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第11頁(yè)

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解：?$(1)$?∵?$a=-1< 0$?

∴圖像開(kāi)口向下

∵?$y=-(x-2)^2+3$?

∴頂點(diǎn)坐標(biāo)是?$(2，$??$3)$?

?$(2)$?∵對(duì)稱軸為直線?$x=2，$?圖像開(kāi)口向下，?$y$?隨?$x$?增大而減小

∴?$x$?的取值范圍為?$x> 2 $?

?$(3)$?∵拋物線的對(duì)稱軸為直線?$x=2，$?滿足?$1< x< 4$?

∴此時(shí)?$y$?的最大值為?$3$?

∵當(dāng)?$x=1$?時(shí)，?$y=2；$?當(dāng)?$x=4$?時(shí)，?$y=-1$?

∴當(dāng)?$1< x< 4$?時(shí)，?$y$?的取值范圍是?$-1< y≤3$?

解：?$(1)$?∵拋物線?$y=a(x-3)^2+2$?經(jīng)過(guò)點(diǎn)?$(1，$??$-2)$?

∴?$-2=a(1-3)^2+2$?

∴?$a=-1 $?

?$(2)$?∵?$y=-(x-3)^2+2$?

∴此函數(shù)的圖像開(kāi)口向下

當(dāng)?$x< 3$?時(shí)，?$y$?隨?$x$?增大而增大，當(dāng)?$x> 3$?時(shí)，?$y$?隨?$x$?增大而減小

∵點(diǎn)?$A(m，$??$y_{1})、$??$B(n，$??$y_{2})(m< n< 3)$?都在該拋物線上

∴?$y_{1}< y_{2}$?

解：?$(1)$?當(dāng)?$m=2$?時(shí)，?$y=- \frac {1}{2} (x-4)^2-1$?

∵點(diǎn)?$A(6，$??$n)$?在該函數(shù)的圖像上

∴?$n=- \frac {1}{2} (6-4)^2-1=-3 $?

?$(2)$?若頂點(diǎn)是?$(2，$??$-1)，$?則?$2m=2①，$??$1-m=-1②$?

由①得?$m=1，$?由②得?$m=2$?

故小明的說(shuō)法錯(cuò)誤

?$(3)$?∵?$P、$??$Q $?兩點(diǎn)的縱坐標(biāo)相等

∴對(duì)稱軸為直線?$x=\frac {a+1+4\ \mathrm {m}-7+a}{2}=a+2m-3$?

∴?$a+2m-3=2m$?

∴?$a=3$?

∴?$P(4，$??$c)$?

∴?$c=- \frac {1}{2} (4-2m) ^2+1-m=-2(m-\frac {7}{4} ) ^2-\frac {7}{8}$?

∵?$-2(m-\frac {7}{4})^2 ≤0$?

∴?$c≤-\frac {7}{8}$?

解：?$(1)$?由表格可知，將點(diǎn)?$(1，$??$2)$?和點(diǎn)?$(2，$??$1)$?代入函數(shù)解析式

得?$\begin{cases}{a(1-2)^2+k=2}\\{a(2-2)^2+k=1}\end{cases}，$?解得?$\begin{cases}{a=1}\\{k=1}\end{cases}$?

∴?$y_{1}=(x-2)^2+1$?

?$(2) $?由題意，得?$y_{2}=(x-2+2)^2+1=x^2+1$?

把點(diǎn)?$A(m，$??$n_{1})、$??$B(m+1，$??$n_{2})$?分別代入?$y_{1}、$??$y_{2}$?的表達(dá)式中，

?$n_{1}=(m-2)^2+1=\ \mathrm {m^2}-4m+5，$??$n_{2}=(m+1)^2+1=\ \mathrm {m^2}+2\ \mathrm {m}+2$?

∴?$n_{1}-n_{2}=(\ \mathrm {m^2}-4m+5)-(\ \mathrm {m^2}+2m+2)=-6m+3$?

當(dāng)?$-6m+3> 0$?時(shí)，?$m< \frac {1}{2}；$?當(dāng)?$-6m+3=0$?時(shí)，?$m=\frac {1}{2}；$?當(dāng)?$-6m+3< 0$?時(shí)，?$m> \frac {1}{2}$?

∴當(dāng)?$m< \frac {1}{2} $?時(shí)，?$n_{1}-n_{2}> 0，$?即?$n_{1}> n_{2}；$?

當(dāng)?$m=\frac {1}{2} $?時(shí)，?$n_{1}-n_{2}=0，$?即?$n_{1}=n_{2}；$?

當(dāng)?$m>\frac {1}{2} $?時(shí)，?$n_{1}-n_{2}< 0，$?即?$n_{1}< n_{2}$?

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