解：?$ (1)$?由圖可知，二次函數(shù)的頂點(diǎn)坐標(biāo)為?$(\frac {5}{2}，$??$-\frac {9}{4})$?
∴二次函數(shù)表達(dá)式為?$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$?軸的交點(diǎn)為?$(4，$??$0)、$??$(1，$??$0) $?
?$(2)$?由圖像可知
①當(dāng)?$x>4$?或?$x<1$?時，?$y>0$?
②當(dāng)?$x= 1$?或?$x=4$?時，?$y=0$?
③當(dāng)?$1＜x＜4$?時，?$y＜0$?
