bonsoir
f (1) = 3 et f( - 3 ) = 2
( 2 - 3 ) / ( - 3 - 1 ) = - 1 /- 4 = 1/4
ax = x / 4
f ( 1) = 3
1 *1/4 + b = 3
1/4 + b = 3
b = 3 - 1/4 = 12/4 - 1/4 = 11 /4
f (x) = x /4 + 11/ 4
d'où
f ( - 3 ) = - 3 * 1/4 + 11 /4 = - 3/4 + 11/4 = 8 /4 = 2
f (x) = x /4 + 11 /4
s'annule en - 11
x - ∞ - 11 + ∞
x /4 + 11 /4 - 0 +
f (x) ≥ 0 [ - 11 ; + ∞ [