Question

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

Answer 1

Gf ∩ oy 
a) f(0) = -8
=> -3a + 1/2 = -8
-6a + 1= -16
-6a= -17
 a= 17/6
f(1)= 1-17/2+1/2= -14/2= -7
f(-1)= -1+17/2+1/2= 16/2=8
f(2)= 2-17/2+1/2= -12/2=-6
=> -7+8-6= -5
b) f(0)= 1/4
-3a +1/2=1/4
-12a + 2=1
-12a=-1
a=1/12
f(1)= 1-1/4+1/2=5/4
f(-1)= -1 + 1/4+1/2= -1/4
f(2)= 2-1/4+1/2= 9/4
=> 5/4-1/4+9/4= 13/4