Find the Vertical, Horizontal and Slant Asymptote
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- เผยแพร่เมื่อ 7 ก.ย. 2024
- 👉 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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I know you dropped this 8 years ago, but thanks anyways man. I got a Pre-Calc test tomorrow I've been studying for, and all I forgot to take notes on was slant asymptotes.
this video came out when i was in fifth grade ..... now i’m in my second semester of college and i use your videos almost every night for my math hw. THANK YOU!
some people still need that intimate teaching, but youtube has allowed a lot of students to skip tutoring for sure
but why dont we use remainder as slant asymptote. ithink thats what she was asking towards the end
THANK YOU SO MUCH, you've literally saved my pre-calc grade. if i hadn't found your videos i probably would've failed my final but i feel so much more comfortable with the material, i can't thank you enough!!! :)
esraa sammme
how was the pre calc final , im here for the same reason lol
LOL hi i honestly don't remember but i go to usc marshall now so don't stress😭@@deathreaper5226
This was 11 years ago!? Man how time flies, I remember watching these videos back in high school, didn't seem that long ago until now
hey it's not simple, keep up the hard work, glad I was able to help you out. Stick with it
Sir, I am very thankful to have teachers online to help with my pre-calculus and I'll make sure to thumb you up. It is indeed a good asset to have people who use TH-cam in a positive manner.
didn't know randy orton could teach maths so well haha xD
now you know, haha
You’re doing gods work on this channel. My teacher just gives us packets that she finds online with fill in the blanks and I don’t learn anything from them. I also don’t get engaged in the lesson. These videos are my savior before tests and thank you for helping me pass my classes :)
well when we multiply (x-2) on both sides we have a couple things happen. On the left side of the equation we are multiply (x-2) by 0 so the left side will be zero. When we multiply (x-2) on the right side we have (x-2) in the numerator so that will divide to one leaving us with 0=2x^2-5x+5 per the video. Am i missing something with your question or did I clear it up. Let me know
I got excited about synthetic division because you taught me how to do that. You are a GREAT teacher. You are a blessing 🙌
This is awesome! Unfortunately, my math class is my first class of the day and I am usually still half asleep... Thanks so much!!!
Mr. McLogan coming in clutch for my Pre-Calc test tomorrow, thanks!
Your videos are great. They’re an amazing supplement to my class and you’re great at teaching
Thanks Eric
How are you such a good teacher?!?!! Saved my ass for a test tomorrow! Thanks Mr. McLogan!!!
Have a test tomorrow and this helped a lot. Thanks
brother , you are one hell of a teacher
Man, I don't even know why people would even pay for tutors anymore. This video really helped, thanks!
god i fucking wish you were my professor your voice is soothing and you explain everything great and it seems like you enjoy teaching math
I do, and am very happy to help you out
sweet so glad I could help you out!
this man is my math teacher and he doesn’t even know my existence
what are you referring too?
really helpful!!! but try to look at the camera
Bro you are the goat
how can i pay attention to these videos but not my own professors lectures?? somethings not adding up LMAOO
Do you have a video where you teach more complicated problems where the vertical asymptote would require a bit more work? As in factoring, grouping, canceling?
of course here you go th-cam.com/play/PL0G-Nd0V5ZMrCy9FP2EeZRJUlawuGJ0xr.html
th-cam.com/play/PL0G-Nd0V5ZMqmhh50eb84f5hePlrT7BrL.html
You are really a semester saver 😭😭❤
i have a question why you multiple with 2. where"s you get this number?
This really helped reinforce my memory for my math final.
awesome! best of luck!
You just saved my grade from tanking. Thank you my brother
God my teacher explains everything so horribly, your videos are so clear and concise...thank you !
Didnt you say in the previous video that if the top x has a higher power than the x in the denominator, then the horizontal asymptote is y=0? but in this video you said it does not exist. could you please clarify. thanks. really like your videos btw.
Ibrahim Ansari Horizontal asymptote is different from slant asymptote
If top power is more than denominator...then there is no horizantal asymptote...but there will be oblique asymptote if degree of numerator=(degree of denominator) +1
such a great word
You are magic, man! I finally understand!
You just saved my algebra grade for the week,thank you soooo much in a matter of 1 minute and 42 seconds you cleared up what i'be been struggling with,Thank you!!
this was so helpful omg
your videos are sooo helpful, thanks
you did synthetic division in the video, but if you were to do long division would you come up with a different answer? because I keep getting 2x+1 (And a remainder of 7 over the divisor) instead of the -2x-1 that you got in the video.
Thank you This was very helpful!
you are very welcome!
Very informative.. love so much
just do long division (x-2)/ 2x^2-5x+5 then we have salt asymtote y= 2x-1
yes
thanks
great explanation
How do you make this look so easy?!?! Every single time I attempt to do this I can't.
chymoney1 a lot of practice, it took me many times to learn this stuff. Struggled just like you over and over. Keep working hard and it will eventually come to you
what if the leading degree of both numerator and denominator is equal? does it have a Vertical asymptote or horizontal asymptote?
the horizontal asymptote will always by y=0 in the scenario, the VA remains the same, what values of x make the denominator equal to zero
oh im sorry, what i meant was if it is a slant or horizontal
love ur vids but u did not show how to solve to horizontal asymptote this time
wish a recommended graphing video was next so i could start graphing it.
Still this video waching.......after 10 years
dont we get 2x^2-6x+ 7 after we mutiply both side by (x - 2) ......x-2 = 2x^2 -5x + 5 ....minus x and plus 2 from both side .
I can't put my finger on it, but I like this guy. Subbed.
appreciate it, and happy to have you
So what is h(x)=(x^2-25)/(x^2+2x-80)? Please tell me 😢
the horz assymptote will be the numerator ld coeffecient / denominator ld coeffecient while vertical asymptote conditions remain the same
Amazing... thank you so much!
you are very welcome Alki!
Hi, I don't know if you still reply to a video from 8 years ago. First off, appreciate the lesson. It was very helpful in understanding asymptotes and finding them.
Question: What if there is a function with no denominator, in that case, there won't be any horizontal asymptote because of the top> bottom rule. But, what about the vertical asymptote, I can't equate 1x^0 =0 to find the same? What happens here?
thirdly, in graph transformations, are the asymptotes to be kept at x=0 and y=0, to begin with, and make changes to them based on the transformations? like for eg: f(x) = 3^(x-1) - 2
here, in the 1st step y =3^x , the horizontal asymptote was at y = 0 and in the next step f(x) = 3 ^(x-1), also the horizontal asymptote was at y=0 and finally, during the f(x) = 3^(x-1) -2 transformation, the horizontal asymptote moves down by 2 units and is at y =-2. Could you please explain this?
I'd greatly appreciate your help. Thanks in advance.
Hi Minus, yes I still try to answer as many comments as I can. You have a lot of questions and I cannot always answer them but will do my best 1.) It depends on the function, if there is no denominator then yes you do not have to worry about the denominator equally zero for a vertical asymptote. However other functions have vertical asymptotes and no denominators, i.e. logarithms, tangent, cotangent, secant, cosecant. 2.) again depends on the asymptotes of the original function, if your function is 1/x then yes the vertical asymptotes are at x=0 and will move based on the transformations. The function you provide does not have a vertical asymptote but a horizontal asymptote at y=0 and since the graph is being shifted down 2 the new horizontal asymptote will be ay y=-2 Hopefully this helps
They told me we can't multiply zero by x-2 in 1:00. Can you help me Mr.McLogan?
you can but you also can think of it as this: if y/x = 0, then y must equal zero because if x were to equal zero, then it would be undefined. its basically the zero product property
How's that not factorable?, If you can use (x-5) and (2x+5) as factors that will give you the middle term which is 5x?
multiply it out, it does not equal the equation i.e. 5 *-5 is -25
@@brianmclogan Oh, crap
you should get the same answer, you are probably making an algebra mistake or setting up the problem wrong.
do you have any videos about rational inequalities?
ummmmm no I do not believe I do, same process, just use test points at the end to determine where to shade
thank you
you are very welcome!
How to find slant if its 4 cm greater than its projection
The distance of slant from the plane is 8 cm
Ohh so you basically ignore the remainder. Is it all the time tho
It wouldn't be 2x^2-6x+7?
How do you find the holes
holes are discontinuities that are divided out
wait this dude doesn't age! I watched another video from 2017 and he looks the exact same 😳
Bro its been 10 years. you okay there?
beauty!!!
Ahhh I need help with finding the ha when it's like x^2+2x over -4x+8 😭😭
Ajel Laqe there is no HA because the degree of the numerator is larger than the denominator therefore there is a SA instead
What was the Horizontal?
there is none
Does he say astmptope?
how many times he gonna say therefore
lol math teachers always talk alittle like robots XD
Is finding the vertical asymptote the same as finding the domain ?
chymoney1 finding the vertical asymptotes helps us determine the domain because at each asymptote the function is undefined. So therefore usually you can say all real numbers except for the x-values where a vertical asymptotes is
Thank you so much, your videos are very helpful.
Why didn't he do quadratic equation to find the x intercept
because the x-intercepts are imaginary, so what is the point on knowing the imaginary intercepts?
goat
aspmtopes lol
istill dont get it
im so lost
🎸
rock on
The adds you placed are encouraging people not to watch your videos. Talk about irony.
I do not place any ads, TH-cam does that, I just allow them
who cares? Its way cheaper to watch an ad than to pay for an overfilled college course where its easy to miss steps. or just get an adblocker like most people do.