bonjour
f (x) = ( x + 3)² - ( 3 x - 1 )²
f (x) = x² + 6 x + 9 - ( 9 x² - 6 x + 1 )
= x² + 6 x + 8 - 9 x² + 6 x - 1
= - 8 x² + 12 x + 7
f (x) = ( x + 3 + 3 x - 1 ) ( x + 3 - 3 x + 1 )
= ( 4 x + 2 ) ( - 2 x + 4 )
f(0) = 7
f ( - 7/3) = - 8 ( - 7/3)² + 12 ( - 7/3) + 7
= - 8 *49/9 - 84/3 + 7
= - 392/9 - 252/9 + 63/9
= - 581/9
f (√3 ) = - 8 ( √3)² + 12 √3 + 7
= - 24 + 12 √3 + 7 = - 17 + 12 √3
f(x) = 0 = tu prends la forme factorisée et un des 2 facteurs est nul
f(x) = 8 = tu calcules le Δ
forme canonique
α = - 12 / - 16 = 3/4
β = - 8 ( 3/4)² + 12 * 3/4 + 7 = - 8*9/16 + 36/4 + 7 = - 72/16 + 144/16 + 112/16
= 184/16 = 23/2
f(x) = - 8 ( x - 3/4)² + 23/2
