zarkam21:

A?

8 months ago
zarkam21:

1 attachment
8 months ago
Vocaloid:

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

8 months ago
zarkam21:

B

8 months ago
Vocaloid:

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

8 months ago
zarkam21:

1 attachment
8 months ago
zarkam21:

B

8 months ago
Vocaloid:

good

8 months ago
zarkam21:

1 attachment
8 months ago
zarkam21:

B

8 months ago
Vocaloid:

good

8 months ago
zarkam21:

1 attachment
8 months ago
zarkam21:

5

8 months ago
Vocaloid:

well done

8 months ago
zarkam21:

1 attachment
8 months ago
Vocaloid:

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

8 months ago
zarkam21:

A and B

8 months ago
Vocaloid:

good

8 months ago
zarkam21:

1 attachment
8 months ago
zarkam21:

D

8 months ago
Vocaloid:

hm. 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?

8 months ago
zarkam21:

A

8 months ago
Vocaloid:

awesome, A

8 months ago
zarkam21:

1 attachment
8 months ago
zarkam21:

B

8 months ago
Vocaloid:

hm, 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)

8 months ago
Vocaloid:

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

8 months ago
zarkam21:

D

8 months ago
zarkam21:

C

8 months ago
Vocaloid:

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

8 months ago
Vocaloid:

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

8 months ago
zarkam21:

A

8 months ago
Vocaloid:

good

8 months ago
zarkam21:

1 attachment
8 months ago
Vocaloid:

well, for the regular y = cot(x) where are the asymptotes normally?

8 months ago
zarkam21:

B and C

8 months ago
Vocaloid:

hm, 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?

8 months ago
zarkam21:

Oh A and D

8 months ago
Vocaloid:

hm, 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?

8 months ago
zarkam21:

C and D

8 months ago
Vocaloid:

is pi/2 a multiple of 2pi?

8 months ago
zarkam21:

No

8 months ago
Vocaloid:

hint - sin(0) = 0

8 months ago
zarkam21:

I know it has to do with x=2πn

8 months ago
Vocaloid:

good, so what do you get when n = 0?

8 months ago
zarkam21:

0

8 months ago
zarkam21:

B

8 months ago
Vocaloid:

good, so 0, 2pi, -2pi, etc. so B+D

8 months ago
zarkam21:

1 attachment
8 months ago
zarkam21:

-7

8 months ago
Vocaloid:

good

8 months ago
zarkam21:

1 attachment
8 months ago
zarkam21:

C

8 months ago
Vocaloid:

well done

8 months ago
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