Vertex of a function

I have no clue how to do this! Please help

Find the vertex of the function
y = 2x^2 + 8x + 1

The vertex of every parabola is defined by the coefficients of the equation. So given ax^2 + bx + c, the vertex is:

(-b/(2a) , c - b^2/(4a) )

In your case a = 2, b = 8, and c = 1. Your vertex is:

(-8/(2*2) , 1 - (8)^2/(4*2)) )

(-8/4 , 1 - 64/8 ) -----> (-2,-7)

The vertex is located at (-2,-7).