To determine which logarithmic function has a y-intercept, we need to check the value of each function at [tex]\( x = 0 \)[/tex]. A y-intercept exists where the function is defined and produces a finite value at [tex]\( x = 0 \)[/tex].
Consider the functions provided:
Option A: [tex]\( f(x) = \log(x + 1) - 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0 + 1) - 1 = \log(1) - 1 = 0 - 1 = -1 \][/tex]
Since this produces a finite value ([tex]\(-1\)[/tex]), this function has a y-intercept.
Option B: [tex]\( f(x) = \log(x) + 1 \)[/tex]
Evaluate at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \log(0) + 1 \][/tex]
However, [tex]\(\log(0)\)[/tex] is undefined. Therefore, this function does not have a y-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]
Since [tex]\(\log(-1)\)[/tex] is undefined in the real number system, this function does not have a y-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]
Since [tex]\(\log(-1)\)[/tex] is undefined in the real number system, this function does not have a y-intercept.
Based on this analysis, the only function that has a y-intercept is:
[tex]\[ \boxed{1} \][/tex]
