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To determine which function has an asymptote at [tex]\( x = 5 \)[/tex] and an [tex]\( x \)[/tex]-intercept of [tex]\( (6, 0) \)[/tex], let's examine each option step-by-step.
### Option A: [tex]\( f(x) = \log (x-5) \)[/tex]
1. Asymptote Analysis:
- The logarithmic function [tex]\( \log (x-5) \)[/tex] is undefined when [tex]\( x-5 \leq 0 \)[/tex], which means at [tex]\( x = 5 \)[/tex].
- Therefore, there is an asymptote at [tex]\( x = 5 \)[/tex].
2. [tex]\( x \)[/tex]-intercept Analysis:
- The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( \log(x-5) = 0 \)[/tex], we get:
[tex]\[ \log(x-5) = 0 \implies x-5 = 10^0 \implies x - 5 = 1 \implies x = 6 \][/tex]
- Hence, the [tex]\( x \)[/tex]-intercept is at [tex]\( (6, 0) \)[/tex].
Both the conditions (asymptote at [tex]\( x = 5 \)[/tex] and [tex]\( x \)[/tex]-intercept at [tex]\( (6, 0) \)[/tex]) are satisfied by this function.
### Option B: [tex]\( f(x) = \log (x+5) \)[/tex]
1. Asymptote Analysis:
- The logarithmic function [tex]\( \log (x+5) \)[/tex] is undefined when [tex]\( x+5 \leq 0 \)[/tex], which means at [tex]\( x = -5 \)[/tex].
- Therefore, there is an asymptote at [tex]\( x = -5 \)[/tex].
2. [tex]\( x \)[/tex]-intercept Analysis:
- The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( \log(x+5) = 0 \)[/tex], we get:
[tex]\[ \log(x+5) = 0 \implies x+5 = 10^0 \implies x + 5 = 1 \implies x = -4 \][/tex]
- Hence, the [tex]\( x \)[/tex]-intercept is at [tex]\( (-4, 0) \)[/tex], not [tex]\( (6, 0) \)[/tex].
This option does not satisfy both conditions.
### Option C: [tex]\( f(x) = \log x - 5 \)[/tex]
1. Asymptote Analysis:
- The logarithmic function [tex]\( \log x \)[/tex] is undefined when [tex]\( x \leq 0 \)[/tex], which means at [tex]\( x = 0 \)[/tex].
- Therefore, there is an asymptote at [tex]\( x = 0 \)[/tex].
2. [tex]\( x \)[/tex]-intercept Analysis:
- The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( \log x - 5 = 0 \)[/tex], we get:
[tex]\[ \log x - 5 = 0 \implies \log x = 5 \implies x = 10^5 \][/tex]
- Hence, the [tex]\( x \)[/tex]-intercept is at [tex]\( (10^5, 0) \)[/tex], not [tex]\( (6, 0) \)[/tex].
This option does not satisfy both conditions.
### Option D: [tex]\( f(x) = \log x + 5 \)[/tex]
1. Asymptote Analysis:
- The logarithmic function [tex]\( \log x \)[/tex] is undefined when [tex]\( x \leq 0 \)[/tex], which means at [tex]\( x = 0 \)[/tex].
- Therefore, there is an asymptote at [tex]\( x = 0 \)[/tex].
2. [tex]\( x \)[/tex]-intercept Analysis:
- The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( \log x + 5 = 0 \)[/tex], we get:
[tex]\[ \log x + 5 = 0 \implies \log x = -5 \implies x = 10^{-5} \][/tex]
- Hence, the [tex]\( x \)[/tex]-intercept is at [tex]\( (10^{-5}, 0) \)[/tex], not [tex]\( (6, 0) \)[/tex].
This option does not satisfy both conditions.
### Conclusion
Only option [tex]\( A \)[/tex], [tex]\( f(x) = \log (x-5) \)[/tex], has an asymptote at [tex]\( x = 5 \)[/tex] and an [tex]\( x \)[/tex]-intercept at [tex]\( (6, 0) \)[/tex].
Therefore, the correct choice is:
[tex]\[ \boxed{A} \][/tex]
### Option A: [tex]\( f(x) = \log (x-5) \)[/tex]
1. Asymptote Analysis:
- The logarithmic function [tex]\( \log (x-5) \)[/tex] is undefined when [tex]\( x-5 \leq 0 \)[/tex], which means at [tex]\( x = 5 \)[/tex].
- Therefore, there is an asymptote at [tex]\( x = 5 \)[/tex].
2. [tex]\( x \)[/tex]-intercept Analysis:
- The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( \log(x-5) = 0 \)[/tex], we get:
[tex]\[ \log(x-5) = 0 \implies x-5 = 10^0 \implies x - 5 = 1 \implies x = 6 \][/tex]
- Hence, the [tex]\( x \)[/tex]-intercept is at [tex]\( (6, 0) \)[/tex].
Both the conditions (asymptote at [tex]\( x = 5 \)[/tex] and [tex]\( x \)[/tex]-intercept at [tex]\( (6, 0) \)[/tex]) are satisfied by this function.
### Option B: [tex]\( f(x) = \log (x+5) \)[/tex]
1. Asymptote Analysis:
- The logarithmic function [tex]\( \log (x+5) \)[/tex] is undefined when [tex]\( x+5 \leq 0 \)[/tex], which means at [tex]\( x = -5 \)[/tex].
- Therefore, there is an asymptote at [tex]\( x = -5 \)[/tex].
2. [tex]\( x \)[/tex]-intercept Analysis:
- The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( \log(x+5) = 0 \)[/tex], we get:
[tex]\[ \log(x+5) = 0 \implies x+5 = 10^0 \implies x + 5 = 1 \implies x = -4 \][/tex]
- Hence, the [tex]\( x \)[/tex]-intercept is at [tex]\( (-4, 0) \)[/tex], not [tex]\( (6, 0) \)[/tex].
This option does not satisfy both conditions.
### Option C: [tex]\( f(x) = \log x - 5 \)[/tex]
1. Asymptote Analysis:
- The logarithmic function [tex]\( \log x \)[/tex] is undefined when [tex]\( x \leq 0 \)[/tex], which means at [tex]\( x = 0 \)[/tex].
- Therefore, there is an asymptote at [tex]\( x = 0 \)[/tex].
2. [tex]\( x \)[/tex]-intercept Analysis:
- The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( \log x - 5 = 0 \)[/tex], we get:
[tex]\[ \log x - 5 = 0 \implies \log x = 5 \implies x = 10^5 \][/tex]
- Hence, the [tex]\( x \)[/tex]-intercept is at [tex]\( (10^5, 0) \)[/tex], not [tex]\( (6, 0) \)[/tex].
This option does not satisfy both conditions.
### Option D: [tex]\( f(x) = \log x + 5 \)[/tex]
1. Asymptote Analysis:
- The logarithmic function [tex]\( \log x \)[/tex] is undefined when [tex]\( x \leq 0 \)[/tex], which means at [tex]\( x = 0 \)[/tex].
- Therefore, there is an asymptote at [tex]\( x = 0 \)[/tex].
2. [tex]\( x \)[/tex]-intercept Analysis:
- The [tex]\( x \)[/tex]-intercept occurs where [tex]\( f(x) = 0 \)[/tex].
- Setting [tex]\( \log x + 5 = 0 \)[/tex], we get:
[tex]\[ \log x + 5 = 0 \implies \log x = -5 \implies x = 10^{-5} \][/tex]
- Hence, the [tex]\( x \)[/tex]-intercept is at [tex]\( (10^{-5}, 0) \)[/tex], not [tex]\( (6, 0) \)[/tex].
This option does not satisfy both conditions.
### Conclusion
Only option [tex]\( A \)[/tex], [tex]\( f(x) = \log (x-5) \)[/tex], has an asymptote at [tex]\( x = 5 \)[/tex] and an [tex]\( x \)[/tex]-intercept at [tex]\( (6, 0) \)[/tex].
Therefore, the correct choice is:
[tex]\[ \boxed{A} \][/tex]
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