The equation of the axis of the parabola 2x^2 | vedclass

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(A) Given equation of the parabola is $2x^2 + 5y - 3x + 4 = 0$. Rearranging the terms,we get $2x^2 - 3x = -5y - 4$. Dividing by $2$,we have $x^2 - \fr

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$x = \frac{3}{4}$

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(A) Given equation of the parabola is $2x^2 + 5y - 3x + 4 = 0$. Rearranging the terms,we get $2x^2 - 3x = -5y - 4$. Dividing by $2$,we have $x^2 - \frac{3}{2}x = -\frac{5}{2}y - 2$. Completing the square on the left side: $x^2 - \frac{3}{2}x + (\frac{3}{4})^2 = -\frac{5}{2}y - 2 + \frac{9}{16}$. $(x - \frac{3}{4})^2 = -\frac{5}{2}y - \frac{23}{16}$. The axis of a parabola of the form $(x - h)^2 = 4a(y - k)$ is given by $x - h = 0$. Therefore,the equation of the axis is $x - \frac{3}{4} = 0$,which implies $x = \frac{3}{4}$.

The equation of the axis of the parabola $2x^2 + 5y - 3x + 4 = 0$ is

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