如图 在平面直角坐标系中 抛物线$y=-x^{2}+bx+c$与$x$轴交于点$A$ $B\left(3,0\right)$ 交$y$轴于点$C\left(0,
1、(1)将点$B\left(3,0\right)$,点$C\left(0,3\right)$代入解析式可得:$\left\{\begin{array}{l}{0=-9+3b+c}\\{c=3}\end{array}\right.$,
2、解得:$\left\{\begin{array}{l}{b=2}\\{c=3}\end{array}\right.$,
3、$\therefore $抛物线的函数表达式为:$y=-x^{2}+2x+3$;
4、$(2)\because y=-x^{2}+2x+3=-\left(x-1\right)^{2}+4$,
5、$\therefore D\left(1,4\right)$,
6、设直线$BD$解析式为$y=kx+n$,
7、由题意可得:$\left\{\begin{array}{l}{4=k+n}\\{0=3k+n}\end{array}\right.$,
8、解得:$\left\{\begin{array}{l}{k=-2}\\{n=6}\end{array}\right.$,
9、$\therefore $直线$BD$的函数表达式为$y=-2x+6$;
10、$(3)$①如图,过点$D$作$DG\bot AB$于$G$,
11、$\therefore DG=4$,$OG=1$,
12、$\therefore BG=2$,
13、$\therefore DB=\sqrt{DG{}^{2}+BG{}^{2}}=\sqrt{16+4}=2\sqrt{5}$,
14、$\because NH\bot x$轴,$DG\bot AB$,
15、$\therefore DG$∥$NH$,
16、$\therefore \angle GDB=\angle DFN$,
17、又$\because \angle NMF=\angle DGH=90^{\circ}$,
18、$\therefore \triangle MNF$∽$\triangle GBD$,
19、$\therefore \frac{MN}{GB}=\frac{NF}{BD}$,
20、$\therefore MN=\frac{2}{2\sqrt{5}}\times NF=\frac{\sqrt{5}}{5}NF$,
21、设点$N$坐标为$(m$,$-m^{2}+2m+3)$,则点$F\left(m,-2m+6\right)$,
22、$\therefore NF=-m^{2}+2m+3-\left(-2m+6\right)=-\left(m-2\right)^{2}+1$,
23、$\therefore MN=-\frac{\sqrt{5}}{5}(m-2)^{2}+\frac{\sqrt{5}}{5}$,
24、$\therefore m=2$时,$MN$有最大值为$\frac{\sqrt{5}}{5}$,
25、$\therefore $点$F\left(2,2\right)$;
26、②如图,连接$AC$,过点$Q$作$PQ\bot CO$,交$AC$于$Q$,
27、$\because $抛物线$y=-x^{2}+2x+3$与$x$轴交于点$A$,$B$两点,
28、$\therefore $点$A\left(-1,0\right)$,
29、$\therefore AO=1$,
30、$\because \tan \angle ACO=\frac{AO}{CO}=\frac{PQ}{CP}$,
31、$\therefore \frac{1}{3}=\frac{PQ}{CP}$,
32、$\therefore PQ=\frac{1}{3}CP$,
33、$\therefore HF+FP+\frac{1}{3}PC=2+PF+PQ$,
34、$\therefore $当点$F$,点$P$,点$Q$三点共线时,$HF+FP+\frac{1}{3}PC$有最小值,
35、$\therefore FP\bot CO$时,$HF+FP+\frac{1}{3}PC$有最小值,
36、$\therefore OP=FH=2$,
37、$\therefore CP=1$.
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