Question

Asking this question for the fourth time on here. Can someone help me find the function for a -tan(x) and a csc(x) graph?

Answer 1

Sure

Answer 2

ok, the first image on the screen is tan(x) right.

Answer 3

it is -tan, yeah @arjund12

Answer 4

yea

Answer 5

i don't get what you're asking for then?

Answer 6

I need to write an equation for the graph. Like, y = a-tan(bx)
But I don't know how to find the period and work the equation for the -tan(x) graph or the csc(x) graph.

Answer 7

the period is the distance between your x intercepts in the case for the tangent fucntion

Answer 8

period measures how long it takes to repeat itself over again

Answer 9

the markings on your graphs are horrendous ... no offense

Answer 10

So the period in this case is 3?

Answer 11

I know, it's from a practice paper my teacher sent out

Answer 12

let me try to detemrine where we are crossing

6 bars is equal to pi/2

therefore

1 bar is equal to pi/12

we have intercepts at +4 bars and -8 bars for a span of 12 bars

the period is 12*pi/12 = pi

Answer 13

Meaning, the period is pi?

Answer 14

yes, the period is pi, the interval at which the function starts to repeat itself has a distance of pi units

Answer 15

Alright, now so far I have \[y = a-\tan(\pi \times x)\]

Answer 16

I was told tangent does not have an amplitude because it does not have a max and min, so the next step is where I'm lost

Answer 17

and that's supposed to be A times negative tangent, not A minus tangent

Answer 18

a little information: tan(x) by default has a period of pi already.

tan(pi * x) is speeding up the period by a factor of pi so its not the same graph

Answer 19

notice that tan(pi x) has a period much faster than tan(x)

Answer 20

Alright I see that difference. So if I don't need the pi in the equation, where do I go next with the equation?

Answer 21

well, tan(x) = 0 when x=0 ... the graph is clearly shifted away from, or up from where it needs to normally be

Answer 22

its been moved 4 bars to the right, 4*pi/12 = pi/3

tan(x-pi/3) is equal to 0 when x=pi/3 gets us at least centered back correctly

Answer 23

and amplitude does affect tan(x)

Answer 24

3tan(pi/4) is not equal to tan(pi/4) is it?

Answer 25

now we can find the amplitude, by using a reference point

a tan(x1-pi/3) = y1

a = y1/tan(x1-pi/3)

Answer 26

when x=pi/4, y=1 appear to be a point we have to hit

a = 1/tan(pi/4-pi/3)

a = cot(-pi/12)

Answer 27

bah, we might need a few reference points to play with

a - b tan(x1-pi/3) = y1
a - b tan(x2-pi/3) = y2

and solve the system for a and b

Answer 28

I'm sorry I'm really slow with this. When you're typing x1 and y1, are you typing it 1 * x and 1 * y, or 1 in place of x and 1 in place of y

Answer 29

x1 in this case means x_1 its just a variable of x

Answer 30

\[x_1,y_1\]

Answer 31

we know (pi/4,1) , (pi/2,-1)

Answer 32

a - b tan(pi/4-pi/3) = 1
a - b tan(pi/2-pi/3) = -1


-a + b tan(pi/4-pi/3) = -1
a - b tan(pi/2-pi/3) = -1
-----------------------
b(tan(-pi/12)-tan(pi/6)) = -2

etc ...

a = 1/2 (sqrt(3)-1)
b = 1/2 (3+sqrt(3))

is this spose to be a pretty picture or is it spose to be this convoluted?

Answer 33

I'm sorry, I'm not gonna be able to finish this because I have no idea what I'm doing. I appreciate the help I really do, I'll give you the fan and the medal and all that. Sorry for wasting your time @amistre64

Answer 34

it wasnt a waste, by trying to help others it helps me to remember what its all about. without the practice id be dead and gone along time ago :) thnx tho

Answer 35

a - b tan(pi/12) = 0

a = b tan(pi/12)
.....................
a - b tan(pi/4) = 1

b tan(pi/12) - b tan(pi/4) = 1

b = 1/(tan(pi/12) - tan(pi/4))
a = tan(pi/12)/(tan(pi/12) - tan(pi/4))

theres just no pretty way about this