PLEASE CAN SOMEONE HELP ME WITH 2 TRIG QUESTIONS THAT I HAVE BEEN STUCK ON FOR ABOUT 2 DAYS AND NOBODY SEEMS TO HELP ME!

- anonymous

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- schrodinger

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- anonymous

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- anonymous

work backwards,
determine your amp, period , phase shift, vertical/ horizontal shift

- anonymous

can someone just show me how to solve the first one so that I can do the second one

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## More answers

- anonymous

hey isn't that from con nexus

- anonymous

what

- anonymous

connections academy

- anonymous

no its from my trig class in Texas

- anonymous

._. hmm bai

- anonymous

can u help me or not?

- anonymous

if that problem is really from where you say it is, can you give me a link so I cam find a solution

- anonymous

@Nnesha you helped me before on these problems, can u help me one last time

- anonymous

oh im not looking to graph, im just trying to find a function out of a graph

- anonymous

@jim_thompson5910

- anonymous

Please help @Vocaloid

- jim_thompson5910

#3 has an asymptote, and its range is the set of all real numbers, so the function is either tan or cot

- anonymous

ok

- jim_thompson5910

let's assume it's tan(x)
what is the value of tan(x) when x = pi/4

- anonymous

its 1

- jim_thompson5910

so IF the point (pi/4, 1) was on the graph, then tan(x) might be the answer. But it's not

- jim_thompson5910

something like 4*tan(x) would work
but if we plugged in x = pi/2, then we'd get undefined. The function f(x) = tan(x) is undefined when x = pi/2. So it turns out that tan doesn't even work at all

- jim_thompson5910

is this making sense?

- anonymous

i see, so it has to be cotangent

- jim_thompson5910

what is the value of cot(x) when x = pi/4 ?

- anonymous

the answer is 1, but if I put a 4 behind cotangent: 4cot(pi/4) then it is 4

- jim_thompson5910

good

- anonymous

and it works with the rest as well

- jim_thompson5910

so it appears that 4*cot(x) works. Check the other x coordinates to see if it holds up for the other two points

- jim_thompson5910

yes 4*cot(x) = 0 when x = pi/2
and 4*cot(x) = -4 when x = 3pi/4

- anonymous

ok so is thar it then, the function for this graphy is just y=4cot(x)

- anonymous

that is it then*

- jim_thompson5910

correct

- anonymous

is that how you properly write it down:\[y=4\cot (x)\]

- jim_thompson5910

yes

- jim_thompson5910

or f(x) = 4*cot(x)

- anonymous

how about the second one, that is the really confusing one

- jim_thompson5910

what's the vertical distance from the very peak of one mountain to the very bottom of the valley?
|dw:1438045445818:dw|

- anonymous

1

- jim_thompson5910

hint: look at the points (-pi/6, 10) and (5pi/6, -10)
specifically the y coordinates

- anonymous

ok i looked at them, just a question when you asked me the distance from the peak of the mountain to the bottom, did you already know the answer or were you asking me because the graph was unclear?

- jim_thompson5910

I knew the answer. I wanted to get you thinking on how to solve this problem

- anonymous

oh

- anonymous

sorry im stuck and i dont know where you were going with in the hint

- jim_thompson5910

what is the change from y = 10 to y = -10

- anonymous

20

- jim_thompson5910

cut that in half to get 20/2 = 10
the amplitude is 10 units. This is the distance from the center to the peak

- jim_thompson5910

look at the graph and tell me what the two peak points are

- anonymous

(-1,2) and (4,2)

- jim_thompson5910

how are you getting them? look at the attachment. They give you the points

- anonymous

arent they the two highest points

- jim_thompson5910

they list the points below the graph

- jim_thompson5910

2 of them are peak points

- jim_thompson5910

yeah they are the highest points

- anonymous

\[(4\pi/3,0) and (11\pi/6,10)\]

- jim_thompson5910

there's a point higher than (4pi/3, 0)

- jim_thompson5910

(11pi/6, 10) is one of the peak points

- anonymous

(-pI/6,-10)

- jim_thompson5910

yes, now subtract the x coordinates

- jim_thompson5910

(11pi/6) - (-pi/6) = ???

- anonymous

2pi

- jim_thompson5910

so that is the period. Every 2pi units, the function repeats itself

- anonymous

I see

- jim_thompson5910

let's assume this is a cosine function
when x = 0, cos(x) = 1 and 10*cos(x) = 10
however, the point (0,10) isn't on this function. Instead, it's shifted to the left pi/6 units. So we need to account for the phase shift
phase shift = -pi/6
the negative tells us to go left

- anonymous

are we sure its a cosine function, could it not be something else

- jim_thompson5910

either that or sine

- jim_thompson5910

cosine is easier to work with though

- anonymous

ok

- jim_thompson5910

A = amplitude = 10
T = period = 2pi
P = phase shift = -pi/6
-------------------------------------------------------
The general cosine equation is
y = A*cos(Bx - C) + D
where
|A| is the amplitude
T = 2pi/B is the period
C/B is the phase shift
y = D is the midline. In this case, the midline is y = 0 so D = 0

- anonymous

im confused as to how to write the equation out

- jim_thompson5910

If T = 2pi/B is the period
and we know T = 2pi is the period, then what is the value of B?

- anonymous

1

- anonymous

right?

- jim_thompson5910

yes B = 1

- anonymous

y=10cos(x+2pi)

- jim_thompson5910

A = 10
B = 1
C = unknown
to find C, solve C/B = -pi/6

- anonymous

c=-pi/6

- anonymous

?

- jim_thompson5910

yes, so the function is \[\Large f(x) = 10\cos\left(x + \frac{\pi}{6}\right)\]

- anonymous

uhhhh, Im graphing this equation, but it looks nothing like the graph given

- anonymous

@jim_thompson5910

- jim_thompson5910

what are you using to graph?

- anonymous

www.demos.com

- jim_thompson5910

you mean this?
https://www.desmos.com/calculator

- anonymous

yep

- jim_thompson5910

and you typed in 10*cos(x+pi/6) ?

- anonymous

yes

- jim_thompson5910

are you in radian mode?

- anonymous

yes

- jim_thompson5910

this is what I get

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- anonymous

oh wait it is working now

- anonymous

if its not a problem can I ask you a way simpler question, that will only take 5 minutes

- anonymous

how to we graph this function: \[y=-3\sec(\pi-2x)+5\]

- jim_thompson5910

by using \[\Large \sec(x) = \frac{1}{\cos(x)}\]

- jim_thompson5910

or desmos should be able to handle secant functions

- anonymous

yeah, but i'm not getting a proper graph, im just getting something like this:

##### 1 Attachment

- jim_thompson5910

that's what secant graphs look like. They have asymptotes and those U shapes

- anonymous

oh ok thanks

- jim_thompson5910

yw

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