Question

find vertical asymptote
f(x) = -ln(x + 3)

Explanation:

Step1: Recall log domain rule

The natural logarithm function \( \ln(u) \) is defined when \( u > 0 \). For \( f(x)=-\ln(x + 3) \), set the argument \( x + 3>0 \)? No, wait—vertical asymptote of \( \ln(u) \) occurs where \( u = 0 \) (since as \( u
ightarrow0^+ \), \( \ln(u)
ightarrow-\infty \)). So solve \( x + 3=0 \).

Step2: Solve for x

\( x+3 = 0\implies x=-3 \). Check the domain: \( x + 3>0\implies x>-3 \). As \( x
ightarrow - 3^+ \), \( x + 3
ightarrow0^+ \), so \( \ln(x + 3)
ightarrow-\infty \), and \( -\ln(x + 3)
ightarrow\infty \). Thus, the vertical asymptote is at \( x=-3 \).

Answer:

The vertical asymptote of \( f(x)=-\ln(x + 3) \) is \( x = - 3 \)