\[\boxed{\mathbf{971\ (971).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[y = ax^{2} + bx + c;\ \ (0;10);\ \]
\[\text{\ \ }\left( x_{0};y_{0} \right) = (6;\ - 2)\]
\[10 = 0 + 0 + c \Longrightarrow \ \ c = 10\]
\[- \frac{b}{2a} = 6 \Longrightarrow \ \ b = - 12a\]
\[- 2 = 36a + 6b + 10\]
\[- 2 = 36a + 6 \cdot ( - 12a) + 10\]
\[- 2 = 36a - 72a + 10\]
\[36a = 12\]
\[a = \frac{1}{3} \Longrightarrow \ \ b = - 12 \cdot \frac{1}{3} = - 4\]
\[Ответ:a = \frac{1}{3};\ \ b = - 4;\ \ c = 10.\]