Question

16．一次函數(shù)y=kx+b（k≠0）的圖象如圖所示，則（　�。�

• A． $\left\{\begin{array}{l}{k=-\frac{1}{3}}\\{b=-1}\end{array}\right.$
• B． $\left\{\begin{array}{l}{k=\frac{1}{3}}\\{b=1}\end{array}\right.$
• C． $\left\{\begin{array}{l}{k=3}\\{b=1}\end{array}\right.$
• D． $\left\{\begin{array}{l}{k=\frac{1}{3}}\\{b=-1}\end{array}\right.$

Answer 1

分析 根據(jù)圖象的性質(zhì)，利用待定系數(shù)法得出解析式即可．

解答 解：由圖象可得：$\left\{\begin{array}{l}{b=-1}\\{3k+b=0}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{k=\frac{1}{3}}\\{b=-1}\end{array}\right.$，
故選D

點評 此題考查一次函數(shù)的解析式，關(guān)鍵是利用待定系數(shù)法得出解析式．