已知二次函數(shù)y=x^2+(a+1)x+b當(dāng)x=3時(shí),y=3,并且對(duì)一切實(shí)數(shù)x,都有y大于等于x,設(shè)y1=y-x,試?yán)煤瘮?shù)y1的圖像特征求a、b的值請(qǐng)?jiān)敿?xì)解答,謝謝
熱心網(wǎng)友
將點(diǎn)(3,3)代入y中化簡(jiǎn)得 3a+b+9=0 ①y1=y-x≥0,當(dāng)且僅當(dāng)x=3時(shí),取得y1=0。也就是y1=x^2+ax+b的最低點(diǎn)為(3,0)。故對(duì)稱軸為x=3,-a/2 = 3, == a=-6代入①求得b=9。故a=-6, b=9。