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> (2011•毕节地区)一次函数y=kx+k(k≠0)和反比例函数在同一直角坐标系中的图象大致是()-八年级数学
(2011•毕节地区)一次函数y=kx+k(k≠0)和反比例函数在同一直角坐标系中的图象大致是()-八年级数学
题目简介
(2011•毕节地区)一次函数y=kx+k(k≠0)和反比例函数在同一直角坐标系中的图象大致是()-八年级数学
题目详情
(2011•毕节地区)一次函数y=kx+k(k≠0)和反比例函数
在同一直角坐标系中的图象大致是( )
题型:单选题
难度:中档
来源:不详
答案
C
A、由反比例函数的图象在一、三象限可知k>0,由一次函数的图象过二、四象限可知k<0,两结论相矛盾,故本选项错误;
B、由反比例函数的图象在二、四象限可知k<0,由一次函数的图象与y轴交点在y轴的正半轴可知k>0,两结论相矛盾,故本选项错误;
C、由反比例函数的图象在二、四象限可知k<0,由一次函数的图象过二、三、四象限可知k<0,两结论一致,故本选项正确;
D、由反比例函数的图象在一、三象限可知k>0,由一次函数的图象与y轴交点在y轴的负半轴可知k<0,两结论相矛盾,故本选项错误.
故选C.
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题目简介
(2011•毕节地区)一次函数y=kx+k(k≠0)和反比例函数在同一直角坐标系中的图象大致是()-八年级数学
题目详情
答案
B、由反比例函数的图象在二、四象限可知k<0,由一次函数的图象与y轴交点在y轴的正半轴可知k>0,两结论相矛盾,故本选项错误;
C、由反比例函数的图象在二、四象限可知k<0,由一次函数的图象过二、三、四象限可知k<0,两结论一致,故本选项正确;
D、由反比例函数的图象在一、三象限可知k>0,由一次函数的图象与y轴交点在y轴的负半轴可知k<0,两结论相矛盾,故本选项错误.
故选C.