The Lagrange quadratic polynomial $$p_2(x)=\ell_0(x)\textsf{f}_0(x)+\ell_1(x)\textsf{f}_1(x)+\ell_2(x)\textsf{f}_2(x)$$ So I wanted to prove that $p_2(x)$ can be written in the following form $$p_2(x)=Ax^2+Bx+C$$ with $A,B,C$ arbitrary scalars.
I was trying to use the Lagrange interpolation method but it didn't work.
Any thoughts? Thanks.
Consider an example as follows:
Example: Construct the quadratic Lagrange interpolating polynomial $p_2(x)$ that interpolates the points $~(1,2), (3,4), ~\text{and}~ (5,6)~$.
Answer:
$$P_2(x)=2~\frac{(x−3)(x−5)}{(1−3)(1−5)}+4~\frac{(x−1)(x−5)}{(3−1)(3−5)}+6~\frac{(x−1)(x−3)}{(5−1)(5−3)}$$ $$\qquad=\frac{1}{4}~(x−3)(x−5)−(x−1)(x−5)+\frac{3}{4}~(x−1)(x−3)$$ $$\qquad=\frac{1}{4}~(x^2-8x+15)−(x^2-6x+5)+\frac{3}{4}~(x^2-4x+3)$$ $$\qquad=0~\cdot~x^2+x+1$$
So, the quadratic Lagrange interpolating polynomial, $~p_2~$ through the points $~(x_0,y_0), ~(x_1,y_1), ~\text{and}~ (x_2,y_2)$ as the following function: $$p_2(x)=\ell_0(x)\textsf{f}_0(x)+\ell_1(x)\textsf{f}_1(x)+\ell_2(x)\textsf{f}_2(x)$$ can be written in the following form $$p_2(x)=Ax^2+Bx+C$$where $$\ell_0(x)=\frac{(x−3)(x−5)}{(1−3)(1−5)}=\frac{1}{8}~(x−3)(x−5)$$ $$\ell_1(x)=\frac{(x−1)(x−5)}{(3−1)(3−5)}=-\frac{1}{4}~(x−1)(x−5)$$ $$\ell_2(x)=\frac{(x−1)(x−3)}{(5−1)(5−3)}=\frac{1}{8}~(x−1)(x−3)$$ $$\textsf{f}_0(x)=y_0=2,\qquad \textsf{f}_1(x)=y_1=4,\qquad \textsf{f}_2(x)=y_2=6$$ and $$A=0,\qquad B=1,\qquad C=1$$