You know, for a video made over a decade ago, I appreciate how willing you are to find an easier route. Today's textbooks make this process more complicated than it needs to be and I've always thought the math was a little bit too much work. Sincerely, thank you.
Thank you very much for producing this video. It was incredibly informative and much easier to follow than my Calculus book. Maybe you could help write the examples for a Calculus I book at some point? That would be good.
Hi, Really helpful video. Thank you! Also, could you please show a solved tutorial on how to get the horizontal asymptotes of a function where: a) the highest power of the denominator is lesser than that of the numerator, and b) the highest power of the denominator is greater than that of the numerator, c) functions where there is a natural log of a rational function (like for example ln((x^3 - 3x +2)/(x^2 + 1))
Can you please explain the Vertical asymptote for f(x)= (8x-16)/(x^4-8x^3-16x^2) As I found asymptotes are x=0,4 But @ 0, from left no doubt its negative infinity, but from right its showing 2 behaviors like @ 1 negative infinity, @ 2 zero, @ 3 and above positive infinity, how it is possible to show 2 behaviors with in the range of 0 and 4. Can you please give explanation for the same As soon as possible...
There is a horizontal asymptote at y=0. This is because when you divide x to the other side you get y=6/x and if you would plug in a really really big positive number into x, you would get a very very very small number (but still positive), so we say it is approaching zero. If you would plug in a really really big negative number into it you would also get a very very very small number (but this time negative), so it is still approaching zero. If a function approaches the same number to positive infinity as it does to negative infinity you have a horizontal asymptote at that point (which is a y value since it is horizontal)
I found that there are horizontal and vertical asymptote, both are equal to 0. If i am wrong correct me. I had this for homework, and i was stuck on critical points and searching for minimum and maximum (ps. my professor told us there is only maximum).
Yes there is also a verticle asymptote at 0 as well as a horizontal aymptote at 0, but they are not the same zero. The horizontal asymptote goes from left and right across the graph at y=0 while the verticle asymptote goes up and down the graph at x=0. They are at the same place (the origin) but the way the asymptotes are going makes the difference. to find the minimum and maximum you would take the derivative of your equation. since your equation in 6/x the derivative of that is -6/(x^2). so you would then set the top and bottom equal to zero(separately) to find the critical numbers. in this case the top is -1 and the bottom is x^2.. [These critical numbers are the x values where the original equation is going to have a maximum or a minimum, and also where the tangent line (derivative) is equal to zero. OR when the tangent line does not exist.] If you set -1=0 that doesn't make sense, so there are no critical numbers coming from the top part of the equation. If you set the bottom equal to zero you get x=0 (which is your vertical asymptote so there obviously isn't going to be a maximum or a minimum there). So for your equation there are no minimums or maximums because the tangent line(the derivative) is never horizontal (equal to zero). does that clear it up?
a positive infi. and a negative infi. in limit are conditions for vertical assymtot! vertical assy. occurs two positive infinities or two negative infinities!
Do you know how to factor? Factoring can be confusing sometimes but with practice, it will become one of the easiest thing to do in terms of Calculus. He simply factored the original function (numerator and denominator) and that is what it became. You should have learned about factoring in previous classes but assume that you didn't - I suggest you take some extra time in learning how to factor because it will become essential to your lectures. :)
You know, for a video made over a decade ago, I appreciate how willing you are to find an easier route. Today's textbooks make this process more complicated than it needs to be and I've always thought the math was a little bit too much work. Sincerely, thank you.
Learned more in this 12 minutes than in the 2 hours of reading the book and my instructors notes. Thanks!
Were you in college?
@@justneedlife2001
It could have been high school.
Thank you so much! Such a pleasant voice, style and presentation too! Keep these coming and you may yet replace Salman Khan!
Thomas White mmm ni
You definitely know wtf you’re talking about! This helped me out so much
Awesome video! Almost everyone stops at the first method, so helpful you showed the limit method. Thanks so much!
Years later… this video helped me so much. I love you
Thank you so much for actually showing the vertical asymptotes from both sides; now I understand it clearly!
Really clear and easy to follow, helped me out a great deal. Thanks.
You taught me in 12 minutes, what my maths professor couldn't in 8 classes!!
Great explanation indeed!
Thank you very much! :D
Clear and concise. Your voice makes you sound like the bob ross of math lol
You explained so well. This is just the review I needed for my class. Thank you.
Thank you so much!! My calc teachers doesn’t explain very well, but you did a great job
Thank you very much for producing this video. It was incredibly informative and much easier to follow than my Calculus book. Maybe you could help write the examples for a Calculus I book at some point? That would be good.
Great work man, easy to ubderstand. Very well done thank you!
thank you soooooooooo much helped me a lot more than many books and videos. bless you man
Your voice is so soothing. Omg.
I have a calculus exam later today. Thanks, learned a lot!
omg I'm doing this in precalc and i never understood it until just now--Thanks!
Wowie! Limits in Precal o.o.
best of luck buddy !
Brilliantly explained, thank you very much for helping me with my homework :) Please do more calculus tutorials, you are a great teacher imo.
Very informative video that reinforced my knowledge in limits. Thanks a lot brother!
You're explanation destroyed my teacher's and the homework software we are using. Kudos for knowing your shiz. lol
You just explained the end behavior very great , now i get it
Aced my calc quiz today thanks!
Thanks for the video. I got this wrong on my first test but now I'm prepared for the final!
Very clear and helpful! Thank you so much.
God bless you and your family and your soul you beautiful human being
Thanks so much !! This explanation was so good, clear and easy to follow.. :)
YOURE A LEGEND MATE
Thank you so much, i now feel more prepared for my exam!
Great video, thank you so much!
very straight forward thank you so much
for the first problem, would there be an asymp. at x=-2 if we took the limit of negative infinity?
Nah, it'll still be 2
i love you so much man.
Great video bro, thanks for the help.
Thank you so much sir, this makes things cleaner for me.
god bless you-you are a beautiful soul
Your videos are awesome
Hi,
Really helpful video. Thank you!
Also, could you please show a solved tutorial on how to get the horizontal asymptotes of a function where:
a) the highest power of the denominator is lesser than that of the numerator, and
b) the highest power of the denominator is greater than that of the numerator,
c) functions where there is a natural log of a rational function (like for example ln((x^3 - 3x +2)/(x^2 + 1))
That was an amazing video... I wish ı could find this at morning :( İt was so clear and effective 🙏🏻🙏🏻
How would this work if you have X/X^2. Finding the HA with limits will give you a very different answer which is confusing me.
sir you're great .. respects
I am taking the extra time, hence the fact I'm watching videos like these. Thanks for the explanation though.
thanks man gave me a clearer idea
fantastic!!! thank you so much!
Thank you for your video it help a lot.:)
great video, thanks man!
Great teaching!
Thank you for this video!
did you make any videos on intergration by part ?
thanks you, much appreciate
Thanks for the help!
Can you please explain the Vertical asymptote for f(x)= (8x-16)/(x^4-8x^3-16x^2)
As I found asymptotes are x=0,4
But @ 0, from left no doubt its negative infinity, but from right its showing 2 behaviors like @ 1 negative infinity, @ 2 zero, @ 3 and above positive infinity, how it is possible to show 2 behaviors with in the range of 0 and 4. Can you please give explanation for the same As soon as possible...
What if there is no denominator? or exponents in the function?
Can you please tell me are there any asymptotes in this function: xy=6
There is a horizontal asymptote at y=0. This is because when you divide x to the other side you get y=6/x and if you would plug in a really really big positive number into x, you would get a very very very small number (but still positive), so we say it is approaching zero. If you would plug in a really really big negative number into it you would also get a very very very small number (but this time negative), so it is still approaching zero. If a function approaches the same number to positive infinity as it does to negative infinity you have a horizontal asymptote at that point (which is a y value since it is horizontal)
I found that there are horizontal and vertical asymptote, both are equal to 0. If i am wrong correct me. I had this for homework, and i was stuck on critical points and searching for minimum and maximum (ps. my professor told us there is only maximum).
Yes there is also a verticle asymptote at 0 as well as a horizontal aymptote at 0, but they are not the same zero.
The horizontal asymptote goes from left and right across the graph at y=0 while the verticle asymptote goes up and down the graph at x=0. They are at the same place (the origin) but the way the asymptotes are going makes the difference.
to find the minimum and maximum you would take the derivative of your equation. since your equation in 6/x the derivative of that is -6/(x^2). so you would then set the top and bottom equal to zero(separately) to find the critical numbers. in this case the top is -1 and the bottom is x^2.. [These critical numbers are the x values where the original equation is going to have a maximum or a minimum, and also where the tangent line (derivative) is equal to zero. OR when the tangent line does not exist.]
If you set -1=0 that doesn't make sense, so there are no critical numbers coming from the top part of the equation. If you set the bottom equal to zero you get x=0 (which is your vertical asymptote so there obviously isn't going to be a maximum or a minimum there).
So for your equation there are no minimums or maximums because the tangent line(the derivative) is never horizontal (equal to zero).
does that clear it up?
It was very helpful indeed.
I've always loved Asymptotes for some reason
Very talented
what if we dont know the function, but only have the graph
a positive infi. and a negative infi. in limit are conditions for vertical assymtot! vertical assy. occurs two positive infinities or two negative infinities!
thanks. Very helpful.
you're awesome!!!!
high quality! :)
thank you that was realy helpful
Sure that helps thanks man!
well done! thank you!
thank you this is great
Thank you ❤️😘
really it helps thanks a lot:)
OH MY GAWD THANK YOUUUUUUUUUUUUUUUU!!!!
Very helpful thanks
You rock
tnx you saved me !
Do you know how to factor? Factoring can be confusing sometimes but with practice, it will become one of the easiest thing to do in terms of Calculus. He simply factored the original function (numerator and denominator) and that is what it became. You should have learned about factoring in previous classes but assume that you didn't - I suggest you take some extra time in learning how to factor because it will become essential to your lectures. :)
great help thanks!
thank yoou very much as possible!
Thanks for describing without writing out tables
This video helped so fucking much thank you!
how did u factor the bottom 😑😑
He used difference of squares to factor the bottom.
Thank you
Thank you.
Nice stuffs
Thanks so so so much
really helpfull thanks
U R Legend
yayyyy im the 10,000th sub
Thanks!!!
omg why didn't i find this video 5 hours ago
thanks man
thank you :-)
U sound like Gearhead from Hot Wheels: Velocity X.
good , sir
coulda swore this was a white dude till I saw the hands lol
super helpful video!!
good viid thanks
Thanks
tnx
great
it did
you lost me at 6:00 buddy--have no clue where you got that equation from the original one :/
i love u
a nice video but you didn't graph a thing sir tat's the most difficult thing in the world