To determine the nature of the stationary points of

and sketch its graph.

Solution. The intercepts are at $x=0,$ $f(0)=1$; the equation $x^{2}+x+1=0$ has no real solution so the graph does not intercept the $x$ axis.

We divide and obtain

where $1/(x+1)\rightarrow 0$ as $x\rightarrow\pm\infty$; hence $y=x$ is an asymptote as $x\rightarrow\pm\infty$.

Consider $x$ near to $-1$; now $f(x)\rightarrow\infty$ as $x\rightarrow(-1)+$;

$f(x)\rightarrow-\infty$ as $x\rightarrow(-1)-$; hence $x=-1$ is a vertical asymptote.