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 !