The vertex of the parabola x^2 + 8x + 12y + 4 | vedclass

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Question

(A) Given equation of the parabola is $x^2 + 8x + 12y + 4 = 0$. Completing the square for the $x$ terms: $x^2 + 8x = -12y - 4$ $(x^2 + 8x + 16) = -12y

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$(-4, 1)$

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(A) Given equation of the parabola is $x^2 + 8x + 12y + 4 = 0$. Completing the square for the $x$ terms: $x^2 + 8x = -12y - 4$ $(x^2 + 8x + 16) = -12y - 4 + 16$ $(x + 4)^2 = -12y + 12$ $(x + 4)^2 = -12(y - 1)$ Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,where $(h, k)$ is the vertex: $h = -4$ and $k = 1$. Thus,the vertex is $(-4, 1)$.

The vertex of the parabola $x^2 + 8x + 12y + 4 = 0$ is

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