Mapping Notation Ac0B58

1. **State the problem:** We are given a transformation of the form $(x, y) \to \left(\frac{1}{b} x + h, ay + k\right)$ that transforms the function $y = f(x)$ into $y = 2 + 7f\left(\frac{1}{2} (x + 3)\right)$. We need to find the mapping notation in the form $(x,y) \to (x_\_, y_\_)$. 2. **Analyze the transformation inside the function argument:** The transformed function argument is $\frac{1}{2}(x + 3)$. This matches the form $\frac{1}{b} x + h$ where: $$\frac{1}{b} x + h = \frac{1}{2} x + \frac{3}{2}$$ So, $\frac{1}{b} = \frac{1}{2} \implies b = 2$ and $h = \frac{3}{2}$. 3. **Analyze the transformation outside the function:** The transformed function is $y = 2 + 7 f(\ldots)$ which matches $ay + k$ where: $$ay + k = 7y + 2$$ So, $a = 7$ and $k = 2$. 4. **Write the mapping notation:** Using the values found: $$(x, y) \to \left(\frac{1}{2} x + \frac{3}{2}, 7y + 2\right)$$ In the required notation form $(x,y) \to (x_\_, y_\_)$, the mapping is: "(\frac{3}{2}_2, 2_7)" This means: - For $x$: $x \to \frac{1}{2} x + \frac{3}{2}$ corresponds to $(x_2)$ with $h = \frac{3}{2}$ - For $y$: $y \to 7y + 2$ corresponds to $(y_7)$ with $k = 2$ **Final answer:** "(\frac{3}{2}_2, 2_7)"