Prove that $|f(x)| \le \frac{3}{2}$ when $f(x)=ax^2+bx+c$ Suppose $f(x) = ax^2+bx+c$ where $x \in [-1,1]$. If $f(-1),f(0),f(1)\in [-1,1]$ show that $|f(x)| \le \frac{3}{2}$ $\forall x \in [-1,1]$.
This is how I tried: 
$f(0)=c$ 
$f(1)=a+b+c$ 
$f(-1)=a-b+c$ 
Putting $f(0)=c$ we get $f(1)-f(0)=a+b$, $f(-1)-f(0)=a-b$. Solving for $a$ and $b$ respectively we get $$a=\frac{f(1)+f(-1)-2f(0)}{2}$$ and $$b=\frac{f(1)-f(-1)}{2}$$ Then
$$|f(x)|=|(\frac{f(1)}{2})(x^2+x)+(\frac{f(-1)}{2})(x^2-x)+(f(0))(1-x^2)|$$
$$\le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|$$. Now individually the maximum of $|x^2+x|$ happens at $x=1$ , $|x^2-x|$ at $x=-1$ and $|1-x^2|$ at $x=0$.  So All I get $|f(x)| \le 3$. But I need $\frac{3}{2}$.
Thanks for the help!!

As mentioned above, this has already been answered in Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. with the better bound $5/4$. But it seems that your
proof is almost complete and slightly shorter than (or at least different from) the answers given there.
You already showed that
$$
|f(x)| \le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|
$$
and for $0 \le x \le 1$ the right-hand side is
$$
\frac{x^2+x}{2} -\frac{x^2-x}{2}+1-x^2
 = x + 1 - x^2 = \frac 54 - (x - \frac 12)^2 \le \frac 54 < \frac 32 \quad .
$$
Similarly for $-1 \le x \le 0$.
