Question

(本小題滿分10分)                                                       
已知函數f ( x ) =( a > 0且a ≠1)圖象經過點Q(8, 6).
(Ⅰ) 求a的值,并在直角坐標系中畫出函數f ( x )的大致圖象;
(Ⅱ) 求函數f ( t ) – 9的零點.

Answer 1

(Ⅰ) a = 3.圖像略
(Ⅱ) x =  – 2 , 或x =" 5" .   

（1) 由x =" 8" > 3, 且點Q在函數圖象上得: 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" (負舍去)
得 x =  – 2 , 或x =" 5" .