Question

find the horizontal asymptote
-e^x - 1

Explanation:

Step1: Recall the behavior of exponential function

The exponential function \( y = e^x \) has a horizontal asymptote \( y = 0 \) as \( x \to -\infty \), and it approaches \( +\infty \) as \( x \to +\infty \). For the function \( y=-e^x - 1\), we analyze the limit as \( x\to\pm\infty \).

Step2: Find the limit as \( x\to -\infty \)

As \( x\to -\infty \), \( e^x\to 0 \) (since the exponential function with base \( e>1 \) approaches 0 as the exponent goes to \( -\infty \)). So, we calculate the limit of \( -e^x - 1 \) as \( x\to -\infty \):

$$ \lim_{x\to -\infty}(-e^x - 1)=- \lim_{x\to -\infty}e^x-1=- 0 - 1=-1 $$

Step3: Find the limit as \( x\to +\infty \)

As \( x\to +\infty \), \( e^x\to +\infty \), so \( -e^x\to -\infty \), and then \( -e^x - 1\to -\infty \). So, there is no horizontal asymptote from the right - hand limit in the sense of a finite value, but from the left - hand limit (as \( x\to -\infty \)) we get a horizontal asymptote.

Answer:

The horizontal asymptote of the function \( y = -e^x-1 \) is \( y=-1 \)