Answer
There is a hole at point $(-3,-15)$.
The $x$-intercepts: $(2,0)$ and $(3,0)$.
The $y$-intercept: $(0,6)$.
The vertical asymptote: $x=-1$.
The slant asymptote: $y=x-6$.
See graph
Work Step by Step
Simplify the function:
$$f(x)=\frac{(x-3)(x-2)(x+3)}{(x+3)(x+1)}$$ $$f(x)=\frac{(x-3)(x-2)}{x+1}$$
Equate $x+3$ to $0$:
$$x+3=0$$ $$x=-3$$
$$f(-3)=\frac{(-3-3)(-3-2)}{-3+1}=-15$$
Thus, there is a hole at point $(-3,-15)$.
Equate the numerator to $0$ to find the $x$-intercept:
$$(x-3)(x-2)=0$$ $$x-3=0$$ $$x=3$$ $$x-2=0$$ $$x=2$$
Thus, the $x$-intercepts are at $(2,0)$ and $(3,0)$.
Set $x=0$ to find the $y$-intercept:
$$f(0)=\frac{(0-3)(0-2)}{0+1}=6$$
Thus, the $y$-intercept is at $(0,6)$.
The function is undefined when the denominator is $0$:
$$x+1=0$$ $$x=-1$$
Thus, the vertical asymptote is at $x=-1$.
Since the degree of the numerator is one more than the degree of the denominator, divide the leading coefficient of the numerator by the leading coefficient of the denominator and ignore the remainder to find the slant asymptote is:
$$y=\frac{x^3-2x^2-9x+18}{x^2+4x+3}$$ $$y=x-6+\frac{12}{x+1}$$
Ignoring the remainder, the slant asymptote is $y=x-6$.
The sketch of the graph of the function is as shown.
