Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.