Question

MEDAL!! Determine the period, domain, range, zeros, and asymptotes (if any). y=-3tan(1/2x)

Answer 1

I know the period is 2pi and the range is all real numbers already

Answer 2

@Kainui

Answer 3

So what are you missing and what do you think they might be? Make your best guess and I'll help you get it right!

Answer 4

I am missing domain, asymptotes, and zeros. Asymptotes = 2, pi/2? I really don't know those 3 ones

Answer 5

@OrangeMaster If you don't know an answer, it's better if you don't reply at all.

Answer 6

Lol okay

Answer 7

@Nurali

Answer 8

y = -3tan(x/2).

Period: Since tan(x) has a period of pi, tan(x/2) will have a period of pi / (1/2) = 2(pi).

Domain: tan(x) is undefined when x is: pi/2 + n(pi) where n is an integer ...-2, -1, 0, 1, 2, .... Therefore tan(x/2) will be undefined when x is: pi + 2n(pi) or (2n+1)pi. Domain of tan(x/2) is:
{x|x≠(2n+1)π,n=...,−2,−1,0,1,...}={x|x≠...,−3π,−π,π,3π,...}

Range: tan(x) has the range (-infinity, infinity) and so does tan(x/2)

Zeros: tan(x) has zeros at x = n(pi) where n is an integer: ...-1, 0, 1, ...
Therefore, tan(x/2) has zeros at x = 2n(pi) where n is an integer: ...-1, 0, 1, ... or at x = ..., -4pi, -2pi, 0, 2pi, 4pi, ...

Asymptotes: Vertical asymptotes will be when tan(x/2) goes to +/- infinity. We found those point earlier and excluded them from the domain because tan(x/2) will be undefined at those x values. Those same x values will be the location of the vertical asymptotes. At x = (2n+1)pi where n is an integer ot at x = ..., -3pi, -pi, pi, 3pi, ...

Answer 9

@Nurali the equation is y=-tan(1/2x). so to recap:
Domain= -2,-1,0,1, -3pi, -pi, pi, 3pi
Zeros= -4pi,, -2pi, 0, 2pi, 4pi
Asymptotes= -3pi, =pi, pi, 3i

Is that exactly what I would type?

Answer 10

I mean y=-3tan(1/2x). Is the recap correct @Nurali

Answer 11

@primeralph

Answer 12

Domain is all real numbers.

Answer 13

so the domain and range are both all real numbers for y=-3tan(1/2x)?

Answer 14

Think of a number for which you can't find the tangent or the inverse tangent.

Answer 15

ok that makes sense. what about zeros and asymptotes? I've been trying to get help for almost 2 hours, so thank you!

Answer 16

Solve for x when the equation equals 0.

Answer 17

0, +-2, +-2pi?

Answer 18

There's more. So simply, 2n*pi, where n is an integer.

Answer 19

so 0 +-2npi, n+-2npi?

Answer 20

Just 2npi.

Answer 21

so the answer for zeros is just 2npi?

Answer 22

You have to specify that n is any integer.

Answer 23

so I would exactly say, zeros: "2npi where n is any integer"

Answer 24

Yeah, that works. Put a comma before "where".
And pi -don't write that in words.

Answer 25

Ok :) asymptotes?

Answer 26

Try plotting the graph.

Answer 27

I did already

Answer 28

Then you should see the asymptotes.

Answer 29

I dont? 2pi?

Answer 30

What's an asymptote?

Answer 31

a line that approaches a curve but doesn't meet it

Answer 32

https://www.google.com/search?q=y%3D-3tan(1%2F2x)&rlz=1C1CHFX_enUS524US524&oq=y%3D-3tan(1%2F2x)&aqs=chrome..69i57&sourceid=chrome&espv=210&es_sm=93&ie=UTF-8#q=y%3D-3tan(1%2F2*x)

Answer 33

There's a lot of them in that graph.

Answer 34

I know the graph because I graphed it already. pi/2?

Answer 35

is it 0±2πn,π±2πn

Answer 36

and I specify again that n is any integer?

Answer 37

Again, if you graphed it right, you'd see the correct asymptotes.

Answer 38

what are they then? I seriously don't know. I thought they were my last answer

Answer 39

(2n-1)*pi.

Answer 40

You should really learn this stuff.

Answer 41

can you explain why that is the correct answer for asymptotes?

Answer 42

@primeralph

Answer 43

For the tangent function, asymptotes occur when tan goes to +-infinity as seen in the graph.
This occurs at pi, 3pi, 5pi, and so on.
To model odd consecutive numbers, you have to have 2n-1.

Answer 44

Or 2n+1. Either way.

Answer 45

ohhh but is (2n-1)*pi the right answer? or 2n+1?

Answer 46

Both are right. They both model odd numbers.

Answer 47

should I say (2n+-1)*pi?

Answer 48

You can pick one or both.

Answer 49

oh so (2n+-1)* pi is an acceptable answer then? I just don't want to screw up later on

Answer 50

Actually, I'll just stay with (2n-1)* pi

Answer 51

Okay, good luck.

Answer 52

thanks. the test is next week, so hopefully you were right on these or I just completely learned the wrong way lol

Answer 53

You're in charge of how you learn.

Answer 54

but your answers and explanation were correct right lol?

Answer 55

How about you check them out? If you got a test, you should be able to know a right answer from a wrong one by now.

Answer 56

why would you help me by giving me the wrong answers?

Answer 57

Then why are you asking if they're right?

Answer 58

I'm tired lol thanks for the help!

Answer 59

Good luck. Study well.