Question

ui

Answer 1

looks pretty much correct.

Answer 2

just 3pi

Answer 3

Woops, your phase shift looks off.\[\large\rm y=\tan\left[6\pi x-3\pi\right]\]\[\large\rm y=\tan\left[6\pi\left(x-\frac{1}{2}\right)\right]\]So your phase shift should be C=1/2, ya?

Answer 4

Oh you just label C=3pi,
ok my mistake :) just different lettering maybe hehe

Answer 5

I guess I'm a little confused by your labeling.. hmm
why start at 3pi if your phase shift is 1/2...

Answer 6

Becomes easier to add your fractions when you have the correct shift value:

x1 = 1/2
x2 = 1/2 + 1/12
x3 = 1/2 + 2/12
x4 = 1/2 + 3/12
x5 = 1/2 + 1/3

Answer 7

Yes, that's what I wrote.
You have to factor out the B from each term to get the correct horizontal shift.

Answer 8

C helps us find the horizontal shift,
it isn't the shift itself though.

The horizontal shift is C/B

Answer 9

A=1
B=6pi
C=3pi
D=0

Therefore:
Amplitude=1
Period = 2pi/6pi = 1/3
Phase Shift = 3pi/6pi = 1/2

Answer 10

It's going to make your calculations easier which is nice :)

Answer 11

Oh oh oh oh oh,
we forgot another important detail.

Tangent is periodic in Pi! Not 2Pi!

Answer 12

Period = pi/6pi

Answer 13

Period = 1/6

So that messes up your points as well XD lol

Answer 14

this is getting a little confusing,
maybe you can look at my notes really quick and see if this makes sense.
https://www.twiddla.com/2509337

Answer 15

i don't think you should have any pi's in your intervals or asymptotes.. hmm

Answer 16

yay team \c:/
I think we did it

Answer 17

Why did you edit your question?