bound for $a,b,c$ in $|ax^2+bx+c| \leq 1\;\forall x\in \left[0,1\right]$ 
Consider $a,b,c\in\mathbb{R}$ such that $|ax^2+bx+c|\leq 1\;\forall x\in \left[0,1\right]$. Prove that $|a|\leq 8\;\;,|b| \leq 8$ and $|c| \leq 1$.

My Attempt:
Set $x = 0$ in $|ax^2+bx+c|\leq 1$ to get $|c|  \leq 1$. Similarly set $x = 1$ in $|ax^2+bx+c| \leq 1$ to get $|a+b+c|\leq 1$.
From here, how can I calculate bounds for $a$ and $b$?

Let $f(x)=ax^2+bx+c$.
Hence, $f(0)=c$, $f(1)=a+b+c$ and $f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$.
After solving of this system we obtain:
$|a|=|2f(1)-4f\left(\frac{1}{2}\right)+2f(0)|\leq2|f(1)|+4|f\left(\frac{1}{2}\right)|+2|f(0)|\leq8$,
$|b|=|-f(1)+4f\left(\frac{1}{2}\right)-3f(0)|\leq|f(1)|+4|f\left(\frac{1}{2}\right)|+3|f(0)|\leq8$,
$|c|=|f(0)|\leq1$.
