Question

Sa se determine functia de gradul al II al carei grafic contine punctele A (1;3) B (0;5) si C (-1;11)

Answer 1

$A(1;3),\quad B (0;5), \quad C (-1;11) \\ \\ Consideram f(x) = a\text{x}^2+b\text{x}+c \\ \\ \bullet A(1;3) \in G_f \Rightarrow f(1) = 3 \\\bullet B(0;5) \in G_f \Rightarrow f(0)=5 \\ \bullet C(-1;11)\in G_f \Rightarrow f(-1) = 11 \\ \\ Formam sistem: \\ \\ \left\{ \begin{array}{c}f(1) = 3 \\ f(0)=5 \\f(-1) = 11 \end{array} \right \Rightarrow \left\{ \begin{array}{c}a\cdot(1)^2+b\cdot 1+c = 3 \\ a\cdot (0)^2-b\cdot 0+c = 5 \\ a\cdot(-1)^2+b\cdot(-1)+c=11 \end{array} \right \Rightarrow$

$\Rightarrow \left\{ \begin{array}{c} a+b+c = 3 \\ c = 5 \\ a-b+c = 11\end{array} \right \Rightarrow \left\{ \begin{array}{c} a+b+5 = 3 \\ c = 5 \\ a-b+5 = 11\end{array} \right \Rightarrow \left\{ \begin{array}{c} a+b= -2 \\ c = 5 \\ a-b =6 \end{array} \right \Rightarrow \\ \\ ~\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (adunam ecuatia 1 cu ecuatia 2) \\ \\ \Rightarrow a+a+b-b=-2+6 \Rightarrow 2a=4 \Rightarrow a = \dfrac{4}{2} \Rightarrow a = 2$

$\Rightarrow \left\{ \begin{array}{c} a=2 \\ 2-b = 6 \\ c=5\end{array} \right \Rightarrow \left\{ \begin{array}{c} a=2 \\ -b = 4 \\ c=5\end{array} \right \Rightarrow \left\{ \begin{array}{c} a=2 \\ -b = 4 \\ c=5\end{array} \right \Rightarrow \left\{ \begin{array}{c} a=2 \\ b = -4 \\ c=5\end{array} \right| \\ \\ \\ \Rightarrow \boxed{f(x) = 2\text{x}^2-4\text{x}+5}$