$证明：y=x^2+ax+a-2=(x+\frac {a}2)^2-\frac {a^2}4+a-2$
$∴函数图像的顶点是(-\frac {a}2，-\frac {a^2}4+a-2)$
$-\frac {a^2}4+a-2=-(\frac {a}2-1)^2-1$
$∵不论a取何值，总有-(\frac a{2}-1)^2≤0$
$∴-(\frac {a}2-1)^2-1＜0，即-\frac {a^2}4+a-2＜0$
$∴不论a取何值，函数图像顶点总在x轴的下方$
