- Let $f(x) = k_1x^2 + k_2$, where $k_1$ and $k_2$ are non-zero constants.
$\begin{array}{rcl}
f(2) & = & 59 \\
k_1(2)^2 + k_2 & = & 59 \\
4k_1 + k_2 & = & 59 \ldots \unicode{x2460}
\end{array}$Also,
$\begin{array}{rcl}
f(7) & = & -121 \\
k_1 (7)^2 +k_2 & = & -121 \\
49k_1 +k_2 & = & -121 \ldots \unicode{x2461}
\end{array}$$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
45k_1 & = & -180 \\
k_1 & = & -4
\end{array}$Sub. $k_1=-4$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
4(-4) + k_2 & = & 59 \\
k_2 & = & 75
\end{array}$$\therefore f(x)= -4x^2 + 75$. Hence, we have
$\begin{array}{rcl}
f(6) & = & -4(6)^2 + 75\\
& = & -69
\end{array}$ - By the result of (a), $a=-69$. Consider the graph $y=-4x^2 + 75$. Note that the graph is symmetric about the $y$-axis. Therefore, $a=b=-69$. Hence, the length of the base of $\Delta ABC$
$\begin{array}{cl}
= & 2 \times 69 \\
= & 138
\end{array}$Note that $AB$ is a horizontal line. Therefore, the height of $\Delta ABC$
$\begin{array}{cl}
= & 6- 0 \\
= & 6
\end{array}$Hence, the area of $\Delta ABC$
$\begin{array}{cl}
= & \dfrac{1}{2} \times 138 \times 6 \\
= & 414
\end{array}$
2014-I-13
Ans: (a) $-69$ (b) $414$
