Find parameter a for which.. Find parameter $a$ for which $$\frac{ax^2+3x-4}{a+3x-4x^2}$$ takes all real values for $x \in  \mathbb{R}$


I have equated the function to a real value, say, k
which gets me a quadratic in x. I have then put $D\geq 0$ (since $x \in \mathbb{R}$) which gets me $a \geq -9/16$

How do I proceed further to get other parameter for $a $?
Find $a$ so that the equation 

$$y= \frac{a x^2+3x -4}{-4x^2+3x+a}$$

has a root which is in the domain of the function

The discriminant must be $\geq 0$ for all $y$

$$(9+16a)y^2+\left(4a^2+46\right)y +(9+16a)\geq 0$$ 

This is the case if its discriminant is negative and $9+16a&gt;0$

$$16 (-7+a) (-1+a) (4+a)^2&lt;0$$

which says $a$ must be in the interval $(1,7)$
$ a \geq -9/16$ is not correct.  For $ a \leq -9/16$ the domain of the function $f(x)=\frac{ax^2+3x-4}{a+3x-4x^2}$ is $ \mathbb R.$