Question

list any vertical asymptotes for the function given below.
$f(x)=\frac{1}{x^{2}+6x}$
a. $y = 0$ and $y=-6$
b. $y = 1$
c. $x = 0$ and $x=-6$
d. $x = 0$ and $x = 6$

Explanation:

Step1: Analyze the denominator

For a rational function \(f(x)=\frac{1}{x^{2}+6x}\), vertical asymptotes occur where the denominator is zero (and the numerator is non - zero). Set the denominator \(x^{2}+6x = 0\).
Factor the denominator: \(x(x + 6)=0\).

Step2: Solve for \(x\)

Using the zero - product property \(ab = 0\) implies \(a = 0\) or \(b = 0\).
If \(x=0\), then \(x(x + 6)=0\); if \(x+6=0\) (i.e., \(x=-6\)), then \(x(x + 6)=0\).
The numerator \(1
eq0\) for \(x = 0\) and \(x=-6\).

Answer:

C. \(x = 0\) and \(x=-6\)