For the parabola y=x^2-3x+2,match the items i | vedclass

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(A) The given parabola is $y = x^2 - 3x + 2$. Rewriting it as $(x - \frac{3}{2})^2 = y + \frac{1}{4}$.Comparing with $(x-h)^2 = 4a(y-k)$,we get $h = \

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$A-I, B-II, C-III, D-IV$

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(A) The given parabola is $y = x^2 - 3x + 2$. Rewriting it as $(x - \frac{3}{2})^2 = y + \frac{1}{4}$.Comparing with $(x-h)^2 = 4a(y-k)$,we get $h = \frac{3}{2}$,$k = -\frac{1}{4}$,and $4a = 1 \implies a = \frac{1}{4}$.$1$. $Q$ is the vertex,which is $(h, k) = (\frac{3}{2}, -\frac{1}{4})$. Thus,$B-II$.$2$. $S$ is the focus $(h, k+a) = (\frac{3}{2}, -\frac{1}{4} + \frac{1}{4}) = (\frac{3}{2}, 0)$. Thus,$C-III$.$3$. $Z$ is the intersection of the axis $(x = \frac{3}{2})$ and directrix $(y = k-a = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2})$,so $Z = (\frac{3}{2}, -\frac{1}{2})$. Thus,$D-IV$.$4$. $P$ is an end point of the latus rectum $(h \pm 2a, k+a) = (\frac{3}{2} \pm \frac{1}{2}, 0)$. For $(2, 0)$,we have $A-I$.Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

$A$. $P$	$I$. $(2,0)$
$B$. $Q$	$II$. $(\frac{3}{2}, -\frac{1}{4})$
$C$. $S$	$III$. $(\frac{3}{2}, 0)$
$D$. $Z$	$IV$. $(\frac{3}{2}, -\frac{1}{2})$
	$V$. $(0, \frac{3}{2})$

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