数学函数,这题怎么做??要所有过程,谢谢啦!!
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解:①因为抛物线于x轴有两个交点,所以△>0,即 b²﹣4ac>0 ;
②∵抛物线开口向上,∴a>0,对称轴x=1,即﹣b/2a=1,b=﹣2a········(1)
∴b<0,由图知: f(0)<0,即c<0,abc>0 ;
③由图知f(-2)>0, f(,-1)<0,即4a﹣2b+c>0··········(2),
a﹣b+c<0···········(3)
把(1)式代入(2)得8a+c>0 ;
④由(1)得2a+b=0,∴ 6a+3b=0·········(4),
把(1)式代入(3)得:3a+c<0··········(5),
(4)+(5)得:9a+3b+c<0 。即4个答案都对。选D
②∵抛物线开口向上,∴a>0,对称轴x=1,即﹣b/2a=1,b=﹣2a········(1)
∴b<0,由图知: f(0)<0,即c<0,abc>0 ;
③由图知f(-2)>0, f(,-1)<0,即4a﹣2b+c>0··········(2),
a﹣b+c<0···········(3)
把(1)式代入(2)得8a+c>0 ;
④由(1)得2a+b=0,∴ 6a+3b=0·········(4),
把(1)式代入(3)得:3a+c<0··········(5),
(4)+(5)得:9a+3b+c<0 。即4个答案都对。选D
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