Why D≥0 while finding the range of rational functions 
To find the range of a rational expression $f(x)=y$,
a) We first make a quadratic in $x$ in terms of $y$.
b) Make the discriminant $\Delta ≥ 0$
c) Solve the resulting inequality to get the range of $y$.

I am having trouble in the second step.
I searched online and this is the explanation I got:
For real values of $y$, the given equation has real roots. So $\Delta≥0$
But I don't find this convincing.
Say we had to find the range of $$y= f(x)=  \frac{x^2−3x−4 }{x^2−3x+4}$$
I can solve this question to this step:
$$(y−1)x^2−3(y−1)x+4(y+1)=0 $$

Can you please explain why $\Delta≥0$?

Don't solve the question entirely, just help me with this doubt.
 A: First of all, for $a,b,c\in\mathbb R$ where $a\not=0$, 
$$ax^2+bx+c=0\tag1$$
can be written as
$$(\Delta=)\ b^2-4ac=\left(2ax+b\right)^2$$
Since $(2ax+b)^2\ge 0$ for $x\in\mathbb R$, we can see that
$$\text{$(1)$ has at least one real root $x$}\iff \Delta\ge 0\tag2$$
Now for your example, we have
$$\begin{align}y=f(x)=\frac{x^2-3x-4}{x^2-3x+4}&\iff y(x^2-3x+4)=x^2-3x-4\\&\iff (y-1)x^2-3(y-1)x+4(y+1)=0\tag3\end{align}$$
with $x^2-3x+4=(x-3/2)^2+7/4\gt 0$ and $y-1\not=0$. 
So, $(3)$ is a quadratic equation in $x$ in terms of $y$.
That $y=y_0$ is in the range of $y=f(x)$ means that there exists at least one $x=x_0$ such that $y_0=f(x_0)$. So, there exists at least one $x$ such that $y=f(x)$, i.e. $(3)$ for every $y$ in the range.
Hence, we want to find the condition that $(3)$ has at least one real root $x$. This is $\Delta\ge 0$ from $(2)$.
A: Before finding the range, let's look at the domain of the function. $$y= f(x)=  \frac{x^2−3x−4 }{x^2−3x+4}$$ We do not want the denominator to be zero, and when we find the discriminant of the denominator; it gives complex values. This means that the denominator is never zero for any real number.Hence the domain of the function is $\mathbb{R}$, the set of real numbers.  Hence the equation $(y−1)x^2−3(y−1)x+4(y+1)=0$ means that it has a solution for "every" $x$$\in$$\mathbb{R}$ . Also we need the discriminant to be greater than zero, otherwise the solution would give imaginary values of $x$ but our domain is restricted to real numbers.
A: $y$ is in the range of $g$
iff there is an $x$ such that $y=f(x)$
iff there is an $x$ that solves the quadratic.
A quadratic has a real solution iff its discriminant is nonnegative. 
This argument only works for rational expressions of degree $2$ (= max (deg  numerator, deg denominator)).
A: Hopefully, this explication helps:


Find the range of $f:\mathbb R\to\mathbb R,$ where $$f(x)=  \frac{x^2−3x−4 }{x^2−3x+4}.$$



*

*Let $k$ be in the range of $f.$
Then some real $x$ satisfies $$f(x)=k\\(k−1)x^2−3(k−1)x+4(k+1)=0.$$
Thus, the above equation is quadratic in $x$ with ($k\ne1$ and) a real solution.
So—noting that the quadratic has real coefficients҂—it has a nonnegative discriminant with $k\ne1:$
$$\big[−3(k−1)\big]^2-4\big[k−1\big]\big[4(k+1)\big]\ge0\quad\text{and}\quad
k\ne1\\-\frac{25}7\le k<1.$$


*The above argument is reversible.


*Hence, $k$ is in the range of $f$ if and only if $-\dfrac{25}7\le k<1.$
That is, $$\text{the range of $f$ is}\left[-\frac{25}7,1\right).$$
P.S. ҂A quadratic polynomial with a real root and a non-real coefficient need not have a nonnegative discriminant: $$ix^2-3ix+i$$ has roots $x = \frac12(3\pm\sqrt5)$ and discriminant $-5.$
