Question

fie functia f r-r , f(x)=ax+b determinati functia f stiind ca punctele M(-2;2) si N(1;8) apartin graficului

Answer 1

$\boxed{\text{Metoda I}:}\\ \\ f(x) = ax+b \\ \\ M(-2,2)\in Gf \Rightarrow f(-2) = 2 \Rightarrow -2a+b = 2 \\ N(1;8) \in Gf \Rightarrow f(1) = 8 \Rightarrow a+b = 8 \\ \\\left\{ \begin{array}{ll} -2a+b=2 \\ a+b=8\Big|\cdot 2 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} -2a+b = 2 \\ 2a+2b = 16 \end{array} \right )(+) \Rightarrow \\ \\ \Rightarrow -2a+2b+b+2b = 2+16 \Rightarrow 3b = 18 \Rightarrow b = 6 \\ \\ a+b = 8 \Rightarrow a+6 = 8 \Rightarrow a = 8-6 \Rightarrow a = 2$

$\Rightarrow \boxed{f(x) = 2x+6} \\ \\ \\ \\\boxed{\text{Metoda II:}}\\ \\ Aplicam metoda polinomului de interpolare Lagrange: \\ \\ Punctele prin care trece functia sunt: (-2;2),\quad (1;8). \\ \\ f(x) = 2\cdot \dfrac{x-1}{-2-1}+8\cdot \dfrac{x-(-2)}{1-(-2)} = 2\cdot \dfrac{x-1}{-3}+8\cdot \dfrac{x+2}{3} = \\ \\ = -\dfrac{2(x-1)}{3} +\dfrac{8(x+2)}{3} = \dfrac{-2(x-1)+8(x+2)}{3} = \\ \\ = \dfrac{-2x+2+8x+16}{3} = \dfrac{6x+18}{3} \Rightarrow \boxed{f(x) = 2x+6}$