What is the vertex of the parabola x^2 - 8y - | vedclass

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(B) The given equation of the parabola is $x^2 - x - 8y + 19 = 0$.Completing the square for $x$:$x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}

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$\left( \frac{1}{2}, \frac{75}{32} \right)$

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(B) The given equation of the parabola is $x^2 - x - 8y + 19 = 0$.Completing the square for $x$:$x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 - 8y + 19 = 0$$\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - 8y + 19 = 0$$\left(x - \frac{1}{2}\right)^2 = 8y - 19 + \frac{1}{4}$$\left(x - \frac{1}{2}\right)^2 = 8y - \frac{75}{4}$$\left(x - \frac{1}{2}\right)^2 = 8\left(y - \frac{75}{32}\right)$Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,the vertex $(h, k)$ is $\left(\frac{1}{2}, \frac{75}{32}\right)$.

What is the vertex of the parabola $x^2 - 8y - x + 19 = 0$?

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