Question

Graficul functiei f:R→R, f(x)= x- 3a+1 supra 2   intersecteaza axa Oy in punctul cu ordonata= 10. Sa se calculeze f(1)+ f(-1)+f(2).

Answer 1

Presupun ca functia este: f(x)=x-3a+1/2
Punctul de intersectie este: A(0;10) de unde rezulta: f(0)=10;  3a-1/2=10
Determinam a:   
3a=10+1/2
3a=21/2
a=21/(2·3)
a=7/2
f(x)=x-21/2+1/2=x-20/2=x-10
Am obtinut f(x)=x-10
Calculam:
f(1)=1-10=-9
f(-1)=-1-10=-11
f(2)=2-10=-8
f(1)+f(-1)+f(2)=-9-11-8=-28
Spor la lucru in continuare!