【答案】
分析:(1)把點(diǎn)A坐標(biāo)代入一次函數(shù)求出m的值,即可得解;
(2)先解一元二次方程求出二次函數(shù)的對稱軸,然后根據(jù)對稱軸與點(diǎn)A的坐標(biāo)列出方程組求出a、b的值,即可得解;
(3)利用一次函數(shù)解析式求出點(diǎn)B的坐標(biāo),利用二次函數(shù)解析式求出點(diǎn)D的坐標(biāo),并判斷出△AOB是等腰直角三角形,根據(jù)等腰直角三角形的性質(zhì)求出AB的長度,∠OAB=∠OBA=45°,然后根據(jù)△ABD中沒有45°的角和∠ABD=135°判斷出∠BAC和∠ABD是對應(yīng)角為135°,從而判斷出點(diǎn)C在點(diǎn)A的左邊,再分AC和BD,AC和AB是對應(yīng)邊兩種情況,根據(jù)相似三角形對應(yīng)邊成比例列式求出AC的長度,再求出OC的長度,從而得解.
解答:解:(1)∵一次函數(shù)y=-x+m圖象經(jīng)過點(diǎn)A(-2,0),
∴-(-2)+m=0,
∴m=-2,
∴一次函數(shù)解析式為y=-x-2;
(2)由2x
2-3x-2=0得,x
1=-

,x
2=2,
∴二次函數(shù)y=ax
2+bx-4的對稱軸為直線x=-

,
∴

,
解得

,
∴二次函數(shù)的解析式為y=2x
2+2x-4;
(3)令x=0,一次函數(shù)與y軸的交點(diǎn)B(0,-2),

二次函數(shù)與y軸的交點(diǎn)為D(0,-4),
∴△AOB是等腰直角三角形,BD=-2-(-4)=2,
∴AB=

=2

,∠OAB=∠OBA=45°,
∵△ABD中,∠BAD、∠ADB都不等于45°,∠ABD=180°-45°=135°,
∴∠BAC和∠ABD是對應(yīng)角為135°,
∴點(diǎn)C在點(diǎn)A的左邊,
①AC和BD是對應(yīng)邊時,∵△ADB∽△BCA,
∴

=

=1,
∴AC=BD=2,
∴OC=OA+AC=2+2=4,
點(diǎn)C的坐標(biāo)為(-4,0),
②AC和AB是對應(yīng)邊時,∵△ADB∽△CBA,
∴

=

=

,
∴AC=

AB=

×2

=4,
∴OC=OA+AC=2+4=6,
∴點(diǎn)C的坐標(biāo)為(-6,0),
綜上所述,在x軸上有一點(diǎn)C(-4,0)或(-6,0),使以點(diǎn)A、B、C組成的三角形與△ADB相似.
點(diǎn)評:本題是二次函數(shù)綜合題型,主要考查了待定系數(shù)法求一次函數(shù)解析式,解一元二次方程,待定系數(shù)法求二次函數(shù)解析式,相似三角形對應(yīng)邊成比例的性質(zhì),難點(diǎn)在于(3)要分情況討論.