Question

6.9.4 quiz: more than one vertical asymptote
at which values of x does the function f(x) have a vertical asymptote? check all that apply.
f(x) = \frac{1}{x(x + 6)(x - 1)}
a. -1
b. 6
c. 0
d. 1
e. -6

Explanation:

Step1: Find the values that make the denominator zero

For a rational function \(F(x)=\frac{1}{x(x + 6)(x-1)}\), vertical asymptotes occur at the values of \(x\) that make the denominator equal to zero.
Set \(x(x + 6)(x - 1)=0\).
Using the zero - product property: if \(abc = 0\), then \(a=0\) or \(b = 0\) or \(c=0\).
For \(x=0\), the denominator is \(0\times(0 + 6)\times(0-1)=0\).
For \(x+6=0\) (i.e., \(x=-6\)), the denominator is \((-6)\times(-6 + 6)\times(-6-1)=0\).
For \(x - 1=0\) (i.e., \(x = 1\)), the denominator is \(1\times(1 + 6)\times(1-1)=0\).

Answer:

C. \(0\), D. \(1\), E. \(-6\)