Réponse:
on a f(x) = ax² + bx + c
A(–4 ; 0) , B(1 ; 0) , C(0 ; 4)
♦ pour A(–4 ; 0)
f(-4) = 0 ==> a(-4)² + b(-4) + c = 0
==> a × 16 – 4b + c = 0
==> 16a – 4b + c = 0 (1)
♦ pour B(1 ; 0)
f(1) = 0 ==> a(1)² + b(1) + c = 0
==> a + b + c = 0 (2)
♦ pour C(0 ; 4)
f(0) = 4 ==> a(0)² + b(0) + c = 4
==> c = 4 (3)
(3) dans (1) et (2)
==> (1) : 16a – 4b + 4 = 0
==> (2) : a + b + 4 = 0 } (–16)
______________________
16a + a(–16) – 4b + b(-16) + 4 + 4(–16) = 0
==> 16a –16a –4b –16b + 4 –64 = 0
==> –20b –60 = 0
==> –20b = 60
==> b = 60/(-20)
==> b = –3 (4)
(4) et (3) dans (2) ==> a + (–3) + 4 = 0
==> a –3 + 4 = 0
==> a + 1 = 0
==> a = –1
donc a = –1 ; b = –3 ; c = 4
