What are the asymptotes for # g(x)= (x^3+1)/(x+1)#?

1 Answer
Jul 13, 2015

Answer:

#g(x) = (x^3+1)/(x+1) = ((x+1)(x^2-x+1))/(x+1) = x^2-x+1#

with exclusion #x != -1#

The polynomial #x^2-x+1# has no asymptotes, so neither has #g(x)#

Explanation:

The sum of cubes identity is:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

So

#x^3+1 = x^3+1^3 = (x+1)(x^2-x+1)#

So

#g(x) = (x^3+1)/(x+1) = ((x+1)(x^2-x+1))/(x+1) = x^2-x+1#

with exclusion #x != -1#

The polynomial #x^2-x+1# has no linear asymptotes, so neither has #g(x)#.