Question

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

Answer 1

X= 3 and X=6

Explanation:

I took the test and got it right. I would explain but the guy on top pretty much got it covered.

Answer 2

Rewriting the function


`f(x)= (x+6)/(x^2-9x+18)`

The vertical asymptotes will be at the value of
`x` when the denominator is zero


`x^2-9x+18=0` ⇒ factorising

`(x-3)(x-6)=0`

`x=3` and
`x=6`

The value of
`x` must not be at 3 and 6 because we cannot have the domain of the denominator equal to zero (fraction with zero denominators is undefined)

Hence, vertical asymptote is at
`x=3` and
`x=6`