Question

Sa se determine functia f:IR→IR, f(x)=ax²+ bx+c; a, b, c ∈ IR, stiind ca punctele A(1;3), B(0;5), C(-1;11) apartin graficului functiei f.

Answer 1

Răspuns:

f(x) = 2x²-4x+5

Explicație pas cu pas:

f(x)=ax²+ bx+c

A(1;3), B(0;5), C(-1;11)  ∈ Gf

A(1;3) ∈ Gf <=> f(1) = 3  => a·1²+b·1+c = 3  => a+b+c = 3

B(0;5) ∈ Gf <=> f(0) = 5 => a·0²+b·0+c = 5  => c = 5

C(-1;11) ∈ Gf <=> f(-1) = 11 => a·(-1)²+b·(-1)+c = 11 => a-b+c = 11 =>

a-b+5 = 11 => a = b+6

a+b+c=3 <=> b+6+b+5 = 3  => 2b = -8 => b = -4

a = -4+6 => a = 2

f(x) = 2x²-4x+5