Condition for inverse of quadratic function - alternative solutions

Question

I was helping my friend teacher to assemble a list of exercises to their precalculus students. So I came up with this problem:

Let $f$ be a quadratic function, i.e. $$f(x) = ax^2 + bx + c,$$ where $a,b,c\in \mathbb{R}$ and $a\neq 0$. Prove that if $\exists \,x,y\in \mathcal{D}om_f$ such that $x\neq y$ and $$|2ax + b|=|2ay + b|,$$ then $\not\exists f^{-1}$.

The most efficient way I find to solve this is squaring both sides of the equality and easily conclude that $f(x) = f(y)$, which implies that $f$ is not injective (the plan is to show that $f$ is not an injection).

My question is: are there alternative (and yet interesting for the students) ways to prove it?

(Feel free to give more than one solutions, make observations, generalizations, etc)

Answer 1

Hint: $\mid 2ax+b\mid =\mid 2ay+b\mid$ is equivalent to say that $2ax+b=2ay+b$ or $2ax+b=-2ay-b$.

Suppose $2ax+b=2ay+b$ thus $2a(x-y)=0, a=0$ or $x=y$ contradiction since by hypothesis $a\neq 0, x\neq y$.

$2ax+b=-2ay-b$ i.e $2a(x+y)+2b=0$.

On the other hand, $ax^2+bx+c=ay^2+by+c$ i.e $a(x-y)(x+y)+b(x-y)=0$ i.e $a(x+y)+b=0$ since $x\neq y$.

Answer 2

Assuming they know the derivative:

If a polynomial $f$ is not monotonic in its domain, then $f$ cannot be injective. But, since $f'(x)=2ax+b$ the given condition may be written as $$|f'(x)|=|f'(y)|$$ So, you have two distinct points with the same absolute value derivative. If both are positive, then $$2ax+b=2ay+b \implies a(x-y)=0 \implies a=0 \text{ or } x=y$$ which is a contradiction. The same if both are negative. So one must be negative and one positive, but this implies that there is one point, say $x$ with $f'(x)>0$ and one point, say $y$ with $f'(y)<0$. So, $f$ is not monotonic (draw also a shape).