General algorithm for range of function given by division of two quadratics

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Study topic in textbook for 9th graders

$$f:D\to\mathbb{R},f(x)=\frac{ax^2+bx+c}{a_1x^2+b_1x+c_1},D\subseteq\mathbb{R}$$

Find a general method for determining $\text{Im}(f)$

No calculus allowed

I would like to know whether I am overcomplicating this problem, or I am on the right way.

$$f(x)=y\in\text{Im}(f)$$ $$h(x)=(a_1y-a)x^2+(b_1y-b)x+(c_1y-c)=0$$

Condition: $\exists$ at least one solution $x\in D$

Now, extracting roots of $h$ and conditioning that at least one of the roots belongs to $D$ will make things messy, involving inequalities with square roots etc. so we need to think of another method.

For the sake of simplicity let $D=(\alpha,\beta)$ and then we will generalize for reunion of intervals.

A. $\exists$ exactly one solution $x\in D$

Calculate values of $h$ at the endpoints of $D$

Let $E_1(x)=a_1x^2+b_1x+c_1,E_2(x)=ax^2+bx+c$.

Then

$$h(\alpha)=E_1(\alpha)y-E(\alpha)$$ $$h(\beta)=E_1(\beta)y-E(\beta)$$

Condition: the values of $h$ at endpoints have opposite sign.

$$h(\alpha)h(\beta)<0$$

The quadratic $h(\alpha)h(\beta)=0$ has roots

$$y_1=\frac{E(\alpha)}{E_1(\alpha)}=f(\alpha)$$ $$y_2=\frac{E(\beta)}{E_1(\beta)}=f(\beta)$$

There will be two cases depending on the main coefficient of $h(\alpha)h(\beta)$

Let $y_{min}=min(y_1,y_2),y_{max}=max(y_1,y_2)$

a. $E_1(\alpha)E_1(\beta)>0$

$$y\in(y_{min},y_{max})$$

b. $E_1(\alpha)E_1(\beta)<0$

$$y\in(-\infty,y_{min})\cup(y_{max},\infty)$$

This will eventually be the interval where $f$ is injective, and it seems to be correct as illustrated in this GeoGebra program:

However I have to find the interval where $f$ is not injective and make the reunion.

B. $\exists$ exactly two solutions $x\in D$

$$\alpha<x_1,x_2<\beta\Leftrightarrow\begin{cases}x_1,x_2>\alpha\ & (1) \\ x_1,x_2<\beta & (2)\end{cases}$$

Now for example

$$(1)\Leftrightarrow\begin{cases}x_1-\alpha>0 \\ x_2-\alpha>0\end{cases}\Leftrightarrow\begin{cases}(x_1-\alpha)+(x_2-\alpha)>0 \\ (x_1-\alpha)(x_2-\alpha)>0\end{cases}$$

And I could go on with Vieta's formulas, but it gets messy really fast.

Any suggestion on how to continue this algorithmic problem?

(Thanks for reading)