Ans: A
It is given that $f(x)=(x+h)(x-3) +k$.
$\begin{array}{rcl}
f(0) & = & 1 \\
(0+h)(0-3) +k & = & 1 \\
-3h +k & = & 1 \ldots \unicode{x2460}
\end{array}$
$\begin{array}{rcl}
f(8) & = & 1 \\
(8+h)(8-3) +k & = & 1 \\
40 +5h +k & = & 1 \\
5h +k & = & -39 \ldots \unicode{x2461}
\end{array}$
$\unicode{x2461} -\unicode{x2460}$, we have
$\begin{array}{rcl}
8h & = & -40 \\
h & = & -5
\end{array}$
Sub. $h = -5$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
-3(-5) +k & = & 1 \\
15 +k & = & 1 \\
k & = & -14
\end{array}$
