Answer :
Certainly! Let's determine the vertical asymptotes of the given function [tex]\( f(x) = \frac{14}{(x-5)(x+1)} \)[/tex].
Vertical asymptotes occur where the denominator of the fraction goes to zero, as the function value approaches infinity or negative infinity at these points.
1. Identify the Denominator:
The denominator of [tex]\( f(x) \)[/tex] is [tex]\( (x-5)(x+1) \)[/tex].
2. Set the Denominator Equal to Zero:
To find the vertical asymptotes, we need to solve the equation [tex]\( (x-5)(x+1) = 0 \)[/tex].
3. Solve for [tex]\( x \)[/tex]:
- First, set each factor of the denominator equal to zero separately:
[tex]\[ x - 5 = 0 \quad \text{and} \quad x + 1 = 0 \][/tex]
- Solving these equations gives:
[tex]\[ x = 5 \quad \text{and} \quad x = -1 \][/tex]
These values of [tex]\( x \)[/tex] are where the vertical asymptotes of [tex]\( f(x) \)[/tex] are located. Thus, the vertical asymptotes for the function [tex]\( f(x) = \frac{14}{(x-5)(x+1)} \)[/tex] are at [tex]\( x = 5 \)[/tex] and [tex]\( x = -1 \)[/tex].
Therefore, the correct answer is:
[tex]\[ x = -1 \quad \text{and} \quad x = 5 \][/tex]
Vertical asymptotes occur where the denominator of the fraction goes to zero, as the function value approaches infinity or negative infinity at these points.
1. Identify the Denominator:
The denominator of [tex]\( f(x) \)[/tex] is [tex]\( (x-5)(x+1) \)[/tex].
2. Set the Denominator Equal to Zero:
To find the vertical asymptotes, we need to solve the equation [tex]\( (x-5)(x+1) = 0 \)[/tex].
3. Solve for [tex]\( x \)[/tex]:
- First, set each factor of the denominator equal to zero separately:
[tex]\[ x - 5 = 0 \quad \text{and} \quad x + 1 = 0 \][/tex]
- Solving these equations gives:
[tex]\[ x = 5 \quad \text{and} \quad x = -1 \][/tex]
These values of [tex]\( x \)[/tex] are where the vertical asymptotes of [tex]\( f(x) \)[/tex] are located. Thus, the vertical asymptotes for the function [tex]\( f(x) = \frac{14}{(x-5)(x+1)} \)[/tex] are at [tex]\( x = 5 \)[/tex] and [tex]\( x = -1 \)[/tex].
Therefore, the correct answer is:
[tex]\[ x = -1 \quad \text{and} \quad x = 5 \][/tex]
