Question

find all vertical asymptotes of the following function.
$f(x) = \frac{x^2 - 36}{3x + 6}$

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(x^2 - 36\) as a difference of squares: \(x^2 - 36=(x - 6)(x + 6)\).
Factor the denominator \(3x + 6\) by taking out the common factor: \(3x + 6 = 3(x + 2)\).
So the function becomes \(f(x)=\frac{(x - 6)(x + 6)}{3(x + 2)}\).

Step2: Simplify the function (if possible)

Check for common factors in numerator and denominator. Here, there are no common factors (the numerator has factors \(x - 6\), \(x + 6\) and the denominator has \(3\), \(x + 2\)).

Step3: Find values that make denominator zero

Set the denominator equal to zero and solve for \(x\):
\(3(x + 2)=0\)
Divide both sides by 3: \(x + 2 = 0\)
Subtract 2 from both sides: \(x=-2\).

We need to check if at \(x = -2\), the numerator is non - zero. Substitute \(x=-2\) into the numerator \((x - 6)(x + 6)\):
\((-2-6)(-2 + 6)=(-8)(4)=-32
eq0\).

Answer:

The vertical asymptote is \(x=-2\)