Answer
$$
y=\frac{-x-4}{3x+6}
$$
We obtain that :
Asymptotes: $y=-\frac{1}{3} $ and $ x=-2$
$x$-intercept: $-4$ the value when $y= 0$.
$y$-intercept: $-\frac{2}{3}$ the value when $x= 0$
Work Step by Step
$$
y=\frac{-x-4}{3x+6}
$$
The value $ x=-2$makes the denominator 0, but not the numerator, so the line $ x=-2$ is a vertical asymptote.
To find a horizontal asymptote, let $x$ get larger and we obtain :
$$
y=\lim\limits_{x \to \infty}\frac{-x-4}{3x+6}=-\frac{1}{3}
$$
This means that the line $y=-\frac{1}{3} $ is a horizontal asymptote.
So, we have:
Asymptotes: $y=-\frac{1}{3} $ and $ x=-2$
$x$-intercept: $-4$ the value when $y= 0$.
$y$-intercept: $-\frac{2}{3}$ the value when $x= 0$