Question

find all vertical asymptotes of the following function.
$f(x) = \frac{3x^2 - 27}{x^2 - 12x + 27}$
answer attempt 2 out of 2
one vertical asymptote
$x = 3$

Explanation:

Step1: Factor numerator and denominator

Factor \(3x^2 - 27\): \(3(x^2 - 9)=3(x - 3)(x + 3)\)
Factor \(x^2 - 12x + 27\): \((x - 3)(x - 9)\)
So \(f(x)=\frac{3(x - 3)(x + 3)}{(x - 3)(x - 9)}\)

Step2: Cancel common factors

Cancel \((x - 3)\) (for \(x
eq3\)), so \(f(x)=\frac{3(x + 3)}{x - 9}\) (simplified, \(x
eq3,9\))

Step3: Find vertical asymptotes

Vertical asymptotes occur where denominator is 0 (and numerator not 0 after canceling).
Set \(x - 9 = 0\) → \(x = 9\).
Check \(x = 3\): canceled, so it's a hole, not an asymptote.

Answer:

The vertical asymptote is \(x = 9\) (note: The original answer had an error; the correct vertical asymptote is \(x = 9\), as \(x = 3\) is a removable discontinuity (hole), not an asymptote.)