Question

f(x) = \frac{x^2 - x}{-4x - 12}
\bigcirc vertical asym.: x = 0, x = 1
holes: none
horz. asym.: y = 0
\bigcirc vertical asym.: x = 3
holes: (0, 0)
horz. asym.: none
\bigcirc vertical asym.: x = -3
holes: none
horz. asym.: none
\bigcirc vertical asym.: x = -3
holes: none
horz. asym.: y = 0

Explanation:

Step1: Simplify the function

First, factor the numerator and the denominator. The numerator \(x^2 - x\) factors to \(x(x - 1)\). The denominator \(-4x - 12\) can be factored as \(-4(x + 3)\) (or \(4(-x - 3)\), but factoring out -4 is helpful here). So the function becomes:

$$ f(x)=\frac{x(x - 1)}{-4(x + 3)} $$

There are no common factors between the numerator and the denominator, so there are no holes (holes occur when there are common factors that can be canceled).

Step2: Find Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is not zero, since there are no common factors). Set the denominator equal to zero:

$$ -4(x + 3)=0 $$

Solving for \(x\), we divide both sides by -4: \(x + 3 = 0\), so \(x=-3\). So the vertical asymptote is \(x = -3\).

Step3: Find Horizontal Asymptotes

To find horizontal asymptotes, we compare the degrees of the numerator and the denominator. The numerator \(x^2 - x\) is a degree 2 polynomial (the highest power of \(x\) is 2), and the denominator \(-4x - 12\) is a degree 1 polynomial (highest power of \(x\) is 1). When the degree of the numerator is greater than the degree of the denominator (2 > 1), there is no horizontal asymptote.

Answer:

Vertical Asym.: \(x = -3\)
Holes: None
Horz. Asym.: None (corresponding to the option with these values)