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

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

hey isn't that from con nexus

what

connections academy

no its from my trig class in Texas

._. hmm bai

can u help me or not?

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

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

ok

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

its 1

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

is this making sense?

i see, so it has to be cotangent

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

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

good

and it works with the rest as well

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

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

that is it then*

correct

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

yes

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

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

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

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

oh

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

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

20

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

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

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

arent they the two highest points

they list the points below the graph

2 of them are peak points

yeah they are the highest points

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

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

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

(-pI/6,-10)

yes, now subtract the x coordinates

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

2pi

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

I see

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

either that or sine

cosine is easier to work with though

ok

im confused as to how to write the equation out

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

right?

yes B = 1

y=10cos(x+2pi)

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

c=-pi/6

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

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

what are you using to graph?

www.demos.com

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

yep

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

yes

are you in radian mode?

yes

this is what I get

oh wait it is working now

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

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

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

or desmos should be able to handle secant functions

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

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

oh ok thanks