1 f(x)=(x+1)(6-2x)=6x-2x²+6-2x=-2x²+4x+6
2-f(x)=-2(x-1)²+8=2*(x²-2x+1)+8=-2x²+4x-2+8=-2x²+4x+6=f(x)
3-a) f(x)=0
(x+1=0 ou 6-2x=0
x=-1 ou x=3
b) 4=f(x)
4=-2(x-1)²+8
-4=-2(x-1)²
2=x²-2x+1
x²-2x-1=0
Δ=(-2)²-4*1*(-1)=8
x1=[tex]\frac{-2-\sqrt{8} }{2*1} =\frac{-2-2\sqrt{2} }{2}=-1-\sqrt{2}[/tex]
x2=[tex]\frac{-2+\sqrt{8} }{2*1} =\frac{-2+2\sqrt{2} }{2}=-1+\sqrt{2}[/tex]
4- f(-1)=0
f(5/3)=(5/3+1)(6-2*5/3)
= 8/3*8/3
=64/6=32/3
J'espère que tu as compris, n'hésite pas si tu as des questions.
