Linear algebra, finding polynomial of degree of 2 or less that gives minimal result for certain condition

Question

yesterday I had an exam in linear algebra, and I had the following problem: find a real polynomial of deg $\leq$ $2$ $$p(x)= A+Bx+Cx^2$$ that satisfies $p(0)=0$ and the value of $$|p(1)-1|^2+|p(-1)-1|^2+|p(2)|^2$$ is minimal. $$$$after the test I realized almost no one solved this question on the exam(good thing there was a choice :) ), I would be more than thankful if someone could clarify how to solve it or give a good solution. I'll add and say it's probably related somehow to IP spaces and minimization problems (using the orthogonal projection, triangle equality, or Cauchy-Schwarz inequality in some shape or form)

Answer 1

Since $p(0)=0$, $A=0$.

Now $p(1)=B+C$, $p(-1)=-B+C$ and $p(2)=2B+4C$. So, the objective function is $$\Phi=(B+C-1)^2+(-B+C-1)^2+(2 B+4 C)^2=6 B^2+16 B C+18 C^2-4 C+2$$ Compute the partial derivatives $\Phi'_B$ and $\Phi'_C$; set them equal to zero and you have two linear equations in $(B,C)$. When solved, plug the values in $\Phi$.

Answer 2

Indeed there is a nice geometry way:

we have to mimimize the square of the distance between the points $(x,y,z)$ and $(1,1,0)$ where $x=p(1)=B+C,y=p(-1)=C-B,z=p(2)=2B+4C$ clearly $x,y,z$ lie in the plane $\phi=3x+y-z=0$.

So the minimum distance between the point $(x,y,z)$ and $(1,1,0)$ is the distance between the plane $\phi$ and $(1,1,0)$ $$=\frac{3(1)+1-0}{\sqrt{3^2+1^2+1^2}}$$ and the answer will be the square of this distance which is $\frac{16}{11}$