[tex]\bf \cfrac{x+6}{x^2-9x+18}
\\\\\\
x^2-9x+18=\underline{0}\implies (x-6)(x-3)=0\implies x=
\begin{cases}
6\\
3
\end{cases}[/tex]
vertical asymptotes occur, at zeros of the denominator, so long the value doesn't make the numerator zero.
so, in this case are x = 6 and x = 3.
