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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

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looks good to me i never remember the formula, so i set \(2x-\frac{\pi}{3}=0\) and solve for \(x\)
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
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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Other answers:

|dw:1349892894075:dw|
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
hold on, i did not say what you wrote set it equal to zero \[2x-\frac{\pi}{3}=0\] \[2x=\frac{\pi}{3}\] \[x=\frac{\pi}{6}\]
i meant set \[2x-\frac{\pi}{3}=0\] not \[2x-\frac{x}{3}=0\]
haha okay, i thought = to zero
whew
haha sorry about that
okay so to figure the points, i start at pi/6 and move right by pi/6?
lets go slow because i do not want to confuse you
you know \(\cos(0)=1\) right?
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
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
all you wrote is correct, here is a nice picture http://www.wolframalpha.com/input/?i=+y+%3D+cos+%282x+-+pi%2F3%29+%2B1
OH! i forgot i have to move it up 1
okay so then i have it like this: |dw:1349893756732:dw|
then i move up 1
|dw:1349893880189:dw|
@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)
|dw:1349894136637:dw|
@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
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
how did you do that?
  • phi
you know the period is pi, so every time you increase x by pi you will get the same value as x
sorry, im new to precalculus. all this is super new to me
  • 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
how did you do that? are you asking how to get the lowest point at x= 2/3 pi ?
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
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?
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
|dw:1349897328597:dw|
this is the graph i got, but i got pi/2 as my lowest. did i do something wrong?
  • phi
type y = cos (2x - pi/3) +1 in the google search window, it should graph it
okay im lookin at the graph
so how do i figure the lowest with that graph?
  • 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
cos(2x-pi/3) = -1 okay i get this like u did
  • 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?
i end up getting 2pi/3. and that graph on google is confusing me with different numbers
  • 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
*The graph shows the curve with x in radians.
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
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.
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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