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Answer 1

Area of ABC : AB*AC/2

the maximum of the parabola is reached at x=-4/(2*(-1))=2 hence A is at (2,0) and B is at (2,(-2)^2+4*2+C)=(2,12+C)

C is the second root (x-intersect), which we can find :

determinant1 : D=16-4*(-1)*C=4(4+C) thus the second root is at x=$\frac{-4-\sqrt{4(4+C)}}{-2}=2+\sqrt{4+C}$

Hence the area of the triangle is $AB*AC/2=(4+C)*(2+\sqrt{4+C}-2)/2=(4+C)\sqrt{4+C}/2=32$ hence $(4+C)\sqrt{4+C}=64$ .

We remark that $64=16*4=16*\sqrt{16}$

Hence 4+C=16 thus C=12