2.2 Lösning 2b

\( f(2) = -3\cdot 2^2 + 2\cdot 2 - 12 = -3\cdot 4 + 4 - 12 = -12 + 4 - 12 = -20 \)

\( f(-2) = -3\cdot (-2)^2 + 2\cdot (-2) - 12 = -3\cdot 4 - 4 - 12 = -12 - 4 - 12 = -28 \)

\( \Delta y = \)

\( {\Delta y \over \Delta x} = {f(3) \, - \, f(2) \over 2 - (-2)} = {5\cdot 3 + 23 \, - \, (5\cdot 2 + 23) \over 1} = \)

\( = 15+23 \, - \, (10+23) = 38-33 = 5 \)

\( y = -3\,x^2 + 2\,x - 12 \) i intervallet \( -2 \leq x \,\leq\, 2 \)