急急急~ 一道中考题,望求解~ 三克油~~ 在线等! 谢~~~

2024-12-15 12:34:57
推荐回答(1个)
回答1:

解:(1)y=kx+3,
当x=0时,y=3,
∴D的坐标是(0,3);

(2)y=kx+3,
∵当y=0时,x=-

3
k
,∴OC=-

3
k
,∵

OC
AC
=

1
2
,∴AC=2OC=-

6
k
,∴OA=BP=-

3
k
+(-

6
k
)=-

9
k
,即P的横坐标是-

9
k


代入y=kx+3得:y=-6,
即P(-

9
k
,-6),

∴OB=AP=6,
∵S△DBP=27,
∴S△DOC+S梯形OCPB=27,


1
2
×(-

3
k
)×3+

1
2
×(-

3
k
-

9
k
)×6=27,∴k=-

2
3
,∴P(

27
2
,-6),m=

27
2
×(-6)=-81,即一次函数的解析式是y=-

2
3
x+3,反比例函数的解析式是y=-

81
x


(3)根据图象写出当x>

27
2 时,一次函数的值小于反比例函数的值.(中间的空档时分数线)

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