Answer
$$
y =\frac{9-6 x+x^{2}}{3-x}
$$
There are no asymptotes, but there is a hole at $x=3$.
There is no $x$-intercept, since $3-x$ implies $x=3$
but there is a hole at $x=3$
$y$-intercept: $3$ the value when $x= 0$.
Work Step by Step
$$
y =\frac{9-6 x+x^{2}}{3-x}
$$
To find a asymptote, let
$$
\begin{aligned} y &=\frac{9-6 x+x^{2}}{3-x} \\ &=\frac{(3-x)(3-x)}{3-x} \\ &=3-x, x \neq 3 \end{aligned}
$$
There are no asymptotes, but there is a hole at $x=3$.
There is no $x$-intercept, since $3-x$ implies $x=3$
but there is a hole at $x=3$
$y$-intercept: $3$ the value when $x= 0$.
