Question

arătați că vârfurile parabolelor asociate familiei de funcții fm=mx^2+2(m+1)x+m+2 se află pe dreapta y=x+1​

Answer 1

$f_m = mx^2+2(m+1)x+m+2\\ \\ \text{Trebuie aratat ca varful }V\Big(-\dfrac{b}{2a},-\dfrac{\Delta}{4a}\Big) \in y = x+1,\quad \forall m\in \mathbb{R}\\ \\ \Rightarrow\,\,-\dfrac{\Delta}{4a} = -\dfrac{b}{2a}+1\,\,\Leftrightarrow \\ \\ \Leftrightarrow\,\,-\dfrac{4(m+1)^2-4m(m+2)}{4m} = -\dfrac{2(m+1)}{2m}+1\\ \\ \Leftrightarrow\,\, -\dfrac{4m^2+8m+4-4m^2-8m}{4m} = -\dfrac{m+1}{m}+1\\ \\ \Leftrightarrow\,\, -\dfrac{1}{m} = \dfrac{-(m+1)+m}{m}$

$\Leftrightarrow\,\, -\dfrac{1}{m} = \dfrac{-m-1+m}{m}\quad (A)\quad\quad \mathrm{Q.E.D.}$