Question

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

Answer 1

Rewriting the function

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

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 $x=6$

Answer 2

Answer:

X= 3 and X=6

Step-by-step explanation:

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