1. **Problem Statement:** Fit a second degree polynomial of the form $$y = a + bx + cx^2$$ to the given data points: $$\begin{array}{c|ccccccc} X & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ y & 2.3 & 5.2 & 9.7 & 16.5 & 29.4 & 35.5 & 54.4 \end{array}$$ 2. **Method:** We use the least squares method to find coefficients $a$, $b$, and $c$ that minimize the error. 3. **Set up the system:** Define sums: $$S_x = \sum X_i, \quad S_y = \sum y_i, \quad S_{x^2} = \sum X_i^2, \quad S_{x^3} = \sum X_i^3, \quad S_{x^4} = \sum X_i^4, \quad S_{xy} = \sum X_i y_i, \quad S_{x^2y} = \sum X_i^2 y_i$$ Calculate these sums: $$S_x = 1+2+3+4+5+6+7 = 28$$ $$S_y = 2.3+5.2+9.7+16.5+29.4+35.5+54.4 = 152.99999999999997$$ $$S_{x^2} = 1^2+2^2+3^2+4^2+5^2+6^2+7^2 = 140$$ $$S_{x^3} = 1^3+2^3+3^3+4^3+5^3+6^3+7^3 = 784$$ $$S_{x^4} = 1^4+2^4+3^4+4^4+5^4+6^4+7^4 = 4676$$ $$S_{xy} = 1*2.3 + 2*5.2 + 3*9.7 + 4*16.5 + 5*29.4 + 6*35.5 + 7*54.4 = 811.1$$ $$S_{x^2y} = 1^2*2.3 + 2^2*5.2 + 3^2*9.7 + 4^2*16.5 + 5^2*29.4 + 6^2*35.5 + 7^2*54.4 = 5040.3$$ 4. **Normal equations:** $$\begin{cases} 7a + 28b + 140c = 152.99999999999997 \\ 28a + 140b + 784c = 811.1 \\ 140a + 784b + 4676c = 5040.3 \end{cases}$$ 5. **Solve the system:** Using matrix methods or substitution, the solution is approximately: $$a \approx 0.914$$ $$b \approx -0.571$$ $$c \approx 1.429$$ 6. **Final fitted polynomial:** $$y = 0.914 - 0.571x + 1.429x^2$$ This polynomial best fits the data in the least squares sense.