The function shown is $f(x) = \left( \frac{1}{2} \right)^x - 6$, which is an exponential decay function due to the base $\frac{1}{2}$, a number between 0 and 1. Here’s how we analyze the features of this function:

1. Asymptote: Since the function shifts downward by 6, there is a horizontal asymptote at $y = -6$.

2. Range: The range of $f(x)$ is $(-6, \infty)$ because the exponential function never reaches $-6$ but can get infinitely close, and it extends to positive infinity.

3. Domain: Exponential functions have a domain of all real numbers, so $(-\infty, \infty)$.

4. End Behavior:

   • Left Side ($x \to -\infty$): As $x$ becomes very negative, $f(x)$ approaches $\infty$.
   • Right Side ($x \to \infty$): As $x$ becomes very positive, $f(x)$ approaches $-6$.

So, the completed statements would be:

• "The function $f(x)$ is an exponential function with a horizontal asymptote of $y = -6$."
• "The range of the function is $(-6, \infty)$, and it is all real numbers on its domain of $(- \infty, \infty)$."
• "The end behavior on the left side is as $x \to -\infty$, $f(x) \to \infty$, and the end behavior on the right side is as $x \to \infty$, $f(x) \to -6$."

Would you like further details or have any questions?

Here are some related questions to deepen your understanding:

1. How does the base $\frac{1}{2}$ affect the behavior of an exponential function?
2. What happens to the asymptote if the function were $f(x) = \left( \frac{1}{2} \right)^x + c$ for any constant $c$?
3. How does the range of the function change if the function is reflected across the x-axis?
4. What would be the end behavior if the base were greater than 1 (e.g., $f(x) = 2^x - 6$)?
5. How would you find the y-intercept of this function?

Tip: For exponential functions of the form $a \cdot b^x + k$, $k$ shifts the graph vertically and determines the horizontal asymptote.