We have a quadratic equation $$y=ax^2+bx+c$$
With roots $\alpha,\beta$
Case 1
$\alpha\to\infty$ and $\beta$ is a finite number the equation transforms to $$y_1=bx+c$$
Case 2
$\alpha,\beta\to\infty$ the new equation becomes
$$y_2=c$$
Case 3
$\alpha\to\infty,\beta\to -\infty$
$$y_3=??$$
I figured out how the equation transforms in the first two cases by tinkering around on a graphing calculator but when I had to prove it rigorously using math I treated as a problem in limits but it looks like a problem in multivariable calculus? Since $a,b,c$ are dependent on each other and I haven't learnt that yet.
Is there some other way to figure out these transformations without using math above High School level?
By factoring, your equation is immediately converted to $$y = a(x - \alpha)(x - \beta) $$ which can then be re-expanded to give $$y = ax^2 - a(\alpha+\beta)x + a \alpha\beta $$
As you can see, the constant $a$ is not determined by $\alpha$ and $\beta$, and so this problem is underdetermined. One way to deal with that is to choose $a$ in some appropriate manner, depending on the case we are in.
For instance, in Case 1 we could choose $a = \frac{1}{\alpha}$ and then the equation becomes $$y = \frac{1}{\alpha} x^2 - (1 + \frac{\beta}{\alpha}) x + \beta $$ Now, taking a limit as $\alpha \to +\infty$ the equation becomes $$y = -x + \beta $$ whose only root is $\beta$. Good so far.
In Case 2, we could choose $a = \frac{1}{\alpha\beta}$ and the equation becomes $$y = \frac{1}{\alpha\beta}x^2 - \left(\frac{1}{\beta} + \frac{1}{\alpha}\right) x + 1 $$ Again, taking a limit as $\alpha,\beta \to +\infty$, the equation becomes $$y = 1 $$
We can deal with Case 3 in the same fashion as Case 2, with the exact same outcome: $$y = 1 $$