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