???$解： (2)頂點(diǎn)坐標(biāo)為(\frac {m-1}{2}，\frac {(m+1)2}{4})$???
???$把x=\frac {m-1}{2}代入y= (x+1)2，$???
???$得y=\frac {(m+1)2}{4}$???
???$不論m為何值，該函數(shù)的圖像的頂點(diǎn)都在函數(shù)y= (x +1)2的圖像上$???
???$(3)設(shè)頂點(diǎn)縱坐標(biāo)為z ，$???
???$則z=\frac {(m+1)2}{4} $???
???$當(dāng)m=-1時，z有最小值0$???
???$當(dāng)m=- 2時，z=\frac {1}{4}$???
???$當(dāng)m=3時， z= 4$???
???$所以當(dāng)-2≤m≤3時，0≤z≤4$??
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