Question 706221
{{{f(x)=(x^3+2x^2-8x)/(x^2+1)}}} The denominator {{{x^2+1}}} is never zero, so there are no vertical asymptotes and no holes.
(You need zeros in the denominator for those).
The graph is shown in red below.
{{{graph(300,300,-10,10,-10,10,(x^3+2x^2-8x)/(x^2+1), x+2)}}}
 
EXTRA:
It has a slanted asymptote (shown in green) because
{{{f(x)=(x^3+2x^2-8x)/(x^2+1)=((x^2+1)(x+2)-(9x+2))/(x^2+1)=x+2-(9x+2)/(x^2+1)}}}
and {{{abs((9x+2)/(x^2+1))}}} gets smaller and smaller as {{{x^2}}} grows larger and larger.