1. Problem 11 states: Given a one-to-one function mapping volcano names to ice cream flavors, find the domain and range of the inverse function. 2. The original function's domain is the set of volcano names: $\{\text{Arenal}, \text{Etna}, \text{Krakatoa}, \text{Kilauea}, \text{Mayon}, \text{Pinatubo}\}$. 3. Its range is the set of ice cream flavors: $\{\text{black liquorice}, \text{pistachio}, \text{maple walnut}, \text{tiger tail}, \text{amaretto}, \text{mint chocolate chip}\}$. 4. For the inverse function, the domain and range swap: - Domain of inverse: $\{\text{black liquorice}, \text{pistachio}, \text{maple walnut}, \text{tiger tail}, \text{amaretto}, \text{mint chocolate chip}\}$ - Range of inverse: $\{\text{Arenal}, \text{Etna}, \text{Krakatoa}, \text{Kilauea}, \text{Mayon}, \text{Pinatubo}\}$ 5. Problem 12 states: Apply these transformations to the parent function $y=x^{2}$: - Vertically compressed by $\frac{1}{6}$ - Horizontally compressed by $\frac{1}{2}$ - Horizontally reflected - Vertically translated up 3 units - Horizontally translated left 4 units 6. Step-wise transformation of $y=x^{2}$: - Vertical compression by $\frac{1}{6}$ changes to $y=\frac{1}{6}x^{2}$ - Horizontal compression by $\frac{1}{2}$ means replace $x$ by $2x$, so $y=\frac{1}{6}(2x)^{2} = \frac{1}{6}4x^{2} = \frac{2}{3}x^{2}$ - Horizontal reflection replaces $x$ by $-x$: $y=\frac{2}{3}(-x)^{2} = \frac{2}{3}x^{2}$ (note reflection on $x^{2}$ doesn't change formula) - Horizontal translation left 4 means replace $x$ by $(x+4)$: $y=\frac{2}{3}(x+4)^{2}$ - Vertical translation up 3 units: $y=\frac{2}{3}(x+4)^{2} + 3$ 7. Final transformed equation is: $$ y = \frac{2}{3}(x+4)^{2} + 3 $$