Question

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

Answer 1

The zeros are where the factors are zero, at x=-2 and x=6. The line of symmetry and the x-value of the vertex is halfway between these at (-2+6)/2 = 2. Then the y-value of the vertex is

... f(2) = (2-6)(2+2) = -16

The vertex is (x, y) = (2, -16).