jmc

Let $p(x)=a{x}^2+bx+c.$ Then from the given information, $$\begin{array}{rl}9a-3b+c& =10,\\ c& =1,\\ 4a+2b+c& =5.\end{array}$$Then $9a-3b=9$ and $4a+2b=4,$ which reduce to $3a-b=3$ and $2a+b=2.$ Adding, we get $5a=5,$ so $a=1.$ Then $4+2b=4,$ so $b=0.$ Therefore, $p(x)=\boxed{x^2+1}.$