088]解:(1)法一:由图象可知:抛物线经过原点. 设抛物线解析式为．把.代入上式得:·····················································——青夏教育精英家教网—

088]解:(1)法一:由图象可知:抛物线经过原点. 设抛物线解析式为．把.代入上式得:···················································································· 1分解得······················································································· 3分 ∴所求抛物线解析式为········································································ 4分法二:∵.. ∴抛物线的对称轴是直线．设抛物线解析式为()······························································ 1分把.代入得解得··········································································· 3分 ∴所求抛物线解析式为．····························································· 4分 (2)分三种情况: ①当.重叠部分的面积是.过点作轴于点. ∵.在中... 在中... ∴. ∴．··············································· 6分 ②当.设交于点.作轴于点. .则四边形是等腰梯形. 重叠部分的面积是． ∴. ∴．··········· 8分 ③当.设与交于点.交于点.重叠部分的面积是．因为和都是等腰直角三角形.所以重叠部分的面积是． ∵.. ∴. ∴. ∴ ．················································ 10分 (3)存在 ···································································································· 12分 ··································································································· 14分【查看更多】

题目列表(包括答案和解析)

9、如图是抛物线y=ax²+bx+c的一部分，其对称轴为直线x=1，若其与x轴一交点为B（3，0），则由图象可知，不等式ax²+bx+c＞0的解集是

x＜-1或x＞3

．