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

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).

50- 250 is correct i did that one too

step-by-step explanation:

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)