[tex]f(x) = (m+3)x^2-(m+4)x+m+4 \\ \\ f(x) > 0 \Rightarrow V(x_0,y_0) \\ \\ \boxed{1}\quad m+3\geq0,\quad x_0 \geq -2\\ \\\Rightarrow m \geq-3,\quad -\dfrac{b}{2a}\geq -2 \\ \\\Rightarrow m\geq-3,\quad \dfrac{m+4}{2(m+3)}\geq-2\\\\\Rightarrow m\geq -3,\quad\dfrac{m+4+4m+12}{m+3}\geq 0 \\ \\ \Rightarrow m\geq-3,\quad \dfrac{5m+16}{m+3}\geq 0 \\ \\ \Rightarrow m \geq -3\quad\text{si}\quad \Big(m\leq -\dfrac{16}{5}\quad \text{sau}\quad m \geq - 3\Big)[/tex]
[tex]\Rightarrow m\in [-3,+\infty)\cap \Bigg(\Big(-\infty - \dfrac{16}{5}\Big]\cup [-3,+\infty)\Bigg) \\ \\ \\\Rightarrow m\in [-3,+\infty)[/tex]
[tex]\\\boxed{2}\quad x_0 < -2,\quad \Delta < 0,\quad m+3 \geq 0 \\ \\ \dfrac{m+4}{2(m+3)}< - 2,\quad (m+4)^2 - 4(m+3)(m+4) < 0,\quad m\geq -3 \\ \\\\m\in \Big(-\dfrac{16}{5},-3\Big),\quad (m+4)^2 - 4(m+3)(m+4) < 0,\quad m\geq -3 \\ \\ (Fals)[/tex]
[tex]\\\text{Din }\boxed{1}\,\text{ sau }\,\boxed{2} \Rightarrow m\in [-3,+\infty)[/tex]
