Due to isoparametrics, the same is valid for any $\,f$, resulting in: $$f_2 = \frac{f_1 + f_3}{2} \quad ; \quad f(\xi) = f_2 + \frac{1}{2}(f_3 - f_1) \xi = \frac{1}{2}(1-\xi) f_1 + \frac{1}{2}(1+\xi) f_3$$ So it's my theory that, e.g. with an equidistant grid, quadratic elements aren't useful. But I have a personal bias towards linear elements, admittedly :-)

The one-dimensional Finite Difference stencil / quadratic parent Finite Element is defined geometrically by:

Algebraically, for any (Test) function $\,T(\xi)$ , it is defined by a quadratic interpolation, part of a Taylor series expansion:
$$
T = T(0) + \frac{\partial T}{\partial \xi}(0).\xi + \frac{1}{2} \frac{\partial^2 T}{\partial \xi^2}(0).\xi^2
$$
Because of the mapping $(T_1,T_0,T_2) \to (-1,0,+1)$ it follows that:
$$
T_0 = T(0)\\
T_1 = T(0) - \frac{\partial T}{\partial \xi}(0) + \frac{1}{2} \frac{\partial^2 T}{\partial \xi^2}(0)\\
T_2 = T(0) + \frac{\partial T}{\partial \xi}(0) + \frac{1}{2} \frac{\partial^2 T}{\partial \xi^2}(0)\\
\quad \mbox{ F.E. } \leftarrow \mbox{ F.D. }
$$
Solving these equations is not much of a problem and well-known Finite Difference schemes are recognized:
$$
T(0) = T_0 \\
\frac{\partial T}{\partial \xi}(0) = \frac{T_2-T_1}{2}\\
\frac{\partial^2 T}{\partial \xi^2}(0) = T_1-2T_0+T_2 \\
\quad \mbox{ F.D. } \leftarrow \mbox{ F.E. }
$$
Finite Element shape functions may be constructed as follows:
$$
T = N_0.T_0 + N_1.T_1 + N_2.T_2 = \\
T_0 + \frac{T_2-T_1}{2}\xi + \frac{T_1-2T_0+T_2}{2}\xi^2 =\\
(1-\xi^2)T_0 + \frac{1}{2}(-\xi+\xi^2)T_1 + \frac{1}{2}(+\xi+\xi^2)T_2
\\ \Longrightarrow \quad
\begin{cases}
N_0 = 1-\xi^2 \\
N_1 = (-\xi+\xi^2)/2\\
N_2 = (+\xi+\xi^2)/2 \end{cases}
$$
Isoparametric means that the *same* interpolation will be employed for *any* other function at the element.
The global one-dimensional Cartesian coordinate $\,x\,$ itself may serve as an outstanding example of such another function:
$$
x = x(0) + \frac{\partial x}{\partial \xi}(0).\xi + \frac{1}{2} \frac{\partial^2 x}{\partial \xi^2}(0).\xi^2\\
x = x_0 + \frac{x_2-x_1}{2}.\xi + \frac{1}{2} (x_1-2x_0+x_2).\xi^2\\
x = (1-\xi^2)x_0 + (-\frac{1}{2}\xi+\frac{1}{2}\xi^2).x_1 + (+\frac{1}{2}\xi+\frac{1}{2}\xi^2).x_2
$$
Before trying to establish the inverse transformation - which means solving for $\,\xi$ - we make some substitutions.
First assume that $\,x_1 < x_0 < x_2$ ; then define Left arm $\,L = x_0 - x_1\,$ and Right arm $\,R = x_2-x_0\,$ of the element:

Algebraically:
$$
x = x_0 + \frac{x_2-x_1}{2}.\xi + \frac{1}{2} (x_1-2x_0+x_2).\xi^2 \\ \Longrightarrow \quad
\frac{R-L}{2}\xi^2 + \frac{R+L}{2}\xi + x_0 = x \\ \Longrightarrow \quad
x(\xi) = \frac{R-L}{2}\left[\xi + \frac{R+L}{2(R-L)}\right]^2 - \frac{R+L}{8(R-L)} + x_0
$$
This means that the curve $\,x(\xi)\,$ is a parabola for $\,R\ne L\,$ and *linear* for $\,R=L\,$ . Let's assume in the sequel
that our element is *not* the important special case, which is linear.

If $\,R\ne L$ , then the parabola $\,x(\xi)\,$ is convex and has a minimum for $\,R > L$ , is concave and and has a maximum for $\,R < L$ .
A few numerical experiments shall reveal what's going on.

In the pictures below the **viewport is $\,[-1,+1]\times[0,2]$** .
Curve in *black* is the function $\,x(\xi)\,$ with $\,x_0=1$ , curve in
$\color{blue}{blue}$ is an isoparametric Test function $\,T(\xi) = 1-\xi^2\,$ with $\,x_0=0$ . Thin lines in $\color{green}{green}$ are
for support of the ideas. Regular behavior without any anomalies is displayed first, for $\,R < L$ , $R \approx L$ , $R > L$ :

$\color{red}{Red}$ dots indicate anomalous behavior. Resulting in a Test function which is multi-valued outside the element, ipse est (i.e.):
** not a function at all**. It is seen in the pictures that such is the case if the maximum or minimum of the parabola $\,x(\xi)\,$
is

Therefore, to avoid anomalous behavior, the position of the

**Note.** Anomalous behavior of the kind is also observed in 2-D, e.g. with quadrilateral elements having an obtuse angle
or being self-intersecting. See :
Quadrilateral Finite Elements must be convex and not self-intersecting. But why? .

**Sad remark.** A detailed analysis as the one above requires little more than some simple algebra and elementary geometry.
Yet such an analysis is seldom seen in Finite Element contexts; though the conclusions of such an analysis seem to be important enough.
So one can only guess for the reasons *why it is not done*.