What is the vertex of the parabola 9x^2 - 6x | vedclass

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(A) Given equation: $9x^2 - 6x + 36y + 9 = 0$Rearranging the terms: $9x^2 - 6x = -36y - 9$Divide by $9$: $x^2 - (2/3)x = -4y - 1$Complete the square o

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$(1/3, -2/9)$

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(A) Given equation: $9x^2 - 6x + 36y + 9 = 0$Rearranging the terms: $9x^2 - 6x = -36y - 9$Divide by $9$: $x^2 - (2/3)x = -4y - 1$Complete the square on the left side: $x^2 - (2/3)x + (1/9) = -4y - 1 + (1/9)$$(x - 1/3)^2 = -4y - 8/9$$(x - 1/3)^2 = -4(y + 2/9)$Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,the vertex $(h, k)$ is $(1/3, -2/9)$.

What is the vertex of the parabola $9x^2 - 6x + 36y + 9 = 0$?

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