Summarize the pertinent infornation oblai wed by upplying tto graphing strategy and sketch the grap' of \( y=I(x) \) \[ f(x)=\ln (x+3)-2 \] Select the correct choice below and, it necessary, ill in ine answer box to complote your choice. A. The domain of I is \( (-3, \infty) \). (Type your answer in inteval nolation. Use a comma to separalo answors as nooded.) B. The domain of I is emply. Select the correct choice below and, if necessary, fill in the answer box to corrplale your choice. A. Tho \( x \)-intercept(s) of \( I \) is \( x=e^{2}-3 \). (Type an oxacl answer Uso a comma lo separefe answors as needed.) B. The funclion f has no \( x \)-inlercepls. Select the correct choice beow and, f necessary, fll in the answer box lo complate your choico. A. The \( y \)-intercept ofl is \( y= \) \( \square \) ]. (Type on exact answer. Use a comma lo soparate answers as needod.) B. The function I has no y-inlercepl.

The domain of the function \( f(x) = \ln(x + 3) - 2 \) is derived from the properties of logarithms, which only accept positive arguments. Thus, for \( x + 3 > 0 \), we have \( x > -3 \). This gives us the interval \( (-3, \infty) \). So, the correct choice is A. The \( x \)-intercept occurs when \( f(x) = 0 \). Therefore, we solve \( \ln(x + 3) - 2 = 0 \), which simplifies to \( \ln(x + 3) = 2 \) or \( x + 3 = e^2 \). This results in \( x = e^2 - 3 \). Thus, the correct choice is A. To find the \( y \)-intercept, we evaluate \( f(0) = \ln(0 + 3) - 2 = \ln(3) - 2 \). Therefore, the \( y \)-intercept is at \( y = \ln(3) - 2 \), making the correct choice A as well. Let's wrap this up with a fun sketch of \( y = f(x) \)! Picture this: the graph starts at the point where \( x = -3 \) (but does not touch it, since that's the boundary of our domain), rises steeply, and gently curves upwards to the right. It dances above the \( x \)-axis, crossing at \( x = e^{2} - 3 \), and reflects its \( y \)-intercept at \( y = \ln(3) - 2 \). Happy graphing!