i like how you derive stuff. i see other youtubers just giving formulas and plugging and chugging
@celebrity_rooster74884 жыл бұрын
Exactly. Just write that shit on paper.
@deenessop59116 жыл бұрын
Hey Eddie. About that proof for the equation of the asymptote, I was just wondering that isn't the multiplication rule for limits: that the lim x-> infinity [f(x)] multiplied by the lim x -> infinity [g(x)] NOT the same as the limit of x-> infinity [f(x).g(x)] I've been taught that multiplication rule for limits only holds true for when x-> a constant value. Is that a correct proof then? Crown Academy > DrDu. Gvng gvng. Eskeetit mums Hehexd
@jeejee30913 жыл бұрын
no what Eddie's multiplication rule is the right one
@joshgun53112 жыл бұрын
The angle between the hyperbola asymptotes is 60 degrees, find the eccentricity, how do we do this question?
@LimeriuxasLT10 ай бұрын
hello nice video but sorry not following how u got 1-(a^2/x^2) at 3:48.. cause we divided.... divided what??
@TheDanubeDepleter9 ай бұрын
He divided the entire radicand by x^2/a^2
@NadeemShaikh-ot8yl4 жыл бұрын
This is a very nice trick!! Thanks a lot.
@warrenw68883 жыл бұрын
Isn’t using linear equation the same thing since it’s a straight line?
@christiancarter255 Жыл бұрын
Yes.
@guitarttimman4 жыл бұрын
That's good reasoning.
@kevinding12043 жыл бұрын
ahhhh I would love to like this video so much but the total amount of likes right now is at 314 and I don't want to ruin it :((((((((((
@Sobioytccc3 жыл бұрын
Wonderful explaination 👍
@zizo-ve8ib4 жыл бұрын
May I ask where the rest of this part is because I can't find the video where you do the rest of the derivation
@evajaymusic13223 жыл бұрын
I like it
@ugursoydan81874 жыл бұрын
it was very helpful for me. thank you
@aashishrimal5793 жыл бұрын
Love from nepal
@cavelinguam6444 Жыл бұрын
Reallllly helpful
@aryanbansal6245 жыл бұрын
could you please please arrange the videos it is really hard to watch all hyperbola asymptote videos in order please reply dont ignore it is a humble request
@andeslam73706 жыл бұрын
would like to know how to infer -bx/a after knowing bx/a is one of the asymptotes
@Zosso-16185 жыл бұрын
Not sure how helpful this is a year after, but I'll give it a shot. When you square root something, there is both a positive and negative variant. The same is true for here because when he took the square root of both sides, there is both a positive and negative version. Like Mr. Woo said, you can just take bx/a and slap a negative onto it to get the other asymptote. I hope that helps!