4. (2024·无锡中考)已知二次函数$y=ax^2+x+c$的图象经过点$A(-1,-\dfrac{1}{2})$和点$B(2,1)$.
(1)求这个二次函数的表达式.
(2)若点$C(m+1,y_1),D(m+2,y_2)$都在该二次函数的图象上,试比较$y_1$和$y_2$的大小,并说明理由.
(3)点$P,Q$在直线$AB$上,点$M$在该二次函数图象上.问:在$y$轴上是否存在点$N$,使得以$P,Q,M,N$为顶点的四边形是正方形?若存在,请直接写出所有满足条件的点$N$的坐标;若不存在,请说明理由.

4. (1) 把$A(-1,-\dfrac{1}{2})$,$B(2,1)$代入$y=ax^2+x+c$得$\begin{cases}a-1+c=-\dfrac{1}{2},\\4a+2+c=1,\end{cases}$解得$\begin{cases}a=-\dfrac{1}{2},\\c=1,\end{cases}$
∴这个二次函数的表达式为$y=-\dfrac{1}{2}x^2+x+1$.
(2)
∵$C(m+1,y_1)$,$D(m+2,y_2)$都在该二次函数的图象上,
∴$y_1=-\dfrac{1}{2}(m+1)^2+(m+1)+1$,$y_2=-\dfrac{1}{2}(m+2)^2+(m+2)+1$,
∴$y_1-y_2=-\dfrac{1}{2}(m+1)^2+(m+1)+1-[-\dfrac{1}{2}(m+2)^2+(m+2)+1]=m+\dfrac{1}{2}$.当$m+\dfrac{1}{2}＞0$,即$m＞-\dfrac{1}{2}$时,$y_1＞y_2$;当$m+\dfrac{1}{2}=0$,即$m=-\dfrac{1}{2}$时,$y_1=y_2$;当$m+\dfrac{1}{2}＜0$,即$m＜-\dfrac{1}{2}$时,$y_1＜y_2$.
(3) 存在.点N的坐标为$(0,\dfrac{-15+5\sqrt{41}}{16})$或$(0,\dfrac{-15-5\sqrt{41}}{16})$或$(0,±5)$或$(0,±\dfrac{5}{8})$.
解析:设直线AB的表达式为y=kx+e,把$A(-1,-\dfrac{1}{2})$,$B(2,1)$代入得$\begin{cases}-\dfrac{1}{2}=-k+e,\\1=2k+e,\end{cases}$解得$\begin{cases}k=\dfrac{1}{2},\\e=0,\end{cases}$
∴直线AB的表达式为$y=\dfrac{1}{2}x$.
当PQ为正方形的边时:
①
∵B(2,1),
∴$\tan∠BOE=\dfrac{1}{2}$.如①,过点M作y轴的垂线,垂足为点G,过点P作MG的垂线,垂足为点H,
∵PQ//MN,MG//x轴,
∴∠BOE=∠NMG,
∴$\tan∠NMG=\tan∠BOE=\dfrac{1}{2}$,则MG=2NG.设NG=t,则MG=2t,
∴M(-2t,-2t²-2t+1),
∴点N的纵坐标为-2t²-2t+1+t=-2t²-t+1,即N(0,-2t²-t+1).
∵以P,Q,M,N为顶点的四边形是正方形,
∴∠PMN=90°,PM=MN,
∴∠PMH+∠NMG=90°.
∵∠PMH+∠MPH=90°,
∴∠NMG=∠MPH.
∵∠NMG=∠MPH,∠H=∠MGN,PM=MN,
∴△PHM≌△MGN,
∴PH=MG=2t,HM=NG=t,
∴P(-3t,-2t²+1).把P(-3t,-2t²+1)代入$y=\dfrac{1}{2}x$得$-2t²+1=\dfrac{1}{2}×(-3t)$,解得$t_1=\dfrac{3+\sqrt{41}}{8}$,$t_2=\dfrac{3-\sqrt{41}}{8}$(舍去),
∴$N(0,\dfrac{-15+5\sqrt{41}}{16})$;
②如②,构造Rt△MQG,Rt△NMH,和①同理可得△MQG≌△NMH,$\tan∠MNH=\dfrac{1}{2}$,设NH=GM=2t,则QG=MH=t,
∴M(2t,-2t²+2t+1),N(0,-2t²+t+1),Q(t,-2t²+4t+1).把Q(t,-2t²+4t+1)代入$y=\dfrac{1}{2}x$得$-2t²+4t+1=\dfrac{1}{2}t$,解得$t_1=2$,$t_2=-\dfrac{1}{4}$(舍去),
∴N(0,-5);
③如③,构造Rt△GMN,Rt△HPM,和①同理可得△GMN≌△HPM,$\tan∠GNM=\dfrac{1}{2}$,设GN=HM=2t,则GM=HP=t,
∴M(-2t,-2t²-2t+1),N(0,-2t²-t+1),P(-t,-2t²-4t+1).把P(-t,-2t²-4t+1)代入$y=\dfrac{1}{2}x$得$-2t²-4t+1=-\dfrac{1}{2}t$,解得$t_1=\dfrac{1}{4}$,$t_2=-2$(舍去),
∴$N(0,\dfrac{5}{8})$;
④如④,构造Rt△GMN,Rt△HNP,和①同理可得△GMN≌△HNP,$\tan∠GMN=\dfrac{1}{2}$,设GM=HN=2t,则GN=HP=t,
∴M(2t,-2t²+2t+1),N(0,-2t²+t+1),P(t,-2t²-t+1).把P(t,-2t²-t+1)代入$y=\dfrac{1}{2}x$得$-2t²-t+1=\dfrac{1}{2}t$,解得$t_1=\dfrac{-3+\sqrt{41}}{8}$,$t_2=\dfrac{-3-\sqrt{41}}{8}$(舍去),
∴$N(0,\dfrac{-15+5\sqrt{41}}{16})$;
当PQ为正方形的对角线时:
⑤如⑤,构造矩形HGJI,过点P作PK⊥IJ于点K,易得∠QPK=∠BOE,
∴$\tan∠QPK=\tan∠BOE=\dfrac{1}{2}$.设QK=x,则PK=2x,和①同理可得△PNH≌△MPG≌△QMJ≌△NQI,
∴HN=PG=MJ=IQ,PH=GM=QJ=NI,
∴四边形HGJI为正方形,
∴PK=IJ=2x,
∴IQ=PG=JK=$\dfrac{1}{2}(IJ-QK)=\dfrac{1}{2}x$,则PH=GM=QJ=NI=$\dfrac{3}{2}x$,
∴$\tan∠PMG=\dfrac{PG}{GM}=\dfrac{1}{3}$.设PG=NH=t,则PH=GM=3t,
∴M(2t,-2t²+2t+1),N(0,-2t²+6t+1),P(-t,-2t²+3t+1).把P(-t,-2t²+3t+1)代入$y=\dfrac{1}{2}x$得$-2t²+3t+1=-\dfrac{1}{2}t$,解得$t_1=2$,$t_2=-\dfrac{1}{4}$(舍去),
∴N(0,5);
⑥如⑥,构造Rt△PMH,Rt△NPG,和⑤同理可得△PMH≌△NPG,$\tan∠PNG=\dfrac{1}{3}$,设PG=HM=t,则PH=GN=3t,
∴M(-2t,-2t²-2t+1),N(0,-2t²-6t+1),P(-3t,-2t²-5t+1).把P(-3t,-2t²-5t+1)代入$y=\dfrac{1}{2}x$得$-2t²-5t+1=-\dfrac{3}{2}t$,解得$t_1=\dfrac{1}{4}$,$t_2=-2$(舍去),
∴$N(0,-\dfrac{5}{8})$.
综上,点N的坐标为$(0,\dfrac{-15+5\sqrt{41}}{16})$或$(0,\dfrac{-15-5\sqrt{41}}{16})$或(0,-5)或(0,5)或$(0,\dfrac{5}{8})$或$(0,-\dfrac{5}{8})$.