Graphs of Other Trigonometric Functions 4.6

[Raphealle]

The question/thing says "sketch the graph of the function. Include two full periods."

my problem is with this equation:
y = -1/2 sec X

And I don't think I'm solving it properly. My teacher said find 5 essential points. The beginning and the end and three in between....

[stapel_eliz]

"sketch the graph of the function. Include two full periods."

y = -1/2 sec X

And I don't think I'm solving it properly.

You'll need to upload a scan of your work to some place like ImageBucket, and then post a link here to the graphic, in order for us to view and comment on what you've done.

My teacher said find 5 essential points. The beginning and the end and three in between....

That's usually the best bet, actually.

The secant is the reciprocal of the cosine, so you can work from what you've learned about the cosine wave. The cosine is zero at pi/2 and at (3pi)/2. It is 1 at 0 and 2pi; it is -1 at pi. These are the five "interesting" points; the rest can be viewed as "filler" once you've learned the basic shape of the cosine wave.

Since the secant is the reciprocal of the cosine, it is -1 whenever cosine is -1, it is 1 whenever cosine is 1, and it has a vertical asymptote whenever the cosine is zero.

So look at the cosine wave. Dash in the vertical asymptotes wherever the cosine is zero. Draw the dots at -1 and 1 wherever the cosine takes on those values. Since you know that the cosine is always positive or always negative between its zeroes and is always less than or equal to 1 (in absolute value), you know that the secant has the same signs on those intervals and is always greater than or equal to 1 (in absolute value). Then, between the vertical asymptotes you've drawn, and passing through the dots you've drawn inside each interval, draw the "U" shapes for the secant's graph.

Have fun!