Asked by Poppy
The following transformations š¦ = 3š (1 š„ ā 2š) ā 1 were applied to the parent function f(x) = csc(x). Graph the transformed function for the interval ā4š ⤠š„ ⤠4š. On your graph, label the asymptotes, local max/min, and number each axis. Use mapping notation to show your work for a minimum of 5 key points for full marks.
I'm having a lot of trouble with the local max/min and the 5 key points, could anyone help me? Please and thank you.
I'm having a lot of trouble with the local max/min and the 5 key points, could anyone help me? Please and thank you.
Answers
Answered by
oobleck
csc(x) has asymptotes, with their local branch turning points at min = 1 and max = -1
so, 3f(x-2Ļ)-1 has its turning points at 3-1 and -3-1
Note also that since the period of csc(x) is 2Ļ, shifting right by 2Ļ does not change the graph in any way. So your graph will still cover exactly four periods. The turning points are all just shifted down by 1. There are still no x-intercepts.
The only real key points for csc(x) are the asymptotes at x = kĻ, and the turning points at x = odd multiples of Ļ/2. That is at x = kĻ + Ļ/2
so, 3f(x-2Ļ)-1 has its turning points at 3-1 and -3-1
Note also that since the period of csc(x) is 2Ļ, shifting right by 2Ļ does not change the graph in any way. So your graph will still cover exactly four periods. The turning points are all just shifted down by 1. There are still no x-intercepts.
The only real key points for csc(x) are the asymptotes at x = kĻ, and the turning points at x = odd multiples of Ļ/2. That is at x = kĻ + Ļ/2
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