(1)6+3m<0時,滿足y隨x增大而減小,故m<-2, n為任意值時,滿足條件.
(2)函數(shù)圖像在y軸上的點滿足x=0,當x=0時,y=(6+3m)*0+(n-4)<0,得:n<4;
又因為題目已確定函數(shù)為一次函數(shù),故6+3m≠0,得:m≠-2
結論:當n<4,且m≠-2時,滿足條件.
(3)函數(shù)圖像經(jīng)過原點滿足x=y=0,把(0, 0)點代入函數(shù)得:0=(6+3m)*0+(n-4),得:n=4;
又因為題目已確定函數(shù)為一次函數(shù),故6+3m≠0,得:m≠-2
結論:當n=4,且m≠-2時,滿足條件.
已知一次函數(shù)y=(6+3m)x+(n-4).
已知一次函數(shù)y=(6+3m)x+(n-4).
當m、n為何值時,y隨x的增大而減小?
當m、n為何值時,函數(shù)的圖象與y軸的交點在x軸的下方?
當m、n為何值時,函數(shù)的圖象經(jīng)過原點?
當m、n為何值時,y隨x的增大而減小?
當m、n為何值時,函數(shù)的圖象與y軸的交點在x軸的下方?
當m、n為何值時,函數(shù)的圖象經(jīng)過原點?
數(shù)學人氣:444 ℃時間:2020-04-06 13:48:41
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