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!

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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!

Mathematics
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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

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Other answers:

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
@Nnesha you helped me before on these problems, can u help me one last time
oh im not looking to graph, im just trying to find a function out of a graph
Please help @Vocaloid
#3 has an asymptote, and its range is the set of all real numbers, so the function is either tan or cot
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
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
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
so it appears that 4*cot(x) works. Check the other x coordinates to see if it holds up for the other two points
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
what's the vertical distance from the very peak of one mountain to the very bottom of the valley? |dw:1438045445818:dw|
1
hint: look at the points (-pi/6, 10) and (5pi/6, -10) specifically the y coordinates
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?
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
cut that in half to get 20/2 = 10 the amplitude is 10 units. This is the distance from the center to the peak
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
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
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
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
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?
1
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
yw

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