1. The problem asks for the domain, range, and asymptote of the function $$h(x) = (0.5)^x - 9$$. 2. Recall that for an exponential function $$f(x) = a^x$$ where $$a > 0$$ and $$a \neq 1$$, the domain is all real numbers $$\mathbb{R}$$ because you can raise $$a$$ to any real power. 3. The range of $$a^x$$ is $$y > 0$$ because exponential functions are always positive. 4. When we have $$h(x) = (0.5)^x - 9$$, the graph is the exponential function shifted down by 9 units. 5. Therefore, the domain remains all real numbers $$\mathbb{R}$$. 6. The range shifts down by 9, so the new range is $$y > -9$$. 7. The horizontal asymptote of the original function $$y = (0.5)^x$$ is $$y = 0$$. 8. After shifting down by 9, the asymptote becomes $$y = -9$$. 9. Summarizing: - Domain: $$\{x \mid x \in \mathbb{R}\}$$ - Range: $$\{y \mid y > -9\}$$ - Asymptote: $$y = -9$$