Some sort of iteration involving derivatives

$\text{ Let }f_{n}(x) = a_{n}x^2 + b_{n}x + c_{n}$ $\text{ for }n\in \mathbb{W}\text{(Set of whole numbers)}\text{ with }a_{0} = 1\;,\;b_{0} = 0\;,\;c_{0} = 0$

$\text{ Then according to question, }\newline f_{n+1}(0) = c_{n}\newline f_{n+1}(1) = b_{n}\newline f_{n+1}(2) = 2a_{n}$

$\text{But according to sequence }f_{n}(x)\;,\quad f_{n+1}(x) = a_{n+1}x^2 + b_{n+1}x + c_{n+1}\newline\Rightarrow\hspace{6pt} f_{n+1}(0) = c_{n+1}\newline\hspace{19pt} f_{n+1}(1) = a_{n+1}+b_{n+1}+c_{n+1}\newline\text{and } f_{n+1}(2) = 4a_{n+1}+2b_{n+1} + c_{n+1}$

$\text{Comparing both sets of equations we get}\newline c_{n+1} = c_{n}\hspace{118pt}\dots\;Eq. 1\newline a_{n+1} + b_{n+1} + c_{n+1} = b_{n}\hspace{56pt}\dots\;Eq. 2\newline 4a_{n+1} + 2b_{n+1} + c_{n+1} = 2a_{n}\hspace{40pt}\dots\;Eq. 3$

$\text{Since }c_{0} = 0\;,\;Eq. 1 \text{ implies that }c_{n} = 0 \hspace{15pt}\forall \;n\in \mathbb{W}.\text{ Rewriting }Eq. 2\text{ and }Eq. 3\text{ we get }\newline a_{n+1} + b_{n+1} = b_{n}\hspace{43pt}\dots\;Eq. 4\newline 4a_{n+1} + 2b_{n+1} = 2a_{n}\hspace{27pt}\dots\;Eq. 5$

$\text{Solving }Eq. 4$ $\text{ and }Eq. 5$ $\text{ for }a_{n+1}$ $\text{ and }b_{n+1}$ $\text{ we get}\newline a_{n+1} = a_n - b_n\hspace{25pt}\dots\;Eq. 6\newline b_{n+1} = 2b_n - a_n\hspace{21pt}\dots\;Eq. 7\newline\text{Applying recurrence relation }Eq. 6\text{ and }Eq. 7 \text{ on }Eq.6\text{ we get }\newline a_{n+1} = a_n - b_n\newline \hspace{22pt} = (a_{n-1} - b_{n-1}) - (2b_{n-1} - a_{n-1}) \newline\hspace{22pt} = 2a_{n-1} - 3b_{n-1}\newline\text{Applying reccurence relation for two more times we get}\newline a_{n+1} = 5a_{n-2} - 8b_{n-2}\newline\hspace{22pt} = 13a_{n-3} - 21b_{n-3}\newline\text{We see the Fibonacci sequence in the coefficient of }a \text{ and }b\text{ in the expression of }a_{n+1}$ $\text{ with initial terms }F_0 = 1\;,\;F_1 = 1\text{. That is }\newline a_{n+1} = F_{0}a_n - F_{1}b_n\newline\hspace{22pt}=F_{2}a_{n-1} - F_{3}a_{n-1}\newline\text{and in general }\newline a_{n+1} = F_{2k}a_{n-k} - F_{2k+1}b_{n-k}$ $\text{ for }0\leq k\leq n \hspace{20pt} \dots\;Eq. 8$

$\newline Eq. 8\text{ can be proved as follows: }$

$a_{n+1} = a_n - b_n = F_{0}a_n - F_{1}b_n\;\;\text{(Base case }k = 0\text{ is true)}\newline \text{Let }Eq. 8\text{ be true for }k = l\text{, then }\newline a_{n+1} = F_{2l}a_{n-l} - F_{2l+1}b_{n-l}\newline\hspace{22pt} = F_{2l}(a_{n-l-1} - b_{n-l-1}) - F_{2l+1}(2b_{n-l-1} - a_{n-l-1})\newline\hspace{22pt} = (F_{2l} + F_{2l+1})a_{n-l-1} - (F_{2l} + 2F_{2l+1})b_{n-l-1}\newline\hspace{22pt} = F_{2l+2}a_{n-l-1} - F_{2l+3}b_{n-l-1}\newline\hspace{22pt} = F_{2(l+1)}a_{n-(l+1)} - F_{2(l+1)+1}b_{n-(l+1)}\quad \text{(true for }k=l+1)$

$\newline \text{Thus from mathematical induction }Eq. 8\text{ is true.}$

$\text{Putting }k=n\text{ in }Eq. 8\text{ we get }\newline a_{n+1} = F_{2n}a_{0} - F_{2n+1}b_{0} = F_{2n}\;\forall \;n\in \mathbb{W}$

$\newline \Rightarrow a_n = F_{2n-2} \;\forall\;n\in \mathbb{N}\newline\text{and } b_n = a_n - a_{n+1} = F_{2n-2} - F_{2n} = -F_{2n-1}\;\forall\;n\in \mathbb{N}$

$\text{Roots of }f_{n}(x) = 0 \text{ are }x = 0 \text{ and }x = \large-\frac{b_n}{a_n}.$ $\text{ Hence the equation of axis of symmetry is }x = \large-\frac{b_n}{2a_n}$

$\large\lim_{n\to\infty}(-\frac{b_n}{2a_n}) = \frac{1}{2}\cdot\lim_{n\to\infty}\frac{F_{2n-1}}{F_{2n-2}}=\frac{\varphi}{2} =\frac{\sqrt{5}+1}{4}\;\text{(where }\varphi\text{ is the golden ratio.)}$

$f_n(x)=a_nx^2+b_nx$ , $a_0=1$ , $b_0=0$

$a_{n+1}=a_n-b_n$

$b_{n+1}=2b_n-a_n$

Solve the recurrence relations

$a_n=\frac{\sqrt{5}+5}{10}\left(\frac{3-\sqrt{5}}{2}\right)^n+\frac{5-\sqrt{5}}{10}\left(\frac{3+\sqrt{5}}{2}\right)^n$

$b_n=\frac{\left(3-\sqrt{5}\right)^n-\left(3+\sqrt{5}\right)^n}{2^n\sqrt{5}}$

$\lim\limits_{n\to\infty}a_n=\frac{5-\sqrt{5}}{10}\left(\frac{3+\sqrt{5}}{2}\right)^n$

$\lim\limits_{n\to\infty}b_n=-\frac{\left(3+\sqrt{5}\right)^n}{2^n\sqrt{5}}$

Thus the axis of symmetry approaches

$x=\frac{\sqrt{5}+1}{4}$