1. **Stating the problem:** We are given a table of water temperatures over time and asked to find a quadratic function $f(x) = ax^2 + bx + c$ that models the temperature, where $x$ is time in minutes and $f(x)$ is temperature in °C. 2. **Given data:** \begin{align*} \text{Time } x &: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \\ \text{Temperature } f(x) &: 10, 14.5, 20, 26.5, 34, 42.5, 52, 62.5, 74, 86.5, 100 \end{align*} 3. **Find first differences:** These are $\Delta f(x) = f(x+1) - f(x)$: \begin{align*} 14.5 - 10 &= 4.5 \\ 20 - 14.5 &= 5.5 \\ 26.5 - 20 &= 6.5 \\ 34 - 26.5 &= 7.5 \\ 42.5 - 34 &= 8.5 \\ 52 - 42.5 &= 9.5 \\ 62.5 - 52 &= 10.5 \\ 74 - 62.5 &= 11.5 \\ 86.5 - 74 &= 12.5 \\ 100 - 86.5 &= 13.5 \end{align*} 4. **Find second differences:** These are $\Delta^2 f(x) = \Delta f(x+1) - \Delta f(x)$: \begin{align*} 5.5 - 4.5 &= 1 \\ 6.5 - 5.5 &= 1 \\ 7.5 - 6.5 &= 1 \\ 8.5 - 7.5 &= 1 \\ 9.5 - 8.5 &= 1 \\ 10.5 - 9.5 &= 1 \\ 11.5 - 10.5 &= 1 \\ 12.5 - 11.5 &= 1 \\ 13.5 - 12.5 &= 1 \end{align*} The second difference is constant and equals 1. 5. **Use the formula for quadratic functions from second differences:** For equally spaced $x$ values with step 1, the second difference $\Delta^2 f(x) = 2a$. So, $$2a = 1 \implies a = \frac{1}{2} = 0.5$$ 6. **Find $b$ and $c$ using initial values:** We know: $$f(0) = c = 10$$ Also, $$f(1) = a(1)^2 + b(1) + c = a + b + c = 14.5$$ Substitute $a=0.5$ and $c=10$: $$0.5 + b + 10 = 14.5 \implies b = 14.5 - 10.5 = 4$$ 7. **Final quadratic function:** $$f(x) = 0.5x^2 + 4x + 10$$ This function models the temperature of the water over time. **Answer:** $f(x) = 0.5x^2 + 4x + 10$