Find the vertex of y=x^2+6x+2

Question

I need help!!!!!!

Answer 1

To do this, we need to complete the square.

For a complete square, the constant coefficient must be (6/2)² = 9. We get

y=x² + 6x + 2 = x² + 6x + 9 - 7 = (x+3)² - 7

This is of the vertex form y = a(x - h)² + k with vertex at (h,k) = (-3,-7)