解:
âµä¸æ¬¡å½æ°y=kx+bçå¾è±¡ä¸ç´çº¿y=-2x+5å¹³è¡
æ以k=-2
æ以解æå¼ä¸ºy=-2x+b
å 为y=-2x+bçå¾è±¡è¿è¿ç¹A(1,-1)
æ以-1=-2+b
æ以b=1
æ以解æå¼ä¸ºy=-2x+1
æ¯-xæï¼0å³æè½ä»£å
¥è§£æå¼çè¿æ ·è§£å½x大äºçäº0æ¶,f(x)=x(1+x)
å½xå°äº0æ¶,-x>0,f(-x)=-x(1-x)
åå½æ°f(x)æ¯å®ä¹å¨Rä¸çå¥å½æ°
æ以f(-x)=-f(x)
äºæ¯f(x)=-f(-x)=x(1-x) xå°äº0
å¾[x+(b-k)/2]^2+[(k+2)^2+4c-4]/4a-[(b-k)/2a]^2=0 åªæä¸ä¸ªè§£
å¾[(k+2)^2+4c-4]/4a=[(b-k)/2a]^2
å并为kçå类项å¾
ï¼a-1)k^2+(4a+2b)k+3a+ac-b^2=0
æ ¹æ®é¢æï¼ä¸å¼å¯¹ä»»æå®æ°ké½æç«
é£ä¹å¯å¾ï¼a-1=0
4a+2b=0
3a+ac-b^2=0
解ä¸é¢çæ¹ç¨ç»å¾ï¼a=1,b=-2,c=1
å æ¤äºæ¬¡å½æ°y=x^2-2x+1
温馨提示:答案为网友推荐,仅供参考