Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 83

Solve for a second degree polynomial $F$ such that $F(2)=5$, $F'(2)$ and $F''(2) = 2$

The general equation for a second degree polynomial is $y = ax^2 + bx + c$ and we are
asked to solve for the numerical coefficients $a,b$ and $c$.

Equation 1:

$\begin{equation} \begin{aligned} F(2) &= 5\\ \\ 5 &= a(2)^2 + 2b + c\\ \\ 5 &= 4a + 2b +c \end{aligned} \end{equation}$


Equation 2:

$\begin{equation} \begin{aligned} F'(2) &= 3\\ \\ y &= ax^2 + bx + c\\ \\ y' &= 2ax + b\\ \\ 3 &= 2a(2) + b\\ \\ 3 &= 4a +b \end{aligned} \end{equation}$


Equation 3:

$\begin{equation} \begin{aligned} F''(2) &= 2\\ \\ y &= ax^2 + bx + c\\ \\ y' &= 2ax + b\\ \\ y'' &= 2a\\ \\ 2 &= 2a\\ \\ a &= 1 \end{aligned} \end{equation}$


Using the Equations 1, 2 and 3. We can now get the numerical coefficients of the equation.


$\begin{array}{l} 4a + 2b + c & = 5\\ 4a + b & = 3\\ a & = 1\\ \hline\\ a = 1\\ b = -1\\ c = 3 \end{array}$


Therefore, the required equation is $y = x^2 - x + 3$