Question

Aflati functia de gradul intai a carui grafic trce prin punctele A(3/2;5/2) si B(7;0)

Answer 1

f(3/2)=5/2
 f(7)=0

f(x) =ax+b
 3a/2+b=5/2
 3a+2b=5
 dar -b/a=7 pt ca se anuleaza in 7
deci b=-7a
 a tunci

3a+2*(-7a)=5
3a-14a=5
11a=-5
 a=-5/11
b=-7a=35/11
 f(x)=ax+b=-5x/11+35 /11

verificare
f(7) =-35/11+35/11=0 adevarat
 f(3/2) =-15/22+35/11= -15/22+70/22 =55/22=5/2 adevarat
 problema e bine rezolvata