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.
 A: 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 $1 < m < 7$.
A: Hint
$$f(x)=\frac{mx^2+3x-4}{-4x^2+3x+m}$$
The function $f$ must to be onto. It means that:
$$p=\frac{mx^2+3x-4}{-4x^2+3x+m} \Rightarrow (m+4p)x^2+3x(1-p)-4-pm=0$$
The above equation must have roots for any $p \in \Bbb R$. It means that:
$$\Delta=9(1-p)^2+4(m+4p)(4+pm)\geq 0$$
for any choice of $p$.
$$(9+16m)p^2+(4m^2-46)p+9+16m \geq0$$
Then we have to analyze that new quadratic equation. Once the above expression must be always non negative, then.
$$\Delta'=(4m^2-46)^2-4(9+16m)^2=(m^2-8m-16)(m^2+8m-7)\leq 0$$
Solving the above inequality we have the values of $m$.
A: I don't know whether it's the simplest way but I would do this:
1) Your expanded equation $(m+4y)x^2+(3-3y)x-(4+my)=0$ needs to have real roots for all $y$ so you need to have $\Delta\ge 0$ for all $y$
2) Your $\Delta$ is itself a quadratic inequality in $y$ with positive coefficient, so to have $\Delta\ge 0$ for all $y$ you need $\Delta_y\le 0$
Which gives you $1\le m\le 7$
A: Hint: $$\frac{mx^2+3x-4}{-4x^2+3x+m} = y$$ The denominator shouldn't have any real roots. Thus, $ \delta_{\text{denom}} = 3^2+16m< 0\Rightarrow m<-\frac{9}{16}$
