SOS!一道初三二次函數(shù)題...不難的..好心人幫忙!謝謝了!
SOS!一道初三二次函數(shù)題...不難的..好心人幫忙!謝謝了!
函數(shù)y=x^2-2mx+m^2+m-2
(1)當(dāng)m為何值時,二次函數(shù)圖象經(jīng)過原點(diǎn).
(2)當(dāng)m為何值時,二次函數(shù)圖象關(guān)于y軸對稱.
(3)當(dāng)m為何值時,二次函數(shù)圖象與x軸交點(diǎn)在原點(diǎn)兩側(cè)
注:要有具體一點(diǎn)的過程哦!最好幫忙算出來...謝謝!感激不盡!
優(yōu)質(zhì)解答
1.x=0,y=0
m^2+m-2=0
m=1 or -2
2.-2m=0 m=0
3.另y=0
x^2-2mx+m^2+m-2 =0
x1*x2=m^2+m-2 <0
得-2又判別式=(-2m)^2-4(m^2+m-2)=-4(m-2)>0,m<2
綜上所述 -2