Range of rational function

Let $y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9}$. Rewrite as: $y(2x^2 + 4x + 9) = x^2 + 2x + 4$ $(2y - 1)x^2 + (4y - 2)x + (9y - 4) = 0$

Since $x$ is real, the discriminant $D \ge 0$: $D = (4y - 2)^2 - 4(2y - 1)(9y - 4) \ge 0$ $4(2y - 1)^2 - 4(2y - 1)(9y - 4) \ge 0$ $4(2y - 1)[(2y - 1) - (9y - 4)] \ge 0$ $4(2y - 1)(3 - 7y) \ge 0$ $(2y - 1)(7y - 3) \le 0$

This gives $y \in [\frac{3}{7}, \frac{1}{2}]$.

However, check if $y = \frac{1}{2}$ can be attained: If $y = \frac{1}{2}$, the coefficient of $x^2$ is $2(\frac{1}{2}) - 1 = 0$, giving $0x^2 + 0x + \frac{1}{2} = 0$, which has no real solution. Hence $y = \frac{1}{2}$ is an asymptote and cannot be reached.

Range = $[\frac{3}{7}, \frac{1}{2})$.