Question

What is the vertex of the quadratic function f(x) = (x-6) (x+2)

Answer 1

well, notice, we can pretty much see the x-intercepts if we set f(x) = 0, clearly they're just 6 and -2,


`\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}`

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.