complete the sqaure to rewrite y=x^2-6x+16 in vertex form. then state whether the vertex is a maximum and give its coordinates!

y = (x-3) ^2 +7

Vertex (3,7)  minimum

Step-by-step explanation:

y=x^2-6x+16

The equation of a parabola in vertex form is

y = a(x - h)2 + k , where (h, k) is the vertex and a tells us where it opens up or down  a>0 opens up and a < 0 opens down

First complete the square

Take the coefficient of the x term, divide by 2 and square it

-6 /2 = -3  

(-3)^2 = 9

Take the 16 and split it into 9 and 7

y=x^2-6x+9 +7

y = (x-3) ^2 +7

The vertex is (3,7) and a=1

Since a =1, this opens up and if it opens up, the vertex is a minimum

(3,7) min

Step-by-step explanation:

y = x² - 6x + 16

Since coefficient of x² is 1, which is greater than 0.. it will have a minimum point.

y = x² - 2(x)(3) + 3² - 3² + 16

y = (x - 3)² - 9 + 16

y = (x - 3)² + 7

Vertex: (3,7)

dd

step-by-step explanation:

dd