[tex]f(x) = ax^2+bx+c \\ \\ A(1;-2),\quad B(-2,13) \in Gf\\ \\ Gf\text{ are axa Oy ca axa de simetrie}: \\ \\ \Rightarrow C(-1;-2) \in Gf \\ \\ f(x) = \\ \\ =-2\cdot \dfrac{\Big(x-(-2)\Big)\Big(x-(-1)\Big)}{\Big(1-(-2)\Big)\Big(1-(-1)\Big)}+ (-2)\cdot \dfrac{(x-1)\Big(x-(-2)\Big)}{(-1-1)\Big(-1-(-2)\Big)} + \\ \\ + 13\cdot\dfrac{(x-1)\Big(x-(-1)\Big)}{(-2-1)\Big(-2-(-1)\Big)} \\ \\ = -2\cdot \dfrac{(x+2)(x+1)}{3\cdot 2} +2\cdot \dfrac{(x-1)(x+2)}{2\cdot 1} +13\cdot \dfrac{(x-1)(x+1)}{-3\cdot(-1)} [/tex]
[tex]= -2\cdot \dfrac{(x+2)(x+1)}{3\cdot 2} +2\cdot \dfrac{(x-1)(x+2)}{2\cdot 1} +13\cdot \dfrac{(x-1)(x+1)}{-3\cdot(-1)} \\ \\ = \dfrac{-2\cdot (x+2)(x+1) +2\cdot 3(x-1)(x+2) + 13\cdot 2(x-1)(x+1)}{3\cdot 2} \\ \\ = \dfrac{-2(x^2+3x+2)+6(x^2+x-2)+26(x^2-1)}{6} \\ \\ = \dfrac{(-2x^2+6x^2+26x^2) +(-6x+6x) +(-4-12-26)}{6} \\ \\ = \dfrac{30x^2-42}{6} \\ \\ = \boxed{5x^2-7}[/tex]
Am folosit teorema polinomului de interpolare Lagrange.
Se pare ca daca axa Oy este axa de simetrie, inseamna ca b va fi 0.
Functia nu il va avea pe x, ci doar pe x^2 si termenul liber, intotdeauna.
f(x) s-ar fi redus la f(x) = ax² + c
A(1;-2) ∈ Gf => f(1) = -2 => a + c = -2
B(-2;13) ∈ Gf => f(-2) = 13 => 4a + c = 13
=> 3a = 15 => a = 5 => 5 + c = -2 => c = -7
=> f(x) = 5x² - 7
Deci, intotdeauna cand axa Oy e axa de simetrie a graficului, coeficientul lui x va fi 0.
