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) =
`(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