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At which values of [tex] x [/tex] does the function [tex] F(x) [/tex] have a vertical asymptote? Check all that apply.

[tex]\[ F(x) = \frac{1}{7(x+2)(x+3)} \][/tex]

A. 2
B. 7
C. 3
D. -3
E. -2

Sagot :

GinnyAnsweridn
To determine the values of [tex]\(x\)[/tex] where the function [tex]\(f(x) = \frac{1}{7(x + 2)(x + 3)}\)[/tex] has vertical asymptotes, we need to identify where the denominator of the function is equal to zero. Vertical asymptotes occur at points where the function becomes undefined due to division by zero.

Let's set the denominator equal to zero and solve:

[tex]\[ 7(x + 2)(x + 3) = 0 \][/tex]

First, we can disregard the constant factor [tex]\(7\)[/tex] because a non-zero constant does not affect the points where the product is zero. Thus, we focus on:

[tex]\[ (x + 2)(x + 3) = 0 \][/tex]

We solve this equation by setting each factor to zero:

1. [tex]\(x + 2 = 0\)[/tex]

Solving for [tex]\(x\)[/tex]:

[tex]\[ x = -2 \][/tex]

2. [tex]\(x + 3 = 0\)[/tex]

Solving for [tex]\(x\)[/tex]:

[tex]\[ x = -3 \][/tex]

Therefore, the function [tex]\(f(x)\)[/tex] has vertical asymptotes at [tex]\( x = -2 \)[/tex] and [tex]\( x = -3 \)[/tex].

So, the values of [tex]\(x\)[/tex] at which the function [tex]\(f(x)\)[/tex] has vertical asymptotes are:

- [tex]\( x = -2 \)[/tex]
- [tex]\( x = -3 \)[/tex]

Hence, the correct options are:

D. -3

E. -2
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