已知一次函數(shù)y=(1-2k) x+(2k+1).
①當(dāng)k取何值時,y隨x的增大而增大?
②當(dāng)k取何值時,函數(shù)圖象經(jīng)過坐標(biāo)系原點?
③當(dāng)k取何值時,函數(shù)圖象不經(jīng)過第四象限?
分析:(1)當(dāng)1-2k>0時,y隨x的增大而增大;
(2)當(dāng)2k+1=0時,函數(shù)圖象經(jīng)過坐標(biāo)系原點;
(3)當(dāng)1-2k>0,2k+1≥0,函數(shù)圖象不經(jīng)過第四象限.
解答:解:(1)一次函數(shù)y=(1-2k) x+(2k+1),當(dāng)1-2k>0時,y隨x的增大而增大;即k<
.
(2)一次函數(shù)y=(1-2k) x+(2k+1),當(dāng)2k+1=0時,函數(shù)圖象經(jīng)過坐標(biāo)系原點;即k=-
.
(3)一次函數(shù)y=(1-2k) x+(2k+1),當(dāng)1-2k>0,2k+1≥0,函數(shù)圖象不經(jīng)過第四象限,即-
≤k<
.
點評:熟練掌握一次函數(shù)y=kx+b的性質(zhì).當(dāng)k>0,圖象過第1、3象限;當(dāng)k<0,圖象過第2、4象限;當(dāng)b<0,圖象與y軸的交點在x軸的下方;當(dāng)b>0,圖象與y軸的交點在x軸的上方;當(dāng)k=0,圖象過坐標(biāo)原點.