Let $a,$ $b,$ $c$ be real numbers such that
\[|ax^2 + bx + c| \le 1\]for all $0 \le x \le 1.$  Find the largest possible value of $|a| + |b| + |c|.$
Setting $x = 0,$ we get $|c| \le 1.$  Setting $x = 1,$ we get
\[|a + b + c| \le 1.\]Setting $x = \frac{1}{2},$ we get
\[\left| \frac{a}{4} + \frac{b}{2} + c \right| \le 1.\]Let
\begin{align*}
p &= c, \\
q &= \frac{a}{4} + \frac{b}{2} + c, \\
r &= a + b + c,
\end{align*}so $-1 \le p,$ $q,$ $r \le 1.$  Solving for $a,$ $b,$ and $c,$ we find
\begin{align*}
a &= 2p - 4q + 2r, \\
b &= -3p + 4q - r, \\
c &= p.
\end{align*}Hence, by Triangle Inequality,
\begin{align*}
|a| &= |2p - 4q + 2r| \le |2p| + |4q| + |2r| = 8, \\
|b| &= |-3p + 4q - r| \le |3p| + |4q| + |r| = 8, \\
|c| &= |p| \le 1.
\end{align*}Therefore, $|a| + |b| + |c| = 8 + 8 + 1 = 17.$

Consider the quadratic $f(x) = 8x^2 - 8x + 1.$  We can write
\[f(x) = 8 \left( x - \frac{1}{2} \right)^2 - 1.\]For $0 \le x \le 1,$ $0 \le \left( x - \frac{1}{2} \right)^2 \le \frac{1}{4},$ so $-1 \le f(x) \le 1.$

Therefore, the largest possible value of $|a| + |b| + |c|$ is $\boxed{17}.$