Find range of values of $a,b,c$ such that $ax^2+bx+c$ satisfies given conditions Let $y = f(x) = ax^2+bx+c$ where $a\neq0$. Find the range of values $a,b$ and $c$ such that it satisfies the following:

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*$0 \le f(x) \le 1 \quad \forall \space x \in [0,1]$

What I have found so far?

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*For $x=0$, we get $0 \le c \le 1$


*For $x=1$, we get $0 \le a+b+c \le 1$


*If roots exist, then they should lie outside the interval $[0,1]$.
Therefore, ${{-b - \sqrt{D}}\over{2a}} < 0$ and ${{-b + \sqrt{D}}\over{2a}} >1$. I tried to solve these but didn't get any fruitful results.
EDIT 1: As explained in comments by @insipidintegrator, this is only for distinct roots.


*Vertex of quadratic $f(x)$ is $\big( \frac{-b}{2a}, -\frac{D}{4a}\big)$. If the x-coordinate of the vertex lies between 0 and 1, then y-coordinate should also be between 0 and 1. But I have no idea how to proceed from here.
I want to find some lower/upper limits for $a,b,c$ or any kind of relation between them.  Any solution or hints for solving this problem is greatly appreciated.

You have $0\le f(x) \le 1$ on $[0,1]$ if and only if $f(0)$, $f(1) \in [0,1]$ and $(\ -\frac{b}{2a} \in [0,1] \implies -\frac{\Delta}{4 a} \in [0, 1]\ )$, that is all of the conditions below are satisfied

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*$ c\in [0,1]$


*$a + b + c \in [0,1]$


*$ - \frac{b}{2 a } \not \in [0,1]$ or $-\frac{\Delta}{4 a} \in [0,1]$
