證明：?$y=x^2+ax+a-2=(x+\frac {a}2)^2-\frac {a^2}4+a-2$?
∴函數(shù)圖像的頂點(diǎn)是?$(-\frac {a}2，$??$-\frac {a^2}4+a-2)$?
?$-\frac {a^2}4+a-2=-(\frac {a}2-1)^2-1$?
∵不論?$a$?取何值，總有?$-(\frac a{2}-1)^2≤0$?
∴?$-(\frac {a}2-1)^2-1＜0，$?即?$-\frac {a^2}4+a-2＜0$?
∴不論?$a$?取何值，函數(shù)圖像頂點(diǎn)總在?$x$?軸的下方
