Urgent! How to find asymptotes for a tan graph?

Question

Argh important test tomorrow!
For the equation y= 3tan(2X - pi/3) for pi/6 < X <13pi/6
How do I find the asymptotes? I've tried using the formula
x= {(2k + 1) / n x pi/2 } but I'm not getting the right answers?
Can someone please spell it out to me how to find them?? Thanks!

Answer 1

tan(pi/2+n*pi) for integer n is undefined - infinite. So if you set 2x - pi/3 = pi/2+n*pi and solve for x, you get a formula showing the x for which y goes infinite, and hence defines a vertical asymptote.

It may also help to graph this, by pencil and paper if nothing else is allowed or available.

Answer 2

The tangent asymptotes are at tan(pi/2) and tan(3 pi/2), and odd multiples of pi/2 (plus and minus). That works out to pi/2 plus or minus n pi.

Do this: Set 2x - pi/3 = pi/2 + n pi, and solve for x.

2x = pi/3 + pi/2 + n pi
x = pi/2 (1/3 + 1/2 + n) = pi/2 (2/6 + 3/6 + n)
x = pi (5/12 + n)

Okay. These x values will give you asymptotes. The only thing left is to see which of these x values are in your range.

Try this:

pi/6 < pi (5/12 + n) < 13 pi/6
1/6 < 5/12 + n < 13/6
1/6 - 5/12 < n < 13/6 - 5/12
-3/12 < n < 21/12

n is an integer between 0 and 1, according to that inequality.

So if n is 0 or 1, x = pi (5/12 + n). For n=0, x = 5 pi/12; and for n=1, x = 17 pi/12.

Plug those two x-values in and see what happens:

y = 3 tan [2(5 pi/12) - pi/3] = 3 tan (10 pi/12 - pi/3)
y = 3 tan (5 pi/6 - 2 pi/6) = 3 tan (3 pi/6) = 3 tan(pi/2)
which gives you an asymptote.

For n=1 and x = 17 pi/12:
y = 3 tan [2(17 pi/12) - pi/3] = 3 tan(34 pi/12 - pi/3)
y = 3 tan(17 pi/6 - 2 pi/6) = 3 tan(15 pi/6) = 3 tan (5 pi/2)
y = 3 tan (pi/2)
which is another asymptote.

[Edit: Although I'm sleepy, I just checked the algebra and found no mistake. Your solutions are x = 5 pi/12 and 17 pi/12.]