Question

20．x-2y=2變成直線2x′-y′=4的伸縮變換為$\begin{array}{l}\left\{\begin{array}{l}{x^'}=x\\{y^'}=4y\end{array}\right.\end{array}$．

Answer 1

分析 將直線x-2y=2變成直線2x′-y′=4即直線x′-$\frac{1}{2}$y′=2，橫坐標(biāo)不變，縱坐標(biāo)變?yōu)樵瓉淼?倍，即可得出結(jié)論．

解答 解：直線2x′-y′=4即直線x′-$\frac{1}{2}$y′=2．
將直線x-2y=2變成直線2x′-y′=4即直線x′-$\frac{1}{2}$y′=2，
故變換時(shí)橫坐標(biāo)不變，縱坐標(biāo)變?yōu)樵瓉淼?倍，
即有伸縮變換$\begin{array}{l}\left\{\begin{array}{l}{x^'}=x\\{y^'}=4y\end{array}\right.\end{array}$．
故答案為$\begin{array}{l}\left\{\begin{array}{l}{x^'}=x\\{y^'}=4y\end{array}\right.\end{array}$．

點(diǎn)評 本題考查函數(shù)的圖象變換，判斷橫坐標(biāo)不變，縱坐標(biāo)變?yōu)樵瓉淼?倍，是解題的關(guān)鍵．