Range of a Rational Function

$$\implies x^2(y-1)+x(3-3y)+4y+4=0$$

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?

Given $$\displaystyle y = \frac{x^2-3x-4}{x^2-3x+4}\Rightarrow yx^2-3yx+4y = x^2-3x-4$$

so we get $$\displaystyle (y-1)x^2-3(y-1)x+4(y+1) = 0$$

For real values of $y\;,$ Then given equation has real roots. So $\bf{Discriminant\geq 0}$

So $$9(y-1)^2-16(y-1)\cdot (y+1)\geq 0$$

So we get $$(y-1)\left[9(y-1)-16(y+1)\right]\geq 0$$

so $$(y-1)\left[-7y-25\right]\geq 0$$

So $$\displaystyle (y-1)(y+\frac{25}{7})\leq 0$$

So $$\displaystyle -\frac{25}{7}\leq y\leq 1\Rightarrow y\in \left[-\frac{25}{7}\;,1\right]$$

Now when $y=1\;,$ Then $$\displaystyle 1 = \frac{x^2-3x-4}{x^2-3x+4}\Rightarrow -4=+4\; \bf(Not\; Possible)$$

So we get $y\neq 1$

So Original Range $$\displaystyle -\frac{25}{7}\leq y< 1\Rightarrow y\in \left[-\frac{25}{7}\;,1\right)$$

You can graph this function and see the range from the graph. Read a book on calculus about how to graph a rational function.