If $p(x)$ is a polynomial of degree $n$ such that $$p(-2)=-15,\ p(-1)=1,\ p(0)=7,\ p(1)=9,\ p(2)=13,\ p(3)=25.$$ Then smalest possible value of $n$ is
Options $(a)\; 2\;\;(b)\; 3\;\; (c)\;\; 4\;\; (d)\; 5$
Try: Tracing curve on coordinate axis, it gave one point of intersection Further $p(x)$ must be an odd degree polynomial. And slope of function is not same in each interval. So it is not linear. So it must have least degree $3$.
Can someone explain me if I am doing right? Thanks.
Otherwise please provide solution.