Find the interpolating polynomial for the given points $$(-1,3)\;\;(0,1)\;\;(1,1)\;\;(2,3)$$
Then add the point $(-2,7)$ to the set of your points and explain the result.
The unique interpolating polynomial of the function is given by $$L\left(x\right)=\sum_{i=0}^{3}y_{i}l_{i}\left(x\right)$$
Where $$l_{i}\left(x\right)=\prod_{m=0 \\ m \ne i }^{3 }\frac{x-x_{m}}{x_{i}-x_{m}}$$
So $$L\left(x\right)=\sum_{i=0}^{3}y_{i}l_{i}\left(x\right)=x^2-x+1$$
I also added the point $(-2,7)$ and noticed that the interpolating polynomial is the same as the above, by the properties of $L(x)$, I know that $L(x)-f(x)$ has at least $4$ roots, however I don't know if $L(x)-f(x)$ is of degree at most $3$ ,so this is why I cannot say $L(x)=f(x)$ and conclude that adding more points will not change the interpolating polynomial.
In general, adding more points will change the polynomial. What happened in this case in that the new point is already in the graphic of the initial interpolating polynomial... If you start with a polynomial $p(x)$ and keep adding points of the form $(x_i, p(x_i))$, you will keep getting the same polynomial, and this is due to the uniqueness of the interpolating polynomial.