Question

What is the range of this exponential function?

y=4(0.8)x+2


(1 point)
$$

Answer 1

To determine the range of the function \( y = 4(0.8)^x + 2 \), we can analyze its components.

The function \( 0.8^x \) is an exponential function that approaches 0 as \( x \) increases (because \( 0.8 < 1 \)), and it approaches infinity as \( x \) decreases (for negative \( x \)).

Now, consider the entire function:

1. As \( x \) approaches positive infinity (\( x \to \infty \)):
\[
0.8^x \to 0 \quad \Rightarrow \quad y = 4(0) + 2 = 2
\]

2. As \( x \) approaches negative infinity (\( x \to -\infty \)):
\[
0.8^x \to \infty \quad \Rightarrow \quad y = 4(\infty) + 2 = \infty
\]

Thus, the value of \( y \) can get arbitrarily close to 2 but can never actually reach it (as \( x \) approaches infinity). However, it can take on all values greater than 2.

Therefore, the range of the function is:
\[
(2, \infty)
\]

In conclusion, the range of the function \( y = 4(0.8)^x + 2 \) is \( (2, \infty) \).