The parabola through three points

Text

There is a unique parabola $p(x)=ax^2+bx+c$ with $p(0)=40$, $p(1)=3$ and $p(2)=10$. Find it and compute $p(3)$.

Solution

From $p(0)=40$ we get $c=40$ immediately. Then: $p(1)=a+b+40=3 \ \Rightarrow\ a+b=-37,$ $p(2)=4a+2b+40=10 \ \Rightarrow\ 2a+b=-15.$ Subtracting the first from the second: $a=22$, hence $b=-37-22=-59$. So $p(x)=22x^2-59x+40,$ and $p(3)=22\cdot 9-59\cdot 3+40=198-177+40=\boxed{61}.$