8. Given $O A=(4,3)$, the graph of the function $y=x^{2}+b x+c$ is translated by the vector $O A$ to obtain a graph that is tangent to the line $4 x+y-8=0$ at point $T(1,4)$. Then the original function is ( ).
(A) $y=x^{2}+2 x+1$
(B) $y=x^{2}+2 x$
(C) $y=x^{2}+2 x+2$
(D) $y=x^{2}+2 x-2$
8.D.

The derivative of the function $y=x^{2}+b x+c$ is $y^{\prime}=2 x+b$. If the point of tangency $T^{\prime}(x, y)$ is translated by the vector $O A=(4,3)$ to $T(1,4)$, then $x=1-4=-3, y=4-3=1$, which means $T^{\prime}(-3,1)$.
Given that the slope of the tangent line is -4, and the point of tangency $T^{\prime}(-3,1)$ lies on the graph of the function $y=x^{2}+b x+c$, we have:
$$
\left\{\begin{array}{l}
-4=2 \times(-3)+b, \\
1=(-3)^{2}+b(-3)+c .
\end{array}\right.
$$

Solving these equations gives $b=2, c=-2$. Therefore, $y=x^{2}+2 x-2$.