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 $y = x^2 - x + 3$