To determine where the function [tex]\( F(x) = \frac{1}{x(x+6)(x-1)} \)[/tex] has vertical asymptotes, we need to find the values of [tex]\( x \)[/tex] that make the denominator zero. The function will be undefined at these values, leading to vertical asymptotes.
We start by setting the denominator equal to zero:
[tex]\[ x(x+6)(x-1) = 0 \][/tex]
Now, solve for [tex]\( x \)[/tex] by finding the roots of each factor individually:
1. Setting [tex]\( x = 0 \)[/tex]:
[tex]\[ 0 \cdot (0+6) \cdot (0-1) = 0 \][/tex]
So, [tex]\( x = 0 \)[/tex] is a root.
2. Setting [tex]\( x+6 = 0 \)[/tex]:
[tex]\[ (x+6) = 0 \][/tex]
[tex]\[ x = -6 \][/tex]
So, [tex]\( x = -6 \)[/tex] is another root.
3. Setting [tex]\( x-1 = 0 \)[/tex]:
[tex]\[ (x-1) = 0 \][/tex]
[tex]\[ x = 1 \][/tex]
So, [tex]\( x = 1 \)[/tex] is another root.
Thus, the values of [tex]\( x \)[/tex] that make the denominator zero, and hence where vertical asymptotes occur, are [tex]\( x = 0 \)[/tex], [tex]\( x = 1 \)[/tex], and [tex]\( x = -6 \)[/tex].
The correct answers are:
A. 1
C. 0
E. -6
