What is the oblique asymptote of the function f(x) = the quantity of x squared plus x minus 2, all over x plus 1? y = x + 3 y = x y = x - 1 y = x - 3 @satellite73
i guess using the equation editor is out of the question huh? \[\frac{x^2+x-2}{x+1}\]
should i use it?
O.o
no no i got it the "all over" and the "plus" is kind of cute
lol ur soooo funnyy omg u always make my day when im most upset lol
what you need to do for this one is divide synthetic division is easiest, long division is annoying
yeah i got it but idk which one is the answer http://www.wolframalpha.com/input/?i=%28x%5E2%2Bx-2%29%2F%28x%2B1%29
ok let me show you how to interpret the answer you got from wolf
alright
you see where it says "quotient and remainder" and has \[x^2+x-2 = x × (x+1)+-2\]
yea
oops it has \[x^2+x-2=x(x+1)-2\] that means \[\frac{x^2+x-2}{x+1}=x-\frac{2}{x+1}\]
is it y = x-3?
your slant asymptote is therefore just \(y=x\)
oh oops
no it is just he quotient, ignore the remainer
oh gotcha okayy!
we can do another if you like, make sure you know how to get to the answer from the wolf
yeah that'll be great I always cant seem to get the answer off of wolf cuz it's not that obvious for mw
you got one or you want me to make one up better if you have one, kill two birds
LOL okay i'll look for one
ok i will wait patiently
okay i have one but instead of asking for the oblique asymptote this one is asking for a vertical asymptotes is that okay?
yeah but that is way easier no wolf needed because you dont have to divide anything
go ahead we can do it in five seconds probably with our eyeballs depending on the question
yeah u have to divide in my question
ok post !
What are the vertical asymptotes of the function f(x) = the quantity of 5 x plus 5, all over x squared plus x minus 2?
you crack me up
\[f(x)=\frac{5x+5}{x^2+x-2}\]
how?
the quantity of 5 x plus 5, all over x squared plus x minus 2?
no u said u crack me up lol
it is cute, very literal i like it in any case lets do this quickly, you want the VERTICAL asymptotes right?
LOL okay yeah
we do this by setting the denominator equal to zero and solving for \(x\)
\[x^2+x-2=0\\ (x+2)(x-1)=0\\ x+2=0,x=-2\\ x-1=0,x=1\]
two vertical asymptotes are \(=-2\) and \(x=1\)
i got x = 1 and x = -2
YAYYYYYY!!!
i got it *victory dance* oh yeaahhh! LOL
I was able to find that on wolf!!!!! :D
cheater
LOL jellyyy!
i meant CONGRATULATIONS
xD THANK YOUUUU!!
do i get a medal? no? no? ok xD
yw you done?
ummmm... idk
you don't know if you are done??!
xD IKR.. kinda crazyy but ik im not done it's just that i dont know if i will need u now cuz if i can do others by myself no need to bother u but if im stuck i'll surely bother YA!! hahah
ok i will do some actual work now if you need help just post
Okayyy!! gracias!
ATTENTIONNNNNN!!!!!
I NEED HELP RIGHT AWAAAYYY!
zzzzz
lol WAKE UP!
i think i have the answer to this one but not sure.
ok
Steve can paint a room three times as fast as Billy. When they work together, Steve and Billy can paint a large room in 4 hours. How many hours would it take Billy to paint it by himself? sixteen-thirds <----- MY ANSWER three sixteenths 16 3
@satellite73
hold on a sec
okk
one rate x, the other 3x
for a total of 4x
yesss
4 HOWAAAS
so \[4x\times 4=1\] making ' \[x=\frac{1}{16}\] and \[3x=\frac{3}{16}\]
oh c'moooon I FLIPPED THEMM!! lol
i'm soo dumb LOL
no it is ok, we still have to flip one is 16 hours the other is 16/3 hours
but you want the time for the slower one, so i guess you are a bit dim
try 16
lol so is it 16/3 or 3/16 cuz i'm confused
it it is neither because you want the time for the SLOWER one not the faster one faster one's rate is \(\frac{3}{16}\) takes \(\frac{16}{6}\) hours but the SLOWER ones' rate is\[\frac{1}{16}\] takes \[16\]hours
go with 16
oh wooow how REATRDED xD
yeah just what i was thinking reatrded
OMGG!! LOL noooooooooo xD i swear i can spell LOL
i can spel to
now are you done? i got stuff to do
you should definitely enter a spelling bee LOL
um no u can go though i'll just get help from someone else.
ok if i am here i will help
ok :)
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