Graph the function Identify the vertex and axis of symmetry
f(x)=2x^2 + 4x +1
please explain steps I need to do. I
keep looking at notes and just can't grasp any of it. Failing math not sure how to start again. I think I missed something major.

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  1. as with any quadratic ax^2+bx+c, the vertex lies on the axis of symmetry, at x = -b/2a

    In this case, that is x = -1
    f(-1) = -1
    so the vertex is at (-1,-1)

    Or, you can complete the square to find the vertex:

    f(x) = 2x^2+4x+1
    = 2(x^2+2x)+1
    = 2(x^2+2x+1)+1-2
    = 2(x+1)^2 - 1

    Recall that if y = a(x-h)^2+k then the vertex is at (h,k)

    Here, it is clear that (h,k) = (-1,-1)

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