考點(diǎn):二次函數(shù)圖象與系數(shù)的關(guān)系,二次函數(shù)圖象與幾何變換
專題:數(shù)形結(jié)合
分析:A.由開口向下,可得a<0;又由拋物線與y軸交于負(fù)半軸,可得c<0,然后由對(duì)稱軸在y軸右側(cè),得到b與a異號(hào),則可得b>0,故得abc>0.
B.根據(jù)圖知對(duì)稱軸為直線x=2,即
-=2,得b=-4a,再根據(jù)圖象知當(dāng)x=1時(shí),y<0,即可判斷;
C.由拋物線與x軸有兩個(gè)交點(diǎn),可得b
2-4ac>0;
D.把二次函數(shù)y=ax
2+bx+c化為頂點(diǎn)式,再求出平移后的解析式即可判斷.
解答:解:A.由開口向下,可得a<0;又由拋物線與y軸交于負(fù)半軸,可得c<0,然后由對(duì)稱軸在y軸右側(cè),得到b與a異號(hào),則可得b>0,故得abc>0,故本選項(xiàng)錯(cuò)誤;
B.根據(jù)圖知對(duì)稱軸為直線x=2,即
-=2,得b=-4a,再根據(jù)圖象知當(dāng)x=1時(shí),y=a+b+c=a-4a+c=-3a+c<0,故本選項(xiàng)正確;
C.由拋物線與x軸有兩個(gè)交點(diǎn),可得b
2-4ac>0,故本選項(xiàng)錯(cuò)誤;
D.y=ax
2+bx+c=
a(x+)2+,
∵
-=2,
∴原式=
a(x-2)2+,
∴向左平移2個(gè)單位后所得到拋物線的解析式為
y=ax2+,故本選項(xiàng)錯(cuò)誤;
故選:B.
點(diǎn)評(píng):本題考查了二次函數(shù)圖象與系數(shù)的關(guān)系.二次函數(shù)y=ax2+bx+c(a≠0)系數(shù)符號(hào)由拋物線開口方向、對(duì)稱軸、拋物線與y軸的交點(diǎn)和拋物線與x軸交點(diǎn)的個(gè)數(shù)確定.