Question

I'm trying to figure out the asymptotes for y= -2 sec (pi*x/3) ? with the period being 6. By looking at the graph it kind of looks like 1.5 and 4.5 but I'm confused as to how to find the exact x value?

Answer 1

hey sourwing hows it goin.. do you think you may be able to point me in the right direction here?

Answer 2

what do you think sec(u) is?

Answer 3

I just set pi*x/3 equal to pi/2 and 3pi/2, since the cos of theta is at pi/2 and 3pi/2 when it's zero, but I'm doing something wrong?

Answer 4

@sourwing is wierd

Answer 5

x)

Answer 6

lol @Pooch

@lahernan why did you set xpi/3 = pi/2 and 3pi/2?

Answer 7

I know sec is = 1/cos theta

Answer 8

good, what did you do next?

Answer 9

well, since the function of sec is just the reciprocal of cos I just set pi*x/3=pi/2 to figure the the asymptote is when cos can't be zero, and cos is zero on the unit circle at pi/2 ? but I'm lost somewhere because i know for whatever reason that's not right?

Answer 10

You're on the right track. Except you forgot the sec(x) is a periodic function. Which means has how many asymoptes?

Answer 11

oh yeah. sorry. Infinite. I guess I should of said, just for one period. Period being equal to 6

Answer 12

Do you you have the graph. There are infinitely many parts of the graph that has the period of 6.

Answer 13

umm , i have it on my computer, but i dont know how to post it on here?

Answer 14

What is the domain on the graph?

Answer 15

So how are you supposed to know which period to designate?

Answer 16

domain is all real numbers except for according to what I'm thinking is pi/2+pi?

Answer 17

well yes, But I was preferring the domain on the graph. Is it like x = 0 to x = 6 ?

Answer 18

oh yes. correct.

Answer 19

well, x = 3/2 and x = 9/2 are the vertical asymptotes on that specific range. Unless the questions wants you to find ALL vertical asymptotes

Answer 20

no those are the only ones. but I don't understand how did you get those values?

Answer 21

well if 6 is the period, what do you think the value of the function at x= 1/2 + 6 is? or at 9/2 + 6?

Answer 22

it would be the same value just adding 6.

Answer 23

typo at x = 3/2 + 6

Answer 24

Yes, it would be the same, which means f is also what at 3/2 + 6?

Answer 25

zero

Answer 26

yes, 0 for cos(x) but what about 1/cos(x)?

Answer 27

but i'm trying to think in terms algebraically? as if i had no graph

Answer 28

Yes, i'm trying to help you get all asymtopes.

So what is the value of the original function at x = 3/2 + 6 or at x = 9/2 + 6?

Answer 29

1?

Answer 30

no, we're looking for *vertical asymptotes* yes? which means the function is what at those asymptotes?

Answer 31

oh you mean in radians?

Answer 32

the function is unbounded?

Answer 33

well, x is already in radian. The answer I was looking for is the function is *undefined* at those vertical asymptotes

Answer 34

see thats exactly where I'm mixed up somehow because if the value for 1/cos is undefined at cos =0 why don't I get those two values 3/2 and 9/2 when I set pix/3= pi/2?

Answer 35

well, let me give you a simple function. Say f(x) = 1/(2x). Clearly the function is undefined at x = 0 yes?

Answer 36

sorry bout that. yes

Answer 37

which means x ≠ 0 yes?

Answer 38

correct

Answer 39

Back the the original problem. -2sec(pix/3) means -2/cos(pix/3)

which means cos(pix/3) ≠ 0, yes?

Answer 40

correct.

Answer 41

ok. so it would be negative two divided by cos pi/3 which would be 1/2?

Answer 42

i have to constanly refresh my browser i keep losing connection.

Answer 43

no, -2 is on the numerator so there is no problem.

when you were solving cos(pix/3) ≠ 0, you were really trying to solve
cos(pix/3) = 0. Whatever the solutions are, those are the value x can *not* be yes?

so really x ≠ 3/2 and x ≠ 9/2. But the definition of vertical asymptote is where x can *not* be. Which means vertical asymptotes are at x = 3/2, x = 9/2 yes?

Answer 44

which also means the function is *undefined* at those x values yes? Now the only problem we have left is how to find the other ones since there are infinitely many of them

Answer 45

ok. i see what your saying. So I was trying to solve for the problem when there was none.

Answer 46

man, i think im so used to dealing with pi, i'm not connecting the dots, when i see numbers. thank you for your help . As always, you've shown me the path. Thanks.

Answer 47

so when i asked you what is the valued at 3/2 + 6. You should have said *undefined*, yes?

Answer 48

"":) yea

Answer 49

what do you think the value of the function at x = 3/2 + 2(6) is?

Answer 50

undefined

Answer 51

good, how about at x = 3/2 + 3(6) ?

Answer 52

undefined. Yea i need a personal tutor, just to really get the basics to sink in.
Thanks Sour

Answer 53

no problem. What do you the final answer is?

Answer 54

in other works, where do you think all the asymptotes are at?

Answer 55

*words*, ..

Answer 56

where the function is undefined. thanks man. I'll definitely hit u up next time i get on. TTyl.