A quadratic function $f(x)$ satisfies the inequality $-1 < f(x) < 1$ for $x \in [0, 1]$. What can we say about the range of its coefficients? Let a function $f(x) = ax^2 + bx + c$, where $a, b, c \in R$,  satisfy $-1 \leq f(x) \leq 1$ for all $x \in [0, 1]$ then which of the following conclusions can be made?
A) $|a| \leq 8$
B) $|b| \leq 8$
C) $|c| \leq 1$
D) $|a| + |b| + |c| \leq 17$
I know that we can get $|c| \leq 1$ by substituting $x = 0$. Apart from this I don't have any idea about how to proceed.
 A: We can reduce the range of searching we will need to do by only considering $ \ a \ > \ 0 \ \ , \ $ since the case for $ \ a \ < \ 0 \ $ produces vertical reflections of those parabolas.  As we are interested in the absolute-values of the coefficients, we can omit these redundant parabolas.
It will also be convenient to write the quadratic polynomial in "vertex form" as $ \ f(x) \ = \ a·(x - x_{v})^2 + y_{v} \ \ , \ $ where the coordinates of the vertex have the familiar expressions $ \ x_v \ = \ -\frac{b}{2a} \ $ and $ \ y_{v} \ = \ c - a· x_{v}^2 \ = \ c -  \frac{b^2}{4a}  \ \ . $  We will only need to examine $ \ 0 \ \le \ x_{v} \ \le \ \frac12 \ \ , \ $ since the parabolas for $ \ \frac12 \ \le \ x_{v} \ \le \ 1 \ \   $ are just reflections of the others about the line $ \ x \ = \ \frac12 \ \ . $
Using this form for the quadratic function, we can write function values in terms of the location of the parabola's vertex as
$$ f(x_v) \ \ = \ \ y_{v} \ \ , \ \ f(0 ) \ \ = \ \ a· x_v^2 \ + \ y_{v} \ \  = \ \ f(x_v) \ + \ a· x_v^2 \ \  ,  $$ $$ f(1) \ \ = \ \ a· (x_v - 1)^2 \ + \ y_{v} \ \  = \ \ f(x_v) \ + \ a· (x_v - 1)^2  \ \  .  $$
In this way, we can examine the values of $ \ f(x) \ $ at the endpoints of $ \ [0 \ , \ 1] \ $ simply as a function of the placement of the vertex.  Because the parabolas "open upward", we get the greatest possible range for $ \ a \ $ by positioning the vertex "as low as permitted" and the function values at the interval endpoints "as high as permitted". Thus we have $$ \ f(x_v) \ \ = \ \ -1 \ \ \rightarrow \ \ f(0) \ \  = \ \ 1 \ \ = \ \ a· x_v^2 \ - \ 1 \ \ ,  \ \  f(1) \ \ = \ \ 1 \ \ = \ \  a· (x_v - 1)^2 \ - \ 1 \ \ . \ $$
Over $ \ 0 \ \le \ x_{v} \ \le \ \frac12 \ \ , \ $ both of these functions have their greatest value at $ \ x_v \ = \ \frac12 \ \ , \ $ that being $ \ \frac{a}{4} - 1 \ \ $ and that maximum is equal to $ \ 1 \ $ for $ \ \frac{a}{4} - 1 \ = \ 1 \ \Rightarrow \ a \ = \ 8  \ \ . \ $  (Using smaller values of $ \ f(0) \ $ or $ \ f(1) \ $ would give us smaller values for $ \ a \ \ . \  $ Should we decide to check the interval $ \ \frac12 \ \le \ x_{v} \ \le \ 1 \ \ , \ $ the function for $ \ f(0) \ $ would be the same and the function for $ \ f(1) \ $ becomes $ f(1) \ \ = \ \ a· (1 - x_v)^2 \ - \ 1 \ \ , \  $ which does not alter our conclusion.)
If we now include the neglected case of $ \ a \ < \ 0 \ \ , \ $ we have $ \ | \ a \ | \ \le \ 8 \ \ . $
As you noted, we must have for the $ \ y-$intercept $ \ -1 \ \le \ f(0)  = c \ \le \ 1 \ \Rightarrow \ | \ c \ | \ \le \ 1 \ \ . \ $
EDIT (8/5) -- (Sorry, I wasn't happy with/convinced by the argument I had here earlier.)  Since we set $ \ y_v \ = \ c - \frac{b^2}{4a} \ = \ -1 \ \ , \ $ then $ \ 4ac - b^2 \ = \ -4a \ \Rightarrow \ b^2 \ = \ 4a·(1 + c) \ \ $ for $ \ a \ > \ 0 \ \ . $  Then
$$ -1 \ \le \ c \ \le \ 1 \ \ \Rightarrow \ \ 0 \ \le \ 1 + c \ \le \ 2 \ \ \Rightarrow \ \ 0 \ \le \ b^2 = 4a·(1 + c) \ \le \ 8a \ \ . $$
Hence $ \ b^2 \ \le \ 8·8 \ \Rightarrow \ | \ b \ | \ \le \ 8 \ \ . $
So we can immediately conclude that $ \ | \ a \ | + | \ b \ | + | \ c \ | \ \le \ 17 \ \ , \ $ as Izaak van Dongen remarks.  All four statements are correct.
Note that these bounds for the coefficients cannot pertain all at once, so while we may say that $ \ | \ a + b + c \ | \ \le \ | \ a \ | + | \ b \ | + | \ c \ | \ \le \ 17 \ \   $ (by the triangle inequality), the bound on the absolute-value of the coefficient sum is substantially smaller (in fact, $  \ | \ f(1) \ = \ a + b + c \ | \ \le \ 1 \ ) \ . $
A: Assume first that $a\geq 0$ and apply , by continuty, the assumption to $x=\pm 1$. By continuity
$a\pm b+c  \leq 1$ so that $\vert a \vert +\vert b\vert +c \leq  1$ and $\vert a \vert +\vert b\vert - \vert c \vert \leq 1$
and the same is true if $a <0$ while changing $f$ to $-f$.
As $\vert c\vert <1$ we get  $\vert a \vert +\vert b\vert < 2$, so that
$\vert a \vert <2$, $\vert b \vert <2$, and $\vert a \vert +\vert b\vert + \vert c \vert \leq 5$
