function has a vertical asymptote of x=−3. [It also has a domain of x ≠ −3, but that won't really affect the graph we draw because we already have a vertical asymptote there. This is because the vertical asymptote can't have a point on it (otherwise the function could not have flown off to ±∞ on either side).]
  • Draw graph axes and draw a dashed vertical line at x=−3 to indicate the asymptote is there.
  • At this point, it would help to have more points to get a sense of how the function curves. Make a table of values to help plot some points. Notice that the most "interesting" thing for this function is the asymptote at x=−3 so we want to focus our table around that.
    x
    f(x)
    −9
    −0.166
    −5
    −0.5
    −4
    −1
    −3.5
    −2
    −2.5
    2
    −2
    1
    −1
    0.5
    3
    0.166
    Plot these points on the graph. [As you become more comfortable with graphing functions like this, you will need fewer points because you'll have a better idea of the shape of the function. For now though, plot however many you need to be comfortable with graphing it.]
  • As x → −3, the graph will fly off to ±∞. We need to figure out which direction it will fly off, though. From our table of points, it seems that on the left side, it flies off to − ∞, while on the right side it goes to + ∞. We might want to be sure, though, so we could also figure out where f(−3.01) and f(−2.99) would go. Alternatively, we could also just think about whether or not the function will go up or down based on if the x-value is slightly more or less than x=−3. Once you know which direction it goes, draw a curve that approaches the dashed line, but does not actually cross it. As it gets closer, it gets lower/higher.
  • Finally, think about what happens to the function as x goes very far to the right or left. Because the numerator never changes, a very large x (+ or −) will cause the function output to become very small. Thus, as x goes very far to the right or left, it will get pulled to a very small height. Combine all of these ideas, and draw in all the curves based off your plotted points.
Draw the graph of the rational function below.
g(x) = 15

24−6x
  • Just like we did in the previous questions, begin by identifying the function's domain and the vertical asymptotes. They'll help when we need to draw the graph.
  • First, figure out the function's domain by setting the denominator to 0:
    24−6x = 0
    The function will "break" at x=4, so the domain is x ≠ 4.
  • Next find the vertical asymptotes. Begin by canceling any common terms on the top and bottom:
    15

    24−6x
        =     3·5

    3(8−2x)
        =     5

    8−2x
    Using the new, simplified version, find the location of any vertical asymptotes by setting the denominator to 0:
    8 − 2x = 0
    The function has a vertical asymptote at x=4. [Notice that this is basically what we figured out when we were looking for the domain. While we canceled out a common factor, nothing changed because the common factor was just a number. Still, we want to make sure to check again after canceling factors because it can change what we find for the asymptotes.]
  • Draw graph axes and draw a dashed vertical line at x=4 to indicate the asymptote is there.
  • At this point, it would help to have more points to get a sense of how the function curves. Make a table of values to help plot some points. Notice that the most "interesting" thing for this function is the asymptote at x=4 so we want to focus our table around that.
    x
    f(x)
    −1
    0.5
    2
    1.25
    3
    2.5
    3.5
    5
    4.5
    −5
    5
    −2.5
    6
    −1.25
    9
    −0.5
    Plot these points on the graph. [As you become more comfortable with graphing functions like this, you will need fewer points because you'll have a better idea of the shape of the function. For now though, plot however many you need to be comfortable with graphing it.]
  • As x → 4, the graph will fly off to ±∞. We need to figure out which direction it will fly off, though. From our table of points, it seems that on the left side, it flies off to + ∞, while on the right side it goes to − ∞. We might want to be sure, though, so we could also figure out where f(3.99) and f(4.01) would go. Alternatively, we could also just think about whether or not the function will go up or down based on if the x-value is slightly more or less than x=4. Once you know which direction it goes, draw a curve that approaches the dashed line, but does not actually cross it. As it gets closer, it gets lower/higher.
  • Finally, think about what happens to the function as x goes very far to the right or left. Because the numerator never changes, a very large x (+ or −) will cause the function output to become very small. Thus, as x goes very far to the right or left, it will get pulled to a very small height. Combine all of these ideas, and draw in all the curves based off your plotted points.
Draw the graph of the rational function below.
h(x) = −2x

x3−x2−6x
  • Just like we did in the previous questions, begin by identifying the function's domain and the vertical asymptotes. They'll help when we need to draw the graph.
  • First, figure out the function's domain by setting the denominator to 0:
    x3−x2−6x = 0
    We solve by factoring and get
    x(x+2)(x−3) = 0,
    thus it has "holes" at x=−2,  0,   3. Therefore its domain is x ≠ −2,  0,  3.
  • Next find the vertical asymptotes. Begin by canceling any common terms on the top and bottom:
    −2x

    x3−x2−6x
        =     −2x

    x(x+2)(x−3)
        =     −2

    (x+2)(x−3)
    Using the new, simplified version, find the location of any vertical asymptotes by setting the denominator to 0:
    (x+2)(x−3) = 0
    The function has vertical asymptotes at x=−2,   3. [Notice that this is different from the holes in our domain. At the very end, we'll deal with the fact that there is a hole in the domain at x=0 that doesn't show up in our asymptotes.]
  • Draw graph axes and draw dashed vertical lines at both x=−2 and x=3 to indicate the asymptotes.
  • At this point, it would help to have more points to get a sense of how the function curves. Make a table of values to help plot some points. Notice that the most "interesting" things for this function are the asymptotes at x=−2,  3 so we want to focus our table around them. [Also, we can use the simplified version of the function ([(−2)/((x+2)(x−3))]) to make it easier to compute points. With one exception, it works exactly the same. That exception is at x=0. From earlier, we found that the function does not exist there. Still, we can plot the point in the simplified version to know the best place to put the "hole" and to help us draw the function.]
    x
    f(x)
    −7
    −0.04
    −4
    −0.14
    −3
    −0.33
    −2.5
    −0.73
    −1.5
    0.88
    −1
    0.5
    0
    0.33 / DNE
    1
    0.33
    2
    0.5
    2.5
    0.88
    3.5
    −0.73
    4
    −0.33
    5
    −0.14
    8
    −0.04
    Plot these points on the graph. [As you become more comfortable with graphing functions like this, you will need fewer points because you'll have a better idea of the shape of the function. For now though, plot however many you need to be comfortable with graphing it.]
  • As x → −2 and x→ 3, the graph will fly off to ±∞. We need to figure out which direction it will fly off, though. From our table of points, it seems that to the left of x=−2 it goes to −∞, then on the right of x=−2 it goes to +∞. For x=3, on the left it goes to +∞, while on the right it goes to −∞. We might want to be sure, though, so we could also figure out where f(−2.01), f(−1.99), f(2.99), and f(3.01) would go. Alternatively, we could also just think about whether or not the function will go up or down based on if the x-value is slightly more or less at each of those asymptote locations. Once you know which direction it goes, draw a curve that approaches the dashed line, but does not actually cross it. As it gets closer, it gets lower/higher.
  • To help plot far to the right and left, think about what happens to the function as x goes very far to the right or left. Because the numerator grows so much less than the denominator, a very large x (+ or −) will cause the function output to become very small. Thus, as x goes very far to the right or left, it will get pulled to a very small height.
  • And there's one last thing to do: remember the "hole" in the graph. From the very beginning, we know that the function does not exist at x=0. We indicate by drawing a little empty circle at where the function "would have been" if it had existed.
Explain why f(x) = [(x+7)/(x+7)] does not have a vertical asymptote.
  • To have a vertical asymptote, there needs to be some x value that causes the denominator to become very, very small as the numerator remains comparatively large. In other words, where the denominator goes to 0, the numerator cannot also go to 0.
  • For this function, the only x-value that will make the denominator go to 0 is at x=−7. However, that location will also cause the numerator to go to 0. Therefore this function can not have a vertical asymptote.
  • An alternative way to look at this function is by canceling common factors. For the most part,
    x+7

    x+7
    = 1,
    so f(x) is just a constant function. A constant function can't fly off to ±∞, like a vertical asymptote requires, so f(x) has no vertical asymptote. The only place where f(x) is different than the simplified version is at x=−7 (because that breaks f, [0/0]), but that doesn't affect whether or not it has a vertical asymptote.
The only place where the denominator approaches 0 is at x=−7, but that causes the numerator to also approach 0, so no vertical asymptote is possible.

*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

Rational Functions and Vertical Asymptotes

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
  • Definition of a Rational Function 1:20
    • Examples of Rational Functions
    • Why They are Called 'Rational'
  • Domain of a Rational Function 3:15
    • Undefined at Denominator Zeros
    • Otherwise all Reals
  • Investigating a Fundamental Function 4:50
    • The Domain of the Function
    • What Occurs at the Zeroes of the Denominator
  • Idea of a Vertical Asymptote 6:23
  • What's Going On? 6:58
    • Approaching x=0 from the left
    • Approaching x=0 from the right
    • Dividing by Very Small Numbers Results in Very Large Numbers
  • Definition of a Vertical Asymptote 10:05
  • Vertical Asymptotes and Graphs 11:15
    • Drawing Asymptotes by Using a Dashed Line
    • The Graph Can Never Touch Its Undefined Point
  • Not All Zeros Give Asymptotes 13:02
    • Special Cases: When Numerator and Denominator Go to Zero at the Same Time
    • Cancel out Common Factors
  • How to Find Vertical Asymptotes 16:10
    • Figure out What Values Are Not in the Domain of x
    • Determine if the Numerator and Denominator Share Common Factors and Cancel
    • Find Denominator Roots
    • Note if Asymptote Approaches Negative or Positive Infinity
  • Example 1 18:57
  • Example 2 21:26
  • Example 3 23:04
  • Example 4 30:01

Transcription: Rational Functions and Vertical Asymptotes

Hi--welcome back to Educator.com.0000

Today, we are going to talk about rational functions and vertical asymptotes.0002

The next few lessons are going to be about rational functions and asymptotes; and they are pronounced aa-sim-tohts.0006

They are spelled kind of funny--asymptotes--but we want to know how to pronounce them; they are pronounced aa-sim-tohts, asymptotes.0012

In exploring these ideas, we will see some very interesting behavior; and we will learn why it occurs.0020

But before you start these lessons, it is important that you have a reasonable understanding of polynomials.0024

Polynomials are going to be central to this, as we are about to see in the definition of a rational function.0028

You will need to know their basics; you will need to know how to factor them to find zeroes and roots.0032

And you will also need to know polynomial division.0036

If you haven't watched the previous section of lessons on polynomials, you might find it helpful to watch0038

the three lessons Introduction to Polynomials, Roots/Zeroes of Polynomials, and Polynomial Division;0042

or just the ones that you need specific help with, because those ideas are going to definitely come up as we explore this stuff.0048

Also, it should be mentioned that we would be working only with the real numbers; so we are back to working just with the reals.0054

And while we briefly worked with the complex numbers in the last couple of lessons, it is back to the reals.0059

We are not going to see anything other than real numbers anymore.0063

If we would have solutions in the complex, too bad; we are not going to really care about them right now.0067

We are back to focusing just on the reals; they are still pretty interesting--there is lots of stuff to explore in the reals.0071

You can leave complex numbers to a future, later course in math.0076

All right, first let's define a rational function: a rational function is the quotient of two polynomials.0079

So, f(x) is our rational function; it is equal to n(x) divided by d(x), where n(x) and d(x) are both polynomials, and d(x) is not equal to 0.0086

And by "not equal to 0," I mean that d(x) isn't 0, like straight "this is zero, all the time, forever."0097

So, d(x) = 0, the constant function, just 0, all the time, forever--that is not allowed.0104

But d(x) can have roots; we can say that d(x) can have roots.0115

For example, d(x) could be x2 - 1, where d would be 0 at positive 1 and -1; that is OK.0120

d(x) can have roots, but it is not going to be the constant function of 0 all the time.0130

That is what is not allowed, because then our function would be broken, completely, everywhere.0137

But we are allowed to have slight breaks occasionally, when we end up having roots appear in the denominator.0141

So, it is just a normal polynomial, but not just 0; that would be bad.0147

All right, some examples: g(x) = (2x + 1)/(x2 - 4) or 1/x3 or (x4 - 3x2)/(x + 2).0152

In all of these cases, we have a polynomial divided by a polynomial--as simple as that.0161

You might wonder why they are called rational functions--what is so rational about them?0166

Remember: many lessons back, when we have talked about the idea of sets, we called fractions made up of integers,0170

things like 3/5 or -47/2, the rational numbers, because they seem to be made in a fairly rational, sensible way.0175

Thus, we are using a similar name, because the rational functions are built similarly.0184

Rational numbers are built out of division, and rational functions are built out of division.0188

So, we are using a similar name; cool.0193

Since a rational function f(x) = n(x)/d(x) is inherently built out of the operation of division, we have to watch out for dividing by 0.0197

That is going to be the Achilles' heel of rational functions; and also, it is going to be what makes them so interesting to look at.0205

Dividing by 0 is not defined--it is never defined; so the zeroes, the roots, the places where it becomes 0,0211

of our denominator polynomial, d(x), will break the rational function.0218

The zeroes of d(x) are not in the domain of our function; so a rational function, right here--0222

wherever it ends up being 0, it is not defined there; that is not in the domain.0231

It is not in the set of numbers that we are allowed to use, the domain, because if we plug in a number0236

that makes us divide by 0, we don't know what to do; we just blew up the world, so it is no good.0241

So, we are not going to be allowed to plug in numbers--we are not going to have those in our domain--0246

when d(x) = 0, when we are looking at the zeroes, the roots, of our denominator polynomial.0250

Now, other than these zeroes in the denominator, a rational function is defined everywhere else,0256

because polynomials have all of the real numbers as their domain.0262

Since you can plug any real number into a polynomial, and get something out of it,0265

the only place we will have any issues is where we are accidentally dividing by 0.0269

So, everything other than these locations where we divide by 0--they are all good.0272

The domain of a rational function is all real numbers, except the zeroes of d(x)--all real numbers, with the exceptions of these zeroes in our denominator polynomial.0276

Everything will be allowed, with a very few exceptions for that denominator polynomial's zeroes.0285

All right, to help us understand rational functions better, let's consider this fundamental rational function, 1/x.0290

So, immediately we see that f is not going to be defined at x = 0.0299

So, our domain for this will be everything except x = 0; domain is everything where x is not 0--so most numbers.0303

We see that we can't divide by 0; but everything else would give us an actual thing.0316

So, notice how it behaves near x = 0; it isn't actually going to be defined at x = 0--we will never see that.0320

But as it approaches it, f(x) grows very, very large--see how it is shooting up?0328

We have it going up and down--it is very, very large.0333

What is going on here? Let's look at a different viewing window to get a sense for just how large f(x) manages to become.0338

If we look at a very small x width, only going from -0.5 to +0.5, we manage to see a very large amount that we move vertically.0344

We can go all the way...as we come from the left, we manage to go from near 0 all the way to -100 on this window.0353

And we can go from near 0 all the way up to positive 100 on this window.0363

And in fact, it will keep going; it just keeps shooting up.0368

It shoots all the way out to...well, not out to infinity, because we can't actually hit infinity...but the idea of shooting out to something arbitrarily large.0371

It is going to get as big as we want to talk about it getting out to; so in a way, we say it goes out to infinity.0378

We see this behavior of going out to infinity in many rational functions.0384

If we look at 1/(x + 2), we get it going out to infinity at -2.0389

Notice that, when we plug in -2 into the denominator, we have a 0 showing up; so we see this behavior around zeroes.0395

If we look at 5/(x2 - 9), we see it at -3 and +3.0401

We have these places where it goes out to infinity.0406

In a few moments, we will formally define this behavior, and we will name it a vertical asymptote.0410

But first, let's understand why this strange behavior is occurring.0415

To understand that, let's look, once again, at f(x) = 1/x, our fundamental rational function, with what makes the most basic form of it.0419

Notice that, while x = 0 is not defined (that would blow up the world, because we are dividing by 0--we are not allowed to do that--0428

it just doesn't make sense), everywhere else is going to be defined.0436

So, let's see what happens as x goes to 0: this little arrow--we say that x is going to 0.0441

We are looking as x gets close to 0--not actually at 0; but as x sort of marches towards 0.0446

Let's look at what happens as we go from the negative side.0452

As we plug in -2, we get -0.5, 1/-2; if we plug in -1, we get -1.0455

But as the x becomes very, very small, we see f(x) become very, very large.0462

-0.5 gets us -2; -0.1 gets us -10; -0.01 gets us -100; -0.001 gets us -1000.0467

As we divide a number by a very small number, 1 divided by 1/10 is equal to positive 10, because the fraction will flip up.0477

We are going to end up having to look at its reciprocal, since it is in the denominator again.0489

We are going to see this behavior: as we get really, really small--as we get to these really small numbers that we are dividing by--0493

we will end up having the whole thing become very, very large.0500

We can see this on the positive side, as well, if we go in from the positive side.0504

The negative side is us starting from the negative side and us moving towards the 0.0508

And the positive one is us starting from the positive side, and us moving towards the 0.0514

It is moving from the right, into the left, versus moving from the left into the right.0519

Positive: we look at 2; we get one-half, 1/2; we plug in 1; we get 1/1, so 1.0524

As we get smaller and smaller, we see it begin to get larger and larger.0529

We plug in 0.5, and we will get 2; we plug in 0.1, and we will get 10; we plug in 0.01, and we will get 100; and so on and so forth.0532

As we get in smaller and smaller and smaller numbers, we are going to get larger and larger things.0540

1/(1 divided by 1 million) is a very small number, but it is one of the things that is defined, because it is not x = 0; it is just very close to x = 0.0545

It will cause it to flip up to the top, and we will get 1 million.0558

We will get to any size number that we want.0562

We can't actually divide by 0; but dividing by very small numbers gives large results.0565

This is what is happening: as we get near 0, we go out to infinity, because we are dividing by very, very small numbers.0576

And when you divide by something very, very small, you get a very big thing.0583

How many times does something very, very small go into a reasonable-sized number?0587

It goes in many times; the smaller the chunk that we are trying to see how many times it fits in, the more times it is going to be able to fit in.0592

So, the smaller the thing that we are dividing by, the larger the number we will get out in response.0598

And that is what is going to cause this behavior of "blowing out."0603

With this idea in mind, we will define a vertical asymptote.0607

A vertical asymptote is a vertical line x = a, where as x gets close to a, the size of f,0610

the absolute value of f(x), becomes arbitrarily large--becomes very big.0619

Symbolically, we show this as x → a will cause f(x) to go to infinity or f(x) to go to negative infinity.0624

Informally, we can see a vertical asymptote as a horizontal location, a, where the function blows out to infinity,0633

either positive infinity or negative infinity, as it gets near a.0641

So, it is going to be some location a that, as we get close to it, we go out to infinity.0645

We manage to blow out to infinity as we get close to it; we get very, very large numbers.0656

So, a vertical asymptote is some horizontal location, some vertical line defined by x = a,0661

where, as we get close to it, we blow out to infinity.0668

Maybe we blow up; maybe we blow down; but we are going to blow out to infinity, one way or the other.0671

Vertical asymptotes and graphs; we show the location of a vertical asymptote with a dashed line, as we have been doing previously.0677

This aids in drawing the graph and in understanding the graph later.0683

So, notice: we just use a dashed line here to show us that this is where the vertical asymptote is.0686

So, for the graph of f(x) = 1/(x - 1), we see, right here at x = +1 (because if we plug in x = +1, we would be dividing by 0 there)--0691

so near that location--that we will be dividing by very small numbers, which will cause us to blow out to infinity around that location.0702

We will see a vertical asymptote there; so as our function approaches that vertical asymptote,0708

it ends up having to get very, very large, because it is now dividing by very, very small numbers.0714

Notice that the graph gets very close to the asymptote; but it won't be able to touch the asymptote.0721

So, it will become arbitrarily close to the line, but it never touches the line.0726

Why can't it touch the line? Because it is not defined there.0731

We have 1 here; so if we plug in x = 1, then we have 1/(1 - 1), which is 1/0.0735

We are not allowed to divide by 0, so we are not allowed to do this, which means we are not allowed to plug in x = 1.0743

Our domain is everything where x is not equal to 1; so we are not defined at that thing.0748

Our function is not actually defined on the asymptote; it will become very large--it will become very close to this asymptote.0754

But it won't actually be allowed to get onto the asymptote, because to be on that point,0761

to be on that horizontal location, would require it to be defined there.0765

And it is not defined there, because that would require us to divide by 0, which is not allowed.0768

So, we are not allowed to do that, which means that we are not actually going to be able to get on it.0773

We will never touch the vertical asymptote, because it is not defined.0777

Now, not all 0's give asymptotes; so far, all of the vertical asymptotes we have seen have happened at the location of a 0 in the denominator.0783

And that is going to be the case: all vertical asymptotes will require there to be a 0 in the denominator.0789

But a 0 in the denominator does not always imply an asymptote.0795

We are not going to always see an asymptote.0801

Consider f(x) = x/x; now, normally, x/x is a very boring function; the x's cancel, and we are left with this effectively being just 1,0803

except at one special place: when we look at x = 0, we are given 0/0.0814

Now, 0/0 is definitely not defined, so we are not allowed to do anything there.0820

We have a pretty good sense that it would be going to 1.0825

But when we actually look at that place, it is not defined, because we are looking at 0/0.0828

0/0 is definitely not defined; so there is going to be a hole in the graph.0834

f is a rational function, x/x, a polynomial divided by a polynomial; they are simple linear polynomials, but they are both polynomials.0839

So, f is a rational function; and it is not defined at x = 0; but it still has no asymptote there,0848

because normally, we just have x/x; it is just really a constant function, 1.0855

We see it here; it is going to be 1 all the time, forever, always, except when we look at x = 0,0861

at which point it blips out of existence, because it is not allowed to actually have anything at 0/0, because that is not defined.0868

So, formally, we can't give it a location; so instead, we have to use this hole.0876

We show that there is a missing place there with this empty circle.0880

As opposed to a filled-in circle to show that there is something there, we use an empty circle to show that we are actually missing a location there.0885

It has a hole at that height there, because it would be going there, but it doesn't actually show up there.0891

There is no asymptote there; why does it happen--why do we lack this asymptote?0896

It is because the numerator and the denominator go to 0 at the same time, so the asymptote never happens.0901

Because they are both going to 0, there is nothing to divide out to blow up to a very large number--to blow down to a very large negative number.0906

The asymptote can't happen, because normally they just cancel out to 1/1...well, not 1/1; they cancel out to something over something.0916

But since it is the same something, it becomes just 1.0923

So, if we have .1 divided by .1, well, they cancel out, and we just get 1.0926

If we have .0001 divided by .0001, well, once again, they cancel out, and we have just 1.0930

So, we don't end up seeing this asymptote behavior, because we have to have something divided by very small numbers.0936

But if we have a very small number divided by a very small number, they end up going at the same rate; and so we don't get this asymptote.0942

To find an asymptote, then, that means we first need to cancel out common factors.0949

We need to be able to say, "Oh, x/x means I am going to cancel these out; and I will get it be equivalent to being just 1."0955

We will have to remember that we are forbidden to actually plug in x = 0; but other than that, it is just going to be 1 all the time, forever.0963

So, with our newfound understanding of all these different things that make up rational functions,0971

we can create a step-by-step guide for finding vertical asymptotes.0975

So, given a rational function in the form n(x)/d(x), where n and d are both polynomials,0978

the first thing we want to do is figure out all of the x-values that we aren't going to be allowed to have in our domain--0984

all of our forbidden x-values that would break our function.0990

These are the zeroes of d(x), because they would cause us to divide by 0.0993

Since dividing by 0 is not allowed, then our domain is everything where we won't divide by 0.0997

So, the zeroes of d(x) are not in our domain.1002

Next, we want to determine if n(x) and d(x) share any common factors.1006

So, to do this, you might have to factor the two polynomials.1010

If they share common factors, cancel out those factors.1013

Alternatively, this step can be done by checking that the zeroes to d(x) are not also zeroes to n(x),1016

because for them to have a common factor, where they would both be going to 0 at the same time,1025

then a common factor means that they have the same root.1029

So, if they are both going to be zeroes at the location, that means we will see this effect.1032

So, we can either cancel out the factors, or we can just and make sure that the roots to d(x) are not also roots to n(x).1036

But the easiest way to find roots in the first place is usually to factor.1043

So, you will probably want to factor, for the most part.1046

But occasionally, it will be easier to just see, "If we plug in this number, does it come out to be 0?"1048

The next step: once you cancel out the common factors, you determine any zeroes that are left in the denominator.1053

These are the vertical asymptotes, because once we have canceled out common factors,1059

we don't have to worry about both the top and the bottom going to 0 at the same time.1063

So, any roots that are left are going to have to make very small numbers to blow everything out to infinity.1067

So, anything that is left after canceling out (it is crucial that it is after you cancel out common factors)--1073

whatever is left as zeroes in the denominator after canceling out will give you vertical asymptotes.1080

Finally, an optional step: if we are graphing the function, you want to figure out which way the graph goes:1086

positive infinity or negative infinity on either side of each asymptote.1091

You can do this by using test values that are very close to the asymptote.1095

For example, if we had an asymptote at x = 2, we might want to check out f at 1.99 and f at 2.01.1098

They are very close to the asymptote, so we will have an idea of where it is going.1105

Alternatively, you can also do this in your head by thinking in terms of positive versus negative,1108

and just thinking, "If I was very slightly more or very slightly less," and we will discuss this in the examples;1114

you will see it specifically in Example 3; we will see this idea of "If I was just a little bit over;1118

if I was just a little bit under," and it will make a little bit more sense when we actually do it in practice.1123

But you can always just test a number, use a calculator, and figure out, "Oh, that is what that value would be";1127

"I am clearly going to be positive on this side; I am clearly going to be negative on this side."1132

All right, let's look at some examples.1136

f(x) = (x - 1)/(2x2 + 8x - 10); what is the domain of f?1139

To figure out the domain of f, we need to figure out where our zeroes on the bottom thing are.1145

So, our very first step, where we do factoring: 2x2 + 8x - 10--how are we going to factor this?1155

Well, let's start out by pulling out that 2 that is in the front; it is just kind of getting in our way.1161

So, x2 +...8x, pulling out a 2, becomes 4x; -10, pulling out a 2, becomes -5.1165

We can check this if we multiply back out; we see that we have the same thing.1171

So, 2(x2 + 4x - 5); could we factor this further? Yes, we can.1174

We see that, if we have +5 and -1, that would factor; so 2(x + 5)(x - 1)...sure enough, that checks out.1178

x2 - x + 5x...that changes to 4x...plus 5 times -1...-5...great; that checks out.1186

So, that means we can rewrite (x - 1)/(2x2 + 8x - 10).1193

It is equivalent to saying (x - 1) over the factored version of the denominator, 2(x + 5)(x - 1).1201

So, all of the places to figure out the domain of f, all of the places where our denominator will go to 0, are going to be x at -5 and x at +1.1210

Now, we don't want to allow those; so when x is not -5, and x is not 1, then we are in the domain.1222

We are allowed those values; so our domain is everything where x is not -5 and not positive 1.1228

Great; the next question is where the vertical asymptotes are.1235

To find the vertical asymptotes, we need to cancel any common factors,1241

so that the numerator and the denominator don't go to 0 at the same time.1244

We notice that we have x - 1 here and x - 1 here; so we can cancel these out, and we get 1/[2(x + 5)].1248

So, now we are looking for where the denominator goes to 0.1258

That is going to happen at...when is x + 5 equal to 0? We get x = -5.1263

So, we have a vertical asymptote when the denominator in our new form with the common factors canceled out.1269

So, we will have a vertical asymptote after we have canceled out common factors,1275

when we can figure out that the denominator still goes to 0, at x = -5; great.1280

The second example: Given that the graph below is of f, our function, find a and b if our function f is 3x/[(x - a)(x - b)].1287

So, we notice, first, that we have a vertical asymptote here, and a vertical asymptote here, off of this graph.1297

So, we see that one of our vertical asymptotes happens at -3; the other one happens at positive 2.1302

So, x = -3; x = +2; we have vertical asymptotes.1308

OK, so if x = -3 and x = 2 mean vertical asymptotes, then that means we need to have a factor that will cause a 0 in the denominator at those locations.1315

So, we want a 0 in the denominator at -3; so x + 3 would give us a 0 when we plug in -3.1324

And x - 2 would give us a 0 when we plug in positive 2.1333

Plus 3...we plug in -3 + 3; that gives us a 0 in our denominator.1338

We plug in +2; 2 - 2...that gives us a 0 in our denominator; great.1342

So, that means that (x + 3)(x - 2)...we need those factors in our denominator to be able to have vertical asymptotes.1346

We don't have to worry about it interfering with our numerator, because 3x doesn't have any common factors with (x + 3)(x - 2).1352

They are very different linear factors.1358

At this point, we are ready...we want -a to be + 3, so that means that a must be equal to -3, because it comes out to be positive 3.1361

There is our a; and b...minus b is -2, so it must be that b is equal to positive 2, so it is still actually subtracting.1371

And there are our answers.1380

All right, the third example: Draw the graph of f(x) = 1/(3 - x).1383

It is a pretty simple one, so we don't have to worry about factoring; it has already been factored.1389

We don't have to worry about canceling any common factors; we see that there are very clearly no common factors.1392

So, we can just get right to figuring this out.1398

We have a vertical asymptote; first, we notice that we have a vertical asymptote at x = 3,1401

because when we plug in 3, 3 minus x...we have 3 minus 3, so we would have a 0, so we would be dividing by very small numbers;1409

so we would be dividing 1 by very small numbers at x = 3.1416

That means we will blow out when we are near x = 3; so we have a vertical asymptote there.1421

Let's draw in our graph axes: let's put this at...we will need more to the right, because the interesting thing happens out here.1426

So, here is 1, 2, 3, 4, 5; I will extend that a little bit further: 6, 1, 2, 3, 4, 5, 6.1436

I will go for roughly the same scale; OK.1450

And 1, 2, 3, -1, -2, -3, -1, -2; great, we have drawn out our graph.1459

We know that we have a vertical asymptote at 3; so let's draw in a vertical line.1467

OK, we see our asymptote on our graph; and now we need to draw it in.1475

So, at this point, we realize that we actually probably should have some points; so let's try some points.1481

If we try x =...oh, let's say 4, just one unit to the right of our asymptote; when we plug that in, we will get 1/(3 - 4), which gets us 1/-1.1486

1/-1 is -1; so at 4, we are at -1; so we plot in that point.1500

Let's try one unit to the left, x = 2: we will get 1/(3 - 2), so we will get 1 over positive 1 (3 - 2 is positive 1).1505

We will be at positive 1 when we are at 2.1515

We might still want a little bit more information, so maybe we will also try x = 0.1518

At x = 0, we will get 1/(3 - 0), so we will be at 1/3; so at 0, we are just 1/3 of the way up to our 1.1525

So, we will put in that point there at x = 6; let's try that one.1533

We will be at 1/(3 - 6), so 1/-3; so we will be at -1/3, so once again, we will be just a little bit...one-third of the way out there.1540

All right, now we are starting to get some sense; what happens is that we actually get near that asymptote.1550

Are we going to go positive on the left side, or are we going to go negative on the left side?1556

Are we going to go positive on the right side, or are we going to go negative on the right side?1560

What we can think about is: we can consider just a hair to the left of 3: consider 2.99999.1564

Consider this number; now, we could actually plug it into a calculator, and we could get a value.1575

But we can also just think about this in our head, and say, "1/(3 - 2.99999)...if we have a bunch of 9's,1578

3 - 2.99999, well, is it going to be a positive number, or is it going to be a negative number?"1591

Well, 3 - 2.99999...2.99999 is still smaller than 3, so it is going to come out as a positive thing.1595

We have 1, a positive number, divided by a positive number; we can just think of it as dividing by a small positive number.1601

That means that the whole thing will come out to be a positive thing, when we are on the left side,1608

because we are subtracting by something, but it ends up being just under; so we stay positive.1613

So, we can think in terms of "Are we being positive, or are we being negative?"1620

We are being positive; so we know we are going to go up on this side.1623

If we consider a very small number above 3, 3.00001--if we consider that, it will be 1/(3 - 3.00001).1627

Well, 3 minus something that is just a little bit above it is going to end up being negative.1642

It is going to be a very small number, but it is going to be negative, and that is key.1647

1 over a negative number on the bottom: since it is a positive divided by a negative, the whole thing will stay negative.1649

And so, we will get very large negative numbers; so we will be going down on the right side.1655

So, at 3.00001, we see that we end up being still negative; so we will have to be negative when we are on the right side of our asymptote.1659

On the left side, 2.99999, just to the left of our asymptote at x = 3...we see that we are going to end up being positive.1668

And so, that guides us on how our asymptote should grow.1675

So, at this point, we can draw in this curve; it is going to be growing and growing and growing.1679

And as it gets to the asymptote, it is going to get larger and larger and larger and larger and larger.1684

It will never actually touch the asymptote; it will end up just getting larger and larger and larger and going up and up and up.1689

It never becomes perfectly vertical; it never touches the asymptote; but it is going to blow out to infinity, slowly but surely.1694

It won't ever actually touch infinity, because we can't touch infinity--infinity is just an idea of ever-larger numbers.1700

But it will go out to ever-larger numbers.1706

On the other side, we see a very similar thing; sorry, that should be curved just a little bit more--not quite straight.1708

And it is going to get very, very large, and once again, going to curve out.1717

It never quite touches that vertical asymptote, but we will get very, very, very close.1721

My right side is a little bit closer to the vertical asymptote than it probably should be.1725

But we have a good idea, and that would be acceptable when turning that in on a test or homework.1729

What happens as we go very, very far to the left?1734

Well, let's think about as we plug in some really big negative number, like, say, -100.1736

Well, x = -100...we are going to end up having 1/(3 - -100); those cancel out, and we get a positive.1741

So, we get 1/103, which is a very small number.1749

So, since it is 3 - x, as we plug in very large negative numbers, we are going to effectively get very large positive numbers in our denominator.1752

And so, we will be dividing by very large things; so we will sort of crush down to nothing.1760

As we get larger and larger and larger, we are going slowly approach 0.1765

We will never quite hit 0 (we will talk about this more when we talk about horizontal asymptotes in the next lesson),1769

but we will get very, very close to the 0 height.1773

The same thing happens as we go out to the right: but we will be coming from the negative side,1776

because if we look at a very, very large positive number, like 1/(3 - 100), a positive 100 plugged in for x,1781

we will get 1/-97, so we will be a very small number,1789

but we will be a very small negative number, because we have that negative on the bottom.1793

And so, we will be on the bottom side; we will be below the x-axis.1797

Great; the final example, Example 4: Give a rational function where the graph goes up on both sides of an asymptote.1800

Also give one where it goes down on both sides.1807

So, so far, we have always seen our asymptotes where on one side it goes up, and on the other side it goes down (or maybe the other way).1809

But is it possible to create something where it is going to go up on both sides or going to go down on both sides?1815

Let's look at our first one that we think of when we think of a rational function.1821

We immediately think, "Oh, the fundamental, basic thing that makes up rational functions is 1/x."1825

This is the simplest form we can think of that really shows these ideas we are looking for.1830

1/x ends up giving us a graph that gets very, very large, that gets to the right side, but goes down on the right.1834

And then, the same thing blows down.1845

The issue here with 1/x is: on one side, it is negative; on the left side, we were dividing by negative numbers, so we have negative.1849

On the other side, it is positive; that is why we go up and down.1865

That is what causes up and down effect--well, down and up, if we go negative to positive.1872

But that is why we are seeing this split, where we are going in two opposite ways,1879

because on one side of the vertical asymptote, we are dividing by a negative number.1882

On the other side of the vertical asymptote, we are dividing by a positive number.1885

They are both very small numbers, but that negative versus positive causes this incredible difference in whether you go to -∞ or +∞.1888

So, what we want is positive on both sides, if we are going to go up on both sides.1894

So, what could we create that would be similar to 1/x, but always putting out positive things on the bottom?1903

Let's keep it as 1 over; but what is always positive?1909

Well, instead of dividing by x, let's divide by x2.1912

We can force it to always be positive: 1/x2...x2 is still a polynomial, so it is still a rational function.1916

And if we draw that one in, we are going to see it behaving very similarly on the right side.1922

It will end up being here and here and here as we get to small numbers, as we get below 1.1927

We are going to be getting to dividing by very small numbers, so it is going to blow out to infinity.1932

And we will get crushed down to 0 as it goes to the right.1939

On the left side, though, when we square a negative number, it becomes positive; so we are going to see a mirror image on the left side, as well.1941

So, we are going to see it going up on both sides.1950

If we want to get one that is going to go down on both sides, if we want it to be down on both sides--1953

well, we see 1/x2; we just want it to go down; so let's make it as easy as just flipping it down.1958

We flip it down by making everything negative, so it is -1/x2.1964

And we will see a graph that looks like this; we have managed to make it go down on both sides.1969

So, a simple way to make both sides go in the same direction is by having it be squared, or by having it be negative and then squared...1978

not negative and then squared, because that would be positive; but a negative square,1986

because then it is a square number, times a negative; cool.1989

All right, we will see you at Educator.com later.1992

Next time, we will talk about horizontal asymptotes and have an understanding of why it is getting crushed to 0,1994

and see that it can, in fact, be crushed to value other than 0--pretty interesting; goodbye!1998