Question

9．畫出|$\frac{1}{2}$x-$\frac{\sqrt{3}}{6}$y|+|$\frac{\sqrt{3}}{3}$y|≤1的圖象．

Answer 1

分析 先去絕對值，然后分別畫出相應的圖象．

解答 解：|$\frac{1}{2}$x-$\frac{\sqrt{3}}{6}$y|+|$\frac{\sqrt{3}}{3}$y|≤1，
∴$\left\{\begin{array}{l}{\frac{1}{2}x-\frac{\sqrt{3}}{6}y≥0}\\{\frac{\sqrt{3}}{3}y≥0}\\{\frac{1}{2}x-\frac{\sqrt{3}}{6}y+\frac{\sqrt{3}}{3}y≤1}\end{array}\right.$或$\left\{\begin{array}{l}{\frac{1}{2}x-\frac{\sqrt{3}}{6}y≥0}\\{\frac{\sqrt{3}}{3}y≤0}\\{\frac{1}{2}x-\frac{\sqrt{3}y}{6}-\frac{\sqrt{3y}}{3}≤1}\end{array}\right.$或$\left\{\begin{array}{l}{\frac{1}{2}x-\frac{\sqrt{3}}{6}y≤0}\\{\frac{\sqrt{3}}{3}y≥0}\\{-\frac{1}{2}x+\frac{\sqrt{3}}{6}y+\frac{\sqrt{3}}{3}y≤1}\end{array}\right.$或$\left\{\begin{array}{l}{\frac{1}{2}x-\frac{\sqrt{3}}{6}y≤0}\\{\frac{\sqrt{3}}{3}y≤0}\\{-\frac{1}{2}x+\frac{\sqrt{3}}{6}y-\frac{\sqrt{3}}{3}y≤1}\end{array}\right.$，
即$\left\{\begin{array}{l}{y≤\sqrt{3}x}\\{y≥0}\\{y≤-\sqrt{3}x+2\sqrt{3}}\end{array}\right.$或$\left\{\begin{array}{l}{y≤\sqrt{3}x}\\{y≤0}\\{y≥\frac{\sqrt{3}}{3}x-\frac{2\sqrt{3}}{3}}\end{array}\right.$或$\left\{\begin{array}{l}{y≥\sqrt{3}x}\\{y≥0}\\{y≤\frac{\sqrt{3}}{3}x+\frac{2\sqrt{3}}{3}}\end{array}\right.$或$\left\{\begin{array}{l}{y≥\sqrt{3}x}\\{y≤0}\\{y≥-\sqrt{3}x-2\sqrt{3}}\end{array}\right.$，
其圖象為：

點評 本題考查了函數(shù)圖象的畫法，關鍵是去絕對值，屬于中檔題．