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).

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

college