Se considera functia f : R -R f(x)=x²+ax+b.
Determinati nr.rele a si b pt care graficul functiei f contine punctele A(2,3) si B(-1,0).

f(2)=3=>f(2)=4+2a+b f(-1)=0=>f(-1)=1-a+b Si faci sistem: 2a+b+4=3 a-b-1=0 Si iti da 3a=0=> a=0 ,dupa care inlocuiesti in ecuatia a doua ca sa il afli si pe b care va da b=-1

[tex]f : \mathbb_{R} \rightarrow \mathbb_{R},$ $\quad f(x) =x^2+ax+b \\ \\ A(2,3)\in G_f \quad $si$ \quad B(-1,0) \in G_f \Rightarrow \left\{ \begin{array}{c} f(2) = 3 \\ f(-1) = 0 \end{array} \right \Rightarrow \\ \\ \left\{ \begin{array}{c} 2^2+2a+b = 3 \\(-1)^2-a+b= 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} 4+2a+b = 3 \\1-a+b= 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} 2a+b = 3-4 \\-a+b= -1 \end{array} \right \Rightarrow [/tex]
[tex]\Rightarrow \left\{ \begin{array}{c} 2a+b = -1 \\-a+b= -1|\cdot2 \end{array} \right \Rightarrow \left\{ \begin{array}{c} 2a+b = -1 \\-2a+2b= -2 \end{array} \right $ \Big($ $adunam\Big)$ \\ \\ \Rightarrow 3b = -3 \Rightarrow \boxed{b = -1} \\ \\ -a+b = -1 \Rightarrow -a -1 = -1 \Rightarrow -a = -1+1 \Rightarrow -a = 0 \Rightarrow \boxed{a = 0} [/tex]