<s>[INST] Range of a Rational Function How to find the Range of function  $$f(x)= \frac{x^2-3x-4}{x^2 - 3x +4}$$
I tried to equate the expression to $y$, then cross multiplied 
$$ y= \frac{x^2-3x-4}{x^2 - 3x +4}$$
$$ y(x^2 - 3x +4)= x^2-3x-4 $$
bought the terms to one side so it becomes a quadratic and made Discriminant to zero , but i cant seem to reach anywhere.. Any suggestions? 
 [/INST] $$\implies x^2(y-1)+x(3-3y)+4y+4=0$$
The discriminant
$$=(3-3y)^2-4(y-1)(4y+4)=-(y-1)(7y+25)$$ which needs to be $\ge0$
Now $(x-a)(x-b)\le0, a\le b\implies a\le x\le b$

Alternatively,
$$\dfrac{x^2-3x-4}{x^2-3x+4}=1+\dfrac{x^2-3x-4}{x^2-3x+4}-1=1-\dfrac8{x^2-3x+4}$$
Now $x^2-3x+4=\dfrac{4x^2-12x+16}4=\dfrac{(2x-3)^2+7}4$
Now $0\le(2x-3)^2\le\infty\iff\dfrac74\le x^2-3x+4\le\infty\iff\dfrac47\ge \dfrac1{x^2-3x+4}\ge0$
Can you take it from here?
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