To graph the function [tex]\( y = \log_5(x - 2) - 1 \)[/tex], we need to understand various properties of this logarithmic function and how transformations affect it.
### Domain
The argument of the logarithmic function, [tex]\( x - 2 \)[/tex], must be greater than 0 because the logarithm of a non-positive number is undefined. This gives us:
[tex]\[ x - 2 > 0 \][/tex]
[tex]\[ x > 2 \][/tex]
Thus, the domain of the function is:
[tex]\[ (2, \infty) \][/tex]
### Range
The function [tex]\( y = \log_5(x - 2) - 1 \)[/tex] is a logarithmic function, which can take any real value as the logarithmic function covers all real numbers. Hence, the range is:
[tex]\[ (-\infty, \infty) \][/tex]
### Intervals of Increase
To determine if the function is increasing, let's consider the behavior of the original logarithmic function. The parent function [tex]\( \log_5(x) \)[/tex] is increasing for [tex]\( x > 0 \)[/tex] because the base [tex]\( 5 \)[/tex] is greater than 1.
Since [tex]\( y = \log_5(x - 2) - 1 \)[/tex] is simply a horizontally shifted and vertically shifted version of [tex]\( \log_5(x) \)[/tex]:
- The shift by 2 units to the right does not change the fact that the logarithm increases.
- Subtracting 1 only shifts the graph downwards but does not affect the interval of increase.
Therefore, the function is increasing on its entire domain:
[tex]\[ (2, \infty) \][/tex]
### Intervals of Decrease
The function does not decrease on any interval within its domain. Hence, there are no intervals of decrease.
### Asymptote(s)
Logarithmic functions have a vertical asymptote where their argument equals zero. In this case, the vertical asymptote occurs when:
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]
So, the vertical asymptote is:
[tex]\[ x = 2 \][/tex]
### Summary of the characteristics:
- Domain: [tex]\( (2, \infty) \)[/tex]
- Range: [tex]\( (-\infty, \infty) \)[/tex]
- Intervals of Increase: [tex]\( (2, \infty) \)[/tex]
- Intervals of Decrease: None
- Asymptote(s): [tex]\( x = 2 \)[/tex]
### Graphing the Function
To graph [tex]\( y = \log_5(x - 2) - 1 \)[/tex]:
1. Start with the parent function [tex]\( y = \log_5(x) \)[/tex].
2. Shift the graph to the right by 2 units to get [tex]\( y = \log_5(x - 2) \)[/tex].
3. Shift the graph down by 1 unit to get [tex]\( y = \log_5(x - 2) - 1 \)[/tex].
Your final graph should have:
- A vertical asymptote at [tex]\( x = 2 \)[/tex].
- The graph passing through points like [tex]\( (3, -1) \)[/tex] since [tex]\( \log_5(1) = 0 \)[/tex] implies that [tex]\( y \)[/tex] will be [tex]\( -1 \)[/tex] when [tex]\( x = 3 \)[/tex].
- As [tex]\( x \)[/tex] increases, the function will continue to rise but at a decreasing rate. As [tex]\( x \)[/tex] approaches 2 from the right, the function will decrease towards [tex]\( -\infty \)[/tex].
