lilsis76
  • lilsis76
Determine Amplitude/period/phase shift /graph function. over 1 period. Indicate xintercepts and coordinates of Hi/low points on graph. y = cos (2x - pi/3) +1 y = A ( Bx - C ) so Amplitude is 1 Period: 2pi/B--> 2pi/2 = Pi Okay is my phase shift right? Phaseshift = C/B pi ---- 3 --------------- = pi/6? 2
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
looks good to me i never remember the formula, so i set \(2x-\frac{\pi}{3}=0\) and solve for \(x\)
anonymous
  • anonymous
this also help me remember whether the graph has moved left or right. in this case you get \(x=\frac{\pi}{6}\) so it is shifted to the right
lilsis76
  • lilsis76
oh!, okay so since it was positive it goes right, right? I thought if if was negative it goes right?? i am still confused on how they shift.

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lilsis76
  • lilsis76
|dw:1349892894075:dw|
anonymous
  • anonymous
think of it this way for regular old cosine, you know \(\cos(0)=1\) now you solve \(2x-\frac{\pi}{3}=0\) and you see \(x=\frac{\pi}{6}\) which means instead of \((0,1)\) on the graph, you have \((\frac{\pi}{6},1)\) on the graph, so it is shifted right
anonymous
  • anonymous
hold on, i did not say what you wrote set it equal to zero \[2x-\frac{\pi}{3}=0\] \[2x=\frac{\pi}{3}\] \[x=\frac{\pi}{6}\]
anonymous
  • anonymous
i meant set \[2x-\frac{\pi}{3}=0\] not \[2x-\frac{x}{3}=0\]
lilsis76
  • lilsis76
haha okay, i thought = to zero
anonymous
  • anonymous
whew
lilsis76
  • lilsis76
haha sorry about that
lilsis76
  • lilsis76
okay so to figure the points, i start at pi/6 and move right by pi/6?
anonymous
  • anonymous
lets go slow because i do not want to confuse you
anonymous
  • anonymous
you know \(\cos(0)=1\) right?
lilsis76
  • lilsis76
okay then. well then what i know know is Amp = 1 period 2pi/b--> 2pi/3 phase shift---> 2x-pi/3=0--> 2x=pi/3 divide both by 2--> = pi/6 Goes Right And yes Cos(0)=1 means at the point zero it goes up to 1 which is the highest point
anonymous
  • anonymous
right, so in this case you have to make \(x=\frac{\pi}{6}\) in order to get \(2x-\frac{\pi}{3}=0\) so for your graph, instead of \((0,1)\) you have \((\frac{\pi}{6},1)\) on the graph
anonymous
  • anonymous
all you wrote is correct, here is a nice picture http://www.wolframalpha.com/input/?i=+y+%3D+cos+%282x+-+pi%2F3%29+%2B1
lilsis76
  • lilsis76
OH! i forgot i have to move it up 1
lilsis76
  • lilsis76
okay so then i have it like this: |dw:1349893756732:dw|
lilsis76
  • lilsis76
then i move up 1
lilsis76
  • lilsis76
|dw:1349893880189:dw|
lilsis76
  • lilsis76
@satellite73 Like this? so amplitude is 1 period is Pi Phase Shift is pi/6 I graphed the function the x intercepts: there are not any. the lowest point is (pi/2,0), and the highest point is (pi/6,2)
lilsis76
  • lilsis76
|dw:1349894136637:dw|
lilsis76
  • lilsis76
@swissgirl @marcoduuuh checked the answers in the back of my book. How did they come up with the x intercept of 2pi/3? the hight point i got is (pi/6,2) but how do i get (7pi/6, 2)? the low point of (2pi/3, 0 ) ?
phi
  • phi
this curve just touches the x-axis when y=0 cos (2x - pi/3) +1 = 0 cos(2x- pi/3)= -1 arc cos both sides 2x - pi/3 = pi 2x= pi+pi/3 = 4/3 pi x= 2/3 pi
lilsis76
  • lilsis76
how did you do that?
phi
  • phi
you know the period is pi, so every time you increase x by pi you will get the same value as x
lilsis76
  • lilsis76
sorry, im new to precalculus. all this is super new to me
phi
  • phi
the hight point i got is (pi/6,2) but how do i get (7pi/6, 2)? pi/6 + pi = pi/6 + 6pi/6 = 7pi/6
phi
  • phi
how did you do that? are you asking how to get the lowest point at x= 2/3 pi ?
lilsis76
  • lilsis76
oh....okay, now is that going to be the same step for getting the HIghs on all problems like that? you get the Period and add it too the phase shift? Yes and I cant get the lowest point either, i end up getting pi/2, 0
phi
  • phi
Let's look at How did they come up with the x intercept of 2pi/3? First, you know x-intercept means the x value where the curve touches or crosses the x-axis?
lilsis76
  • lilsis76
yes how did they come up with 2pi/3, because i got... pi/6 pi/3 pi/2 and wait i guess i got the 2pi/3. but okay i know and see my question. how did they get 2pi/3 an point 0, because when i graphed it by hand it shows that 2pi/3 is at 1
lilsis76
  • lilsis76
|dw:1349897328597:dw|
lilsis76
  • lilsis76
this is the graph i got, but i got pi/2 as my lowest. did i do something wrong?
phi
  • phi
type y = cos (2x - pi/3) +1 in the google search window, it should graph it
lilsis76
  • lilsis76
okay im lookin at the graph
lilsis76
  • lilsis76
so how do i figure the lowest with that graph?
phi
  • phi
You see the curve just touches the x-axis. This happens when y=0 we start with y = cos (2x - pi/3) +1 cos (2x - pi/3) +1 = 0 (look for where the curve crosses or touches the x-axis) use algebra to solve for x: write -1 on both sides of the = cos(2x -pi/3) +1 -1 = 0-1 simplify to get cos(2x-pi/3) = -1
lilsis76
  • lilsis76
cos(2x-pi/3) = -1 okay i get this like u did
phi
  • phi
cos(2x-pi/3) = -1 take the inverse cosine of both sides 2x -pi/3 = acos(-1) type in google acos(-1)= you get pi radians (3.14159....) or 180 degrees. you now have 2x -pi/3 = pi can you finish?
lilsis76
  • lilsis76
i end up getting 2pi/3. and that graph on google is confusing me with different numbers
phi
  • phi
The graph is show the curve with x in radians. the region from x=0 to 2pi (6.283...) is the region your question is asking about. you see 2 peaks in this interval. Decimals make it hard to match these spots with x= pi/6 and x= 7pi/6 but using a calculator or google we see x= 0.523... and x=3.665...
phi
  • phi
*The graph shows the curve with x in radians.
lilsis76
  • lilsis76
oh....okay, i understand that now, thank you. for your help. Imma close this question and fix my homework. Ill be back later. so thank you
phi
  • phi
OK. For graphing, I would pick a few x values and use a calculator to find the corresponding y value. Then plot that point. After a few points you can sketch in the curve.
lilsis76
  • lilsis76
okay, i would have to try that, I probably wont beable to do that on a quiz tomorrow. she is not allowing calculators. but if i know the steps i know i can do the question or problem

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