1. **State the problem:** We have the function $f(x) = 10 \left(\frac{1}{5}\right)^x$. We need to find its domain, range, asymptotes, and the average rate of change from $x = -2$ to $x = 2$. 2. **Domain:** The function is an exponential function with base $\frac{1}{5} > 0$, so it is defined for all real numbers. \[ \text{Domain} = (-\infty, \infty) \] 3. **Range:** Since $\left(\frac{1}{5}\right)^x > 0$ for all $x$, and it is multiplied by 10, the function values are always positive. \[ \text{Range} = (0, \infty) \] 4. **Asymptote:** For exponential decay functions of the form $a b^x$ with $0 < b < 1$, the horizontal asymptote is $y=0$. \[ \text{Asymptote: } y = 0 \] 5. **Average rate of change formula:** \[ \text{Average rate of change} = \frac{f(b) - f(a)}{b - a} \] where $a = -2$ and $b = 2$. 6. **Calculate $f(-2)$:** \[ f(-2) = 10 \left(\frac{1}{5}\right)^{-2} = 10 \times 5^2 = 10 \times 25 = 250 \] 7. **Calculate $f(2)$:** \[ f(2) = 10 \left(\frac{1}{5}\right)^2 = 10 \times \frac{1}{25} = \frac{10}{25} = \frac{2}{5} \] 8. **Calculate average rate of change:** \[ \frac{f(2) - f(-2)}{2 - (-2)} = \frac{\frac{2}{5} - 250}{4} = \frac{\frac{2}{5} - \frac{1250}{5}}{4} = \frac{\frac{2 - 1250}{5}}{4} = \frac{\frac{-1248}{5}}{4} = \frac{-1248}{5} \times \frac{1}{4} = \frac{-1248}{20} \] 9. **Simplify the fraction:** \[ \frac{-1248}{20} = \frac{\cancel{-1248}^{\div 4}}{\cancel{20}^{\div 4}} = \frac{-312}{5} \] 10. **Final answer:** \[ \text{Average rate of change} = -\frac{312}{5} = -62.4 \] This means the function decreases on average by 62.4 units per unit increase in $x$ from $-2$ to $2$.