题目内容
已知y=y1+y2,y1与![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_ST/0.png)
(1)求y与x的函数关系式和x的取范围;
(2)当x=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_ST/1.png)
【答案】分析:根据题意可设y1=k1
,y2=
,所以y=k1
+
;又因为当x=1时,y=-12;当x=4时,y=7,所以将点代入解析式即可得到方程组,解方程即可求得y与x的函数关系式.根据已知可得x>0.将x=
代入函数解析式,即可求得y的值.
解答:解:(1)设y1=k1
,y2=
,则y=k1
+
;
∵当x=1时,y=-12;当x=4时,y=7.
∴
.
解得:
.
∴y与x的函数关系式为y=4
-
,
∵x≥0,x2≠0,
∴x的取范围为x>0;
(2)当x=
时,
y=4×
-
=-254.
∴y的值为-254.
点评:此题考查了待定系数法求函数的解析式,解题的关键是根据题意设得符合要求的解析式,将x与y的取值代入解析式即可求得.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/4.png)
解答:解:(1)设y1=k1
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/8.png)
∵当x=1时,y=-12;当x=4时,y=7.
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/9.png)
解得:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/10.png)
∴y与x的函数关系式为y=4
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/12.png)
∵x≥0,x2≠0,
∴x的取范围为x>0;
(2)当x=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/13.png)
y=4×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/14.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103194614072519159/SYS201311031946140725191024_DA/15.png)
∴y的值为-254.
点评:此题考查了待定系数法求函数的解析式,解题的关键是根据题意设得符合要求的解析式,将x与y的取值代入解析式即可求得.
![](http://thumb2018.1010pic.com/images/loading.gif)
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