Answer :
To find the [tex]\( y \)[/tex]-intercept of a logarithmic function, we need to compute the value of the function when [tex]\( x = 0 \)[/tex]. Let's evaluate each option one by one.
### Option A: [tex]\( f(x) = \log(x + 1) - 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 + 1) - 1 = \log(1) - 1 \][/tex]
Since [tex]\( \log(1) = 0 \)[/tex]:
[tex]\[ f(0) = 0 - 1 = -1 \][/tex]
Thus, this function has a [tex]\( y \)[/tex]-intercept at [tex]\( (0, -1) \)[/tex].
### Option B: [tex]\( f(x) = \log x + 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0) + 1 \][/tex]
The logarithm of zero, [tex]\( \log(0) \)[/tex], is undefined because the logarithm function is only defined for positive real numbers. Hence, this function does not have a [tex]\( y \)[/tex]-intercept.
### Option C: [tex]\( f(x) = \log(x - 1) + 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 - 1) + 1 = \log(-1) + 1 \][/tex]
The logarithm of a negative number, [tex]\( \log(-1) \)[/tex], is also undefined for real numbers. Therefore, this function does not have a [tex]\( y \)[/tex]-intercept.
### Option D: [tex]\( f(x) = \log(x - 1) - 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 - 1) - 1 = \log(-1) - 1 \][/tex]
The logarithm of a negative number, [tex]\( \log(-1) \)[/tex], is undefined for real numbers. Thus, this function does not have a [tex]\( y \)[/tex]-intercept.
### Conclusion
After evaluating each option, we see that only option A, [tex]\( f(x) = \log(x + 1) - 1 \)[/tex], has a [tex]\( y \)[/tex]-intercept. The other functions involve logarithms of zero or negative numbers, which are not defined in the real number system.
Therefore, the correct answer is:
[tex]\[ \boxed{A} \][/tex]
### Option A: [tex]\( f(x) = \log(x + 1) - 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 + 1) - 1 = \log(1) - 1 \][/tex]
Since [tex]\( \log(1) = 0 \)[/tex]:
[tex]\[ f(0) = 0 - 1 = -1 \][/tex]
Thus, this function has a [tex]\( y \)[/tex]-intercept at [tex]\( (0, -1) \)[/tex].
### Option B: [tex]\( f(x) = \log x + 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0) + 1 \][/tex]
The logarithm of zero, [tex]\( \log(0) \)[/tex], is undefined because the logarithm function is only defined for positive real numbers. Hence, this function does not have a [tex]\( y \)[/tex]-intercept.
### Option C: [tex]\( f(x) = \log(x - 1) + 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 - 1) + 1 = \log(-1) + 1 \][/tex]
The logarithm of a negative number, [tex]\( \log(-1) \)[/tex], is also undefined for real numbers. Therefore, this function does not have a [tex]\( y \)[/tex]-intercept.
### Option D: [tex]\( f(x) = \log(x - 1) - 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 - 1) - 1 = \log(-1) - 1 \][/tex]
The logarithm of a negative number, [tex]\( \log(-1) \)[/tex], is undefined for real numbers. Thus, this function does not have a [tex]\( y \)[/tex]-intercept.
### Conclusion
After evaluating each option, we see that only option A, [tex]\( f(x) = \log(x + 1) - 1 \)[/tex], has a [tex]\( y \)[/tex]-intercept. The other functions involve logarithms of zero or negative numbers, which are not defined in the real number system.
Therefore, the correct answer is:
[tex]\[ \boxed{A} \][/tex]
