Find the equation of the curve I've been given the following math question; however, I don't understand the wording of what it's asking.
if $ \dfrac{d^2y}{dx^2} = y''= \dfrac{dy'}{dx} = 3x^2 + x $ for any point on a curve and the tangent line at $ (1,-2) $ is $ y = 2 - 4x $, find the equation of the curve.

We integrate $y''$ to get $y'$. So, $y'=x^3+\frac{1}{2}x^2+c_{1}$, where $c_{1}$ is the constant of integration (to be found).
Recall that $y'$ is the "slope function", $m$, of the tangent line. In this case, $m=-4$. That is, $y'(x=1)=-4$. We use this condition to find the value of $c_{1}$: $$y'(1)=1+\frac{1}{2}+c_{1}=-4\quad\Rightarrow\quad c_{1}=-\frac{11}{2}$$
So, $y'=x^3+\frac{1}{2}x^2-\frac{11}{2}$. Now integrate $y'$ to get $y$ (which is the equation of the curve). So, $y=\frac{1}{4}x^4+\frac{1}{6}x^3-\frac{11}{2}x+c_{2}$, where $c_{2}$ is the constant of integration (to be found).
Recall that we have a point that lies on the curve, namely, $(x,y)=(1,-2)$. This means that $y(x=1)=-2$. So, $$y(1)=\frac{1}{4}+\frac{1}{6}-\frac{11}{2}+c_{2}=-2\quad\Rightarrow\quad c_{2}=\frac{37}{12}$$
So, $y=\frac{1}{4}x^4+\frac{1}{6}x^3-\frac{11}{2}x+\frac{37}{12}$.
