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(A) Since the axis of the parabola is parallel to the $X$-axis,the equation of the parabola is of the form $x = ay^2 + by + c$  $(i)$.Given that the p

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$\left(\frac{5}{2}, 0\right)$

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(A) Since the axis of the parabola is parallel to the $X$-axis,the equation of the parabola is of the form $x = ay^2 + by + c$  $(i)$.Given that the points $(0, -1)$,$(6, 1)$,and $(-2, -3)$ lie on the parabola,we have:$0 = a - b + c$  $(ii)$$6 = a + b + c$  $(iii)$$-2 = 9a - 3b + c$  $(iv)$Subtracting $(ii)$ from $(iii)$,we get $2b = 6$,so $b = 3$.Substituting $b = 3$ into $(ii)$ and $(iii)$,we get $a + c = 3$ and $a + c = 3$ (consistent).Substituting $b = 3$ into $(iv)$,we get $-2 = 9a - 9 + c$,so $9a + c = 7$.Solving $a + c = 3$ and $9a + c = 7$,we subtract the first from the second to get $8a = 4$,so $a = \frac{1}{2}$.Then $c = 3 - \frac{1}{2} = \frac{5}{2}$.The equation is $x = \frac{1}{2}y^2 + 3y + \frac{5}{2}$.To find the point where it cuts the $X$-axis,set $y = 0$:$x = \frac{1}{2}(0)^2 + 3(0) + \frac{5}{2} = \frac{5}{2}$.Thus,the point is $\left(\frac{5}{2}, 0\right)$.

If a parabola having its axis parallel to the $X$-axis passes through the points $(0, -1)$,$(6, 1)$,and $(-2, -3)$,then the point at which this parabola cuts the $X$-axis is

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