The vertex of the parabola $y^{2} + 6 x - 2 y + 13 = 0$ is

(1, - 1)

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(- 2, 1)

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(\frac{3}{2}, 1)

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(- \frac{7}{2}, 1)

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Solution

The correct option is B $(- 2, 1)$
Given equation can be rewritten as
$(y - 1)^{2} = - 6 (x + 2)$
$∴$ Vertex is $(- 2, 1)$

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Similar questions

Solve for x and y:

$\frac{1}{2 x} + \frac{1}{3 y} = 2, \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6} (x \neq 0, y \neq 0) .$

(- 2, 1)

y^{2} - 4 x - 2 y - 3 = 0

(2, 1)

y^{2} = 6 x

(2, 1)

y^{2} - 4 x - 2 y - 3 = 0

(1, - 1)

(2, 1)

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