Question

Identify the asympatote

Answer 1

\[y=1/x+1-2\]

Answer 2

@Kreshnik Im sorry

Answer 3

...@daja2fly I'm sorry too! but I'm not going to help you !

Answer 4

you know it better ! bye .

Answer 5

@Mindy1234 do u no or not

Answer 6

when 1/(x+1) = undefined, then x=? is the vertical asymptote, if i remember it correctly...

Answer 7

0

Answer 8

0?

Answer 9

whats your question

Answer 10

i didn't ask anything...

Answer 11

so how do i figure this out

Answer 12

when 1/(x+1) = undefined, x+1 =0 , solve x and you'll get the vertical asymptote

Answer 13

ok

Answer 14

-1

Answer 15

-1 callisto

Answer 16

x=-1 it should be..

Answer 17

hello

Answer 18

\[\Large y=\frac{1}{x+1}-2=\frac{-2x-1}{x+1}\]

Horizontal asymtote...



\[\LARGE y=\lim_{x\to \infty}\frac{-2x-1}{x+1}=\lim_{x\to\infty}\frac{\cfrac{-2x}{x}-\cfrac{1}{x}}{\cfrac{x}{x}+\cfrac1x}\]

\[\LARGE =\lim_{x\to\infty}\frac{-2-0}{1+0}=-2\]
\[\LARGE \boxed{y=-2}\]

Vertical Asymptote...

\[\LARGE x=-1\]


Not vertical , not horizontal, something in the middle asymptote (I don't know how to say it...not american...but I guess you know what I mean.)

\[\LARGE y=kx+b\]

\[\Large k=\lim_{x\to \infty}\frac{f(x)}{x}=\lim_{x\to\infty}\cfrac{\cfrac{-2x-1}{x+1}}{x}\]
\[\Large \lim_{x\to \infty}\frac{-2x-1}{x^2+x}=0\]

So \[\LARGE k=0\] and there's no asymtote like this



When I was about to finish it... google crashed and I was about to brake my PC.. (here I wrote it again. hope you understand it HOW TO DO !) . @daja2fly