One way to do it is to expand the quadratic, then complete the square to write it in vertex form:
[tex](x-4)(x+2)=x^2-2x-8=(x-1)^2-9[/tex]
Then we get the vertex right away, (1, -9).
Alternatively, if you know about the parity/symmetry of parabolas, you know that the vertex lies on a line between its roots. In this case, we know [tex]x=4[/tex] and [tex]x=-2[/tex] are the roots to this quadratic. The line [tex]x=1[/tex] falls in the middle of these two points (if you're unsure as to why, take the average of the roots: (4 - 2)/2 = 1). So we know the [tex]x[/tex]-coordinate of the vertex, and the [tex]y[/tex]-coordinate is [tex]f(1)=(1-4)(1+2)=-9[/tex], so we again get (1, -9).
