❖ Finding All Asymptotes of a Rational Function: Vertical, Horizontal, and Slant/Oblique ❖

[@Joe230577]

Haven't been to school in six years and this first semester back in college has been a lot easier thanks to your videos. You're a good man for doing these and I appreciate everyone. Please keep doing such great work!

[@jeydkabigas5193]

Im grade 11 damn and we take this as a gen math and my strand is humss

[@NicholasJMV1981]

Be careful. Even if the powers of the numerator don't equal the denominator, there could still be a HA (the denom) has a larger power than the numerator (y=0). I think you misspoke.

[@MandaPanda3100]

Why can't this guy be teaching my college algebra class?

[@zepar6076]

+MandaPanda3100 go contact him and pay him a small salary of billion dollar each month(:

[@projectjt3149]

He is teaching in colleges. His website mentioned he used to teach in Vanderbilt University and still teaches in a community college.

[@STONEAJE]

Your videos are amazing because they explain everything properly but do so in a short amount of time. Thanks!

[@dondigga1757]

Guys I think one should also consult a textbook before or after watching a video. Patrick here made a huge mistake saying "horizontal asymptote exists only when the degree of the numerator is equal to the degree of the denominator, whereas horizontal asymptote exists when the degree of the numerator is is LESS THAN or EQUAL TO that of the denominator... but we all make mistakes, he was mistaken there.

[@OMaMaRMY]

Loafus Kramwell lmao ikr

[@adriancodmw3]

You have literally saved my life like a thousand times. Thanks

[@jcd0402]

Thank you so much. You are a great savior, and a great person for taking your'e every day time to video tape this videos. It took me less than 5 second for everything to make sense. Yesterday in school i was in so much pain for not being able to answer the questions asked to me . Now i now what vertical, horizontal, and obliqe asymptotes are. Now let's see how i go in my test tomorrow. I will post my grade here hope i pass. Again thank you have a nice day.

[@recoil33]

Test in 2 days, your videos are great for revision trying to stay one step ahead.Thanks you so much Patrick Jones.

[@phoenixwings6441]

Thank you so much. I will spread the word. My school has "lecture" videos posted online about this, they don't come close to how well you explain it. Please keep making videos!

[@Triginta37Septem]

I've been watching these videos every night this year and just realised you write with your left hand... ^^ Anyways, thank you so much!

[@craig5716]

Thanks Patrick! Clear voice and neat writing, thumbed up.

[@dewaningram6009]

You are really a life saver. I am making an A in my calculus class because of your material.

[@jewcreww]

i have a test tomo...THANK YOU for this video you just saved me so much time spent staying up late trying to figure this out

[@allykelly6048]

Why didn't i see this weeks ago! I've been stressing over my teachers lecture videos because I don't understand her and I understand it all in 10 mins from your videos!

[@lazyboy395]

for the sake of perfectioninsm, its a 7 min video

[@lazyboy395]

lazyboy395 just sayin

[@allykelly6048]

Do you feel better?

[@twothirdsostrich]

Just found your videos, and you've managed to explain this way better than my professor ever did. Thank you so much for posting. I'll def be subscribing for more! :)

[@95Darkie]

wow ur a great help in helping me to achieve good result in my pure mathematics subject.

[@patrickjmt]

@leafspie2020 you are very welcome!

[@heidyfgt]

this is my second time taking college algebra and i always thought it was "asymptope" lol

[@adithigudipati2962]

My teacher calls it asmatote

[@patrickjmt]

keep watching, subscribe, and spread the word :)

[@shaidasabr5272]

Hello dear patric i have some questions how can i contact with you?! If its possible can send for us your email

[@I54D0R4]

I swear to God, you are saving me right now. Boo, you don't understand, thank you so much!

[@DarkDreamer31]

there he goes, patrick saves me again.. can't thank you enough..

[@desyanuspatiallo4959]

i'm nothing in calculus thanks for uploading your video God bless you forever Sir!

[@patrickjmt]

happy i could help :) spread the word about the videos! :)

[@Dat-plati]

Where did you get the 0x from?

[@osteetheman]

i have pre-calc exam in 3 days that im freaking out for. my prof just talks too dang fast and doesnt explain things step by step. thank you for these videos. ill be watching this one alot over the new few days

[@matthewwroblewski8752]

As always, thanks Patrick

[@samsonbb8279]

your are the best teacher!! thanks man :)

[@thekodajay09]

Definitely just bombed my test on this material just this morning, ahhh! Where was this about a week ago? Haha, thanks for the upload though, for real. :)

[@MrJakeepooh]

Man, that was an excellent explanation. Very clear and easy to follow. All hail hydra! Der, I mean youtube!!

[@vinsanity_52]

Great vid! You're an amazing teacher/tutor!

[@patrickjmt]

@thekodajay09 lots of other examples of this by me already out there :) sorry about the test!

[@mryip06]

It seems that you missed a case When the degree of the denominator is greater than that of the nominator, there exists horizontal asymptote which is y=0. Right?

[@frodo_banggin]

that was what i thinking, too

[@patrickjmt]

@daggerGforcefahrer how could there be any others? i assume it is clear if you stop and think about it

[@VIR092]

This one was certainly very interesting.

[@patrickjmt]

@HanDynasty26349 ha, good timing! :) hope it helps

[@hamsterball.0]

I'm going to quit math, this is my final message good b ye

[@cialee76]

you have saved my life! ♥

[@bboty7344]

THANK YOU MAN < YOU HELPEDME ALOT

[@narminaskerzade3171]

i have found your videos just some days ago,and i have learned more from your videos than on math lectures.Thank you so much! :) I have a midterm exam tomorrow,and you really saved me from getting "F"(well,i hope i won't get it )) And I have a question,what if the function is not the ratio? how we find asymptotes in that case?

[@missgodessful]

In his other video, "shortcut to find horizontal asymptotes of rational functions" he also says that if the degree of the denominator is greater than the degree of the numerator y=0 is the asymptote... if the degree of the numerator=the degree of the denominator, there is a horizontal asymptote at the ratio of the leading coefficients.

[@hannahschneiderman898]

Couldn't you use Synthetic Division instead of Long Division? I find it faster and it takes up much less space on the paper.

[@christianondo9637]

I think he just also wanted to show long division since it is applicable to divisors of all degrees.

[@09inyourbasement]

you are a beautiful and very smart human being thank you SO much.

[@Sparkylove0219]

wow you are literally perfect

[@UPproductionZ]

Actually, I don't think 'literally' can be used that way.

[@Sparkylove0219]

you know what i mean!! :)

[@Sparkylove0219]

AnUnprofessionalProduction You know what i mean!! :)

[@brwaomer]

Thanks bro it really helped

[@shinnsohai]

wootz thanks alot for this video helping alot!

[@joonchild.9393]

My math savior.

[@HeavySpinna415]

These were fun back in pre cal

[@nickturner5721]

Yes, it's possible! Try graphing y=(1/(x-2))-x^2+4 :)

[@suprateekat5338]

This is awesome! A trick to learn in maths (y)

[@BarbaraAbrabra]

So, if there is oblicual aymptotes there are no horizontal???

[@marekc861]

Yes, good luck

[@Lazyskins82]

Thank you so much!!

[@darkcreepyangel5455]

if the number is odd, do we really have to put square root?

[@frodo_banggin]

was that suppose to be top x degree less than bottom x degree so to have a horizontal asymptotes? and suppose when top x degree equal to bottom x degree then it should be approaching to (coefficient of top x / coefficient of bottom x) as x → +infinity?

[@TheSlimyDog]

Another way that I seem to remember horizontal asymtotes/oblique asymtotes is just by using the same formula for both and saying that is y=0x+c then it's horizontal.

[@Zachbjj]

Good god, you're helpful.

[@michaelkoenig7812]

"Horizontal asymptotes only exist if the degree of the numerator is equal to the degree of the denominator." ... FALSE! They only exist if the degree of the numerator is less than or equal to the degree of the denominator. If the degree of the numerator is less than the degree of the denominator, the asymptote is always going to be y=0. Otherwise, it will be the ratio of the coefficients of the leading terms.

[@KingBarney]

Yeah, he made a mistake there. :P

[@barrydavis331]

It will be the ratio of LCs + vertical translation.

[@bonbonpony]

What about implicit equations that cannot simply be rewritten as a rational function and divided?

[@XxStopherLixX]

PAHAHAH. I have my final in like an hour thanks dude.

[@pinkpencil07]

holy shift! look at the asymptote on that mother function!

[@ernestochvv]

You're awesome!

[@prathambabaria5834]

What did you major in?

[@juanpedro19840914]

Excellent!

[@howardbaxter2514]

There is a horizontal asymptote if the denominator has a higher degree than the numerator. And you can only have a linear slant/ oblique asymptote if the numerator has one degree higher than the denominator but you can still have a oblique asymptote if the numerator has a degree that is 2 or more degrees higher than the denominator. I am learning this in Pre-Cal right now.

[@GiftWraper]

wow

[@garywhite8274]

"And you can only have a linear slant/ oblique asymptote if the numerator has one degree higher than the denominator but you can still have a oblique asymptote if the numerator has a degree that is 2 or more degrees higher than the denominator." You've made 2 conflicting statements here. I believe that only the first is accurate (i.e. "you can only have a linear slant/ oblique asymptote if the numerator has one degree higher than the denominator").

[@patrickjmt]

asymptotes by definition are lines.

[@Yamnkelaniyonela]

Very much helpful :)

[@Shashikant-iu6xx]

find all the asymptotes of the following curve show that three asymptotes meet the curve again in three point which lie on a straight line, find the equation of the line: 3x²+2x²y-7xy²+2y³-14xy+7y²+4x+5y=0

[@epruitt]

Thank you!!

[@darkcreepyangel5455]

I've watched 6 videos already in solving asymptotes! but all of you are using different kinds of methods! huhu which one should I use? our test is next week. And my brain is shaking!!!!!!!

[@GooseTheFlick]

How's that going for you?

[@janicaalexandraalvarez4277]

true true different methods.. ...different from my teacher's method!!

[@projectjt3149]

Does the "degree of a polynomial" still come into play when there's a constant in the numerator and some power of x in the denominator, i.e. 1/x?

[@patrickjmt]

1/x is no longer a polynomial, but a rational function. However, if you look at the ratio of the degrees (numerator/denominator) you can still glean some info

[@thanossmith1126]

I wish this guy is my professor

[@neverenough1415]

is there anywhere that the rules are written down

[@epifanioinfante695]

so what is the domain and range?

[@jadem.7227]

Does your last name sound like 'burger'? Because I think I remember your voice and handwriting from 6th grade... You were the video guy from my textbook.

[@patrickjmt]

+Jade M. My last name is flurger

[@jadem.7227]

+patrickJMT have you ever written for Holt?

[@ramichoeb2743]

Please please someone help! Why does he add the 0x while doing the long division? i get everything but I dont understand the adding of 0x !

[@ramichoeb2743]

is it because we have 2 terms ?? the ( x+3 ) ?

[@hoodjedi7115]

Rami choeb so that it lines up right in decending order. he didnt want to skip the 0x because it would throw off the problem.

[@DocHolidayKY99]

In division of N/D or simply Long division , etc... You have to fill in the missing degrees respectively. If you have X^3 +1/X you fill in the Numerator respectively going down the Degrees of X. So X^3 + 0X^2 +0X +1 Otherwise it will mess up your whole division.

[@Nic_0404]

Got my AH maths nab today at school, desperation lead me here. Hope this technique works...

[@patrickjmt]

hope it works? is all of this stuff made up and fake?

[@Levantinelyjan]

what grade would you learn this? is this calculus?

[@shahzaibchauhan8692]

Can u pls tell d name playlist of asymptotes in ur accnt?

[@Gamefreak0101]

Why did you use long division? Synthetic division is much easier (and faster too)! Great video by the way.

[@indigopeach]

What about curvilinear asymptotes?

[@YTD07]

your the best......

[@alyssafavela1096]

how do we identify any holes in the graph?

[@neverenough1415]

THANK YOU

[@yourtypicalniceguy]

I LOVE YOUUUUUUUUUUUUUUUUUUUUUUUUUUU

[@kichigan1]

So the asymptote of the denominator is the value of x which would make the expression undefined?

[@TerryTWeiss]

+Juan Mendez Yes, which is why the main function has to avoid it.

[@GuerreroF5]

huh? but eh 5 and x squared were not in perenthesies so how did u do so

[@songran258]

thank you!!!!

[@dinahk8386]

can u make a tutorial graphing the rational function by its asymptotes

[@ahuman9113]

You could synthetically divide too..its much easier, but only if the factor is linear..and in this case it is

[@terrytownsend5404]

I would say that the numerator is a difference of squares, but not perfect squares. Yes, it factors as you indicated, but I'm just being technical with your explanation.

[@reginahunstock3968]

So how do I find horizontal asymptotes?

[@juanpedro19840914]

True. But a curved line is still a line, so I will take your answer as a "yes".

[@7ommus]

isn't there another case where the degree of the denominator is greater than the degree of the numerator then y=0 ?

[@phendformyself]

Good video, but why didn't you use synthetic division? And can't there be horizontal asymptotes when the degree of the denominator is larger than the numerator?

[@FritzyNB]

You can use either long division or synthetic, it is up to the person. However if you're dividing by a square you can't use synthetic. Also I believe if the degree of the numerator is less than the degree of the denominator then it can't have a horizontal asymptote.

[@garywhite8274]

FritzyNB If the degree of the numerator is less than the degree of the denominator then the horizontal asymptote is y=0. Watch his "Shortcut to Find Horizontal Asymptotes of Rational Functions" video. I'm not sure why he states otherwise in this video.

[@garywhite8274]

FritzyNB Also, this example is not dividing by a square but rather by x^1 + 3 so synthetic division is entirely appropriate here.

[@minhaoling3056]

For long division part , the numerator should be shifted to the right to fulfill the degree of x

[@SalemInsomniac]

[@SteelersFans99]

Are Oblique and Slant Asymptotes the same thing?

[@howardbaxter2514]

Yep

[@liverpoolks]

can u give me solution for this function 2x in square -5x+4/3x-6 ???

[@killsarawr]

woah weird, we just did this today in class

[@Bruja989]

does he have a video graphing everything?

[@yangjela11]

wow helpful!

[@MrGuywhodoesnothing]

but i just learned last week that there can be parabolic asymptotes when the degree of the numerator is exactly 2 more than in the denominator