Precal

Explain  how to graph the following functions:cosb

y = -3cos(-2x+pi/6) +3 and  y = -3cos(2x-pi/6). how are they equal/different? how do you explain the differences and or similarities?

First 5 people (per period) that correctly answer this will get 2 extra points in your next ch test.

29 thoughts on “Precal

  1. The ONLY difference between this two graphs are that the second given equation of the graph has a reflection over the x-axis AND the first given graph has NO reflections BUT has a vertical shift of 3 units UP. Thats about it. And the why is that the – inside the parenthesis of the first given graph must be factored out and as that happens the factored out negative moves in front of the cosine which makes the 2 negatives cancel out so there would be no reflections. 🙂

  2. The first equation does not have a reflection on the x-axis because the negative on the -3 and the -2x cancel each other out. So if you factor out the negative that would make the first equation y = 3cos(2x-pi/6) +3. So you have the same graph as the second one except it doesn’t have a reflection on the x-axis and the first one has a vertical shift of 3 units up.

  3. The first graph has an amplitude of 3 and starts at (0,-6) because it is translated up on the y-axis 3 units and then is translated on the x-axis. They both have the similarities of the same amplitude and phase shift. Their differences are that the second one is not translated up/ down any units, and is not reflected on the x-axis and the second graph’s period is positive instead of negative.

    To Graph 1st one:
    amplitude=3
    period=-6pi
    p.s=pi/12
    reflection on x-axis
    translate up 3 units

    To graph 2nd one:
    amplitude=3
    period=6pi
    p.s=pi/12
    no reflection
    no translation

    -Kyler Palacio Period 4

  4. -3cos (-2x + pi/6) + 3
    Period: set 2x equal to 2pi = pi
    Amplitude: 3
    Phase shift: set (-2x + pi/6) equal to 0 = pi/12
    Reflection on the x-axis because of the negative (-3)
    Translate 3 units up y-axis
    Endpoints: pi/12 to (pi/12 – pi) = -11pi/12
    Keypoints: (pi/12 , 0) (-pi/6 , 3) (-5pi/12 , 6) (-2pi/3 , 3) (-11pi/12 , 0)

    -3cos (2x – pi/6)
    Period: pi
    Amplitude: 3
    Phase shift: pi/12
    Reflection on x-axis
    Endpts: pi/12 to (pi/12 + pi) = 13pi/12
    Keypoints: (pi/12 , -3) (pi/3 , 0) (7pi/12 , 3) (5pi/6 , 0) (13pi/12 , -3)

    Both graphs have a period of pi (the negative makes no difference because periods are always positive), an amplitude of 3 (amplitudes are not negative), a reflection on the x-axis (negative in front of 3), and a phase shift of pi/12 (negatives cancel on graph 1). Graph 1 has a vertical translation because of the +3, and the endpoints of both graphs are different because of the negative in front of 2x (graph 1).

  5. they both share an amplitude of 3, as well as they both share a period of pie, except that one of them is negative, and both of them have a phase shift of a pi/12, as well as the first has a vertical shift of 3. As well as a reflection on the x- axis, in front the of the negative 3.

  6. Graph 1:
    Amplitude: 3
    Period: pi
    P.S: pi/12
    Go up 3 units on the y-axis because of the +3
    Reflection on the x-axis because of the (-) in front of the 3
    End points: pi/12 -> -11pi/12
    Key points : (-11pi/12,0) (-2pi/3,3) (-5pi/6,6) (-pi/6,3) (pi/12,0)

    Graph 2:
    Amplitude: 3
    Period: pi
    P.S: pi/12
    Reflection on the x- axis
    End points: pi/12 -> 13pi/12
    Key points : (pi/12,-3) (pi/3,0) (7pi/12,3) (5pi/6,0) 13pi/12,-3)

    The difference is that the first graph has a translation on the y-axis because of the (+3) and they have different end points because there is a (-) in front of the (2x) on graph 1. They are similar because both graphs have an amplitude of 3, a period of pi, a reflection on the x-axis, and a phase shift of pi/12.

  7. For the first graph y = -3cos(-2x+pi/6) +3
    Reflection on x-axis
    three units up on the y-axis
    Amplitude: 3
    Period: -2x=2pi; period will be -pi
    P.Shift: -2x= -pi/4; p.shift will be pi/12
    Endpoints: pi/12 – 12pi/12 = -11pi/12; the endpoints are pi/12 and -11pi/12
    Points in graph are: (-11pi/12 , 0) , (-2pi/3 , 3) , (-5pi/12 , 6) , (-pi/6 , 3) , and (pi/12 ,0).

    For the second graph, y = -3cos (2x-pi/6)
    reflection on x-axis (same as graph one)
    amplitude is 3 (same as graph one)
    period is pi
    p.shift: is pi/12 (same as graph one)
    endpoints: pi/12 + 12pi/12= 13pi/12; endpoints are pi/12 and 13pi/12
    points in the graph are: (pi/12 , -3) , (pi/3 , 0) , (7pi/12 , 3) , (5pi/6 , 0) , and (13pi/12 , -3)

    The graphs y = -3cos(-2x+pi/6) +3 and y = -3cos (2x-pi/6) both have an amplitude of 3,they both have a p.shift of pi/12, and they both have a period of pi. Also, both graphs have a reflection on the x-axis. The differences they share are their endpoints. Also, on graph one, there is a translation of 3 units up on the y-axis. Graph two does not have any translations going up or down.

  8. First graph:
    The negative sign in front of the 3 signifies a reflection over the x axis
    For the Period: -2x = 2pi
    x = -pi
    Amplitude= 3
    The max/min of the graph are increased by 3 because of the “+3” at the end.
    Phase Shift = pi/12
    Endpoints = from pi/12 to (pi/12 – pi) = -11pi/12
    Keypoints = (pi/12 , 0) (-pi/6 , 3) (-5pi/12 , 6) (-2pi/3 , 3) (-11pi/12 , 0)

    Second Graph:
    Reflection over x as in the first graph
    Period: 2x = 2pi
    x = pi
    Amplitude = 3
    Phase Shift: just like in the first graph, pi/12
    Endpoints: from pi/12 to (pi/12 + pi) = 13pi/12
    Keypoints: (pi/12 , -3) (pi/3 , 0) (7pi/12 , 3) (5pi/6 , 0) (13pi/12 , -3)

    The differences between the two graphs are shift in the y-axis, and the period. Since the 1st graph has a +3 at the end, it’s amplitude goes up by 3. The 1st graph also has a negative period, which gives it different endpoints and key points than those of the 2nd graph.

    The similarities are in the phase shift, reflection, amplitude, and cosine.

    Timur Katsnelson
    Period 4

  9. To ammend what i had previously posted, this would be the difference in the information.

    Amplitude= 3
    x = -pi
    The maximum or the min would be plus 3
    Phase Shift = pi/12
    Endpoints = from pi/12 to (pi/12 – pi) = -11pi/12
    K.P = (pi/12 , 0) (-pi/6 , 3) (-5pi/12 , 6) (-2pi/3 , 3) (-11pi/12 , 0)

    Second Graph:

    Reflection on the x-axis
    Period: pi
    Amplitude = 3
    x = pi
    Phase Shift: pi/12
    Endpoints: from pi/12 to (pi/12 + pi) = 13pi/12
    K.P: (pi/12 , -3) (pi/3 , 0) (7pi/12 , 3) (5pi/6 , 0) (13pi/12 , -3)

    As well as they both share an amplitude of 3, as well as they both share a period of pie, except that one of them is negative, and both of them have a phase shift of a pi/12, as well as the first has a vertical shift of 3. As well as a reflection on the x- axis, in front the of the negative 3.

    Matthew Chin
    Per 3

  10. y=-3cos(-2x+pi/6)+3
    y=3cos(2x-pi/6)
    Similarities: both share the amplitude of 3 and period of pi
    Differences: the phase change in the first equation is positive while the second one is negative. the first equation has a reflection over the x-axis from the first look by the negative before the “2x” and the 3 cancel out so there is no reflection. also, the first graph has a vertical shift of 3.

  11. Graph 1:
    -3cos(-2+pi/6)+3
    Period: set 2x equal to 2pi=pi
    Altitude: 3
    Phase Shift: set (-2+ pi/6) equal to 0 which equals pi/2
    Reflection on the x-axis since there is a negative number (-3)
    Translate 3 units up on the y-axis
    Endpoints: pi/12 to (pi/12-pi)= -11pi/12
    Keypoints: (pi/12, 0) (-5pi/12,6) (-2pi/3,3) (-11pi/12,0)

    Graph 2:
    -3cos(2x-pi/6)
    Period: pi
    Amplitude: 3
    Phase Shift: pi/2
    Reflection on the x-axis
    Endpoints: pi/12 to (pi/12+pi)=13pi/12
    Keypoints: (pi/12,-3) (pi/3,0) (7pi/12,3) (5pi/6,0) (13pi/12, -3)

    Both of these graphs have a period of pi and the negative isn’t included because the period and amplitude are always positive. They also both have amplitude of 3 and a phase shift of pi/12 which the negatives cancel when on the graph. The graph has one vertical translation because there is a positive 3; and the endpoints of both graphs are different because there is a negative in graph 1 in front of the 2x which makes them change. Also the reflection on the x-axis is another similarity. Another difference between the two graphs are the shift in the y-axis.

    Adar Levy; Period 4

  12. For the first equation:
    y=-3cos(-2x+ pi/6) +3

    A: 3
    P: pi
    PS: pi/12
    endpoints: -11pi/12, pi/12 (because you subtract the period from the ps)
    The amplitude came from the number next to the cosine, the period is 2x=2pi, the ps came from setting the numbers in the parenthesis equal to zero (-2x+ pi/6=0)

    Because the graph is reflected on the x axis and translated up 3 units, the key points become : (-11pi/12,0) (-2pi/3, 3) (-5pi/12, 6) (-pi/6, 3) (pi/12, 0)

    For the second equation:
    y=-3cos(2x-pi/6)

    A: 3
    P: pi
    PS: pi/12
    Endpoints: pi/12 and 13pi/12
    These are all found the same way from the first graph, except you add the pi to the ps.

    Since the graph is only being reflected on the x axis, and is not being translated, the key points become : (pi/12,-3) (pi/3, 0) (7pi/12,3) (5pi/6,0) (13pi/12, -3)

    Similarities: both are cosine graphs, both are being reflected on the x axis, and they both have the same amplitude, period, and phase shift.

    Differences: the first graph has a vertical translation of 3 units, they have different endpoints and key points because the first graph is not only being shifted to the left but also moved up 3 units and the second graph is only being shifted to the right.

    Negin Fadaee Period 4

  13. For the first given equation:
    y = -3cos(-2x+pi/6) +3
    Amp=3
    Per= -pi
    PS= pi/2
    Key points= (-11pi/12, 0) (-5pi/12, 6) (-pi, 3) (pi/12, 0) (-2pi/3, 3)
    VS is up 3
    No reflection because the -3 and the -2x cancel out.

    Second graph
    y = -3cos(2x-pi/6)
    Reflection on x axis
    Per== 2x=2pi > x=pi
    Amp=3
    Key points = (pi/3, 0) (pi/12, -3) (7pi/12, 3) (5pi/6, 0) (13pi/12, -3)
    No VS
    PS= pi/2

    The similarities between the two graphs is that they have the same amplitude, and phase shift. Their differences include their periods, and that the second graph involves a reflection on the x-axis, and the first does not. Also, the first graph moves up the y-axis 3 units but the second graph does not.

    Ashley period 3

  14. Graph 1
    period:pi
    amplitude:3
    p.s.:pi/12
    key points:(pi/12,0),(-pi/6,3),(-5pi/12,6),(-2pi/3 ,3),(-11pi/12,0)
    reflection on the x-axis and a vertical shift of +3

    Graph 2
    period:pi
    amplitude:3
    p.s.:pi/12
    key points:(pi/12,-3),(pi/3,0),(7pi/12,3),(5pi/6,0),(13pi/12 ,-3)
    refection on the x-axis and no vertical shift

    These graphs are similar because the period, amplitude, and the phase shift are the same but are different because of the key points and that the first graph has a vertical shift of +3.

  15. Similarities:

    -same amplitude
    -same P.S.

    Differences:

    -first graph moves vertically by 3
    -periods (first one is negative pi and the second one is pi)
    -second graph has a reflection on the x-axis

    1.

    y = -3cos(-2x+pi/6) +3

    No reflection
    Amp= 3
    Per= -pi
    PS= pi/12
    Key points= (-11pi/12, 0) (-5pi/12, 6) (-pi, 3) (pi/12, 0) (-2pi/3, 3)
    Vertical Shift of 3

    2.

    y = -3cos(2x-pi/6)

    Reflection on x axis
    Amp= 3
    Per= pi
    PS= pi/12
    Key points = (pi/3, 0) (pi/12, -3) (7pi/12, 3) (5pi/6, 0) (13pi/12, -3)

    Chloie Bunye; Period 3

  16. Graph 1:
    Amplitude: 3
    Period: -π
    Phase Shift: π/12
    Endpoints: π/12 to -11π/12
    Vertical Shift: +3
    Reflection on X-Axis

    Graph 2:
    Amplitude: 3
    Period: – π
    Phase Shift: π/12
    Endpoints: π/12 to 13 π/12
    Reflection on X-Axis

    The similarities these two graphs share is the amplitude, the period, and the phase shift are the same. Both graphs also have a reflection on the X-Axis. The differences between these two graphs is that they have different endpoints and Graph 1 has a vertical shift of +3 while Graph 2 does not.

  17. Graph #1 and #2 similarities: Amplitude-3, Period-pi/12
    Graph #1 has: phase shift-up 3 units, (this graph doesn’t reflect because the two negatives cancel out)
    Graph #2 has: reflection on the X-axis,

    Hien Luong, Period 3

  18. The first equation has a reflection on the y axis and the x axis while the second equation only has a reflection on the x axis. Also the first equation has a positive phase shift while the second equation has a negative one. And finally, the first equation is translated 3 units up while the second equation remains in the same spot.

  19. First, you would need to get the amplitude from the number in front of the cos, which always because amplitude can only be positive. Next, you need to x equal to 2pi and solve to find the period. Then, you need to set the entire thing in the parentheses equal to zero and solve for x to get the phase shit. To get the key values, you need to find the midpt. of the first and last points, then the midpt. of the first value and first midpt. you found and the midpt. for the last value and first midpt. Then you need to see if the equation is being translated up or down. For example, y=cos x, the first and last keypts would be 0 and 2pi and the midpt of that is pi. The midpt. between 0 and pi would be pi/2 and the midpt between pi and 2pi would be 3pi/2.

    First equation:
    amplitude = 3
    per. = pi
    phase shift = pi/12
    Translate up 3 units
    keypts. = (pi/12,0) (pi/3,3) (7pi/12,6) (5pi/6, 3) (13pi/12,0)

    The second equation is similar with the only difference being that it isn’t translated up 3 units. The keypts. for that one would be (pi/12,-3) (pi/3,0) (7pi/12,3) (5pi/6, 0) (13pi/12,-3)

    • My bad didn’t see the reflection on the first one yea they are different because one of them reflects in both x and y and on is only x and both of them translate in opposite directions

  20. y = -3cos(-2x+pi/6) +3 and y = -3cos(2x-pi/6)
    Both these equations have the same amplitude. The amplitude is calculated by looking at the number behind the cosine.
    Amp for both = 3
    the period for both= pi
    and the phase shift is also the same = -pi/12
    The major differences between the equations is the the first equation all the points are being translated up by 3 units, and that the first graph has no reflection on the x-axis.
    First graph:
    Amplitude: 3
    Period: pi
    P.S: pi/12
    and then it is translated +3 units.

    Second graph:
    period:pi
    amplitude:3
    P.S.:pi/12
    It is also reflected on the x-axis

  21. First graph y=-3cos(-2x+pi) +3
    amplitude= 3
    periold=pi
    reflection on the x- axis
    translation of 3 units up
    p.s.= pi/12

    Second Graph y=-3cos(2x-pi/6)
    amplitude=3
    period= pi
    reflection on the x-axis
    p.s.= pi/12

    The only difference is the first graph is translated 3 units up since with the reflections cos(angle)=cos(-angle)

  22. Graph 1:
    -3cos(-2+pi/6)+3
    Period: pi
    Altitude: 3
    Phase Shift: pi/2
    Reflection on the x-axis since there is a negative number (-3)
    Translate 3 units up on the y-axis
    Endpoints: pi/12 to (pi/12-pi)= -11pi/12
    Keypoints: (pi/12, 0) (-5pi/12,6) (-2pi/3,3) (-11pi/12,0)

    Graph 2:
    -3cos(2x-pi/6)
    Period: pi
    Amplitude: 3
    Phase Shift: pi/2
    Reflection on the x-axis
    Endpoints: pi/12 to (pi/12+pi)=13pi/12
    Keypoints: (pi/12,-3) (pi/3,0) (7pi/12,3) (5pi/6,0) (13pi/12, -3)

    Both of the graphs have the same period which is pi. But the negative is not included because amplitude and period are positive. They also both have a reflection on the x- axis. A difference would be that graph 1 has a vertical translation while graph 2 does not. Also they both have a amp. of 3 and a phase shift of pi/2. Graph 1 has a vertical translation because there is a +3 at the end. The endpoints of the graphs are different because in graph 1 there is a negative in front of the 2x.

    Period 3

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