Complete the square: g(x)=x^2-16x+64-64+2=(x-8)^2-62. Let y=g(x), then y+62=(x-8)^2. The vertex is at (8,-62) because when x=8, y=-62. If x<8, y+62>0, so y>-62; if x>8, y>-62. Therefore y=-62 is the lowest point, or vertex, of the parabola, because all values of x, apart from x=8, make y greater than -62.