Find the equation of the curve $y''=2x+1$

Question

I am stuck on the last step of the following equation:

If $y''=2x+1$ and there is a stationary point at $(3,2)$ find the equation of the curve.

So far I have the following:

If $y''=2x+1$

$y'=x^2+x+c$

then use the given point to solve for $c$ as follows:

$0=3^2+3+c$

$c=-12$

so, if $y'=x^2+x-12$

$y= \frac{x^3}{3}+\frac{x^2}{2}-12x+c$

now I know to find the final equation of the curve I need to find the last $c$ but am stumped on how to do so? any help would be appreciated!

Answer 1

It would be better to tackle the problem this way: Integrating $y''(x)=2x+1$ twice to obtain \begin{align}y'(x)&=x^2+x+c\\y(x)&=\frac{1}{3}x^3+\frac{1}{2}x^2+cx+d\end{align} From $y'(3)=0$ you get $c=-12$ and from $y(3)=2$ you get $\frac{49}{2}$.