\[y = ax^{2} + bx - 4;\ \ \ \]
\[y(1) = 0;\ \ \ y(4) = 0.\]
\[1)\ (x - 1)(x - 4) = 0;\]
\[x^{2} - 4x - x + 4 = 0;\]
\[x^{2} - 5x + 4 = 0;\]
\[y = - x^{2} + 5x - 4.\]
\[2)\ y^{'}(x) = - \left( x^{2} \right)^{'} + (5x)^{'} - (4)^{'} =\]
\[= - 2x + 5 - 0 = 5 - 2x.\]
\[3)\ Точка\ экстремума:\]
\[5 - 2x = 0\]
\[2x = 5\]
\[x = \frac{5}{2};\]
\[y\left( \frac{5}{2} \right) = - \frac{25}{4} + \frac{25}{2} - 4 =\]
\[= \frac{- 25 + 50 - 16}{4} = \frac{9}{4} = 2,25.\]
\[Ответ:\ \ 2,25.\]