如图抛物线y=-x2+5x+k经过点C(4,0)与x轴交于另一点A,与y轴交于点B. (1)求AC的长; (2)求出△ABC的面积.
问题描述:
如图抛物线y=-x2+5x+k经过点C(4,0)与x轴交于另一点A,与y轴交于点B.
(1)求AC的长;
(2)求出△ABC的面积.
答
(1)设A(x1,0),
∵C(4,0),
∴x1+4=5,
∴x1=1,
∴AC=|1-4|=3;
(2)∵x1=1,点C(4,0),
∴4x1=-k,x1=-
=1,k=-4,k 4
∴B点坐标为(0,-4),
∴S△ABC=
AC•|k|,1 2
=
×3×4,1 2
=6.
故答案为:AC=3,S△ABC=6.