Parabolic Interpolation with three data points and measurement noise The question is as follows:

Have$$z_i = y(x_i) + w_i,\quad i = 1, 2, 3$$ where $x_1<x_2<x_3$ and
$z_1<z_2>z_3$. $\space$ Assume the unknown function $y(x)$ can be
approximated with parabolic interpolation.

*

*Find the location of the estimate of the maximum point $y(x)$ $$\hat x_m = f(x_1,x_2,x_3,z_1,z_2,z_3)$$


*Assume $$E[w] = 0; \space \space E[ww^T]=P$$ find the variance associated with the estimate $\hat x_m$ using first-order series
expansion.

$$\\$$
I was able to solve part 1 by considering the model:
$$ z(i) = a{x_{(i)}}^2+bx_{(i)}+c+w$$
Then with three measurements, can write the measurement equation as:
$$\begin{pmatrix}z_1 \\\ z_2 \\\ z_3\end{pmatrix} =
\begin{pmatrix} {x_1}^2 & x_1 & 1\\\ {x_2}^2 & x_2 & 1\\\ {x_3}^2 & x_3 & 1 \end{pmatrix} \begin{pmatrix} a \\\ b \\\ c \end{pmatrix} + \begin{pmatrix} 1\\\ 1\\\ 1\end{pmatrix}w
$$
Denote this as $$Z = XA+W$$
The parabola parameter matrix $A$ can then be written as $$A = X^{-1}Z - X^{-1}W$$
where $\hat A$ can be found as
$$ \hat A = E[A] = X^{-1}Z - X^{-1}E[W]$$
Its terms can be found through matrix operations assuming $E[W]=0$. The maximum point of $y(x)$ occurs at $-\frac{b}{2a}$, which can be written as
$$\hat x_m = -\frac{b}{2a}=
\frac{z_1({x_3}^2-{x_2}^2) + z_2({x_1}^2-{x_3}^2) + z_3(-{x_1}^2+{x_2}^2)}
{2(z_1(x_3-x_2) + z_2(x_1-x_3) + z_3(-x_1+x_2))}$$
But I can't figure out how to find the variance of the estimate. I thought about finding the covariance matrix of $\hat A$, which would give me the variance of $a, b$ and $c$. Assuming the terms are independent, I can calculate the variance of $\hat x$. But the covariance would be
$$E[(\hat A-\bar A)(\hat A-\bar A)^T]$$
but since $$\hat A = \bar A = E[A]$$
the covariance matrix would just be $0$ which doesn't makes sense.
Would really appreciate it if someone could point me in the right direction. Thanks in advance!
 A: I denote vectors by small letters. The estimated coefficients are given by
\begin{align}
\hat{\mathbf{a}}=\mathbf{X}^{-1}\mathbf{z},
  \end{align}
whereas the true coefficients are given by
\begin{align}
\mathbf{a}=\mathbf{X}^{-1}(\mathbf{z}-\mathbf{w}),
  \end{align}
where
\begin{align}
 \mathbf{X}=\left[\begin{array}{ccc}x_1^2&x_1&1\\x_2^2&x_2&1\\x_3^2&x_3&1\end{array}\right],~
 \mathbf{a}=\left[\begin{array}{c}a\\b\\c\end{array}\right],~
 \mathbf{z}=\left[\begin{array}{c}z_1\\z_2\\z_3\end{array}\right],~
 \mathbf{w}=\left[\begin{array}{c}w_1\\w_2\\w_3\end{array}\right].
\end{align}
We see that $\text{var}(\mathbf{z})=\text{var}(\mathbf{w})$. A variance of the obtained parameters $\hat{\mathbf{a}}$ is given by
\begin{align}
   \text{var}(\hat{\mathbf{a}})=\text{var}(\mathbf{X}^{-1}\mathbf{z})=\mathbf{X}^{-1}\text{var}(\mathbf{z})\mathbf{X}^{-\text{T}}=\mathbf{X}^{-1}\underbrace{\text{var}(\mathbf{w})}_{=P\mathbf{I}}\mathbf{X}^{-\text{T}}=P\mathbf{X}^{-1}\mathbf{X}^{-\text{T}},
  \end{align}
where we have used formula $\text{var}(\mathbf{Ay})=\mathbf{A}\text{var}(\mathbf{y})\mathbf{A}^{\text{T}}$ and
$\text{var}(\mathbf{w})=P$. Therefore we get
\begin{align}
 \text{var}(a)&=P(\mathbf{X}^{-1}\mathbf{X}^{-\text{T}})_{11},\\
 \text{var}(b)&=P(\mathbf{X}^{-1}\mathbf{X}^{-\text{T}})_{22}.
\end{align}
Finally, we would like to evaluate $\text{var}\left(-\frac{2a}{b}\right)$, which is not possible directly, but only approximately. By the use of the first order Taylor expansion $-\frac{2a}{b}\approx -\frac{2a_0}{b_0}-\frac{2}{b_0}a+\frac{2a_0}{b_0^2}b$ and the formula $\text{var}(a\mathbf{x}+b)=a^2\text{var}(\mathbf{x})$ we obtain the following approximation
\begin{align}
 \text{var}(\hat{x}_{\text{m}})=\text{var}\left(-\frac{2a}{b}\right)\approx\,& \text{var}\left(-\frac{2}{b_0}a+\frac{2a_0}{b_0^2}b\right)=\frac{4}{b_0^2}\text{var}(a)+\frac{4a_0^2}{b_0^4}\text{var}(b)= \nonumber\\
 &\frac{4P}{b_0^2}\left[(\mathbf{X}^{-1}\mathbf{X}^{-\text{T}})_{11}+\frac{a_0^2}{b_0^2}(\mathbf{X}^{-1}\mathbf{X}^{-\text{T}})_{22}\right].
\end{align}
