6.79. $f(x)=\frac{x^{2}+4 x+1}{x+2}$.
Solution. We find the vertical asymptotes at the point $x=-2$:

$$
\lim _{x \rightarrow-2-0} \frac{x^{2}+4 x+1}{x+2}=\frac{(-2-0)^{2}+4(-2-0)+1}{(-2-0)+2}=\frac{-3}{-0}=+\infty
$$

as we approach the point $x=-2$ from the left, the graph of the function tends to $+\infty, x=-2$ is a vertical asymptote as $x \rightarrow-2$ from the left,

$$
\lim _{x \rightarrow-2+0} \frac{x^{2}+4 x+1}{x+2}=\frac{(-2+0)^{2}+4(-2+0)+1}{(-2+0)+2}=\frac{-3}{+0}=-\infty,
$$

as we approach the point $x=-2$ from the right, the graph of the function tends to $-\infty, x=-2$ is a vertical asymptote as $x \rightarrow-2$ from the right. We will find the oblique asymptote as $x \rightarrow-\infty$:

$$
\begin{aligned}
& k_{1}=\lim _{x \rightarrow-\infty} \frac{f(x)}{x}=\lim _{x \rightarrow-\infty} \frac{x^{2}+4 x+1}{x(x+2)}=1 \\
& b_{1}=\lim _{x \rightarrow-\infty}[f(x)-k x]=\lim _{x \rightarrow-\infty}\left(\frac{x^{2}+4 x+1}{x+2}-x\right)=\lim _{x \rightarrow-\infty} \frac{2 x+1}{x+2}=2
\end{aligned}
$$

$y=x+2$ is the oblique asymptote as $x \rightarrow-\infty$. Similarly, we find

$$
\begin{aligned}
k_{2} & =\lim _{x \rightarrow+\infty} \frac{f(x)}{x}=\lim _{x \rightarrow+\infty} \frac{x^{2}+4 x+1}{x(x+2)}=1 \\
b_{2} & =\lim _{x \rightarrow+\infty} \frac{2 x+1}{x+2}=2
\end{aligned}
$$

and $y=x+2$ is the oblique asymptote as $x \rightarrow+\infty$.