Answer :
Let's analyze the function [tex]\( f(x) = \ln(x-5) \)[/tex] step by step to determine its graph's behavior accurately.
1. Domain of the function:
- The natural logarithm function, [tex]\( \ln(y) \)[/tex], is defined only for [tex]\( y > 0 \)[/tex].
- Therefore, [tex]\( \ln(x-5) \)[/tex] is defined only when [tex]\( x-5 > 0 \)[/tex] or [tex]\( x > 5 \)[/tex].
- So, the domain of the function is [tex]\( x > 5 \)[/tex].
2. Vertical asymptote:
- As [tex]\( x \)[/tex] approaches 5 from the right-hand side (i.e., [tex]\( x \to 5^+ \)[/tex]), [tex]\( x-5 \)[/tex] approaches 0 from the positive side.
- Therefore, [tex]\( \ln(x-5) \to -\infty \)[/tex] as [tex]\( x \to 5^+ \)[/tex].
- Thus, there is a vertical asymptote at [tex]\( x = 5 \)[/tex].
3. Key point on the graph:
- Let's find a specific point the graph passes through by substituting particular [tex]\( x \)[/tex] values.
- For [tex]\( x = 6 \)[/tex], [tex]\( f(6) = \ln(6-5) = \ln(1) = 0 \)[/tex].
- So, the graph goes through the point [tex]\( (6,0) \)[/tex].
4. Behavior of the graph:
- As [tex]\( x \)[/tex] increases from slightly more than 5 to larger values, the rate of increase in [tex]\( \ln(x - 5) \)[/tex] slows down.
- This is because logarithmic functions increase rapidly at first and then increase more gradually as the input becomes larger.
Putting this all together:
- The function increases rapidly when [tex]\( x \)[/tex] is just slightly greater than 5 (as [tex]\( \ln(x-5) \)[/tex] transitions from negative infinity quickly towards higher values).
- It passes through the point [tex]\( (6,0) \)[/tex].
- After passing this point, it continues to increase, but the rate of increase slows down gradually.
Thus, the best descriptor for the graph of [tex]\( f(x) = \ln(x-5) \)[/tex] is:
d. It increases rapidly, goes through the point [tex]\( (6,0) \)[/tex], then increases gradually.
So, the correct answer is d.
1. Domain of the function:
- The natural logarithm function, [tex]\( \ln(y) \)[/tex], is defined only for [tex]\( y > 0 \)[/tex].
- Therefore, [tex]\( \ln(x-5) \)[/tex] is defined only when [tex]\( x-5 > 0 \)[/tex] or [tex]\( x > 5 \)[/tex].
- So, the domain of the function is [tex]\( x > 5 \)[/tex].
2. Vertical asymptote:
- As [tex]\( x \)[/tex] approaches 5 from the right-hand side (i.e., [tex]\( x \to 5^+ \)[/tex]), [tex]\( x-5 \)[/tex] approaches 0 from the positive side.
- Therefore, [tex]\( \ln(x-5) \to -\infty \)[/tex] as [tex]\( x \to 5^+ \)[/tex].
- Thus, there is a vertical asymptote at [tex]\( x = 5 \)[/tex].
3. Key point on the graph:
- Let's find a specific point the graph passes through by substituting particular [tex]\( x \)[/tex] values.
- For [tex]\( x = 6 \)[/tex], [tex]\( f(6) = \ln(6-5) = \ln(1) = 0 \)[/tex].
- So, the graph goes through the point [tex]\( (6,0) \)[/tex].
4. Behavior of the graph:
- As [tex]\( x \)[/tex] increases from slightly more than 5 to larger values, the rate of increase in [tex]\( \ln(x - 5) \)[/tex] slows down.
- This is because logarithmic functions increase rapidly at first and then increase more gradually as the input becomes larger.
Putting this all together:
- The function increases rapidly when [tex]\( x \)[/tex] is just slightly greater than 5 (as [tex]\( \ln(x-5) \)[/tex] transitions from negative infinity quickly towards higher values).
- It passes through the point [tex]\( (6,0) \)[/tex].
- After passing this point, it continues to increase, but the rate of increase slows down gradually.
Thus, the best descriptor for the graph of [tex]\( f(x) = \ln(x-5) \)[/tex] is:
d. It increases rapidly, goes through the point [tex]\( (6,0) \)[/tex], then increases gradually.
So, the correct answer is d.