Find the approximation for the interpolation of $f(x)$ by a polynomial of second degree 
Assume that $f(x)$ has a minimum in the interval $x_{n-1}\leq x\leq x_{n+1}$ where $x_k=x_0+kh$, $k$ being an integer. Show that the interpolation of $f(x)$ by a polynomial of second degree yields the approximation $$f_n-\frac{1}{8}\left[\frac{(f_{n+1} - f_{n-1})^2}{f_{n+1} -2f_{n}+f_{n-1}}\right],$$ $f_k=f(x_k)$.

I know that the error term for second degree polynomial approximation
$$error \leq \frac{M_{3}}{(3)!}\max_{x\in[x_{k-1},x_{k+1}]}|(x-x_{k-1})(x-x_{k})(x-x_{k+1})|$$
where $$M_3=\max_{\xi\in[x_{k-1},x_{k+1}]}|f^3(\xi)|$$
Please help me to solve the problem.

The function is approximated by a second order polynomial. Let:
$$
f(x) = a(x-x_n)^2+b(x-x_n)+c
$$
Specify for the points $x_k$ with $k=n-1,n,n+1$ ,
with $x_{n+1}-x_n=h$ and $x_{n-1}-x_n=-h$ :
$$
\begin{array}{l}
f_{n-1} = a h^2 - b h + c \\
f_{n} = c \\
f_{n+1} = a h^2 + b h + c
\end{array}
$$
Subtract the first equation from the last one:
$$
f_{n+1} - f_{n-1} = 2 b h \quad \Longrightarrow \quad b = \frac{f_{n+1} - f_{n-1}}{2h}
$$
Substitute this and $\,c\,$ into the last equation:
$$
f_{n+1} = a h^2 + \frac{f_{n+1} - f_{n-1}}{2h} h + f_{n}
\quad \Longrightarrow \quad a = \frac{f_{n+1} -2 f_n + f_{n-1}}{2h^2}
$$
The minimum of $f(x)$ - find it e.g. by differentiation - is obtained for $x-x_n = -b/(2a)$ .
And the minimum is:
$$
f(x_n-b/(2a)) = a\left[-\frac{b}{2a}\right]^2 + b\left[-\frac{b}{2a}\right] + c = c-\frac{b^2}{4a}
$$
Now substitute the values found for $\{a,b,c\}$ and you're done.
EDIT. Combine the condition $x_{n-1}\leq x\leq x_{n+1}$ with the place of the minimum:
$$
x-x_n = -b/(2a) = -\frac{(f_{n+1}-f_{n-1})/(2h)}{2(f_{n+1} -2 f_n + f_{n-1})/(2h^2)}
\quad \Longrightarrow \\
x_{n-1} \leq x_n-\frac{f_{n+1}-f_{n-1}}{f_{n+1} -2 f_n + f_{n-1}} \frac{1}{2} h \leq x_{n+1}
\quad \Longrightarrow \\
-h \leq \frac{f_{n+1}-f_{n-1}}{f_{n+1} -2 f_n + f_{n-1}} \frac{1}{2} h \leq +h
\quad \Longrightarrow \\
\left| f_{n+1}-f_{n-1} \right| \leq 2\left|f_{n+1} -2 f_n + f_{n-1}\right|
$$
Unless $f$ is a constant, this ensures that the denominator is nonzero in:
$$
f_n-\frac{1}{8}\left[\frac{(f_{n+1} - f_{n-1})^2}{f_{n+1} -2f_{n}+f_{n-1}}\right]
$$
Also note that still there can be a maximum instead of a minimum, all depending on the sign of the denominator (which is $\sim$ the discretization of the second order derivative). From the question as it is formulated therefore can be concluded that $f_{n+1} -2f_{n}+f_{n-1}$ is positive, or:
$$ f_n < \frac{f_{n+1}+f_{n-1}}{2} $$
So $f_n$ is smaller than the mean of its neighbors. If $f_n$ is itself the minimum, then its neighbors are equal: $f_{n-1}=f_{n+1}$ . If not, then the minimum is $< f_n$ .
