Question

find the horizontal asymptote
$f(x) = 3^{x + 1}$

Explanation:

Step1: Recall Exponential Function Asymptote

For an exponential function \( y = a^x + k \) (or transformed forms), the horizontal asymptote is determined by the limit as \( x \to -\infty \) (for exponential growth/decay). The general form \( f(x)=b^{x + c}+d \) has horizontal asymptote \( y = d \) when considering the limit of \( b^{x + c} \) as \( x\to-\infty \) (since \( b^{x + c}\to0 \) for \( b>1 \) as \( x\to-\infty \)). Here, \( f(x)=3^{x + 1} \) can be rewritten as \( f(x)=3\times3^{x} \).

Step2: Analyze Limit as \( x\to-\infty \)

As \( x\to-\infty \), \( 3^{x}\to0 \) (because the base \( 3>1 \), and exponential functions with base \( > 1 \) approach 0 as \( x\to-\infty \)). Then \( f(x)=3\times3^{x}\to3\times0 = 0 \) as \( x\to-\infty \). So the horizontal asymptote is the line \( y = 0 \).

Answer:

The horizontal asymptote of \( f(x)=3^{x + 1} \) is \( y = 0 \)