Question

Graficului functiei f:R->R,f(x)=x-(3a+1)/2 interesează axa Oy in punctul cu ordonata:a)10;b)-3;c)-8;d)1/4.Sa se calculeze f(1)+f(-1)+f(2) in fiecare caz.

Answer 1

Gf∩Oy =f(0)=-(3a+1)/2

a) -3a-1=20
-3a=19
a=-19/3
f(x)=x+10
f(1) +f(-1) +f(2)=1+10-1+10+2+10=2+3*10=32
observam ca f(1) +f(-1) +f(2) =2+3 *b,
, unde b este termenul liber, chiarcel dat in cerinta, ptca f(0)=b



c) -3a-1=-16
-3a=-15
a=5
 f(x) =x-8

f(-1) +f(1)+f(2)=2-3*6=2-24=-22


d) -3a-1=1/2
-3a=3/2
 a=-1/2
f(x)= x+1/4
f(1) +f(-1) +f(2)=2+3*1/4=2+3/4=11/4

b) -3a-1=-3
3a+1=3
3a=2
a=2/3
f(x) =x-3
 f(1)+f(-1)=f(2) =2+3*(-3)=2-9=-7