已知關(guān)于x的方程mx2-(3m-1)x+2m-2=0.
(1)求證:無論m取任何實數(shù)時,方程恒有實數(shù)根;
(2)若關(guān)于x的二次函數(shù)y=mx2-(3m-1)x+2m-2的圖象與x軸兩交點間的距離為2時,求拋物線的解析式;
(3)在直角坐標(biāo)系xoy中,畫出(2)中的函數(shù)圖象,結(jié)合圖象回答問題:當(dāng)直線y=x+b與(2)中的函數(shù)圖象只有兩個交點時,求b的取值范圍.
【答案】
分析:(1)本題中,二次項系數(shù)m的值不確定,分為m=0,m≠0兩種情況,分別證明方程有實數(shù)根;
(2)設(shè)拋物線與x軸兩交點的橫坐標(biāo)為x
1,x
2,則兩交點之間距離為|x
1-x
2|=2,再與根與系數(shù)關(guān)系的等式結(jié)合變形,可求m的值,從而確定拋物線的解析式;
(3)分三種情況:只與拋物線y
1有兩個交點,只與拋物線y
2有兩個交點,直線過拋物線y
1、y
2的交點,觀察圖象,分別求出b的取值范圍.
解答:解:(1)分兩種情況討論.
①當(dāng)m=0時,方程為x-2=0,x=2.
∴m=0時,方程有實數(shù)根.
②當(dāng)m≠0時,則一元二次方程的根的判別式
△=[-(3m-1)]
2-4m(2m-2)
=9m
2-6m+1-8m
2+8m=m
2+2m+1
=(m+1)
2≥0,
∴m≠0時,方程有實數(shù)根.
故無論m取任何實數(shù)時,方程恒有實數(shù)根.
綜合①②可知,m取任何實數(shù),方程mx
2-(3m-1)x+2m-2=0恒有實數(shù)根;
(2)設(shè)x
1,x
2為拋物線y=mx
2-(3m-1)x+2m-2與x軸交點的橫坐標(biāo),
則x
1+x
2=
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,x
1x
2=
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.
由|x
1-x
2|=
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=
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=
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=
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=|
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|.
由|x
1-x
2|=2,得|
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|=2,
∴
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=2或
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=-2.
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∴m=1或m=-
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.
∴所求拋物線的解析式為y
1=x
2-2x,
y
2=-
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(x-2)(x-4).
其圖象如右圖所示:
(3)在(2)的條件下y=x+b與拋物線
y
1,y
2組成的圖象只有兩個交點,結(jié)合圖象求b的取值范圍.
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,
當(dāng)y
1=y時,得x
2-3x-b=0,有△=9+4b=0得b=-
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.
同理
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,△=9-4(8+3b)=0,得b=-
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.
觀察圖象可知,
當(dāng)b<-
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,或b>-
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直線y=x+b與(2)中的圖象只有兩個交點;
由
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,
當(dāng)y
1=y
2時,有x=2或x=1.
當(dāng)x=1時,y=-1.
所以過兩拋物線交點(1,-1),(2,0)的直線為y=x-2.
綜上所述可知:當(dāng)b<-
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或b>-
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或b=-2時,
直線y=x+b與(2)中圖象只有兩個交點.
點評:本題具有較強的綜合性,考查了一元二次方程的根的情況,二次函數(shù)與對應(yīng)的一元二次方程的聯(lián)系,討論一次函數(shù)與二次函數(shù)圖象交點的情況.