Question

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

Explanation:

Step1: Recall Horizontal Asymptote Rules

For exponential functions \( y = e^{kx}+c \), we analyze limits as \( x \to \pm\infty \).

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

As \( x \to -\infty \), \( 3x \to -\infty \), so \( e^{3x} \to 0 \) (since exponential function with negative exponent approaches 0). Then \( f(x)=e^{3x}+1 \to 0 + 1=1 \).

Step3: Analyze Limit as \( x \to \infty \)

As \( x \to \infty \), \( 3x \to \infty \), so \( e^{3x} \to \infty \), thus \( f(x)=e^{3x}+1 \to \infty \). So the horizontal asymptote is determined by the limit as \( x \to -\infty \).

Answer:

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