1. **State the problem:** We want to find a quadratic function of the form $$y = ax^2 + bx + c$$ that best fits the points $(-1,8)$, $(5,-4)$, and $(7,8)$ using quadratic regression. 2. **Recall the quadratic regression formula:** Given points $(x_i, y_i)$, the quadratic regression finds coefficients $a$, $b$, and $c$ such that the sum of squared residuals is minimized. Since we have exactly three points, the quadratic function passing through all three points can be found by solving the system: $$ \begin{cases} a(-1)^2 + b(-1) + c = 8 \\ a(5)^2 + b(5) + c = -4 \\ a(7)^2 + b(7) + c = 8 \end{cases} $$ 3. **Write the system explicitly:** $$ \begin{cases} a - b + c = 8 \\ 25a + 5b + c = -4 \\ 49a + 7b + c = 8 \end{cases} $$ 4. **Subtract the first equation from the second and third to eliminate $c$:** $$ \begin{cases} (25a + 5b + c) - (a - b + c) = -4 - 8 \\ (49a + 7b + c) - (a - b + c) = 8 - 8 \end{cases} $$ which simplifies to $$ \begin{cases} 24a + 6b = -12 \\ 48a + 8b = 0 \end{cases} $$ 5. **Simplify the equations:** $$ \begin{cases} 4a + b = -2 \\ 6a + b = 0 \end{cases} $$ 6. **Subtract the first from the second to solve for $a$:** $$ (6a + b) - (4a + b) = 0 - (-2) \\ 2a = 2 \\ a = 1 $$ 7. **Substitute $a=1$ into $4a + b = -2$ to find $b$:** $$ 4(1) + b = -2 \\ 4 + b = -2 \\ b = -6 $$ 8. **Substitute $a=1$, $b=-6$ into the first original equation to find $c$:** $$ 1 - (-6) + c = 8 \\ 1 + 6 + c = 8 \\ 7 + c = 8 \\ c = 1 $$ 9. **Final quadratic function:** $$ y = 1x^2 - 6x + 1$$ This quadratic function fits the points exactly.