1. **State the problem:** We analyze the features of the function $$f(x) = -\left(\frac{1}{2}\right)^x + 5$$ including its asymptote, range, domain, monotonicity, and end behavior. 2. **Recall the general form and rules:** The base function is $$g(x) = \left(\frac{1}{2}\right)^x$$ which is an exponential decay function with base between 0 and 1. 3. **Analyze the transformation:** The function is reflected vertically by the negative sign, then shifted upward by 5 units. 4. **Find the horizontal asymptote:** Since $$g(x) \to 0$$ as $$x \to \infty$$, then $$f(x) = -g(x) + 5 \to -0 + 5 = 5$$. So the horizontal asymptote is $$y = 5$$. 5. **Determine the range:** Since $$g(x) > 0$$ for all real $$x$$, $$-g(x) < 0$$. Thus, $$f(x) = -g(x) + 5 < 5$$. Also, as $$x \to -\infty$$, $$g(x) \to \infty$$, so $$f(x) \to -\infty + 5 = -\infty$$. Therefore, the range is $$(-\infty, 5)$$. 6. **Domain:** Exponential functions are defined for all real numbers, so domain is $$(-\infty, \infty)$$. 7. **Monotonicity:** Since $$g(x)$$ is decreasing, $$-g(x)$$ is increasing. Adding 5 does not change monotonicity. So $$f(x)$$ is increasing on $$(-\infty, \infty)$$. 8. **End behavior:** - As $$x \to -\infty$$, $$g(x) \to \infty$$, so $$f(x) \to -\infty$$. - As $$x \to \infty$$, $$g(x) \to 0$$, so $$f(x) \to 5$$ from below. **Final answer:** - Horizontal asymptote: $$y = 5$$ - Range: $$(-\infty, 5)$$ - Domain: $$(-\infty, \infty)$$ - Increasing on entire domain - End behavior: $$f(x) \to -\infty$$ as $$x \to -\infty$$, and $$f(x) \to 5$$ as $$x \to \infty$$.