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

vertex = (2, - 16 )

Step-by-step explanation:

The vertex is positioned midway between the zeros

Given

f(x) = (x - 6)(x + 2)

To find the zeros equate f(x) to zero, that is

(x - 6)(x + 2) = 0

Equate each factor to zero and solve for x

x - 6 = 0 ⇒ x = 6

x + 2 = 0 ⇒ x = - 2

Thus

= = = 2

Substitute x = 2 into f(x) for corresponding y- coordinate

f(2) = (2 - 6)(2 + 2) = (- 4)(4) = - 16

vertex = (2, - 16 )

no this is a nonlinear negative

34.5 cm

step-by-step explanation:

generally for a circular segment of angle, θ° and radius r

arc length

= (θ/360) x circumference

= (θ/360) x 2πr

in our case, θ = 60°, π=3.142 and r = 33 cm

substitute into above formula

arc length

= (60/360) x 2(3.142)( 33) = 34.5 cm