Răspuns:
[tex]m = \frac{-1}{4}[/tex]
Explicație pas cu pas:
[tex]\displaystyle f_m(x) = mx^2 - (2m+1)x + (m+1) \\ \\ V\Big(-\frac{b}{2a}, \frac{-\Delta}{4a}\Big)\\ \\ \frac{-b}{2a} = \frac{2m+1}{2m} = 1 + \frac{1}{2m}\\ \\ \Delta = (2m+1)^2 - 4m(m+1) = 4m^2 + 4m + 1 - 4m^2 - 4m = 1\\ \\ \frac{-\Delta}{4a} = \frac{-1}{4m}\\ \\ V\Big(1+\frac{1}{2m}, \frac{-1}{4m}\Big) \in d \iff 2\Big(1+\frac{1}{2m}\Big) - 3\Big(\frac{-1}{4m}\Big) + 5 = 0\\ \\ 2 + \frac{1}{m} + \frac{3}{4m} + 5 = 0\Bigg | \cdot m\\ \\ 2m + 1 + \frac{3}{4} + 5m = 0\Bigg | \cdot 4\\ \\ 28m + 4 + 3 = 0\\ \\ 28m + 7 = 0\\ \\ m = \frac{-7}{28}\\ \\ \boxed{m = \frac{-1}{4}}[/tex]
