Question

Find the second-degree polynomial P(x)=ax^2+bx+c such that P(1)=4, P'(1)=2, and P"(1)=6

I don't even know where to start..

Answer 1

start by writing the derivatives

Answer 2

\[P'(x)=2ax+b\]
\[P''(x)=2a\]

Answer 3

now you are told that \(P''(1)=6\) which tells you that \(2a=6\) and so \(a=3\)

Answer 4

that makes \(P'(x)=6x+b\) and again you are told that \(P'(1)=2\)
this gives \(P'(1)=2=6\times 1+b\) i.e. \(6+b=2\)
solve for \(b\)
etc