Question

15．根據(jù)如圖所示的函數(shù)y=f（x）的圖象，寫出函數(shù)的解析式．

Answer 1

分析 根據(jù)直線的兩點(diǎn)式方程求出函數(shù)的解析式即可．

解答 解：當(dāng)-3≤x≤-1時(shí)，設(shè)f（x）=ax+b，則$\left\{\begin{array}{l}{f（-3）=1}\\{f（-1）=-2}\end{array}\right.$，即$\left\{\begin{array}{l}{-3a+b=1}\\{-a+b=-2}\end{array}\right.$，解得a=-$\frac{3}{2}$，b=-$\frac{7}{2}$，此時(shí)f（x）=-$\frac{3}{2}$x-$\frac{7}{2}$，
當(dāng)-1≤x≤1時(shí)，設(shè)f（x）=ax+b，則$\left\{\begin{array}{l}{f（-1）=-2}\\{f（1）=1}\end{array}\right.$，即$\left\{\begin{array}{l}{-a+b=-2}\\{a+b=1}\end{array}\right.$，解得a=$\frac{3}{2}$，b=-$\frac{1}{2}$，此時(shí)f（x）=$\frac{3}{2}$x-$\frac{1}{2}$，
當(dāng)1＜x＜2時(shí)，f（x）=1，
即f（x）=$\left\{\begin{array}{l}{-\frac{3}{2}x-\frac{7}{2}，}&{-3≤x≤-1}\\{\frac{3}{2}x-\frac{1}{2}，}&{-1＜x≤1}\\{1，}&{1＜x＜2}\end{array}\right.$．

點(diǎn)評 本題主要考查函數(shù)解析式的求解，利用待定系數(shù)法是解決本題的關(guān)鍵．