For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Rational Functions and Vertical Asymptotes
 A rational function is the quotient of two polynomials:
where N(x) and D(x) are polynomials and D(x) ≠ 0.f(x) = N(x) D(x)
,  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 xvalues 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 xvalue, 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 xvalues 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

 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:
We see that the only place where the function "breaks" (is not defined) is at x=5. Thus, x=5 is the only xvalue not allowed. Therefore any x value other than 5 is in the domain.x−5=0 Domain: x ≠ 5  A vertical asymptote is some specific xvalue that, as x approaches it, the function grows very large (positive or negative). This happens when the numerator is nonzero 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:
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.]7 x−5  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:
The above denominator still only has a zero at x=5, as we figured out before. Thus, x=5 is a vertical asymptote.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:
We see that the only place where the function "breaks" (is not defined) is at x=−3. Thus, x=−3 is the only xvalue not allowed. Therefore any x value other than −3 is in the domain.3x+9=0 Domain: x ≠ −3  A vertical asymptote is some specific xvalue that, as x approaches it, the function grows very large (positive or negative). This happens when the numerator is nonzero 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:
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.]17x 3x+9  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:
The above denominator still only has a zero at x=−3, as we figured out before. Thus, x=−3 is a vertical asymptote.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:
We can find the solutions to this equation by factoring.x^{2}−2x−24=0  (x+4)(x−6)=0, so the function "breaks" (is not defined) at x=−4 and x=6. These are the only two xvalues not allowed. Therefore any x value other than them is in the domain.
Domain: x ≠ −4, 6  A vertical asymptote is some specific xvalue that, as x approaches it, the function grows very large (positive or negative). This happens when the numerator is nonzero 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:
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.]3 x^{2}−2x−24= 3 (x+4)(x−6)  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:
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.3 (x+4)(x−6)

 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:
We can find the solutions to this equation by factoring.x^{3}+5x^{2}−14x=0  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 xvalues not allowed. Therefore any x value other than them is in the domain.
Domain: x ≠ −7, 0, 2  A vertical asymptote is some specific xvalue that, as x approaches it, the function grows very large (positive or negative). This happens when the numerator is nonzero 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:
We see that there is a common factor: (x+7), so we cancel this factor out before trying to find the vertical asymptote(s).2x+14 x^{3}+5x^{2}−14x= 2(x+7) x(x−2)(x+7)
[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 xlocation of −7.]2 x(x−2)  Once any common factors have been canceled, find what zeros (roots) are left in the denominator:
The above denominator only has zeros at x=0 and x=2. These are the vertical asymptotes.2 x(x−2)
 A vertical asymptote is some specific xvalue 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.
 A vertical asymptote is some specific xvalue 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.

 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:
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).]x+3 = 0  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.
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.]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  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 xvalue 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.

 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:
The function will "break" at x=4, so the domain is x ≠ 4.24−6x = 0  Next find the vertical asymptotes. Begin by canceling any common terms on the top and bottom:
Using the new, simplified version, find the location of any vertical asymptotes by setting the denominator to 0:15 24−6x= 3·5 3(8−2x)= 5 8−2x
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.]8 − 2x = 0  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.
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.]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  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 xvalue 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.

 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:
We solve by factoring and getx^{3}−x^{2}−6x = 0
thus it has "holes" at x=−2, 0, 3. Therefore its domain is x ≠ −2, 0, 3.x(x+2)(x−3) = 0,  Next find the vertical asymptotes. Begin by canceling any common terms on the top and bottom:
Using the new, simplified version, find the location of any vertical asymptotes by setting the denominator to 0:−2x x^{3}−x^{2}−6x= −2x x(x+2)(x−3)= −2 (x+2)(x−3)
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.](x+2)(x−3) = 0  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.]
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.]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  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 xvalue 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.
 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 xvalue 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,
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.x+7 x+7= 1,
*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 screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Introduction
 Definition of a Rational Function
 Domain of a Rational Function
 Investigating a Fundamental Function
 Idea of a Vertical Asymptote
 What's Going On?
 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
 Vertical Asymptotes and Graphs
 Not All Zeros Give Asymptotes
 How to Find Vertical Asymptotes
 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
 Example 2
 Example 3
 Example 4
 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
Precalculus with Limits Online Course
Transcription: Rational Functions and Vertical Asymptotes
Hiwelcome 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 aasimtohts.0006
They are spelled kind of funnyasymptotesbut we want to know how to pronounce them; they are pronounced aasimtohts, 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 interestingthere 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, foreverthat 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 x^{2}  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)/(x^{2}  4) or 1/x^{3} or (x^{4}  3x^{2})/(x + 2).0152
In all of these cases, we have a polynomial divided by a polynomialas simple as that.0161
You might wonder why they are called rational functionswhat 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 definedit 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 here0222
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 numberswe are not going to have those in our domain0246
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 0they 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 0so 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 = 0we will never see that.0320
But as it approaches it, f(x) grows very, very largesee how it is shooting up?0328
We have it going up and downit 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/(x^{2}  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 0we are not allowed to do that0428
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 arrowwe say that x is going to 0.0441
We are looking as x gets close to 0not 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 smallas we get to these really small numbers that we are dividing by0493
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 onehalf, 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 reasonablesized 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 largebecomes 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 locationthat 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 largeit 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 filledin 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 happenwhy 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 numberto 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 stepbystep 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 xvalues that we aren't going to be allowed to have in our domain0984
all of our forbidden xvalues 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)/(2x^{2} + 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: 2x^{2} + 8x  10how 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, x^{2} +...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(x^{2} + 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
x^{2}  x + 5x...that changes to 4x...plus 5 times 1...5...great; that checks out.1186
So, that means we can rewrite (x  1)/(2x^{2} + 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...onethird 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, 3.00001if 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 infinityinfinity is just an idea of everlarger numbers.1700
But it will go out to everlarger numbers.1706
On the other side, we see a very similar thing; sorry, that should be curved just a little bit morenot 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 xaxis.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 effectwell, 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 x^{2}.1912
We can force it to always be positive: 1/x^{2}...x^{2} 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 sides1953
well, we see 1/x^{2}; 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/x^{2}.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 0pretty interesting; goodbye!1998
1 answer
Last reply by: Professor SelhorstJones
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?