Let $f(x)=\frac{3 x-6}{x^{2}+8 x+15}=\frac{(3 x-6)}{(x+5)(x+3)}$ Find: 3) $x$ intercepts at the point(s) 4) Vertical asymptotes at $x=-3$ 5) Horizontal asymptote at $y=-5$

Step 1 ：First, we simplify the function $f(x)=\frac{3 x-6}{x^{2}+8 x+15}=\frac{3(x-2)}{(x+5)(x+3)}$

Step 2 ：To find the x-intercepts, we set $f(x)$ equal to zero and solve for $x$. This gives us $3(x-2)=0$ which simplifies to $x=2$

Step 3 ：The vertical asymptote is given by the values of $x$ that make the denominator of the function equal to zero. In this case, we have $x+5=0$ and $x+3=0$, which gives us $x=-5$ and $x=-3$. However, the problem only asks for the vertical asymptote at $x=-3$, so we ignore $x=-5$

Step 4 ：The horizontal asymptote is given by the limit of the function as $x$ approaches infinity. In this case, as $x$ approaches infinity, the function $f(x)$ approaches $-5$

Step 5 ：Therefore, the x-intercept is at the point $(2,0)$, the vertical asymptote is at $x=-3$, and the horizontal asymptote is at $y=-5$