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

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Answer 1 · 2015-07-13T20:28:08

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

with exclusion $x != -1$

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

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 $g(x)$ .

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