Locate the points A and B
$y = y$
$x^2 = 2x + 8$
$x^2 - 2x - 8 = 0$
$x = 4 \text{ and } -2$
For x = 4
$y = 4^2 = 16$
The point is (4, 16)
For x = -2
$y = (-2)^2 = 4$
The point is (-2, 4)
Using the formula for the area polygon by coordinates
$A = \dfrac{1}{2}\left| \begin{matrix} x_1 & x_2 & x_3 & \cdots & x_n & x_1 \\ y_1 & y_2 & y_3 & \cdots & y_n & y_1 \end{matrix} \right|$
Rotate counterclockwise from point A
$A = \dfrac{1}{2}\left| \begin{matrix} x_A & x_C & x_B & x_A \\ y_A & y_C & y_B & y_A \end{matrix} \right|$
$A = \dfrac{1}{2}\left| \begin{matrix} -2 & x & 4 & -2 \\ 4 & y & 16 & 4 \end{matrix} \right|$
$A = \frac{1}{2} \left[ (-2y + 16x + 16) - (4x + 4y - 32) \right]$
$A = \frac{1}{2}(12x - 6y + 48)$
$A = 6x - 3y + 24$
Substitute y = x2 of the parabola
$A = 6x - 3x^2 + 24$
Differentiate then equate to zero to maximize the area
$\dfrac{dA}{dx} = 6 - 6x = 0$
$x = 1$
When x = 1
$y = 1^2 = 1$
Thus, point C is at (1, 1). answer: B