To find the values of [tex]\( x \)[/tex] at which the function [tex]\( F(x) = \frac{1}{7(x+2)(x+3)} \)[/tex] has vertical asymptotes, we need to determine where the denominator is equal to zero. This occurs when:
[tex]\[ 7(x + 2)(x + 3) = 0 \][/tex]
First, note that the constant coefficient (7) does not affect the values of [tex]\( x \)[/tex] at which the product equals zero. Therefore, we focus on solving:
[tex]\[ (x + 2)(x + 3) = 0 \][/tex]
Next, find the values of [tex]\( x \)[/tex] that satisfy each factor being zero.
1. [tex]\( x + 2 = 0 \)[/tex]
[tex]\[
x = -2
\][/tex]
2. [tex]\( x + 3 = 0 \)[/tex]
[tex]\[
x = -3
\][/tex]
Thus, the function [tex]\( F(x) \)[/tex] has vertical asymptotes at [tex]\( x = -2 \)[/tex] and [tex]\( x = -3 \)[/tex].
So, from the given options:
- A. 7
- B. 2
- C. -3
- D. 3
- E. -2
The correct answers are:
C. -3 and E. -2
