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Graph the function. Identify the vertex and axis of symmetry. f(x)= 2x^2+ 2x- 1
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In this case the axis of symmetry is the line along the vertex because this is a parabola. We can tell that it is a parabola because it is a polynomial of degree 2. Since the sign on the highest degree term is positive then we can tell that opens upward. I find the vertex by completing the square so that you get \[2(x^2+1-1/2)\] completinging the square gives you \[2((x+1/2)^2 -1/2+1/2)\] Distrubute and simplify \[2(x+1/2)^2\] the general form of a parabola is \[a(x-h)^2 + k\] where the vertex is (h,k) so the vertex in this problem is (-1/2,0) where the axis of symmetry is x=-1/2
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