tag:blogger.com,1999:blog-30226356.post5606741895148024860..comments2024-02-16T23:32:12.073-08:00Comments on The Exponential Curve: Algebra 2: Horizontal Shift and Review/STAR ProblemsDan Wekselgreenehttp://www.blogger.com/profile/08696028020767073620noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-30226356.post-22467506433833740632008-11-23T16:17:00.000-08:002008-11-23T16:17:00.000-08:00It's late for this time through, but...I start wit...It's late for this time through, but...<BR/><BR/>I start without f(x) notation...<BR/><BR/>Circles and absolute value are great places to begin.<BR/><BR/>(x - h)^2 + (y - k)^2 = r^2<BR/>As you calculate distance to the point, you subtract. Kids know it. I bet even kids not from New York can find the distance from 42nd St to 125th St. <BR/><BR/>Time playing with circles is time well spent.<BR/><BR/>Then y = a|x - h| + k<BR/>Easy to graph. use y = |x| as reference, then run through, for example<BR/><BR/>y = 2|x|<BR/>y = -|x|<BR/>y = ½|x|<BR/>y = |x| + 1<BR/>y = |x| - 4<BR/>y = |x + 2| <BR/>y = |x - 3|<BR/>Really, more practice is better.<BR/>"Discover" the effect of modifying a, h, and k.<BR/><BR/>and then the hook back to circles:<BR/><BR/>y - k = a|x - h|<BR/><BR/>Subtraction is more the normal mode of things.<BR/><BR/>Then, maybe, y = a[f(x - h)] + k.<BR/><BR/>But that's fairly abstract notation, and very very hard without a smooth transition.Anonymousnoreply@blogger.com