a)
[tex]f_m(x)=(m+1)x^2+(2m+3)x+m+4 \\\\ x_V =-\dfrac{2m+3}{2m+2}= - \dfrac{2m+2+1}{2m+2}= -1-\dfrac{1}{2m+2}\\ \\y_V = \dfrac{-\Delta}{4a} = -\dfrac{(2m+3)^2-4(m+1)(m+4)}{4(m+1)} =\\=-\dfrac{4m^2+12m+9-4(m^2+5m+4)}{4(m+1)} = \\= -\dfrac{4m^2+12m+9-4m^2-20m-16}{4(m+1)} = - \dfrac{-8m-7}{4(m+1)} = \dfrac{8m+7}{4(m+1)}\\ \\ =\dfrac{8m+8-1}{4m+4} = 2-\dfrac{1}{4m+4} = \dfrac{1}{2}\cdot (4-\dfrac{1}{2m+2}) =[/tex]
[tex]=\dfrac{1}{2}\cdot (-1-\dfrac{1}{2m+2}+5) = \dfrac{1}{2}\cdot(x_V+5) = \dfrac{1}{2}\cdot x_V + \dfrac{5}{2} \\\\ \Rightarrow y_V = \dfrac{1}{2}\cdot x_V + \dfrac{5}{2} \Rightarrow V(x_V,y_V) \in y = \dfrac{1}{2}\cdot x+\dfrac{5}{2} \\\\ \text{Varfurile parabolelor apartin unei drepte }\Rightarrow \text{Varfurile sunt coliniare.}[/tex]
b)
[tex]f_m(x)=(m+1)x^2+(2m+3)x+m+4 \\ \\ f_m(a) = b\quad (\text{b independent de m}) \\ \\ (m+1)a^2+(2m+3)a+m+4 = b \\a^2m+a^2+2am+3a+m+4 = b \\m(a^2+2a+1)+a^2+4 = b \\m(a+1)^2+a^2+3a+4 = b \\ \\ \Rightarrow (a+1)^2 = 0 \Rightarrow a = -1\\ a^2+3a+4 = b \Rightarrow 1-3+4 = b \Rightarrow b = 2 \\ \\ \Rightarrow (-1,2)\text{ este puctul comun al parabolelor}[/tex]
