Question

dentify the vertical asymptotes of f(x) = quantity x plus 6 over quantity x squared minus 9x plus 18.

Answer 1

So the function we're dealing with is

f(x) = $\frac{x+6}{x^{2}-9x+18 }$

Factorising we get, $x^{2} -9x+18=0$
$(x-3)(x-6)=0$

Therefore, the vertical asymptotes are at x = 3 and x = 6