What is the vertex of the parabola y^2 + 6x - | vedclass

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(B) Given the equation of the parabola: $y^2 + 6x - 2y + 13 = 0$Rearranging the terms to complete the square for $y$:$y^2 - 2y = -6x - 13$Adding $1$ t

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$(-2, 1)$

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(B) Given the equation of the parabola: $y^2 + 6x - 2y + 13 = 0$Rearranging the terms to complete the square for $y$:$y^2 - 2y = -6x - 13$Adding $1$ to both sides:$y^2 - 2y + 1 = -6x - 13 + 1$$(y - 1)^2 = -6x - 12$$(y - 1)^2 = -6(x + 2)$Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex:Here,$h = -2$ and $k = 1$.Therefore,the vertex is $(-2, 1)$.

What is the vertex of the parabola $y^2 + 6x - 2y + 13 = 0$?

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