So we want to find the vertical asymptotes of the function:
[tex]f(x)=\frac{x+6}{x^2-9x+18}[/tex]
Remember that there's a vertical asymptote at points of x where the denominator of the rational function is 0.
So, equal the denominator to zero:
[tex]x^2-9x+18=0[/tex]
And solve this equation for x as follows:
We could factor
[tex]\begin{gathered} x^2-9x+18=0 \\ (x-6)(x-3)=0 \end{gathered}[/tex]
Now, notice that any of both factors could be zero, so:
[tex]\begin{cases}x-6=0\to x=6 \\ x-3=0\to x=3\end{cases}[/tex]
Therefore, the function has vertical asymptotes at x=6 and x=3.
