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