2011-I-11

For $f(-2)=28$, we have

$\begin{array}{rcl}
f(-2) & = & k_1(-2)^2+k_2(-2) \\
28 & = & 4k_1-2k_2 \\
14 & = & 2k_1-k_2 ~\ldots \unicode{x2460}
\end{array}$

For $f(6)=-36$, we have

$\begin{array}{rcl}
f(6) & = & k_1(6)^2 +k_2(6) \\
-36 & = & 36k_1 + 6k_2 \\
-6 & = & 6k_1 + k_2~\ldots \unicode{x2461}
\end{array}$

$\unicode{x2460} + \unicode{x2461}$, we have

$\begin{array}{rcl}
8 & = & 8k_1 \\
k_1 & = & 1
\end{array}$

Sub. $k_1=1$ into $\unicode{x2461}$, we have

$\begin{array}{rcl}
-6 & = & 6(1) +k_2 \\
k_2 & = & -12
\end{array}$

Therefore, $f(x)= x^2 -12x$.

1. Note that the coordinates of the vertex are $(6,k)$.

   Hence by the result of (a), we have

   $\begin{array}{rcl}
   k & = & \dfrac{4(1)(0)-(12)^2}{4(1)} \\
   & = & -36
   \end{array}$

2. Sub. $x=10$ into $y=3(x-6)^2-36$, we have

   $\begin{array}{rcl}
   y & = & 3(10-6)^2 -36 \\
   & = & 12
   \end{array}$

   Therefore, the coordinates of $A$ are $(10,12)$.

   Sub. $x=10$ into $f(x)=x^2-12x$, we have

   $\begin{array}{rcl}
   y & = & 10^2-12(10) \\
   & = & -20
   \end{array}$

   Therefore, the coordinates of $D$ are $(10,-20)$.

   Sub. $y=-20$ into $f(x)=x^2-12x$, we have

   $\begin{array}{rcl}
   -20 & = & x^2-12x \\
   x^2-12x+20 & = & 0 \\
   (x-10)(x-2) & = & 0
   \end{array}$

   Therefore, $x=2$ or $x=10$. Hence, the coordinates of $C$ are $(2,-20)$. Therefore, we have

   $\begin{array}{rcl}
   AD & = & 12- (-20) \\
   & = & 32
   \end{array}$

   and

   $\begin{array}{rcl}
   CD & = & 10 – 2 \\
   & = & 8
   \end{array}$

   Hence, the area of the rectangle $ABCD$

   $\begin{array}{cl}
   = & 8 \times 32 \\
   = & 256
   \end{array}$