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

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Question

(B) Given the equation of the parabola: $y^2 + 6x - 2y + 13 = 0$. Rearrange the terms to group $y$ variables: $y^2 - 2y = -6x - 13$. Complete the squa

Vedclass · Accepted Answer

$(-2, 1)$

Vedclass · Answer

(B) Given the equation of the parabola: $y^2 + 6x - 2y + 13 = 0$. Rearrange the terms to group $y$ variables: $y^2 - 2y = -6x - 13$. Complete the square for the $y$ terms: $(y^2 - 2y + 1) = -6x - 13 + 1$. This simplifies to: $(y - 1)^2 = -6x - 12$. Factor out $-6$ on the right side: $(y - 1)^2 = -6(x + 2)$. 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 $(-2, 1)$.

Find the vertex of the parabola $y^2 + 6x - 2y + 13 = 0$.

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