【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k的取值進行一一驗證,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軸的交點坐標,再分別求出直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/0.png)
x+b經(jīng)過點A、B時的b的取值,進而求出其取值范圍.本題第二問是難點,主要是不會借助計算淘汰不合題意的k值.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/images1.png)
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=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/1.png)
.
當k=1時,方程2x
2+4x+k-1=0有一個根為零;
當k=2時,x
1•x
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/2.png)
,方程2x
2+4x+k-1=0沒有兩個不同的非零整數(shù)根;
當k=3時,方程2x
2+4x+k-1=0有兩個相同的非零實數(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k=3時,二次函數(shù)為y=2x
2+4x+2,把它的圖象向下平移8個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,則A(-3,0),B(1,0).
依題意翻折后的圖象如圖所示.
當直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/3.png)
x+b經(jīng)過A點時,可得b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/4.png)
;
當直線y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/5.png)
x+b經(jīng)過B點時,可得b=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/6.png)
.
由圖象可知,符合題意的b(b<3)的取值范圍為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/7.png)
<b<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/8.png)
.
(3)依圖象得,要圖象y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/9.png)
x+b(b小于k)與二次函數(shù)圖象有兩個公共點時,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y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/10.png)
x+b過(1,0)時,有一個交點,此時b=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/11.png)
.
當y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/12.png)
x+b過(-3,0)時,有三個交點,此時b=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/13.png)
.而在此中間即為兩個交點,此時-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/14.png)
<b<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/15.png)
.
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y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/16.png)
x+b與y=-2(x-1)(x+3)(-3<x<1)相切時,y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/17.png)
x+b與這個二次函數(shù)圖象有三個交點,若直線再向上移,則只有兩個交點.
因為b<3,而y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/18.png)
x+b(b小于k,k=3),所以當b=3時,將y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/19.png)
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)圖象相交只有兩個公共點.這種情況故舍去.
綜上,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/20.png)
<b<
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103101221051868954/SYS201311031012210518689004_DA/21.png)
.
點評:考查知識點: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.不錯的題目,難度不大,綜合性強,考查面廣,似乎是一個趨勢或熱點.