Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real 
Find the interval in which $m$ lies so that the expression $\frac{mx^2+3x-4}{-4x^2+3x+m}$ can take all real values, $x$ being real.    

I don't know how to proceed with this question. I have equated this equation with $y$ to obtain a quadratic equation: $(m+4y)x^2+(3-3y)x-(4+my)=0$. Now I have no idea as to how I can find the answer. A small hint will be helpful.

Momo's answer is correct except for the values $m=1$ and $m=7$.

Let $f$ be given by
$$f(x)=\frac{mx^2+3x-4}{-4x^2+3x+m}\;\;$$
If $m = 1$, then 
\begin{align*}
f(x)&=\frac{x^2+3x-4}{-4x^2+3x+1}\\[4pt]
&=-\frac{x^2+3x-4}{4x^2-3x-1}\\[4pt]
&=-\frac{(x-1)(x+4)}{(x-1)(4x+1)}\\[4pt]
&=-\frac{x+4}{4x+1},\;\;x\ne 1\\[4pt]
\end{align*}
which misses the values$\;y=-1/4\;$and$\;y=-1$.

If $m = 7$, then
\begin{align*}
f(x)&=\frac{7x^2+3x-4}{-4x^2+3x+7}\\[4pt]
&=-\frac{7x^2+3x-4}{4x^2-3x-7}\\[4pt]
&=-\frac{(x+1)(7x-4)}{(x+1)(4x-7)}\\[4pt]
&=-\frac{7x-4}{4x-7},\;\;x\ne -1\\[4pt]
\end{align*}
which misses the values$\;y=-7/4\;$and$\;y=-1$.

Hence the correct answer is $1 < m < 7$.
