Given that the image of quadratic function y = x2 + BX + C is shifted 2 unit lengths to the left and 3 unit lengths to the up, we get the quadratic function y = x2-8x + 10. (1) find the values of B and C; (2) if the intersection of the function in question (1) and the X axis is a and B, try to determine a point P on the image below the X axis, so that the area of △ PAB is the largest. Can you find the area of △ PAB?

(1) Y = x2-8x + 10 = (x-4) 2-6, translate the parabola y = (x-4) 2-6 two unit lengths to the right and three unit lengths to the down, and the analytical formula of the parabola is y = (X-6) 2-9 = x2-12x + 27, so B = - 12, C = 27; (2) solve the equation x2-12x + 27 = 0, then X1 = 3, X2 = 9, so AB = 9-3 = 6

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