Those extremas..

Firstly we can obtain one equation of $a$ and $b$ by plugging in the given point: $2 = a(-1)^3+b(-1)^2-x+2 \\ 1 = -a+b.$ We can find another equation of $a$ and $b$ by realizing that the point $(-1,0)$ is on the derivative: $\begin{aligned} \frac{dy}{dx} =& 3ax^2 + 2bx+1 \\ 0 = & 3a (-1)^2 +2b(-1) +1 \\ -1 = & 3a-2b. \end{aligned}$ Now we have the system formed by $\begin{aligned} 1 = & -a +b \\ -1 = & 3a -2b. \end{aligned}$ Which has the solution point $(a,b) = (1,2)$ . So we have $\frac{a}{b} = \boxed{\frac{1}{2}}$ .