4. [2025合肥庐阳区三模]已知抛物线$y = x^2 + bx + c$与$x$轴交于$O(0,0),A(6,0)$两点，顶点为$B$.
(1)求该抛物线的表达式.
(2)如图，$C$点坐标$(3,3)$，$D$为抛物线对称轴上一动点，过点$D$的直线$EF$平行$x$轴交抛物线于$E,F$两点(点$E$在点$F$的左侧).
①若$BD + EF = 8$，求点$E$的坐标.
②若以$EF$为边构造矩形$EFGH(G,H$在线段$AC,OC$上$)$，求该矩形周长的最大值.

(1)由题意可得 $\begin{cases}c = 0,\\36 + 6b + c = 0,\end{cases}$ 解得 $\begin{cases}b = -6,\\c = 0,\end{cases}$
$\therefore$ 抛物线的表达式为 $y = x^2 - 6x$.
(2)$\because y = x^2 - 6x = (x - 3)^2 - 9$, $\therefore$ 顶点 $B$ 的坐标为 $(3,-9)$.
设 $E(m,m^2 - 6m)$, 则 $D(3,m^2 - 6m)$,
$\therefore BD = m^2 - 6m - (-9) = m^2 - 6m + 9, DE = 3 - m$,
$\therefore EF = 2DE = 6 - 2m$.
①$\because BD + EF = 8, \therefore m^2 - 6m + 9 + 6 - 2m = 8$,
解得 $m = 7$(舍去)或 $m = 1, \therefore$ 点 $E$ 的坐标为 $(1,-5)$.
②设直线 $OC$ 的表达式 $y = kx$.
$\because C(3,3), \therefore 3 = 3k$, 解得 $k = 1$,
$\therefore$ 直线 $OC$ 表达式为 $y = x$,
$\therefore H(m,m), \therefore EH = m - (m^2 - 6m) = 7m - m^2$.
设矩形 $EFGH$ 周长为 $C$, 则 $C = 2(EF + EH) = 2(6 - 2m + 7m - m^2) = -2m^2 + 10m + 12 = -2(m - \frac{5}{2})^2 + \frac{49}{2}(0 ＜ m ＜ 6)$,
$\therefore$ 当 $m = \frac{5}{2}$ 时, $C$ 的最大值为 $\frac{49}{2}$.

5. [2026杭州滨江区段考]如图,在平面直角坐标系中,菱形OABC的顶点A在x轴正半轴上,顶点C的坐标为$(4,3)$,D是抛物线$y=-x^2 +6x$上一点,且在x轴上方,则$△ BCD$面积的最大值是多少?

$\because C(4,3), \therefore OC = \sqrt{4^2 + 3^2} = 5$.
$\because$ 四边形 $OABC$ 是菱形, $\therefore BC = OC = 5$.
设 $D(x,-x^2 + 6x)$,
则 $S_{△ BCD} = \frac{1}{2}BC · (y_D - y_C) = \frac{1}{2} × 5 × (-x^2 + 6x - 3) = -\frac{5}{2}(x - 3)^2 + 15$.
$\because -\frac{5}{2} ＜ 0, \therefore S_{△ BCD}$ 有最大值, 为 $15$.