$解：? (1)?由圖可知，二次函數(shù)的頂點坐標(biāo)為?(\frac {5}{2}，??-\frac {9}{4})?$
$∴二次函數(shù)表達式為?y= (x-\frac {5}{2})2-\frac {9}{4}?$
$令?y=0，??(x-\frac {5}{2})2-\frac {9}{4}=0?$
$解得?{x}_1=4，??{x}_2=1?$
$∴二次函數(shù)與?x?軸的交點為?(4，??0)、??(1，??0) ?$
$?(2)?由圖像可知$
$①當(dāng)?x\gt 4?或?x\lt 1?時，?y\gt 0?$
$②當(dāng)?x= 1?或?x=4?時，?y=0?$
$③當(dāng)?1＜x＜4?時，?y＜0?$
