Question

Which of the following are vertical asymptotes of the function y = 3cot(1/2x) - 4?

Answer 1

A. x = 0
B. x = pi/2
C. x = +- 2pi
D. x = 3pi

Answer 2

What part os giving you trouble with this?

Answer 3

i just cant solve it. hate trig.

Answer 4

If you plan on taking calculus, learn to love trig.

How far do you get before you get stuck?

Answer 5

i dont know how to even start it.

Answer 6

OK, well, do you know what a vertical asymtote is? Lets start there.

Answer 7

no

Answer 8

AH! OK, that is probably the real problem. A vertical asymptote is a line the range of the equation does not cross. So it goes from \(-\infty\) to \(\infty\) and never gets crossed. Where this happens with cotangent is when the bottom part is 0. Know how cotangent is cosign over sign? That is the bottom part I mean. When the sign part would be 0, the equation becomes undefined and there is a vertical asymptote.

Answer 9

Well, sine... LOL. Not sign. Mistyped there, but you should get the point.

Answer 10

so the answer is A?

Answer 11

That is pne answer. This is probably multiple selectables. There is more than one answer.

Answer 12

Test each of the points in say a calculator. Any place you get an error for an answer is it. That is one way. Knowing the unit circle is another.

Answer 13

i dont have a calc left it at skool

Answer 14

ah. Now, one question, is that \(\frac{1}{2}x\) or \(\frac{1}{2x}\) just to be sure I am looking at the question correctly.

Answer 15

first one

Answer 16

So we could call it x/2. So, any place \(\sin{\frac{x}{2}}=0\). If you look carefully, nothing else about that problem matters. The -4 shifts the graph down four, but does not change the period. The 3 stretches it, but does not change the period. So all that matters is when \(\sin{\frac{x}{2}}=0\).

Answer 17

wait i m calculating

Answer 18

np. If you have the unit circle, you can just try those four points and see what they become.

Answer 19

Just remember it is \(\frac{x}{2}\). That DOES change the period!

Answer 20

so how should i try solving this in the calulator

Answer 21

Well, lets take a point. Say that \(\frac{\pi}{2}\). We put it into \(\frac{x}{2}\) That gives us:\[\frac{\frac{\pi}{2}}{2}\implies \frac{\pi}{4}\]
So, is \(\sin(\frac{\pi}{4})=0\)?

Answer 22

0.75....

Answer 23

So that is NOT a vertical asymptote. Now do the same thing for the next point. So on and so forth. Any point that has a sine of 0 means the cotangent would be undefined and would be a vertical asymptote.

Answer 24

so C and A?

Answer 25

Yep. And here is a reference with your points and a few others graphed. The vertical asymptotes are where the cotangent gets close, but never crosses.
https://www.desmos.com/calculator/swir7owcxw

Answer 26

tysm

Answer 27

I hope the graph makes it a little clearer what this is all about. Asymptotes, vertical and otherwise, become more important as you go on in math.

Answer 28

On the graph, you can remove the extra lines I put in and just see the points you are concerned with. It was just an easy way to visualize it for you.

Answer 29

ok thx

Answer 30

Well, don't have too much fun!