Question

如圖，拋物線y=x²﹣2x﹣3與x軸交A、B兩點（A點在B點左側(cè)），直線l與拋物線交于A、C兩點，其中C點的橫坐標(biāo)為2．                                                                                                    

（1）求A、B兩點的坐標(biāo)及直線AC的函數(shù)表達式；                                      

（2）P是線段AC上的一個動點，過P點作y軸的平行線交拋物線于E點，求線段PE長度的最大值．                  

                                                        

　                                                                                                      

                                                                                                          

Answer 1

【考點】二次函數(shù)綜合題．                                                                     

【專題】壓軸題；動點型．                                                                     

【分析】（1）令y=0，解x²﹣2x﹣3=0，可得AB的坐標(biāo)；將C的橫坐標(biāo)代入，易得其縱坐標(biāo)，結(jié)合A的坐標(biāo)，可ACC的方程；                                                                                               

（2）設(shè)出P點的橫坐標(biāo)，表示出P、E的坐標(biāo)，可得PE長度的表達式，進而根據(jù)x的取值范圍可得線段PE長度的最大值．                                                                       

【解答】解：（1）令y=0，解得x₁=﹣1或x₂=3，                                          

∴A（﹣1，0）B（3，0）；                                                                    

將C點的橫坐標(biāo)x=2代入y=x²﹣2x﹣3                                                     

得y=﹣3，                                                                                         

∴C（2，﹣3），                                                                               

設(shè)直線AC的解析式是y=kx+b，                                                              

把A（﹣1，0），C（2，﹣3）代入得：，                                  

解得：k=﹣1，b=﹣1，                                                                            

∴直線AC的函數(shù)解析式是y=﹣x﹣1；                                                    

                                                                                                          

（2）設(shè)P點的橫坐標(biāo)為x（﹣1≤x≤2）（注：x的范圍不寫不扣分）               

則P、E的坐標(biāo)分別為：P（x，﹣x﹣1），E（x，x²﹣2x﹣3）                       

∵P點在E點的上方，PE=（﹣x﹣1）﹣（x²﹣2x﹣3）=﹣x²+x+2=﹣（x﹣）²+，                 

∴當(dāng)時，PE的最大值=．