A?

1 year agothat's not quite right if b = the coefficient of x, and period = 2pi/b what would b need to be?

1 year agoB

1 year agogood, period = pi/2 = 2pi/b so b = 4 and B is the correct sol'n

1 year agoB

1 year agogood

1 year agoB

1 year agogood

1 year ago5

1 year agowell done

1 year agohonestly I'd just graph these, some of these comparisons are a pain to work w/ by hand

1 year agoA and B

1 year agogood

1 year agoD

1 year agohm. check the period again since the period of a tan function is normally pi then the coefficent b would be given by period = pi/b what is b?

1 year agoA

1 year agoawesome, A

1 year agoB

1 year agohm, that's not quite, remember sec has a range from (-infinity,-1) U (1,infinity) so adding 2 would make the graph cross (0,3) not (0,2)

1 year agoas a hint, to make it cross (0,2) you would have to do a vertical stretch by 2

1 year agoD

1 year agoC

1 year agoit can't be a csc function (notice where the asymptotes are) a vertical shift by 2 would be multiplying the entire function by 2

1 year agotake the sec function, apply a vertical stretch by 2, what's the only possibility?

1 year agoA

1 year agogood

1 year agowell, for the regular y = cot(x) where are the asymptotes normally?

1 year agoB and C

1 year agohm, I got something a little different cot(x) = cos(x)/sin(x) so cot(x) takes its asymptotes when sin(x) = 0, what values of x would that be?

1 year agoOh A and D

1 year agohm, that's not quite right sin(x) takes its 0 values when x = n * pi, so every multiple of pi is an asymptote for cot(x) however, the function in the problem is cot(x/2) not cot(x), so the function gets stretched horizontally and takes its asymptote every 2pi so which answers would be included?

1 year agoC and D

1 year agois pi/2 a multiple of 2pi?

1 year agoNo

1 year agohint - sin(0) = 0

1 year agoI know it has to do with x=2πn

1 year agogood, so what do you get when n = 0?

1 year ago0

1 year agoB

1 year agogood, so 0, 2pi, -2pi, etc. so B+D

1 year ago-7

1 year agogood

1 year agoC

1 year agowell done

1 year ago