# [SOLVED] Find the first TWO non-negative asymptotes and the first negative asymptote

• Mar 26th 2009, 11:45 AM
lsnyder
[SOLVED] Find the first TWO non-negative asymptotes and the first negative asymptote
Find the first TWO non-negative asymptotes and the first negative asymptote of the graph of https://webwork.uncc.edu/webwork2_fi...22b824af51.png in radians

smallest non-negative asymptote: https://webwork.uncc.edu/webwork2_fi...9019052b81.png

second non-negative asymptote: https://webwork.uncc.edu/webwork2_fi...9019052b81.png
first negative asymptote: https://webwork.uncc.edu/webwork2_fi...9019052b81.png

okay, i know the first one is pi/5

but the other two i do not know how to find. can some one please help me?

but i keep thinkin the other two answers are x=+-(pi/10). (which is wrong)
• Mar 26th 2009, 02:54 PM
sinewave85
Quote:

Originally Posted by lsnyder
Find the first TWO non-negative asymptotes and the first negative asymptote of the graph of https://webwork.uncc.edu/webwork2_fi...22b824af51.png in radians

smallest non-negative asymptote: https://webwork.uncc.edu/webwork2_fi...9019052b81.png

second non-negative asymptote: https://webwork.uncc.edu/webwork2_fi...9019052b81.png
first negative asymptote: https://webwork.uncc.edu/webwork2_fi...9019052b81.png

okay, i know the first one is pi/5

but the other two i do not know how to find. can some one please help me?

but i keep thinkin the other two answers are x=+-(pi/10). (which is wrong)

Hi, Isnyder. The trick to this is to realize that what you are doing is simply solving for those values of x that make cot undefined (they cause division by 0): $-\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi$

Let $\cot\left(x - \frac{\pi}{5}\right) = \cot{\theta}$ so that $x - \frac{\pi}{5} = \theta$

So the first non-negative asymptote is at $\frac{\pi}{5}$ as you guessed:

$x - \frac{\pi}{5} = 0$

$x = \frac{\pi}{5}$

And the second non-negative asymptote:

$x - \frac{\pi}{5} = \frac{\pi}{2}$

$x = \frac{\pi}{2} + \frac{\pi}{5} = \frac{2\pi + 5\pi}{10} = \frac{7\pi}{10}$

Lastly, the first negative asymptote:

$x - \frac{\pi}{5} = -\frac{\pi}{2}$

$x = -\frac{\pi}{2} + \frac{\pi}{5} = \frac{-3\pi}{10}$
• Mar 26th 2009, 03:06 PM
lsnyder
actually the last two answers are 6pi/5 and -4pi/5.

because you add pi instead of pi/2. because the pi is smaller.

but if it wasn't for you saying all that, I would of never have gotten the right answer. Thank you! (Bigsmile)
• Mar 26th 2009, 03:18 PM
sinewave85
Quote:

Originally Posted by lsnyder
actually the last two answers are 6pi/5 and -4pi/5.

because you add pi instead of pi/2. because the pi is smaller.

but if it wasn't for you saying all that, I would of never have gotten the right answer. Thank you! (Bigsmile)

I am fairly sure that $\pi > \frac{\pi}{2}$.

$\pi \approx 3.14159$

$\frac{\pi}{2} \approx 1.57080$

• Mar 26th 2009, 03:25 PM
lsnyder
Quote:

Originally Posted by sinewave85
I am fairly sure that $\pi > \frac{\pi}{2}$.

$\pi \approx 3.14159$

$\frac{\pi}{2} \approx 1.57080$

(pi/5)+pi = 6pi/5 rather than (pi/5)+(pi/2)=7pi/10
6pi/5 < 7pi/10

same for the negative

(pi/5)-pi =-4pi/5 rather than (pi/5)-(pi/2) = -3pi/10
-4pi/5 <-3pi/10

also, it came up right in my homework when i typed it in.
• Mar 26th 2009, 03:36 PM
sinewave85
Quote:

Originally Posted by lsnyder

(pi/5)+pi = 6pi/5 rather than (pi/5)+(pi/2)=7pi/10
6pi/5 < 7pi/10

same for the negative

(pi/5)-pi =-4pi/5 rather than (pi/5)-(pi/2) = -3pi/10
-4pi/5 <-3pi/10

also, it came up right in my homework when i typed it in.

$\frac{6\pi}{5} \approx 3.7699$ is actually greater than (>) $\frac{7\pi}{10} \approx 2.19911$
$\frac{-4\pi}{5} \approx -2.51327$ is less that $\frac{-3\pi}{10} \approx-0.942478$, but that means that it is further away from 0. The absolute value of the former is greater than the latter.