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

Case 3

$\alpha\to\infty,\beta\to -\infty$


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: $$y = 1 $$

  • $\begingroup$ Why can we assign values to $a$ which are dependent on $\alpha,\beta$ wouldn't doing so mean that it's valid only for a specific case? $\endgroup$ – user659291 Apr 15 at 2:50
  • $\begingroup$ You've essentially done the same thing, but you called it "transforming the equation". So you could similarly ask yourself: Why can you transform the equation? $\endgroup$ – Lee Mosher Apr 15 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy