How to Find Slant and Vertical Asymptotes

👉 Learn how to find the slant/oblique asymptotes of a function. A slant (oblique) asymptote usually occurs when the degree of the polynomial in the numerator is higher than the degree of the polynomial in the denominator. To find the slant asymptote of a rational function, we divide the numerator by the denominator using either long division or synthetic division. The quotient obtained when the numerator is divided by the denominator is the slant asymptote of the function.
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Organized Videos:
✅ Find the Asymptotes of Rational Functions
• Find the Asymptotes of...
✅ Find the Vertical and Horizontal Asymptotes of a Rational Function y=0
• Find the Vertical and ...
✅ Asymptotes of Rational Functions | Learn About
• Asymptotes of Rational...
✅ Find the Asymptotes of a Rational Function with Trig
• Find the Asymptotes of...
✅ Find the Asymptotes and Holes of a Rational Function
• Find the Asymptotes an...
✅ Find the Slant Asymptotes of the Rational Function
• Find the Slant Asympto...
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[@jitdas99]

I plotted the function in Grapher app and didn't show x=1 as an vertical asymptote. This may be due to the fact that we can factorise the numerator and denominator and simplify the expression.

[@brianmclogan]

that is correct, it is actually a hole of the function

[@brianmclogan]

there are actually a lot of us out there, making great videos. Glad you found me and mine and that they work for you!

[@thespicehoarder]

There needs to be more teachers like you out there. Thank you so much!!

[@elismith4040]

I can not thank you enough for all of your videos. You are a legend and I hope that youtube pays you well for efforts. Thank you for making learning awesome

[@brianbotello342]

every math teacher I've had has been bald by choice or by misfortune, just a coincidence

[@brianmclogan]

use and abuse the videos, they are for you!

[@angelmendez-rivera351]

(x^3 - 1)/(x^2 - x) = (x^2 + x + 1)·(x - 1)/[x·(x - 1)]. Because of the common factor of x - 1 in the numerator and denominator, there is no vertical asymptote there. This is a hole. As for the oblique asymptote, this is best found by doing division. When you remove the hole at 1, you get (x^2 + x + 1)/x = x + 1 + 1/x. The polynomial part is x + 1, and so the oblique asymptote is given by y = x + 1.

[@Jezebel1115]

I wish you were my teacher...oops you are. Thanks for the videos! Get it you are, bcuz I watch your videos.:)

[@laurenalongi5439]

Thanks!

[@brianmclogan]

You are very welcome!!

[@kealilasok5005]

Lol the as before this was a “study pod” video that began with “are you really still looking on youtube for math videos? Are those 6 year old videos really still working?? Try study pod.. blah blah” and like yes these videos do help lol

[@augustayarteh7524]

I have to factor the denominator and equal it to zero. Even the numerator. I have to factor them and equal to 0 after that i have to solve them to get my VA. I love your explanation. Professor you are doing great job sir.

[@sharakam846]

يجب تحليل كل من البسط و المقام ثم الاختصار ثم ايجاد المقاربات العمودية

[@danielivey562]

Someone coming over the intercom to announce that students are to be marked tardy if they are not back in their classrooms after a certain number of minutes.... We've heard it once or twice before. XD

[@emresagdik8160]

x=1 is a gap it is not asymptote.