Question

Last one, tan(x/2) I have to draw a graph with three sequences
asymtopes:
period:
x intercepts:

Answer 1

I usually deal with tangent and cotangent graphs in a special way. So, on the normal tangent graph, do you know where the asymptotes would be?

Answer 2

I did it towards the beginning of the semester so I don't remember

Answer 3

So no sorry

Answer 4

Tangent would have asymptotes at -pi/2 and pi/2 and then those asymptotes would repeat every pi. So basically, the pattern/graph of tangent is defined between each set of asymptotes. So my strategy for graphing tangent is to take the angle of tangent and set up an inequality between -pi/2 and pi/2. SO in this case, your angle is x/2, which means I would do this:

-pi/2 < x/2 < pi/2

From here I solve for x, which gives me

-pi/4 < x < pi/4

Doing this does multiple things for me.
1. It tells me where asymptotes are
2. The distance between these asymptotes is the period of the function. Knowing the period and the location of asymptotes lets me knoe where all the asymptotes of the function are
3. This also accounts for any phase shifts if there are any.

So the distance between -pi/4 and pi/4 is pi/2. This means the period of the function is pi/2 and that all asymptotes can be found by adding pi/2 to the location of any known asymptote.

So that takes care of asymptote placement. Im sure you recall the general shape of the tangent graph. Unless a graph is shifted up or down (which it isnt in this case), any x-intercept is in between any 2 asymptotes. So with this information, I could graph this:



Before I add in the last detail, does that all make sense?