Calculus 3 - by Vlad Vasilescu (W)

Since $f(x)$ passes through $(2,8)$ , this means that:

$\begin{aligned} f(2) & = 8 \\ \frac {2^2+2a+b}{2-1} & = 8 \\ 4 + 2a+b & = 8 \\ \implies 2a+b & = 4 & ...(1) \end{aligned}$

Since the tangent of $f(x)$ at $x=2$ is parallel with $y = -3x+1$ , this means that:

$\begin{aligned} \frac {df(x)}{dx} \bigg|_{x=2} & = -3 \\ \frac {(2x+a)(x-1)-(x^2+ax+b)(1)}{(x-1)^2} \bigg|_{x=2} & = -3 \\ \frac {(4+a)(1)-(4+2a+b)}1 & = - 3 \\ -a-b & = - 3 \\ \implies a+b & = 3 & ...(2) \end{aligned}$

$\begin{aligned} (1)-(2): \quad a + 0 & = 4-3 \\ \implies a & = 1 \\ (2): \quad 1+ b & = 3 \\ \implies b & = 2 \end{aligned}$

Therefore, $\overline{ab} = \boxed{12}$ .