Question

help please

Answer 1

@Vocaloid

Answer 2

if you can recall, which graph is symmetric about the y-axis, cosine or sine? you can look these functions up on google images if you are having trouble remembering

Answer 3

its hazy..

Answer 4

which one is symmetric on the y-axis?

Answer 5

cosx

Answer 6

good

if we look at the problem, it is just cos(x) with a vertical shrink by 1/2 so the answer is 1/2 cosx (first choice)

Answer 7

the midline of this function is just the horizontal line y = 0 so the first point can be (pi/2,0) since that's where 2cos(x) = 0

the second point needs to be on the max or min, so 2cos(x) has its max when theta = 0, so it crosses (0,2)

Answer 8

the function basically looks like this, but instead of maxing out/minimum at 1 or -1 it reaches 2 and -2

Answer 9

first we determine the length of one of its cycles

Answer 10

then it's just 1/period

Answer 11

1/2?

Answer 12

thats still 1

Answer 13

pi?

Answer 14

yup so frequency = 1/ period = 1/pi

Answer 15

0.318?

Answer 16

weell it doesn't say what format the answer needs to be in, so if they let you enter 1/pi then do that, but if not, then go ahead and round to 0.318

Answer 17

it has to be another cos function based on its shape (symmetry to the y-axis)

it can't be 2cos(x) because the max/min are still 1 not 2 so cos(1/2x) is our only possiblility

Answer 18

D?

Answer 19

first we draw a horizontal line from peak-to-peak to get our period

Answer 20

about what is the length of this horizontal line? (it's a little off but use your best estimate)

Answer 21

uh....

Answer 22

it starts at -pi and goes to 0, how long is this?

Answer 23

5?

Answer 24

think about how we measure the length of a horizontal line

it's (right point - left point)

so 0 - (-pi) = ?

Answer 25

pi?

Answer 26

good so the period = pi = your answer

Answer 27

thanks