Bonjour !
f(x) = ax + b
1) f(5) = 10 et f(10) = 11. Sachant que a = (y2-y1)/(x2-x1) : (y c'est le résultat de f(x) btw)
Donc a = (11-10)/(10-5) = 1/5 = 0.2
Maintenant pour calculer b :
on sait que f(x) = ax+b. On sait a, donc f(x) = 0.2x+b
Or f(5) = 10. On peut donc dire que f(5) = 0.2*5 + b = 10
On a donc une équation : 0.2*5 + b = 10
<=> 1 + b = 10
<=> b = 10-1 = 9
Donc f(x) = 0.2x + 9
2) f(-7) = 4 et f(4) = -7.
a = (y2-y1)/(x2-x1) = (-7 - 4)/ (4 - (-7)) = (-11)/(11) = -1
f(x) = ax+b = -1*x+b
Or f(-7) = -1*(-7)+b = 4
Donc -1*(-7)+b = 4
<=> 7+b = 4
<=> b = 4-7 = -3
Donc f(x) = -1*x - 3 = -x-3
3) f(-8) = 0 et f(-1) = 13
a = (13-0)/(-1 -(-8)) = 13/7
f(x) = (13/7) * x + b
f(-8) = (13/7)*(-8) + b = 0
<=> -104/7 + b = 0
<=> b = 104/7
f(x) = (13/7)*x + 104/7
4) f(-1) = -5 et f(2) = -9
a = (-9-(-5))/(2-(-1)) = -4/3
f(x) = -4/3*x + b
f(-1) = -4/3*(-1) + b = -5
<=> -4/3*(-1)+b = -5
<=> 4/3 + b = -5
<=> b = -5-4/3 = -19/3
f(x) = -4/3*x -19/3
Voilà !
