6.(2025·博山区模拟)如图，抛物线$y=ax^2+bx+c$与$x$轴交于点$A(3,0),B(1,0)$，与$y$轴交于点$C$，且$OA=OC$。
(1)求抛物线的函数解析式；
(2)如图，在直线$AC$上方的抛物线上存在一点$M$，使得$S_{△ AMC}=6$，求出点$M$的坐标；
(3)若$P$是该抛物线上位于直线$AC$下方的一动点，从点$C$沿抛物线向点$A$运动(点$P$与点$A$不重合)，点$D$在抛物线的对称轴上，$Q$是平面内任意一点，当$B,P,D,Q$四点构成的四边形为正方形时，请直接写出点$Q$的坐标。

6. 解:(1)$\because A(3,0)$,$OA=OC$,$\therefore C(0,3)$,
将$A(3,0),B(1,0),C(0,3)$代入$y=ax^2+bx+c$,得
$\begin{cases}9a+3b+c=0,\\a+b+c=0,\\c=3,\end{cases}$
解得$\begin{cases}a=1,\\b=-4,\\c=3,\end{cases}$
$\therefore$抛物线的函数解析式为$y=x^2-4x+3$.
(2)设直线$AC$的函数解析式为$y=mx+n$,
将点A,C的坐标分别代入,得
$\begin{cases}3m+n=0,\\n=3,\end{cases}$
解得$\begin{cases}m=-1,\\n=3,\end{cases}$
$\therefore$直线$AC$的函数解析式为$y=-x+3$.
如答图①,过点M作$MN// y$轴,交AC于点N,
设$M(a,a^2-4a+3)$,
则$N(a,-a+3)$,
$\therefore MN=a^2-4a+3-(-a+3)=a^2-3a$,
$\therefore S_{△ AMC}=S_{△ CMN}-S_{△ AMN}=\dfrac{1}{2}MN· OA$,
即$\dfrac{1}{2}(a^2-3a)× 3=6$,
解得$a=4$或$a=-1$,
$\therefore$点M的坐标为$(4,3)$或$(-1,8)$.
(3)$\because y=x^2-4x+3=(x-2)^2-1$,
$\therefore$抛物线的对称轴为直线$x=2$,顶点坐标为$(2,-1)$.
如答图②,过点P作$PG⊥$对称轴于点G,作$PF⊥ x$轴于点F,过点Q作$QJ⊥ x$轴于点J,
$\therefore ∠ PFB=∠ PGD=90°=∠ FPG$.
设$P(x,x^2-4x+3)$,$D(2,t)$,
$\because$四边形PBQD是正方形,
$\therefore ∠ BPD=90°$,$PB=PD$,
$\therefore ∠ BPF=90°-∠ BPG=∠ GPD$.
在$△ DPG$和$△ BPF$中,$\begin{cases}∠ PGD=∠ PFB,\\∠ GPD=∠ FPB,\\PD=PB,\end{cases}$
$\therefore △ DPG≌△ BPF(\mathrm{AAS})$,$\therefore DG=BF$,$PG=PF$,
$\therefore \begin{cases}1-x=t-x^2+4x-3,\\x^2-4x+3=2-x,\end{cases}$
解得$\begin{cases}x=\dfrac{3-\sqrt{5}}{2},\\t=\sqrt{5}\end{cases}$或$\begin{cases}x=\dfrac{3+\sqrt{5}}{2},\\t=-\sqrt{5}\end{cases}$(舍去),
$\therefore PF=x^2-4x+3=\dfrac{\sqrt{5}+1}{2}$,$BF=1-x=\dfrac{\sqrt{5}-1}{2}$,
同理可得$BJ=PF=\dfrac{\sqrt{5}+1}{2}$,$QJ=BF=\dfrac{\sqrt{5}-1}{2}$,
$\therefore OJ=1+\dfrac{\sqrt{5}+1}{2}=\dfrac{\sqrt{5}+3}{2}$,$\therefore Q(\dfrac{\sqrt{5}+3}{2},\dfrac{\sqrt{5}-1}{2})$.
如答图③,过点P作$PF⊥ x$轴于点F,$PG⊥$对称轴于点G,过点Q作$QJ⊥ x$轴于点J,
同理可得$△ PBF≌△ PDG$,
$\therefore PF=PG$,$BF=DG$,$\therefore \begin{cases}-x^2+4x-3=x-2,\\x-1=x^2-4x+3-t,\end{cases}$
解得$\begin{cases}x=\dfrac{3-\sqrt{5}}{2},\\t=\sqrt{5}\end{cases}$(舍去)或$\begin{cases}x=\dfrac{3+\sqrt{5}}{2},\\t=-\sqrt{5},\end{cases}$
$\therefore P(\dfrac{3+\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2})$,
同理可得$PF=BJ=\dfrac{\sqrt{5}-1}{2}$,$QJ=BF=x-1=\dfrac{\sqrt{5}+1}{2}$,
$\therefore OJ=1-BJ=1-\dfrac{\sqrt{5}-1}{2}=\dfrac{3-\sqrt{5}}{2}$,
$\therefore Q(\dfrac{3-\sqrt{5}}{2},\dfrac{-1+\sqrt{5}}{2})$;
如答图④,当P为抛物线的顶点$(2,-1)$,点Q,A重合时,记对称轴与x轴的交点为H,
此时$D(2,1)$,$Q(3,0)$,
$\therefore HB=HQ=HP=HD$,且$DP⊥ AB$,
此时四边形PBDQ是正方形,$\therefore Q(3,0)$;
如答图⑤,当P为抛物线的顶点$(2,-1)$,D为对称轴与x轴的交点,
此时,$D(2,0)$,$P(2,-1)$,$\therefore Q(1,-1)$;
如答图⑥,过点P作$PF⊥ x$轴于点F,
易证$△ PFB≌△ BED$,$\therefore PF=BE$,
$\therefore x^2-4x+3=2-1$,$\therefore x^2-4x+2=0$,
解得$x=2-\sqrt{2}$或$x=2+\sqrt{2}$(舍去),
$\therefore DE=BF=1-x=\sqrt{2}-1$,
$\therefore D(2,\sqrt{2}-1)$,易得$Q(3-\sqrt{2},\sqrt{2})$;
如答图⑦,过点P作$PG⊥$对称轴于点G,过点B作$BF⊥ PG$交GP的延长线于点F,易证$△ BFP≌△ PGD$,
$\therefore PF=DG$,$BF=PG$,
$\therefore -x^2+4x-3=2-x$,
解得$x=\dfrac{5-\sqrt{5}}{2}$或$x=\dfrac{5+\sqrt{5}}{2}$(舍去),
$\therefore DG=PF=x-1=\dfrac{3-\sqrt{5}}{2}$,
$\therefore P(\dfrac{5-\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2})$,$D(2,2-\sqrt{5})$,
易得$Q(\dfrac{1+\sqrt{5}}{2},\dfrac{3-\sqrt{5}}{2})$.
综上所述,点Q的坐标为$(\dfrac{\sqrt{5}+3}{2},\dfrac{\sqrt{5}-1}{2})$或$(\dfrac{3-\sqrt{5}}{2},\dfrac{-1+\sqrt{5}}{2})$或$(3,0)$或$(1,-1)$或$(3-\sqrt{2},\sqrt{2})$或$(\dfrac{1+\sqrt{5}}{2},\dfrac{3-\sqrt{5}}{2})$.

7.(2025·崇川区期末)已知二次函数$y=\frac{1}{2}mx^2 + 2x - 2m + 4$（$m$为常数，且$m≠0$）.
(1)当$m=1$时，求该二次函数的图象的顶点坐标；
(2)直线$y=2x+b$与该二次函数的图象交于$A(x_1,y_1),B(x_2,y_2)$两点，若当$m＞0$时，有$x_1＜-2,x_2＞2$，求$b$的取值范围；
(3)顺次连接$A(-2,0),B(8,0),C(8,12),D(-2,12)$，得到矩形$ABCD$，若该二次函数的图象与矩形$ABCD$有三个公共点，请直接写出$m$的取值范围.

7. 解:(1)当$m=1$时,$y=\dfrac{1}{2}x^2+2x+2=\dfrac{1}{2}(x+2)^2$,
$\therefore$该二次函数的图象的顶点坐标为$(-2,0)$.
(2)联立,得$\dfrac{1}{2}mx^2+2x-2m+4=2x+b$,
整理,得$mx^2-4m-2b+8=0$,
则$x_1+x_2=0$,$x_1x_2=-4-\dfrac{2b}{m}+\dfrac{8}{m}$.
$\because x_1＜-2$,$x_2＞2$,$\therefore x_1-2＜-4＜0$,$x_2-2＞0$,
$\therefore (x_1-2)(x_2-2)＜0$,$\therefore x_1x_2-2(x_1+x_2)+4＜0$,
$\therefore -\dfrac{2b}{m}+\dfrac{8}{m}＜0$.
$\because m＞0$,$\therefore b＞4$.
(3)$\because y=\dfrac{1}{2}mx^2+2x-2m+4=\dfrac{1}{2}m(x+\dfrac{2}{m})^2+4-2(m+\dfrac{1}{m})$,
$\therefore$抛物线的顶点坐标为$(-\dfrac{2}{m},4-2m-\dfrac{2}{m})$,
当$m＞0$时,$(\sqrt{m}-\dfrac{1}{\sqrt{m}})^2=m-2+\dfrac{1}{m}\ge0$,
$\therefore m+\dfrac{1}{m}\ge2$,
$\therefore 4-2(m+\dfrac{1}{m})\le0$,
当$x=-2$时,$y=\dfrac{1}{2}m(-2+\dfrac{2}{m})^2+4-2(m+\dfrac{1}{m})=0$,
$\therefore$抛物线$y=\dfrac{1}{2}m(x+\dfrac{2}{m})^2+4-2(m+\dfrac{1}{m})$始终经过点$A(-2,0)$.
当$m=\dfrac{1}{m}$,即$m=1$时,$4-2(m+\dfrac{1}{m})=0$,
此时,抛物线的顶点坐标为$(-2,0)$,抛物线与矩形ABCD有2个公共点,如答图①,

当$0＜m＜1$时,该抛物线与矩形ABCD有2个公共点,如答图②;
当$m＞1$时,抛物线$y=\dfrac{1}{2}mx^2+2x-2m+4$与矩形ABCD恰好有3个公共点,如答图③.
当$m＜0$时,当抛物线$y=\dfrac{1}{2}mx^2+2x-2m+4$经过点$C(8,12)$时,$\dfrac{1}{2}m×64+2×8-2m+4=12$,解得$m=-\dfrac{4}{15}$,
此时抛物线$y=\dfrac{1}{2}mx^2+2x-2m+4$与矩形ABCD恰好有3个公共点,如答图④;
当$4-2(m+\dfrac{1}{m})=12$时,$m=-2\pm\sqrt{3}$,
当$m=-2+\sqrt{3}$时,抛物线过点$(8,30\sqrt{3}-40)$,
此时抛物线$y=\dfrac{1}{2}mx^2+2x-2m+4$与矩形ABCD恰好有3个公共点,如答图⑤,

当$m=-2-\sqrt{3}$时,抛物线过点$(8,-30\sqrt{3}-40)$,此时抛物线与矩形ABCD有2个交点,不符合题意,舍去.
综上所述,该二次函数的图象与矩形ABCD有三个公共点时,$m$的取值范围为$m＞1$或$m=-2+\sqrt{3}$或$m=-\dfrac{4}{15}$.