Question

Identify the vertical asymptotes of f(x)=x+6 over x^2-9x+18

Answer 1

$\bf \cfrac{x+6}{x^2-9x+18} \\\\\\ x^2-9x+18=\underline{0}\implies (x-6)(x-3)=0\implies x= \begin{cases} 6\\ 3 \end{cases}$

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.