Prove that if $\dfrac{ax^2+2bx+c}{\alpha x^2+2\beta x+\gamma} (\alpha\neq 0)$ has $3$ inflection points, then all of them lie on one line?

Question

Prove that if the rational function $f(x)=\dfrac{ax^2+2bx+c}{\alpha x^2+2\beta x+\gamma} (\alpha\neq 0)$ has three inflection points, then all of them lie on one line? (All the parameters are real numbers.

It's an exercise problem. And there is a hint of this problem:

Firstly consider the case $a=0, \alpha=1$. Prove that if $b=0$ or $x^2+2\beta x+\gamma=0$ has real roots, then $f(x)$ would not have $3$ inflection points.

Here's my question:

1. I don't know how to prove the case “$x^2+2\beta x+\gamma=0$ has two real roots”.

2. How to prove the general case if we have prove the case $a=0,\alpha=1$?

Answer 1

For your second question, notice that ... $$ \begin{align} f(x)&=\dfrac{ax^2+2bx+c}{\alpha x^2+2\beta x+\gamma} \\&=\frac a \alpha \Bigg( 1+\dfrac{ \big( \frac ba - \frac \beta\alpha \big)x + \big( \frac ca - \frac \gamma\alpha \big) }{ x^2+\frac\beta\alpha x + \frac\gamma\alpha} \Bigg ) \\ & = \frac a \alpha \big (1+g(x) \big) \end{align}$$

where $g(x)$ is a rational function satisfying the requirements of your hint.