well, notice, we can pretty much see the x-intercepts if we set f(x) = 0, clearly they're just 6 and -2,
[tex]\bf \stackrel{f(x)}{0}=(x-6)(x+2)\implies
\begin{cases}
0=x-6\implies &6=x\\
0=x+2\implies &-2=x
\end{cases}[/tex]
now, this is a quadratic equation, since it only has 2 solutions at most, recall your fundamental theorem of algebra.
so, the vertex will be right in middle of those two x-intercepts.
what's between -2 and 6?, well is +2 or just 2, so x = 2.
so we know the x-coordinate for the vertex is the halfway point of 2, what's the y-coordinate?
f(x) = y = (2 - 6)(2 + 2)
y = ( - 4)( 4 )
y = -16
thus the vertex is at, well you already know.
