Question

20．x-2y=2变成直线2x′-y′=4的伸缩变换为$\begin{array}{l}\left\{\begin{array}{l}{x^'}=x\\{y^'}=4y\end{array}\right.\end{array}$．

Answer 1

分析 将直线x-2y=2变成直线2x′-y′=4即直线x′-$\frac{1}{2}$y′=2，横坐标不变，纵坐标变为原来的4倍，即可得出结论．

解答 解：直线2x′-y′=4即直线x′-$\frac{1}{2}$y′=2．
将直线x-2y=2变成直线2x′-y′=4即直线x′-$\frac{1}{2}$y′=2，
故变换时横坐标不变，纵坐标变为原来的4倍，
即有伸缩变换$\begin{array}{l}\left\{\begin{array}{l}{x^'}=x\\{y^'}=4y\end{array}\right.\end{array}$．
故答案为$\begin{array}{l}\left\{\begin{array}{l}{x^'}=x\\{y^'}=4y\end{array}\right.\end{array}$．

点评 本题考查函数的图象变换，判断横坐标不变，纵坐标变为原来的4倍，是解题的关键．