【答案】
分析:(1)綜合根的判別式及k的要求求出k的取值;
(2)對(duì)k的取值進(jìn)行一一驗(yàn)證,求出符合要求的k值,再結(jié)合拋物線平移的規(guī)律寫出其平移后的解析式;
(3)求出新拋物線與x軸的交點(diǎn)坐標(biāo),再分別求出直線y=
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x+b經(jīng)過點(diǎn)A、B時(shí)的b的取值,進(jìn)而求出其取值范圍.本題第二問是難點(diǎn),主要是不會(huì)借助計(jì)算淘汰不合題意的k值.
解答:
解:(1)由題意得,△=16-8(k-1)≥0.
∴k≤3.
∵k為正整數(shù),
∴k=1,2,3;
(2)設(shè)方程2x
2+4x+k-1=0的兩根為x
1,x
2,則
x
1+x
2=-2,x
1•x
2=
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.
當(dāng)k=1時(shí),方程2x
2+4x+k-1=0有一個(gè)根為零;
當(dāng)k=2時(shí),x
1•x
2=
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,方程2x
2+4x+k-1=0沒有兩個(gè)不同的非零整數(shù)根;
當(dāng)k=3時(shí),方程2x
2+4x+k-1=0有兩個(gè)相同的非零實(shí)數(shù)根-1.
綜上所述,k=1和k=2不合題意,舍去,k=3符合題意.
當(dāng)k=3時(shí),二次函數(shù)為y=2x
2+4x+2,把它的圖象向下平移8個(gè)單位得到的圖象的解析式為y=2x
2+4x-6;
(3)設(shè)二次函數(shù)y=2x
2+4x-6的圖象與x軸交于A、B兩點(diǎn),則A(-3,0),B(1,0).
依題意翻折后的圖象如圖所示.
當(dāng)直線y=
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x+b經(jīng)過A點(diǎn)時(shí),可得b=
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;
當(dāng)直線y=
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x+b經(jīng)過B點(diǎn)時(shí),可得b=-
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.
由圖象可知,符合題意的b(b<3)的取值范圍為
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<b<
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.
(3)依圖象得,要圖象y=
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x+b(b小于k)與二次函數(shù)圖象有兩個(gè)公共點(diǎn)時(shí),顯然有兩段.
而因式分解得y=2x
2+4x-6=2(x-1)(x+3),
第一段,當(dāng)y=
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x+b過(1,0)時(shí),有一個(gè)交點(diǎn),此時(shí)b=-
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.
當(dāng)y=
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x+b過(-3,0)時(shí),有三個(gè)交點(diǎn),此時(shí)b=
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.而在此中間即為兩個(gè)交點(diǎn),此時(shí)-
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<b<
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.
第二段,將平移后的二次函數(shù)的圖象在x軸下方的部分沿x軸翻折后,
開口向下的部分的函數(shù)解析式為y=-2(x-1)(x+3). 顯然,
當(dāng)y=
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x+b與y=-2(x-1)(x+3)(-3<x<1)相切時(shí),y=
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x+b與這個(gè)二次函數(shù)圖象有三個(gè)交點(diǎn),若直線再向上移,則只有兩個(gè)交點(diǎn).
因?yàn)閎<3,而y=
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x+b(b小于k,k=3),所以當(dāng)b=3時(shí),將y=
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x+3代入二次函數(shù)y=-2(x-1)(x+3)整理得,
4x
2+9x-6=0,△>0,所以方程有兩根,那么肯定不將有直線與兩截結(jié)合的二次函數(shù)圖象相交只有兩個(gè)公共點(diǎn).這種情況故舍去.
綜上,-
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<b<
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.
點(diǎn)評(píng):考查知識(shí)點(diǎn):一元二次方程根的判別式、二次函數(shù)及函數(shù)圖象的平移與翻折,最后還考到了與一次函數(shù)的結(jié)合等問題.不錯(cuò)的題目,難度不大,綜合性強(qiáng),考查面廣,似乎是一個(gè)趨勢或熱點(diǎn).