F(x)=x^2-6x+11 how can you find the vertex of this quadratic function?

(3,2)

Step-by-step explanation:

a = 1

b = 6

c = 11

Vertex of x = 6/(2*1)

6/(2*1) = 3

Back to f(x)=x^2-6x+11

Substitute x with 3

3^2-6(3)+11

9-18+11 = 2

y = 2

(3,2)

Step-by-step explanation:

Vertex of x-coordinate = -b/2a

b = -6 and a = 1

Vertex of x-coordinate = -(-6)/(2×1)

= 6/2

= 3

Vertex of y-coordinate = f(3)

= 3^2 - 6(3) + 11

= 9 - 18 + 11

= 2

answer: translate the circles so they share a common center point and dilate circle y by a scale factor of 3.