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Given the equation of the parabola
[tex]y=x^2+2x-1[/tex]which can be written as
[tex]\begin{gathered} y=x^2+2x+1-1-1 \\ =(x+1)^2-2 \end{gathered}[/tex]Comparing with
[tex]y=a(x-h)^2+k[/tex]gives a = 1, h = -1 and k = -2.
The vertex is (h, k) = (-1, -2).