# Finding horizontal and vertical asymptotes | Rational expressions | Algebra II | Khan Academy

Voiceover: We have F of X
is equal to three X squared minus 18X minus 81, over
six X squared minus 54. Now what I want to do in this video is find the equations for the horizontal and vertical asymptotes and I encourage you to
pause the video right now and try to work it out on your own before I try to work through it. I’m assuming you’ve had a go at it. Let’s think about each of them. Let’s first think about
the horizontal asymptote, see if there at least is one. The horizontal asymptote
is really what is the line, the horizontal line that F of X approaches as the absolute value of X approaches, as the absolute value
of X approaches infinity or you could say what does F of X approach as X approaches infinity and what does F of X approach as X approaches negative infinity. There’s a couple of ways
you could think about it. Let me just rewrite the
definition of F of X right over here. It’s three X squared minus 18X minus 81. All of that over six X squared minus 54. Now there’s two ways you
could think about it. One you could say, okay, as X as the absolute value of X becomes larger and larger and larger, the highest degree terms in the numerator and the denominator are going to dominate. What are the highest degree terms? Well the numerator you
have three X squared and in the denominator
you have six X squared. As X approaches, as
the absolute value of X approaches infinity, these two terms are going to dominate. F of X is going to become
approximately three X squared over six X squared. These other terms are going to matter less obviously minus 54 isn’t
going to grow at all and minus 18X is going to grow much slower than the three X squared, the highest degree terms are
going to be what dominates. If we look at just those terms then you could think of
simplifying it in this way. F of X is going to get closer and closer to 3/6 or 1/2. You could say that there’s
a horizontal asymptote at Y is equal to 1/2. Another way we could
have thought about this if you don’t like this whole little bit of hand wavy argument that
these two terms dominate is that we can divide the
numerator and the denominator by the highest degree or X
raised to the highest power in the numerator and the denominator. The highest degree term is
X squared in the numerator. Let’s divide the numerator
and the denominator or I should say the highest degree term in the numerator and the
denominator is X squared. Let’s divide both the numerator and denominator by that. If you multiply the numerator
times one over X squared and the denominator
times one over X squared. Notice we’re not changing the value of the entire expression,
we’re just multiplying it times one if we assume
X is not equal zero. We get two. In our numerator, let’s
see three X squared divided by X squared is going to be three minus 18 over X minus 81 over X squared and then all of that over six X squared times one over X squared,
this is going to be six and then minus 54 over X squared. What’s going to happen? If you want to think in terms of if you want to think of limits as something approaches infinity. If you want to say the limit as X approaches infinity here. What’s going to happen? Well this, this and that
are going to approach zero so you’re going to approach 3/6 or 1/2. Now, if you say this X
approaches negative infinity, it would be the same thing. This, this and this approach zero and once again you approach 1/2. That’s the horizontal asymptote. Y is equal to 1/2. Let’s think about the vertical asymptotes. Let me write that down right over here. Let me scroll over a little bit. Vertical asymptote or possibly asymptotes. Vertical maybe there is more than one. Now it might be very tempting to say, “Okay, you hit a vertical asymptote” “whenever the denominator equals to zero” “which would make this
rational expression undefined” and as we’ll see for this case that is not exactly right. Just making the denominator
equal to zero by itself will not make a vertical asymptote. It will definitely be a place where the function is undefined but by itself it does not
denominator right over here so we can factor it out. Actually let’s factor out the numerator and the denominator. We can rewrite this as F of
X is equal to the numerator is clearly every term
is divisible by three so let’s factor out three. It’s going to be three times X squared minus six X minus 27. All of that over the denominator each term is divisible by six. Six times X squared minus 9 and let’s see if we can
factor the numerators and denominators out further. This is going to be F of
X is equal to three times let’s see, two numbers,
their product is negative 27, their sum is negative six. Negative nine and three seem to work. You could have X minus
nine times X plus three. Just factor the numerator
over the denominator. This is the difference of
squares right over here. This would be X minus
three times X plus three. When does the denominator equal zero? The denominator equals zero when X is equal to positive three or X is equal to negative three. Now I encourage you to pause
this video for a second. Think about are both of
these vertical asymptotes? Well you might realize that the numerator also equals zero when X is
equal to negative three. What we can do is actually
simplify this a little bit and then it becomes a little bit clear where our vertical asymptotes are. We could say that F of X, we could essentially divide the numerator and denominator by X plus three and we just have to key, if we want the function to be identical, we have to keep the [caveat]
that the function itself is not defined when X is
equal to negative three. That definitely did
make us divide by zero. We have to remember that but that will simplify the expression. This exact same function is going to be if we divide the numerator and denominator by X plus three, it’s going to be three times X minus nine over six times X minus three for X does not equal negative three. Notice, this is an identical definition to our original function and I have to put this
qualifier right over here for X does not equal negative three because our original function is undefined at X equals negative three. X equals negative three is
not a part of the domain of our original function. If we take X plus three
out of the numerator and the denominator, we have to remember that. If we just put this right over here, this wouldn’t be the same function because this without
the qualifier is defined for X equals negative three but we want to have the
exact same function. You’d actually have a
point in discontinuity right over here and now we could think about
the vertical asymptotes. Now the vertical asymptotes
going to be a point that makes the denominator equals zero but not the numerator equals zero. X equals negative three
made both equal zero. Our vertical asymptote,
I’ll do this in green just to switch or blue. Our vertical asymptote is going to be at X is equal to positive three. That’s what made the
denominator equal zero but not the numerator
so let me write that. The vertical asymptote
is X is equal to three. Using these two points of information or I guess what we just figured out. You can start to attempt
to sketch the graph, this by itself is not going to be enough. You might want to also plot a few points to see what happens I
guess around the asymptotes as we approach the two
different asymptotes but if we were to look at a graph. Actually let’s just do it for fun here just to complete the
picture for ourselves. The function is going to
look something like this and I’m not doing it at scales. That’s one and this is
1/2 right over here. Y equals 1/2 is the horizontal asymptote. Y is equal to 1/2 and we have a vertical asymptote that X is equal to positive three. We have one, two … I’m going to do that in blue. One, two, three, once again
I didn’t draw it to scale or the X and Y’s aren’t on the same scale but we have a vertical
asymptote just like that. Just looking at this we don’t know exactly what the function looks like. It could like something like this and maybe does something like that or it could do something like that or it could do something
like that and that or something like that and that. Hopefully you get the idea here and to figure out what it does, you would actually want
to try out some points. The other thing we want
to be clear is that the function is also not defined at X is equal to negative three. Let me make X equals negative three here. One, two, three, so
the function might look and once again I haven’t
tried out the points. It could look something like this, it could look something
where we’re not defined at negative three and then it goes something like this and maybe does something like that or maybe it does something like that. It’s not defined at negative three and this would be an asymptote right now so we get closer and closer and it could go something like that or it goes something like that. Once again, to decide
which of these it is, you would actually want
to try out a few values. I encourage you to, after this video, try that out on yourself and try to figure out
what the actual graph of this looks like. ## 62 thoughts on “Finding horizontal and vertical asymptotes | Rational expressions | Algebra II | Khan Academy”

1. greywolf424 says:

You must mean x can not be +3 since that is the only vertical asymptote, x = -3 works just fine

2. BelievinSP says:

You don't have anything on  Power Series, Ordinary, Singular and Regular Singular Points or Fourier Series.  It'd be great if you could do a bunch on this.  Deciphering DiPrima and Boyce's book is a real pain.

3. Amare Tefera says:

I love the tutorial of khan Acady

4. Kepa García de Latorre says:

A parenthesis is missing in the denominator ;)…. great video, as usual!

5. AlfaHanen1 says:

You are one of the best in the World. Thank you for your idea. It helps to understand the world we living in, and hop that we one day can live without War IN Harmony and Peace ONE DAY. Knowledge is the key for that.

6. mrspaz007 says:

congrats on 4,200th video 😉
You've saved many lives including mine!
Thank you!!

7. ⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻ says:

grats on your 4,200 video !

8. xandrora says:

9. Salvador Allende says:

school at its finest! thanks Mr Khan

10. Nisha says:

this problem was a bit more in depth compared to the other tutorials found on youtube but I understood it because of you so thank you!!

11. TheEighthAct says:

Vertical asymptotes at 4:24

12. Mohammed Al-Bashiri says:

Thank you

13. Bahroz Haleem says:

but wouldn't you have either a hole or a Horizontal Asymtote in an equation ???????????????? please answer?????

14. Griffin Hudson says:

Is there sound to this

15. SIBUSISO MASHALABA says:

i love this thing damn

16. Prasanna Kumar says:

Thank You. Great explanation.

17. TheEasyInvestor says:

Great vids

18. Sun-Wukong says:

Great video, but I couldn't find it anywhere on the actual site no matter where I looked.

19. deviltzu says:

Why consider what happens when x approaches negative infinity?

20. Thomas Phu says:

This was extreme helpful but watching video makes me want to swallow my spit when i hear him talk lol

21. Blahblahblah14898 says:

not really helpful when you describe how the first terms will "Dominate". What is that supposed to mean? Please refrain from using buzzwords. I don't have 100% what you are talking about when you talk in this more secluded language.

22. Brent Jones says:

Oh I hate this stuff! I can't do it. I hate math!

23. Gurinder Singh says:

OMG KHAN ACADEMY THANK YOU SOOOOO MUCH FOR MAKING EDUCATIONAL VIDEOS. I GOT 11/10 ON MY ALGEBRA 2 TEST, CONCEPTUAL QUESTIONS PART DUE TO YOUR VIDEOS. I WATCHED YOUR VIDEO AND I AM REALLY HAPPY. THANK YOU SOOO MUCH.

24. Sarah Kelly says:

lost me when he got to vertical asymptotes….

25. Kevin Lin says:

Clearly you understand the content but you take forever to explain one question. A good tutor is able to get their point across to their students. An excellent tutor can get their point across both effectively and quickly.

26. arman sesar says:

it actually looks like -3 is a vertical asymptote once you graph the original function

27. iliriacum666 says:

The same voice in most of Khan's video…you should be a genius knowing all subjects!

28. chris choi says:

finally!! thanks for the tutorial.

29. 케일라Kaylah says:

Thank you!!! I have a calculus packet due and a test on Friday, this is a lifesaver!

30. Holly M says:

Thank you so much!! My teacher failed to explain this concept to our class, so I've sent everyone a link to this video. Really, thank you SO much!

31. Miggy Lumaniog says:

for the vertical asymptote, why not just make the denominator's equation equal to zero like: 6x^2-54=0 then do math. it's much more easier.

32. Sara Mahmood says:

thanks

33. Amir Saif says:

vertical asymptote =-3 not 3 and thank U

34. Jay Dan says:

So what is the significance of vertical asymptotes in practical terms?

35. Brandon Carrillo says:

yo this helped me out a lot, teacher confuses me prob cause I zone out easily

36. Connor S. says:

He teaches an easy topic in the most complicated way possible

37. Aznumi25 says:

1:34 cringe

38. Rosa s says:

if someboy doesnt have any concept of this subject,the explanation seems so insufficient.
I couldnt get the points and failed to follow

39. Prevent says:

I dont understand how they can draw on the computer

40. samady ou says:

Wow Really Nice Tutorial, It Really Helped with my Business Cal problem Thank You Khan Academy!

41. Owen Wu says:

wait do you simplify the equation before using the degree rule to determine the horizontal asymptote?

42. Vape King says:

THANK YOU!!!

43. M W says:

So clear – thank you!!!

44. Sheharyar Bhatti says:

Why -3 is not in the domain of given equation? You should have explained this.

45. Andrew Ku says:

horizontal asymptotes: what happens if the highest degree terms are not the same on the numerator and denominator?

46. Doug Clement says:

These two terms are going to dominate….WTF does that mean?

47. Linus Bao says:

How do u find the y value of the whole discountnity, if i plug neg 3 into x and solve for y it gives my undef

48. xSuper_Nova says:

Actually your explanation is complex a bit but it shows all the steps and why this and that become the answer, I suggest you making a video of a lesson with different explanations for different skill learning level and i think this will be very helpful. Thanks a lot for your efforts 👍👏👏👏

49. Danno [insert last name here] says:

so we would say that there is a hole in the graph at x= -3
right?

50. Scepticul says:

51. Kenny Borek says:

52. Josh Competente says:

Can he ever pronounce numerator and denominator right?

53. Marcelo Ampie says:

I love khanacademy and everything, but i feel this could have been explained easier and faster

54. Owen_newO says:

Sal Khan is a life saver. Much better teacher than Ms. Cornett, who decided it was better to play 2 truths and a lie rather than go over the summer review packet on the first day of school 🙂 And We've got an exam tomorrow :DDDDD. THANKS MS CORNETT

55. Beti A. says:

falling out of focus with sal is never ok. thats like day and night.

56. Sunny Kin says:

Khan Academy is the best one as far as I know. Thanks.

57. Alexandra Trigg says:

I have no clue about the vocabulary but I get the concept so I guess I'm okay.

58. Omar Omax says:

59. Colby Roach says:

Thank you Sal for singlehandedly saving my math grade for the last 3 years

60. Mets7chris says:

Any videos on how to graph something like this?

61. ZingerFlame says:

This is not algebra 2 bro

62. Daniel Leslie says:

Is this Algebra 2 or Precalculus?