Question

16．一次函数y=kx+b（k≠0）的图象如图所示，则（　　）

• 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 根据图象的性质，利用待定系数法得出解析式即可．

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

点评 此题考查一次函数的解析式，关键是利用待定系数法得出解析式．