A function $f:\mathbb{R}\to \mathbb{R}$ is define by, $f(x) = \frac{\alpha x^2 + 6x - 8}{\alpha + 6x - 8x^2}$Find the interval of values of $\alpha$ for which $f$ is onto.
So, here is what I did
Let $f(x)$ be $y$ $y(\alpha + 6x -8x^2) = \alpha x^2 + 6x - 8$ $ (\alpha + 8y)x^2 + 6(1-y)x -(8+\alpha y) = 0 $ Since, $x$ is valid for all Real Numbers, $ \:36(1-y)^2 + 4(8+ \alpha y)(8y + \alpha) \ge0 $ $y^2(9+8\alpha) + y(46 + \alpha^2) + (9+8\alpha) \ge0 \tag{1}$ For $(1)$ to hold for each $y\in\mathbb{R}$, $\:9 + 8\alpha >0 \implies \alpha > -\frac{9}{8}$
But that's all I have been able to do and in my book an another condition is given,$\:(46 + \alpha^2)^2 - 4(9 + 8\alpha)^2\le0$.
I am not able to understand why is the author using this condition, If anyone of you could explain me what the condition has to do with the $\textbf{equation } (1)$, and also if someone could explain me the graphical implication of the condition.
Thank You.