3. (2026·济南期末)如图,抛物线$y=-\dfrac{3}{4}x^2 + bx + c$经过点$C(0,3)$,交$x$轴于点$A,B$(点$A$在点$B$左侧),点$A(-1,0)$,连接$BC$,直线$y=kx+2(k＞0)$与$y$轴交于点$D$,与$BC$上方的抛物线交于点$E$,与$BC$交于点$F$.
(1)求抛物线的表达式.
(2)抛物线上是否存在点$E$使得$\dfrac{EF}{DF}=3$? 若存在,请求出此时点$E$的坐标;若不存在,请说明理由.
(3)第一象限内抛物线上是否存在一点$P$,使得$△ BCO$中有一个锐角与$∠ PCB$相等? 若存在,请直接写出点$P$的横坐标;若不存在,请说明理由.

备用图

3. (1)将$C(0,3)$和$A(-1,0)$代入$y=-\frac{3}{4}x^2+bx+c$,得$\begin{cases}c=3,\\-\frac{3}{4}-b+c=0,\end{cases}$解得$\begin{cases}b=\frac{9}{4},\\c=3,\end{cases}$ $\therefore$ 抛物线的表达式为$y=-\frac{3}{4}x^2+\frac{9}{4}x+3$.
(2)存在,$E(2,\frac{9}{2})$,理由如下:
由题意,点$E$在$y$轴的右侧,作$EG// y$轴,交$BC$于点$G$,$\because EG// y$轴,$\therefore △ EGF ∽ △ DCF$,$\therefore \frac{EF}{DF}=\frac{EG}{CD}$.$\because$ 直线$y=kx+2\ (k＞0)$与$y$轴交于点$D$,$\therefore D(0,2)$,$\therefore CD=3-2=1$,$\therefore \frac{EF}{DF}=EG=3$.令$y=0$,得$-\frac{3}{4}x^2+\frac{9}{4}x+3=0$,解得$x_1=-1$,$x_2=4$,$\therefore A(-1,0)$,$B(4,0)$.设$BC$所在直线的表达式为$y=mx+n\ (m≠0)$,将$B(4,0)$,$C(0,3)$代入上述表达式得$\begin{cases}0=4m+n,\\3=n,\end{cases}$解得$\begin{cases}m=-\frac{3}{4},\\n=3,\end{cases}$$\therefore BC$的表达式为$y=-\frac{3}{4}x+3$.设$E(t,-\frac{3}{4}t^2+\frac{9}{4}t+3)$,则$G(t,-\frac{3}{4}t+3)$,其中$0＜t＜4$,$\therefore EG=-\frac{3}{4}t^2+\frac{9}{4}t+3-(-\frac{3}{4}t+3)=-\frac{3}{4}(t-2)^2+3=3$,解得$t_1=t_2=2$,$\therefore -\frac{3}{4}t^2+\frac{9}{4}t+3=-\frac{3}{4}×2^2+\frac{9}{4}×2+3=\frac{9}{2}$,$\therefore$ 点$E$的坐标为$(2,\frac{9}{2})$.
(3)存在,点$P$的横坐标为$\frac{47}{18}$或3. 解析:①当$∠ PCB = ∠ OCB$时,,过点$O$作$ON ⊥ BC$于点$N$,延长$ON$至$H$,使$NH=NO$,连接$CH$交抛物线于点$P$,过点$H$作$HT ⊥ x$轴于点$T$,$\because NH=NO$,$ON ⊥ BC$,$\therefore BC$是$OH$的垂直平分线,$\therefore CH=CO=3$,$\therefore ∠ PCB = ∠ OCB$,$\therefore$ 点$P$为所求点.在$\mathrm{Rt}△ BCO$中,$BC=\sqrt{OC^2+BO^2}=5$,$\because ∠ OCN = ∠ BCO$,$∠ CNO = ∠ COB$,$\therefore △ OCN ∽ △ BCO$,$\therefore \frac{OC}{BC}=\frac{ON}{OB}$,$\therefore \frac{3}{5}=\frac{ON}{4}$,$\therefore ON=\frac{12}{5}$,$\therefore OH=2ON=\frac{24}{5}$.$\because ∠ CON + ∠ OCN = 90°$,$∠ NOB + ∠ CON = 90°$,$\therefore ∠ OCN = ∠ NOB = ∠ PCB = ∠ OCB$,$△ OCN ∽ △ HOT$,$\therefore \frac{OC}{HO}=\frac{ON}{HT}$,$\therefore \frac{3}{\frac{24}{5}}=\frac{\frac{12}{5}}{HT}$,$\therefore HT=\frac{96}{25}$,$\therefore OT=\sqrt{OH^2-HT^2}=\frac{72}{25}$,$\therefore H(\frac{72}{25},\frac{96}{25})$.设直线$CH$的表达式为$y=m_1x+n_1$,把$H(\frac{72}{25},\frac{96}{25})$,$C(0,3)$代入,得$\begin{cases}n_1=3,\\\frac{72}{25}m_1+n_1=\frac{96}{25},\end{cases}$解得$\begin{cases}n_1=3,\\m_1=\frac{7}{24},\end{cases}$ $\therefore$ 直线$CH$的表达式为$y=\frac{7}{24}x+3$.联立$\begin{cases}y=\frac{7}{24}x+3,\\y=-\frac{3}{4}x^2+\frac{9}{4}x+3,\end{cases}$得$-\frac{3}{4}x^2+\frac{9}{4}x+3=\frac{7}{24}x+3$,解得$x_1=0$(不符合题意),$x_2=\frac{47}{18}$,$\therefore$ 点$P$的横坐标为$\frac{47}{18}$;
②当$∠ PCB = ∠ OBC$时,,则$CP// x$轴,则点$P$,$C$关于抛物线对称轴对称,$\because$ 对称轴为直线$x=-\frac{\frac{9}{4}}{2×(-\frac{3}{4})}=\frac{3}{2}$,$C(0,3)$,$\therefore$ 点$P$的横坐标为$\frac{3}{2}+\frac{3}{2}=3$.综上所述,点$P$的横坐标为$\frac{47}{18}$或3.

4. (2026·烟台月考)如图,已知抛物线$y=ax^2+bx+3(a≠0)$与x轴交于点A(1,0)和点B(-3,0),与y轴交于点C.
(1)求抛物线的表达式.
(2)如图①,对称轴上是否存在点E,使△ACE周长最小?求出此时点E的坐标和周长的最小值.
(3)如图②,点F为第二象限抛物线上一动点,连接AF交BC于点D,$k=S_{△FDC}:S_{△ADC}$,是否存在点F,使k取最大值?如果存在,求出此时点F的坐标和k的最大值;若不存在,请说明理由.
(4)如图③,已知点M是抛物线对称轴上一点,点N是平面内一点,点P是第二象限抛物线上一点,点Q是线段BC上一点,$PQ // y$轴,当线段PQ取得最大值时,是否存在点M,N使得四边形QAMN是菱形?若存在,直接写出点M的坐标,若不存在,请说明理由.

4. (1)将点$A(1,0)$和点$B(-3,0)$代入$y=ax^2+bx+3(a≠0)$得$\begin{cases}a+b+3=0,\\9a-3b+3=0,\end{cases}$解得$\begin{cases}a=-1,\\b=-2,\end{cases}$$\therefore y=-x^2-2x+3$.
(2)$\because y=-x^2-2x+3=-(x+1)^2+4$,$\therefore$ 对称轴为直线$x=-1$,当$x=0$时,$y=3$,则$C(0,3)$,设直线$BC$的表达式为$y=kx+3$,将点$B(-3,0)$代入得$0=-3k+3$,解得$k=1$,$\therefore$ 直线$BC$的表达式为$y=x+3$.$\because A,B$关于$x=-1$对称,,连接$BC$交直线$x=-1$于点$E$,连接$AE,AC$,$\because AC+EA+EC=AC+EB+EC≥ AC+BC$,$\therefore$ 当$B,E,C$三点共线时,$△ ACE$的周长最小,则点$E$即为所求,$\therefore$ 当$x=-1$时,$y=-1+3=2$,则$E(-1,2)$.$\because A(1,0)$,$B(-3,0)$,$C(0,3)$,$\therefore AC=\sqrt{1^2+3^2}=\sqrt{10}$,$BC=\sqrt{3^2+3^2}=3\sqrt{2}$,$\therefore △ ACE$的周长为$AC+AE+CE=AC+BC=\sqrt{10}+3\sqrt{2}$.
(3),过点$F$作$FH// x$轴交$BC$的延长线于点$H$,设$F(t,-t^2-2t+3)$,$\because FH// x$轴,$H$的纵坐标等于$F$的纵坐标,$\therefore H$的横坐标为$x=y-3=-t^2-2t+3-3=-t^2-2t$,即$H(-t^2-2t,-t^2-2t+3)$,$\therefore FH=-t^2-2t-t=-t^2-3t$.$\because A(1,0)$,$B(-3,0)$,$\therefore AB=4$.$\because FH// x$轴,$\therefore △ DFH ∽ △ DAB$.$\therefore \frac{FD}{AD}=\frac{FH}{AB}$.$\because k=S_{△ FDC}:S_{△ ADC}=FD:AD$,$\therefore k=\frac{FH}{AB}=-\frac{1}{4}(t^2+3t)=-\frac{1}{4}(t+\frac{3}{2})^2+\frac{9}{16}$,$\therefore$ 当$t=-\frac{3}{2}$时,即$F(-\frac{3}{2},\frac{15}{4})$时,$k$取得最大值$\frac{9}{16}$.
(4)点$M$的坐标为$(-1,\frac{3\sqrt{2}}{2})$或$(-1,-\frac{3\sqrt{2}}{2})$. 解析:依题意,设$P(m,-m^2-2m+3)$,则$Q(m,m+3)$,$\therefore PQ=-m^2-2m+3-(m+3)=-m^2-3m=-(m+\frac{3}{2})^2+\frac{9}{4}$,则当$P(-\frac{3}{2},\frac{15}{4})$时,线段$PQ$取得最大值,则$Q(-\frac{3}{2},\frac{3}{2})$.又$A(1,0)$,$\therefore AQ=\sqrt{(1+\frac{3}{2})^2+(\frac{3}{2})^2}=\frac{\sqrt{34}}{2}$.$\because$ 点$M$是抛物线对称轴上一点,设$M(-1,n)$,依题意,四边形$QAMN$是菱形,$\therefore AM=AQ$,$\therefore \sqrt{(1+1)^2+n^2}=\frac{\sqrt{34}}{2}$,解得$n=\pm\frac{3\sqrt{2}}{2}$,$\therefore$ 点$M$的坐标为$(-1,\frac{3\sqrt{2}}{2})$或$(-1,-\frac{3\sqrt{2}}{2})$.