Question

How do you find the asymptotes of a trigonometric function?
Say the function is f(x)=tan2x
and the asymptotes are from x=0 to x=pi.
I know how to solve for asymptotes, I just don't know what to do when they give me a "range"

Answer 1

well... a range just means " what vertical asymptotes occur between here and here"

Answer 2

and tangent function has infinite vertical asymptotes, since its period is based on it

Answer 3

infinite? i don't see that as a solution

Answer 4

the question asks *where the asymptotes are

Answer 5

heheh... I said, that tangent itself, has infinite since it runs infintely

Answer 6

well.... the asymptotes will lie where the period for tangent starts and end

so for tan(2x) for example the period for that function will be of \(\bf tan({\color{brown}{ 2}}x)\qquad period\implies \cfrac{\pi}{{\color{brown}{ 2}}}\)

Answer 7

period of a trig function, is the "multiplier" of the angle, dividing by the "regular period" of it

tangent has a regular period of \(\pi\)

Answer 8

mmm.. I know what the period is. but how would i use it to find where the asymptotes are?? do i add pi/2b to it for one of them?

Answer 9

yeap

Answer 10

so i get 5pi/4 for the first asymptote... but is it at x=0??

Answer 11

hmmm

Answer 12

the period for this one is \(\bf tan({\color{brown}{ 2}}x)\qquad period\implies \cfrac{\pi}{{\color{brown}{ 2}}}\)

not \(\bf \cfrac{\pi}{4}\)

Answer 13

so, there's one at 0
then \(\bf\cfrac{\pi}{2}, \cfrac{\pi}{2}+\cfrac{\pi}{2}\)

Answer 14

so i add pi/2 with pi/4?

Answer 15

because pi/2b is pi/4

Answer 16

anyhow

you'd go from one period
to the next
to the next

at every period stop, the x-coordinate is the vertical asymptote

Answer 17

i don't exactly follow. can i just show you the possible solutions to them? and you can explain to me which one is correct? working backwards?

Answer 18

1: x=0, x=pi/4
2: x=pi/4, x=3pi/4
3. x=pi/2, x=3pi/2

Answer 19

... better to graph the function to see where the asymptotes are...