Answer
$$
y=\frac{6-3x}{4x+12}
$$
We obtain that :
Asymptotes: $y=-\frac{3}{4} $ and $ x=-3$
$x$-intercept: $2$ the value when $y= 0$.
$y$-intercept: $\frac{1}{2}$ the value when $x= 0$
Work Step by Step
$$
y=\frac{6-3x}{4x+12}
$$
The value $ x=-3$makes the denominator 0, but not the numerator, so the line $ x=-3$ is a vertical asymptote.
To find a horizontal asymptote, let $x$ get larger and we obtain :
$$
y=\lim\limits_{x \to \infty}\frac{6-3x}{4x+12}=-\frac{3}{4}
$$
This means that the line $y=-\frac{3}{4} $ is a horizontal asymptote.
So, we have:
Asymptotes: $y=-\frac{3}{4} $ and $ x=-3$
$x$-intercept: $2$ the value when $y= 0$.
$y$-intercept: $\frac{1}{2}$ the value when $x= 0$