anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
work backwards, determine your amp, period , phase shift, vertical/ horizontal shift
anonymous
  • anonymous
can someone just show me how to solve the first one so that I can do the second one

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anonymous
  • anonymous
hey isn't that from con nexus
anonymous
  • anonymous
what
anonymous
  • anonymous
connections academy
anonymous
  • anonymous
no its from my trig class in Texas
anonymous
  • anonymous
._. hmm bai
anonymous
  • anonymous
can u help me or not?
anonymous
  • 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
  • anonymous
@Nnesha you helped me before on these problems, can u help me one last time
anonymous
  • anonymous
oh im not looking to graph, im just trying to find a function out of a graph
anonymous
  • anonymous
@jim_thompson5910
anonymous
  • anonymous
Please help @Vocaloid
jim_thompson5910
  • 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
  • anonymous
ok
jim_thompson5910
  • jim_thompson5910
let's assume it's tan(x) what is the value of tan(x) when x = pi/4
anonymous
  • anonymous
its 1
jim_thompson5910
  • 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
  • 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
  • jim_thompson5910
is this making sense?
anonymous
  • anonymous
i see, so it has to be cotangent
jim_thompson5910
  • jim_thompson5910
what is the value of cot(x) when x = pi/4 ?
anonymous
  • anonymous
the answer is 1, but if I put a 4 behind cotangent: 4cot(pi/4) then it is 4
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • anonymous
and it works with the rest as well
jim_thompson5910
  • 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
  • jim_thompson5910
yes 4*cot(x) = 0 when x = pi/2 and 4*cot(x) = -4 when x = 3pi/4
anonymous
  • anonymous
ok so is thar it then, the function for this graphy is just y=4cot(x)
anonymous
  • anonymous
that is it then*
jim_thompson5910
  • jim_thompson5910
correct
anonymous
  • anonymous
is that how you properly write it down:\[y=4\cot (x)\]
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
or f(x) = 4*cot(x)
anonymous
  • anonymous
how about the second one, that is the really confusing one
jim_thompson5910
  • 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
  • anonymous
1
jim_thompson5910
  • jim_thompson5910
hint: look at the points (-pi/6, 10) and (5pi/6, -10) specifically the y coordinates
anonymous
  • 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
  • jim_thompson5910
I knew the answer. I wanted to get you thinking on how to solve this problem
anonymous
  • anonymous
oh
anonymous
  • anonymous
sorry im stuck and i dont know where you were going with in the hint
jim_thompson5910
  • jim_thompson5910
what is the change from y = 10 to y = -10
anonymous
  • anonymous
20
jim_thompson5910
  • 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
  • jim_thompson5910
look at the graph and tell me what the two peak points are
anonymous
  • anonymous
(-1,2) and (4,2)
jim_thompson5910
  • jim_thompson5910
how are you getting them? look at the attachment. They give you the points
anonymous
  • anonymous
arent they the two highest points
jim_thompson5910
  • jim_thompson5910
they list the points below the graph
jim_thompson5910
  • jim_thompson5910
2 of them are peak points
jim_thompson5910
  • jim_thompson5910
yeah they are the highest points
anonymous
  • anonymous
\[(4\pi/3,0) and (11\pi/6,10)\]
jim_thompson5910
  • jim_thompson5910
there's a point higher than (4pi/3, 0)
jim_thompson5910
  • jim_thompson5910
(11pi/6, 10) is one of the peak points
anonymous
  • anonymous
(-pI/6,-10)
jim_thompson5910
  • jim_thompson5910
yes, now subtract the x coordinates
jim_thompson5910
  • jim_thompson5910
(11pi/6) - (-pi/6) = ???
anonymous
  • anonymous
2pi
jim_thompson5910
  • jim_thompson5910
so that is the period. Every 2pi units, the function repeats itself
anonymous
  • anonymous
I see
jim_thompson5910
  • 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
  • anonymous
are we sure its a cosine function, could it not be something else
jim_thompson5910
  • jim_thompson5910
either that or sine
jim_thompson5910
  • jim_thompson5910
cosine is easier to work with though
anonymous
  • anonymous
ok
jim_thompson5910
  • 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
  • anonymous
im confused as to how to write the equation out
jim_thompson5910
  • 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
  • anonymous
1
anonymous
  • anonymous
right?
jim_thompson5910
  • jim_thompson5910
yes B = 1
anonymous
  • anonymous
y=10cos(x+2pi)
jim_thompson5910
  • jim_thompson5910
A = 10 B = 1 C = unknown to find C, solve C/B = -pi/6
anonymous
  • anonymous
c=-pi/6
anonymous
  • anonymous
?
jim_thompson5910
  • jim_thompson5910
yes, so the function is \[\Large f(x) = 10\cos\left(x + \frac{\pi}{6}\right)\]
anonymous
  • anonymous
uhhhh, Im graphing this equation, but it looks nothing like the graph given
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
what are you using to graph?
anonymous
  • anonymous
www.demos.com
jim_thompson5910
  • jim_thompson5910
you mean this? https://www.desmos.com/calculator
anonymous
  • anonymous
yep
jim_thompson5910
  • jim_thompson5910
and you typed in 10*cos(x+pi/6) ?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
are you in radian mode?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
this is what I get
anonymous
  • anonymous
oh wait it is working now
anonymous
  • anonymous
if its not a problem can I ask you a way simpler question, that will only take 5 minutes
anonymous
  • anonymous
how to we graph this function: \[y=-3\sec(\pi-2x)+5\]
jim_thompson5910
  • jim_thompson5910
by using \[\Large \sec(x) = \frac{1}{\cos(x)}\]
jim_thompson5910
  • jim_thompson5910
or desmos should be able to handle secant functions
anonymous
  • anonymous
yeah, but i'm not getting a proper graph, im just getting something like this:
jim_thompson5910
  • jim_thompson5910
that's what secant graphs look like. They have asymptotes and those U shapes
anonymous
  • anonymous
oh ok thanks
jim_thompson5910
  • jim_thompson5910
yw

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