To determine the values of [tex]\( x \)[/tex] at which the function [tex]\( F(x) = \frac{3}{x(x-5)(x+1)} \)[/tex] has vertical asymptotes, we need to identify where the denominator of the function is equal to zero. Vertical asymptotes occur at values of [tex]\( x \)[/tex] that make the denominator zero, as this causes the function to approach infinity.
Let's set the denominator of [tex]\( F(x) \)[/tex] equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x(x-5)(x+1) = 0 \][/tex]
This equation is satisfied when any factor is zero, so we separate the equation accordingly:
1. [tex]\( x = 0 \)[/tex]
2. [tex]\( x - 5 = 0 \implies x = 5 \)[/tex]
3. [tex]\( x + 1 = 0 \implies x = -1 \)[/tex]
Therefore, the function [tex]\( F(x) \)[/tex] has vertical asymptotes at [tex]\( x = 0 \)[/tex], [tex]\( x = 5 \)[/tex], and [tex]\( x = -1 \)[/tex].
Given the options:
A. -5 (Incorrect, as substituting -5 does not make the denominator zero)
B. 0 (Correct, as shown by [tex]\( x = 0 \)[/tex])
C. -1 (Correct, as shown by [tex]\( x = -1 \)[/tex])
D. 3 (Incorrect, as substituting 3 does not make the denominator zero)
E. 5 (Correct, as shown by [tex]\( x = 5 \)[/tex])
F. 1 (Incorrect, as substituting 1 does not make the denominator zero)
Thus, the values of [tex]\( x \)[/tex] at which the function [tex]\( F(x) \)[/tex] has vertical asymptotes are:
[tex]\[ \boxed{0, \, -1, \, 5} \][/tex]
