Answer :
To determine which logarithmic function has a [tex]$y$[/tex]-intercept, we need to evaluate each function at [tex]\( x = 0 \)[/tex]. The [tex]$y$[/tex]-intercept of a function is the value of the function when [tex]\( x = 0 \)[/tex].
### Option A: [tex]\( f(x) = \log(x + 1) - 1 \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 + 1) - 1 \][/tex]
2. Simplify:
[tex]\[ f(0) = \log(1) - 1 \][/tex]
3. Recall that [tex]\(\log(1) = 0\)[/tex]:
[tex]\[ f(0) = 0 - 1 = -1 \][/tex]
Thus, the [tex]$y$[/tex]-intercept for Option A is [tex]\(-1\)[/tex].
### Option B: [tex]\( f(x) = \log(x) + 1 \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0) + 1 \][/tex]
2. Note that [tex]\(\log(0)\)[/tex] is undefined.
As [tex]\(\log(0)\)[/tex] is undefined, Option B does not have a [tex]$y$[/tex]-intercept.
### Option C: [tex]\( f(x) = \log(x - 1) + 1 \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 - 1) + 1 \][/tex]
2. Simplify:
[tex]\[ f(0) = \log(-1) + 1 \][/tex]
3. Note that [tex]\(\log(-1)\)[/tex] is undefined.
As [tex]\(\log(-1)\)[/tex] is undefined, Option C does not have a [tex]$y$[/tex]-intercept.
### Option D: [tex]\( f(x) = \log(x - 1) - 1 \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 - 1) - 1 \][/tex]
2. Simplify:
[tex]\[ f(0) = \log(-1) - 1 \][/tex]
3. Note that [tex]\(\log(-1)\)[/tex] is undefined.
As [tex]\(\log(-1)\)[/tex] is undefined, Option D does not have a [tex]$y$[/tex]-intercept.
### Conclusion
The only logarithmic function that has a [tex]$y$[/tex]-intercept is Option A: [tex]\( f(x) = \log(x + 1) - 1 \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{A} \][/tex]
### Option A: [tex]\( f(x) = \log(x + 1) - 1 \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 + 1) - 1 \][/tex]
2. Simplify:
[tex]\[ f(0) = \log(1) - 1 \][/tex]
3. Recall that [tex]\(\log(1) = 0\)[/tex]:
[tex]\[ f(0) = 0 - 1 = -1 \][/tex]
Thus, the [tex]$y$[/tex]-intercept for Option A is [tex]\(-1\)[/tex].
### Option B: [tex]\( f(x) = \log(x) + 1 \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0) + 1 \][/tex]
2. Note that [tex]\(\log(0)\)[/tex] is undefined.
As [tex]\(\log(0)\)[/tex] is undefined, Option B does not have a [tex]$y$[/tex]-intercept.
### Option C: [tex]\( f(x) = \log(x - 1) + 1 \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 - 1) + 1 \][/tex]
2. Simplify:
[tex]\[ f(0) = \log(-1) + 1 \][/tex]
3. Note that [tex]\(\log(-1)\)[/tex] is undefined.
As [tex]\(\log(-1)\)[/tex] is undefined, Option C does not have a [tex]$y$[/tex]-intercept.
### Option D: [tex]\( f(x) = \log(x - 1) - 1 \)[/tex]
1. Substitute [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 - 1) - 1 \][/tex]
2. Simplify:
[tex]\[ f(0) = \log(-1) - 1 \][/tex]
3. Note that [tex]\(\log(-1)\)[/tex] is undefined.
As [tex]\(\log(-1)\)[/tex] is undefined, Option D does not have a [tex]$y$[/tex]-intercept.
### Conclusion
The only logarithmic function that has a [tex]$y$[/tex]-intercept is Option A: [tex]\( f(x) = \log(x + 1) - 1 \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{A} \][/tex]
