\[f(x)\]
satisfies \[f(0)=a, \: f(1)=b, \: f(2)=c\]
The first of these is satisfied by the Lagrange polynomial
\[f|_1(x)=\frac{x-1)(x-2)}{(0-1)(0-2)} = \frac{x^2-3x+2}{2}a\]
The second of these is satisfied by the Lagrange polynomial
\[f|_2(x)=\frac{x-0)(x-2)}{(1-0)(1-2)} = (-x^2+2x)b\]
The third of these is satisfied by the Lagrange polynomial
\[f|_3(x)=\frac{x-0)(x-1)}{(2-0)(2-1)}c = \frac{x^2-x}{2}c\]
Adding these gives the approximation over the interval
\[[0,2]\]
:\[\begin{equation} \begin{aligned} f(x) & \simeq f_1(x)a+f_2(x)b+f_3(x)c \\ &=\frac{x^2-3x+2}{2}a + (-x^2+2x)b +\frac{x^2-x}{2}c \\ &=x^2(a/2-b+c/2)+x(-3a/2+2b-c/2)+a \end{aligned} \end{equation}\]
This expression is only equal to zero if each coefficient is zero i.e.
\[a/2-b+c/2=-3a/2+2b-c/2=a=0\]
Since
\[a=0\]
, \[-b+c/2=2b-c/2=0\]
The only possibility is that
\[b=c=0\]
.Then these Lagrange polynomials are linearly independent.