Question

find all vertical asymptotes of the following function.
$f(x) = \frac{x^2 - 36}{3x + 6}$
answer attempt 1 out of 2
no vertical asymptotes
no vertical asymptotes
one vertical asymptote
two vertical asymptotes

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator. The numerator \(x^2 - 36\) is a difference of squares, so it factors as \((x - 6)(x + 6)\). The denominator \(3x + 6\) can be factored as \(3(x + 2)\). Wait, no, wait: \(3x + 6 = 3(x + 2)\)? Wait, no, original denominator is \(3x + 6\), factor out 3: \(3(x + 2)\)? Wait, no, wait the numerator is \(x^2 - 36=(x - 6)(x + 6)\), denominator is \(3x + 6 = 3(x + 2)\)? Wait, no, that's not right. Wait, no, let's check again. Wait, the function is \(f(x)=\frac{x^2 - 36}{3x + 6}\). Let's factor numerator: \(x^2 - 36=(x - 6)(x + 6)\). Denominator: \(3x + 6 = 3(x + 2)\)? Wait, no, \(3x + 6 = 3(x + 2)\)? Wait, no, \(3x + 6 = 3(x + 2)\) is correct. Wait, but wait, maybe I made a mistake. Wait, no, the denominator is \(3x + 6\), so factoring out 3 gives \(3(x + 2)\). The numerator is \((x - 6)(x + 6)\). Now, check for common factors. There are no common factors between the numerator and the denominator. Wait, no, wait: numerator has \((x + 6)\), denominator has \(3(x + 2)\). So no common factors. Wait, but to find vertical asymptotes, we set the denominator equal to zero, provided that the numerator is not also zero at that point. So denominator is zero when \(3x + 6 = 0\), which gives \(x = -2\). Now, check the numerator at \(x = -2\): numerator is \((-2)^2 - 36 = 4 - 36 = -32
eq 0\). Wait, but wait, maybe I factored wrong. Wait, no, let's re-express the function. Wait, maybe I made a mistake in factoring. Wait, \(x^2 - 36=(x - 6)(x + 6)\), denominator \(3x + 6 = 3(x + 2)\). So the function is \(\frac{(x - 6)(x + 6)}{3(x + 2)}\). Now, vertical asymptotes occur where the denominator is zero and the numerator is not zero. So denominator zero at \(x = -2\), numerator at \(x = -2\) is \((-2 - 6)(-2 + 6)=(-8)(4)=-32
eq 0\). Wait, but wait, the original function: wait, maybe I made a mistake. Wait, no, let's check again. Wait, the denominator is \(3x + 6\), so solving \(3x + 6 = 0\) gives \(x = -2\). Now, check if the numerator is zero at \(x = -2\): \(x^2 - 36 = (-2)^2 - 36 = 4 - 36 = -32
eq 0\). So there should be a vertical asymptote at \(x = -2\). Wait, but the options are No Vertical Asymptotes, One Vertical Asymptote, Two Vertical Asymptotes. Wait, but maybe I made a mistake. Wait, wait, let's re-express the function. Wait, maybe the numerator and denominator have a common factor. Wait, no, numerator is \((x - 6)(x + 6)\), denominator is \(3(x + 2)\). No common factors. So vertical asymptote at \(x = -2\), so one vertical asymptote. Wait, but wait, the initial simplification: wait, maybe I factored the denominator wrong. Wait, \(3x + 6 = 3(x + 2)\), correct. Numerator \((x - 6)(x + 6)\), correct. So denominator zero at \(x = -2\), numerator not zero there, so vertical asymptote at \(x = -2\). So one vertical asymptote.

Step2: Determine vertical asymptotes

Vertical asymptotes occur when the denominator is zero and the numerator is not zero at that point. We found that the denominator is zero at \(x = -2\), and the numerator at \(x = -2\) is \(-32
eq 0\). So there is one vertical asymptote.

Answer:

One Vertical Asymptote