Sure, let's analyze the function [tex]\( y = \log_{1/2}(x + 9) + 4 \)[/tex]:
### Domain
For the logarithmic function [tex]\( \log_{1/2}(x + 9) \)[/tex] to be defined, the argument [tex]\( x + 9 \)[/tex] must be greater than 0:
[tex]\[ x + 9 > 0 \][/tex]
[tex]\[ x > -9 \][/tex]
Thus, the domain of the function is:
[tex]\[ (-9, \infty) \][/tex]
### Range
The range of a logarithmic function with any base is all real numbers because a logarithmic function can produce any real number output. Therefore:
[tex]\[ (-\infty, \infty) \][/tex]
### Intervals of Increase/Decrease
Since the base of the logarithm is [tex]\( \frac{1}{2} \)[/tex], which is less than 1, the function [tex]\( \log_{1/2}(x + 9) \)[/tex] is a decreasing function. Adding 4 does not change the nature of increase or decrease; it just shifts the entire function up by 4 units. Thus, the function decreases over its entire domain:
- There are no intervals where the function is increasing.
- The function is decreasing on the entire domain:
[tex]\[ (-9, \infty) \][/tex]
### Asymptote(s)
The vertical asymptote occurs where the argument of the logarithm approaches zero (i.e., [tex]\( x + 9 = 0 \)[/tex]):
[tex]\[ x = -9 \][/tex]
Thus, the vertical asymptote of the function is at:
[tex]\[ x = -9 \][/tex]
### Summary
- Domain: [tex]\( (-9, \infty) \)[/tex]
- Range: [tex]\( (-\infty, \infty) \)[/tex]
- Intervals of increase: None
- Intervals of decrease: [tex]\( (-9, \infty) \)[/tex]
- Asymptote(s): [tex]\( x = -9 \)[/tex]
