Question

Help with a trig function?

Answer 1

R(x) = -3tan(1/2x)

1) What kind of reflection does the function experience?
2) What is the vertical stretch factor?
3) What is the horizontal stretch factor?
4) What is the period?
5) What are the equations of the asymptotes?
6) What are the zeros?

My answers:
1) A reflection across the x-axis, a vertical reflection.
2) 3
3) 2
4) 2pi

I'm not sure on 5 and 6

Answer 2

Ooo answers 1 through 4 look good! \(\large\rm yay!\)

Answer 3

:) awesome!

For 6, I put x = 0 + 2pin and 0 - 2npi but I'm not sure if they're right...

Answer 4

I feel like 5 is something similar. There is an asymptote at pi and every 2pi after that.

Answer 5

So normally your asymptotes are at pi/2, 3pi/2, ...
or more generally at pi/2 + n(pi)

Your horizontal stretch is what's messing with that, ya?
2(pi/2 +n(pi)) = pi + 2n(pi)

Ya seems like you've got the right idea :)

Answer 6

For zeros, same idea 0 + n(pi) becomes 2(0+n(pi)) = 0+2n(pi)

Answer 7

When you make a horizontal stretch, eeeeverything is stretched horizontally:
~the period,
~the space between the zeros,
~the space between the asymptotes.

Answer 8

So the equations for the asymptotes are
x = pi + 2n(pi) and x = pi - 2n(pi)
And the zeros of the function are:
0 + 2n(pi)
Thanks for being Ze-best (hey, I tried)

Answer 9

haha nice :D

If n is a NATURAL NUMBER, then yes, you need two separate equations like that.
If instead you let n, or k or something, be an INTEGER,
then that covers both positive and negatives.

\(\large\rm x=\pi+2k\pi,\qquad\qquad k\in\mathbb Z\)
same as
\(\large\rm x=\pi\pm2n\pi,\qquad\qquad n\in\mathbb N\)

Answer 10

I see. Thanks zep :)

Answer 11

If those weird symbols are confusing,
then you can ignore them.

I'm not sure what type of math you've dAbbled in at this point.
Eyyy I still got it!

Answer 12

Haha that was good! xD

The symbols look somewhat familiar from my adventures Khan Academy, but I couldn't tell you what they mean.