For the parabola y^2+6y-2x+5=0:(I) The vertex | vedclass

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(B) Given parabola is $y^2+6y-2x+5=0$. Completing the square for $y$: $y^2+6y+9-9-2x+5=0$ $(y+3)^2-4-2x=0$ $(y+3)^2=2(x+2)$. Comparing with $(y-k)^2=4

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$I$ is true,$II$ is false

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(B) Given parabola is $y^2+6y-2x+5=0$. Completing the square for $y$: $y^2+6y+9-9-2x+5=0$ $(y+3)^2-4-2x=0$ $(y+3)^2=2(x+2)$. Comparing with $(y-k)^2=4a(x-h)$,we get vertex $(h, k) = (-2, -3)$. Thus,statement $I$ is true. For the directrix,$4a=2 \implies a=\frac{1}{2}$. The equation of the directrix is $x = h-a$. $x = -2 - \frac{1}{2} = -2.5$,or $2x+5=0$. Since the given directrix in statement $II$ is $y+3=0$,statement $II$ is false.

For the parabola $y^2+6y-2x+5=0$:
$(I)$ The vertex is $(-2,-3)$
$(II)$ The directrix is $y+3=0$
Which of the following is correct?

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For the parabola $y^2+6y-2x+5=0$:$(I)$ The vertex is $(-2,-3)$$(II)$ The directrix is $y+3=0$Which of the following is correct?