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=[tex]\frac{-4-\sqrt{4(4+C)}}{-2}=2+\sqrt{4+C}[/tex]
Hence the area of the triangle is [tex]AB*AC/2=(4+C)*(2+\sqrt{4+C}-2)/2=(4+C)\sqrt{4+C}/2=32[/tex] hence [tex](4+C)\sqrt{4+C}=64[/tex] .
We remark that [tex]64=16*4=16*\sqrt{16}[/tex]
Hence 4+C=16 thus C=12
