Answer :
To determine which equation could represent the function [tex]\( n \)[/tex] given the conditions, we need to carefully consider how the transformation would affect the rational function [tex]\( m \)[/tex].
### Given Information:
1. The graphs of rational functions [tex]\( m \)[/tex] and [tex]\( n \)[/tex] have the same vertical asymptotes.
2. Both functions have a single [tex]\( x \)[/tex]-intercept at [tex]\( x = 5 \)[/tex].
#### Step-by-Step Analysis:
1. Vertical Asymptotes:
- The vertical asymptotes suggest that the unknown [tex]\( n(x) \)[/tex] should preserve the same factors in the denominator as [tex]\( m(x) \)[/tex] to ensure the asymptotic behavior is the same.
2. [tex]\( x \)[/tex]-Intercept at [tex]\( x = 5 \)[/tex]:
- The [tex]\( x \)[/tex]-intercept for a function [tex]\( f(x) \)[/tex] is found by setting [tex]\( f(x) = 0 \)[/tex]. For [tex]\( n(x) \)[/tex] to have its [tex]\( x \)[/tex]-intercept at [tex]\( x = 5 \)[/tex], we need [tex]\( n(5) = 0 \)[/tex].
#### Evaluating the Options:
Option A: [tex]\( n(x) = 5 m(x) \)[/tex]
- Scaling [tex]\( m(x) \)[/tex] by a factor of 5 will not affect the location of the [tex]\( x \)[/tex]-intercept but it also won't change the vertical asymptotes. However, this does not fix the [tex]\( x \)[/tex]-intercept directly to [tex]\( x = 5 \)[/tex]. It maintains the given [tex]\( x \)[/tex]-intercept of [tex]\( m(x) \)[/tex] but doesn't solve the problem as required.
Option B: [tex]\( n(x) = m(x) + 5 \)[/tex]
- Adding 5 to [tex]\( m(x) \)[/tex] shifts the graph vertically upwards or downwards, which alters the intercepts and affects the graph's shape overall. This would change the [tex]\( x \)[/tex]-intercept location, which does not meet the required condition.
Option C: [tex]\( n(x) = m(x - 5) \)[/tex]
- This option represents a horizontal shift of the function [tex]\( m(x) \)[/tex] to the right by 5 units. To confirm, substituting [tex]\( x = 5 \)[/tex]:
[tex]\[ n(5) = m(5 - 5) = m(0) \][/tex]
For [tex]\( n(x) \)[/tex] to have an [tex]\( x \)[/tex]-intercept at [tex]\( x = 5 \)[/tex]:
[tex]\[ n(5) = 0 \implies m(0) = 0 \][/tex]
Therefore, the intercept at [tex]\( x = 5 \)[/tex] for [tex]\( n \)[/tex] would hold if [tex]\( m \)[/tex] originally had an intercept at [tex]\( x = 0 \)[/tex]. This transformation ensures the [tex]\( x \)[/tex]-intercept is set at [tex]\( x = 5 \)[/tex] appropriately.
Option D: [tex]\( n(x) = m(x + 5) \)[/tex]
- This option would move the function [tex]\( m(x) \)[/tex] left by 5 units. Substituting [tex]\( x = 5 \)[/tex]:
[tex]\[ n(5) = m(5 + 5) = m(10) \][/tex]
Which means that for [tex]\( n(x) \)[/tex] to be zero at [tex]\( x = 5 \)[/tex], [tex]\( m(10) \)[/tex] would need to be zero, which does not ensure that [tex]\( x = 5 \)[/tex] is the intercept in the context provided.
Considering these options, the correct answer is:
[tex]\[ \boxed{C} \][/tex]
This option correctly adjusts the input of [tex]\( m(x) \)[/tex] in a manner that ensures both the vertical asymptotes remain the same and the [tex]\( x \)[/tex]-intercept is specifically located at [tex]\( x = 5 \)[/tex].
### Given Information:
1. The graphs of rational functions [tex]\( m \)[/tex] and [tex]\( n \)[/tex] have the same vertical asymptotes.
2. Both functions have a single [tex]\( x \)[/tex]-intercept at [tex]\( x = 5 \)[/tex].
#### Step-by-Step Analysis:
1. Vertical Asymptotes:
- The vertical asymptotes suggest that the unknown [tex]\( n(x) \)[/tex] should preserve the same factors in the denominator as [tex]\( m(x) \)[/tex] to ensure the asymptotic behavior is the same.
2. [tex]\( x \)[/tex]-Intercept at [tex]\( x = 5 \)[/tex]:
- The [tex]\( x \)[/tex]-intercept for a function [tex]\( f(x) \)[/tex] is found by setting [tex]\( f(x) = 0 \)[/tex]. For [tex]\( n(x) \)[/tex] to have its [tex]\( x \)[/tex]-intercept at [tex]\( x = 5 \)[/tex], we need [tex]\( n(5) = 0 \)[/tex].
#### Evaluating the Options:
Option A: [tex]\( n(x) = 5 m(x) \)[/tex]
- Scaling [tex]\( m(x) \)[/tex] by a factor of 5 will not affect the location of the [tex]\( x \)[/tex]-intercept but it also won't change the vertical asymptotes. However, this does not fix the [tex]\( x \)[/tex]-intercept directly to [tex]\( x = 5 \)[/tex]. It maintains the given [tex]\( x \)[/tex]-intercept of [tex]\( m(x) \)[/tex] but doesn't solve the problem as required.
Option B: [tex]\( n(x) = m(x) + 5 \)[/tex]
- Adding 5 to [tex]\( m(x) \)[/tex] shifts the graph vertically upwards or downwards, which alters the intercepts and affects the graph's shape overall. This would change the [tex]\( x \)[/tex]-intercept location, which does not meet the required condition.
Option C: [tex]\( n(x) = m(x - 5) \)[/tex]
- This option represents a horizontal shift of the function [tex]\( m(x) \)[/tex] to the right by 5 units. To confirm, substituting [tex]\( x = 5 \)[/tex]:
[tex]\[ n(5) = m(5 - 5) = m(0) \][/tex]
For [tex]\( n(x) \)[/tex] to have an [tex]\( x \)[/tex]-intercept at [tex]\( x = 5 \)[/tex]:
[tex]\[ n(5) = 0 \implies m(0) = 0 \][/tex]
Therefore, the intercept at [tex]\( x = 5 \)[/tex] for [tex]\( n \)[/tex] would hold if [tex]\( m \)[/tex] originally had an intercept at [tex]\( x = 0 \)[/tex]. This transformation ensures the [tex]\( x \)[/tex]-intercept is set at [tex]\( x = 5 \)[/tex] appropriately.
Option D: [tex]\( n(x) = m(x + 5) \)[/tex]
- This option would move the function [tex]\( m(x) \)[/tex] left by 5 units. Substituting [tex]\( x = 5 \)[/tex]:
[tex]\[ n(5) = m(5 + 5) = m(10) \][/tex]
Which means that for [tex]\( n(x) \)[/tex] to be zero at [tex]\( x = 5 \)[/tex], [tex]\( m(10) \)[/tex] would need to be zero, which does not ensure that [tex]\( x = 5 \)[/tex] is the intercept in the context provided.
Considering these options, the correct answer is:
[tex]\[ \boxed{C} \][/tex]
This option correctly adjusts the input of [tex]\( m(x) \)[/tex] in a manner that ensures both the vertical asymptotes remain the same and the [tex]\( x \)[/tex]-intercept is specifically located at [tex]\( x = 5 \)[/tex].
