Bonjour,
il suffit de développer
-2(x-5)(x-1)= -2(x²-5x-x+5)= -2(x²-6x+5)= -2x²+12x-10
-2((x-3)²-4)= -2[ (x²-3x-3x+9) - 4 ]= -2 [ (x²-6x+9)-4 ]= -2x²-12x-18+8= -2x²-12x-10
f(x)= 8-2(x-3)²
8-2(x-3)² ≤ 8
-2(x-3)² ≤ 8-8
-2(x-3)² ≤ 0
x-3= 0 => x= 3
x 3
f(x) + + 0 - -
S= [3 ; + ∞ [