Question

(本小题满分10分)                                                        

已知函数f ( x ) =( a > 0且a ≠1)图象经过点Q(8, 6).

(Ⅰ) 求a的值,并在直角坐标系中画出函数f ( x )的大致图象;

(Ⅱ) 求函数f ( t ) – 9的零点.

Answer

(Ⅰ) a = 3.图像略

(Ⅱ) x =  – 2 , 或x = 5 .    

【解析】（1) 由x = 8 > 3, 且点Q在函数图象上得: 6 = ( 8 – 5 ) 2 – a , 解得a = 3.                             

      得f ( x ) =                                                                                                                                                        

(2) 由f (x ) = 9, 得 3  – x = 9或(x – 5)2 – 3  = 9,

解得: x =  – 2 , 或x = 5 (负舍去)

得 x =  – 2 , 或x = 5 .