1. **Problem Statement:** We are given the values of an investment over 4 years: $$f(1) = 10970.46, \quad f(2) = 12962.46, \quad f(3) = 14964.46, \quad f(4) = 16962.46$$ We want to confirm the function type and find the slope (common difference) of the function. 2. **Identify the function type:** The values increase by approximately the same amount each year, suggesting a linear function of the form: $$f(x) = mx + b$$ where $m$ is the slope (rate of change) and $b$ is the initial value (intercept). 3. **Calculate the common difference (slope) $m$:** Calculate the differences between successive values: $$12962.46 - 10970.46 = 1992$$ $$14964.46 - 12962.46 = 2002$$ $$16962.46 - 14964.46 = 1998$$ The differences are approximately equal, so the slope $m \approx 2000$. 4. **Find the intercept $b$:** Using the point $(1, 10970.46)$ and the slope $m = 2000$, substitute into the linear equation: $$10970.46 = 2000 \times 1 + b$$ Solve for $b$: $$b = 10970.46 - 2000 = 8970.46$$ 5. **Write the linear function:** $$f(x) = 2000x + 8970.46$$ 6. **Interpretation:** The investment increases by about 2000 each year, starting from an initial value of approximately 8970.46 at year 0. **Final answer:** $$\boxed{f(x) = 2000x + 8970.46}$$