f(x)=(x-4)√(x^2-8x+32).
f'(x)=(x-4)(2x-8)/2√x^2-8x+32)+√(x^2-8x+32) = ((x-4)^2+x^2-8x+32)/√(x^2-8x+32)=
2(x-4)^2/√(x^2-8x+32).
When f'(x)=0, there is zero gradient marking a turning point.
f'(x)=0 implies (x-4)^2=0 so x=4.
f(4)=0, so there is a turning point at (4,0).
f"(x)=((4x-16)√(x^2-8x+32)-(x-4)^2/√(x^2-8x+32))/(x^2-8x+32)
Clearly at x=4, f"(4)=0, indicating a point of inflexion.
To find the min and max we need to evaluate f(-3)=(-7)√(9+24+32)=-7√65 and f(11)=7√(121-88+32)=7√65.
Max=56.44 and min=-56.44 approx. The line of symmetry is x=4, the point of inflexion (4,0).