Question

AL45 UPT 2019. Materie de cls 9

Answer 1

$\text{Rezolvare clasele a 9}$

$y = ax^2+bx+c \\ \\ f(x) = ax^2+bx+c \\ f'(x) = 2ax+b \\ \\ f'(4) = 0 \Rightarrow 8a+b = 0 \\ f(4) = 2 \Rightarrow 16a+4b+c = 2\\ f(2) = 0 \Rightarrow 4a+2b+c = 0 \\ \\ \Rightarrow 12a+2b = 2 \\ 8a+b = 0\Big|\cdot 2 \Rightarrow 16a+2b = 0 \\ \Rightarrow 4a = -2 \Rightarrow a = -\dfrac{1}{2}\\ \\ \Rightarrow b = 4 \Rightarrow -2+8+c = 0 \Rightarrow c = -6\\ \\ abc = \dfrac{-6\cdot 4}{-2} = 12$

$\text{Rezolvare clasa a 9-a:} \\ \\ f(x) = ax^2+bx+c \\ \\V(4,2) \\ \\ \Rightarrow \dfrac{-b}{2a} = 4,~~\dfrac{-\Delta}{4a} = 2 \Rightarrow b = -8a,\quad \dfrac{4ac-b^2}{4a} = 2 \Rightarrow \\ \\ \Rightarrow 4ac-b^2 = 8a \Rightarrow 4ac - 64a^2 = 8a \Rightarrow 4a(c-16a) = 8a \Rightarrow \\ \Rightarrow c-16a = 2 \Rightarrow c = 2+16a \\ \\ f(2) = 0 \Rightarrow 4a+2b+c = 0 \Rightarrow 4a-16a+2+16a = 0 \Rightarrow$

$\Rightarrow 4a = -2 \Rightarrow a = -\dfrac{1}{2} \Rightarrow c = 2-8 = -6 \Rightarrow b = 4 \\ \\ \Rightarrow abc = -\dfrac{1}{2}\cdot 4\cdot (-6) = 12$