(1)∵抛物线的顶点为A(1,4),
∴设抛物线的解析式y=a(x-1)
2+4,
把点B(0,3)代入得,a+4=3,
解得a=-1,
∴抛物线的解析式为y=-(x-1)
2+4;
(2)点B关于x轴的对称点B′的坐标为(0,-3),
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/64380cd7912397ddfb7ba3b65a82b2b7d0a28705?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
由轴对称确定最短路线问题,连接AB′与x轴的交点即为点P,
设直线AB′的解析式为y=kx+b(k≠0),
则
,
解得
,
∴直线AB′的解析式为y=7x-3,
令y=0,则7x-3=0,
解得x=
,
所以,当PA+PB的值最小时的点P的坐标为(
,0).