Question 613572
How do I find the vertex for 
x^2+6x+1?
First, put equation in standard form by completing the square
y=(x^2+6x+9)+1-9
y=(x+3)^2-8
This is an equation of a parabola that opens upwards.
Its standard form: (x-h)^2+k, (h,k)=(x,y) coordinates of the vertex
For given equation: x^2+6x+1
vertex: (-3,-8)