Pls help me with this:
Find vertical, horizontal, oblique asymptote for the following rational function

G(x) = x^3-8/2x-x^2

Question

Pls help me with this:
Find vertical, horizontal, oblique asymptote for the following rational function

G(x) = x^3-8/2x-x^2

anonymous · Answer

vertical asymptotes are obtained by equating the denominator is equated to zero..

Horizontal asymptote have a rule:

for any function, $$f(x)= (ax ^{n}+.....)/ (bx ^{m}+...........)$$   where n and m are the highest power in numerator and denominators respectively.. then,

1. if n<m then y=0 is the horizontal asymptote
2. if n=m then  y= a/b is the horizontal asymptote
3. if n>m then there is no horizontal asymptote 

And for the oblique asymptote you have to perform the long division.. whatever you get in the quotient, that quantity equated to zero will be your oblique asymptote

You have all what you need.. good luck now.. tell me if you encounter problems..

anonymous · Answer

I get -4-x as the oblique asymptotes, no horizontal asymptotes. But this answer is wrong. That's why I'm so confusing now.

anonymous · Answer

sorry the oblique asymptote are obtained by equating the quotient to y (not zero).. pardon my typo in earlier reply..


Still the quotient is -x.. so the oblique asymptote must be y=-x..

tell me if you aren't satisfied yet

anonymous · Answer

What I found is G(x) = -x-4 + 4/-x. Does this mean that -x-4 is the oblique asymptotes? 

Could you guide me through this? I don't really get it yet....Thanks for your help!