Find the vertex of the following parabola. write your answer in the form: (x, y) with no spaces. (so a vertex of 2,1 would be written as (2,show all your work.

Vertex (-3,1)

Step-by-step explanation:

y = x^2 + 6x + 10          Put brackets around the first two terms

y = (x^2 + 6x) + 10        Take 1/2 the linear term (6) and square it. Put that inside  the brackets. Always add.

y = (x^2 + 6x + (6/2)^2 ) + 10 Since you added in side the brackets, subtract after the 10

y = (x^2 + 6x + 9) + 10 - 9     The 3 terms inside the brackets are a perfect square. Combine the two terms outside the brackets.

y = (x + 3)^2 + 1

The vertex is at (the value that makes x + 3 go to zero, and the number outside the brackets)

(-3,1)

Check

The check is to ask Desmos to graph the equation. That graph is shown below. Notice that the lowest point is(-3,1). Notice that the graph and my answer do agree.

expression #3 is not equivalent. the others are.