For example, given $\overrightarrow{O A}=(4,3)$, the graph of the function $y=x^{2}+b x+c$ is translated by the vector $\overrightarrow{O A}$ to obtain a graph that is tangent to the line $4 x+y-8=0$ at point $T(1,4)$. Find the original function's equation.
Analysis According to the translation formula, find the coordinates of the point corresponding to the tangent point.
Solution The derivative of the function $y=x^{2}+b x+c$ is $y^{\prime}=2 x+b$. Let the tangent point $T^{\prime}(x, y)$ be translated by the vector $\overrightarrow{O A}=(4,3)$ to $T(1,4)$, then
$$
\left\{\begin{array}{l}
x=1-4=-3 \\
y=4-3=1
\end{array}\right.
$$

So $T^{\prime}(-3,1)$.
Since the slope of the tangent line is -4, the tangent point $T^{\prime}(-3,1)$ lies on the graph of the function $y=x^{2}+b x+c$, so
$$
\left\{\begin{array}{l}
-4=2 \times(-3)+b \\
1=(-3)^{2}+b \cdot(-3)+c .
\end{array}\right.
$$

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