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答案：解析 (1)將A(-1,0),B(4,0)代入$y=ax^{2}+bx-4$,得$\begin{cases}a-b-4=0,\\16a+4b-4=0,\end{cases}$解得$\begin{cases}a=1,\\b=-3,\end{cases}$∴ $y=x^{2}-3x-4$.
(2)存在.∵ $y=x^{2}-3x-4=(x-\frac{3}{2})^{2}-\frac{25}{4}$,∴ 拋物線L的頂點(diǎn)坐標(biāo)為$(\frac{3}{2},-\frac{25}{4})$,對稱軸為直線$x=\frac{3}{2}$,∴ 拋物線L'的頂點(diǎn)坐標(biāo)為$(-\frac{3}{2},\frac{25}{4})$,∴ 拋物線L'的解析式為$y=-(x+\frac{3}{2})^{2}+\frac{25}{4}=-x^{2}-3x+4$,∵ 點(diǎn)A'、B'與點(diǎn)A、B關(guān)于原點(diǎn)對稱,∴ A'(1,0),B'(-4,0).
設(shè)$M(\frac{3}{2},m)$,$N(n,-n^{2}-3n+4)$,
①當(dāng)AA'為平行四邊形的對角線時(shí),$n+\frac{3}{2}=-1+1$,解得$n=-\frac{3}{2}$,∴ $N(-\frac{3}{2},\frac{25}{4})$.
②當(dāng)AM為平行四邊形的對角線時(shí),$-1+\frac{3}{2}=1+n$,解得$n=-\frac{1}{2}$,則$-n^{2}-3n+4=-(-\frac{1}{2})^{2}-3×(-\frac{1}{2})+4=\frac{21}{4}$,∴ $N(-\frac{1}{2},\frac{21}{4})$.
③當(dāng)AN為平行四邊形的對角線時(shí),$-1+n=1+\frac{3}{2}$,解得$n=\frac{7}{2}$,則$-n^{2}-3n+4=-(\frac{7}{2})^{2}-3×\frac{7}{2}+4=-\frac{75}{4}$,∴ $N(\frac{7}{2},-\frac{75}{4})$.
綜上所述,點(diǎn)N的坐標(biāo)為$(-\frac{3}{2},\frac{25}{4})$或$(-\frac{1}{2},\frac{21}{4})$或$(\frac{7}{2},-\frac{75}{4})$.

答案：
解析 (1)∵ 二次函數(shù)$y=-x^{2}+bx+c$的圖象與x軸交于點(diǎn)A(-4,0),與y軸交于點(diǎn)C(0,4),∴ $\begin{cases}-16-4b+c=0,\\c=4,\end{cases}$解得$\begin{cases}b=-3,\\c=4,\end{cases}$∴ 該二次函數(shù)的解析式為$y=-x^{2}-3x+4$.
(2)由題意得$P(t,-t^{2}-3t+4)$,∵ A(-4,0),C(0,4),∴ OA=OC=4, ∴ ∠OAC=∠OCA=45°.∵ ∠PCA=45°,∴ ∠PCO=90°,∴ PC⊥OC,∵ OA⊥OC,∴ PC//OA.

∴ 點(diǎn)P的縱坐標(biāo)為4,∴ $-t^{2}-3t+4=4$,解得t=0或t=-3,∵ t≠0,∴ t=-3.
(3)①令y=0,則$-x^{2}-3x+4=0$,解得x=-4或x=1,∵ 點(diǎn)B在x軸正半軸上,∴ B(1,0),∴ OB=1.當(dāng)點(diǎn)P在AC的上方,即-4＜t＜0時(shí),過點(diǎn)P作PD⊥OA于點(diǎn)D,

則$PD=-t^{2}-3t+4$,$OD=0-t=-t$,∴ AD=t-(-4)=t+4,∴ $S=S_{△PAD}+S_{梯形PDOC}=\frac{1}{2}AD·PD+\frac{1}{2}(PD+OC)·OD=\frac{1}{2}(t+4)(-t^{2}-3t+4)+\frac{1}{2}(-t^{2}-3t+4+4)×(-t)=-2t^{2}-8t+8$.當(dāng)點(diǎn)P在AC的下方,即0＜t＜1時(shí),過點(diǎn)P作PE⊥OC于點(diǎn)E,

則PE=t,∴ $S=S_{△OAC}+S_{△OPC}=\frac{1}{2}OA·OC+\frac{1}{2}OC·PE=\frac{1}{2}×4×4+\frac{1}{2}×4t=2t+8$.
綜上,$S=\begin{cases}-2t^{2}-8t+8(-4＜t＜0),\\2t+8(0＜t＜1).\end{cases}$
②當(dāng)-4＜t＜0時(shí),$S=-2t^{2}-8t+8=-2(t+2)^{2}+16$,∵ -2＜0,∴ 當(dāng)t=-2時(shí),S有最大值,為16,∵ |-4-(-2)|=|0-(-2)|,∴ 當(dāng)t=-4或t=0時(shí),$-2(t+2)^{2}+16=8$,∴ 8＜S≤16.當(dāng)0＜t＜1時(shí),S=2t+8,∴ 8＜S＜10.
函數(shù)$S=\begin{cases}-2t^{2}-8t+8(-4＜t＜0),\\2t+8(0＜t＜1)\end{cases}$的大致圖象如圖所示,由圖象可知,當(dāng)8＜S＜10時(shí),存在3個(gè)符合條件的點(diǎn)P;當(dāng)10≤S＜16時(shí),存在2個(gè)符合條件的點(diǎn)P;當(dāng)S=16時(shí),存在1個(gè)符合條件的點(diǎn)P.