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Finding the amplitude and period of a function

Find the amplitude and period of the following functions and graph one cycle.

  1. y = 2 sin ( 1 4 x )
  2. y = −3 sin ( 2 x + π 2 )
  3. y = cos x + 3

We will solve these problems according to the models.

  1. y = 2 sin ( 1 4 x )   involves sine, so we use the form
    y = A sin ( B t + C ) + D

    We know that   | A |   is the amplitude, so the amplitude is 2. Period is   2 π B , so the period is

    2 π B = 2 π 1 4       = 8 π

    See the graph in [link] .

    Graph of y=2sin(1/4 x) from 0 to 8pi, which is one cycle. The amplitude is 2, and the period is 8pi.
  2. y = −3 sin ( 2 x + π 2 )   involves sine, so we use the form
    y = A sin ( B t C ) + D

    Amplitude is   | A | , so the amplitude is   | 3 | = 3. Since   A   is negative, the graph is reflected over the x -axis. Period is   2 π B , so the period is

    2 π B = 2 π 2 = π

    The graph is shifted to the left by   C B = π 2 2 = π 4   units. See [link] .

    Graph of y=-3sin(2x+pi/2) from -pi/4 to 3pi/2, one cycle. The amplitude is 3, and the period is pi.
  3. y = cos x + 3   involves cosine, so we use the form
    y = A cos ( B t ± C ) + D

    Amplitude is   | A | ,   so the amplitude is 1. The period is   2 π .   See [link] . This is the standard cosine function shifted up three units.

    Graph of y=cos(x) + 3 from -pi/2 to 5pi/2. The amplitude and period are the same as the normal y=cos(x), but the whole graph is shifted up on the y-axis by 3.
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What are the amplitude and period of the function   y = 3 cos ( 3 π x ) ?

The amplitude is   3 , and the period is   2 3 .

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Finding equations and graphing sinusoidal functions

One method of graphing sinusoidal functions is to find five key points. These points will correspond to intervals of equal length representing   1 4   of the period. The key points will indicate the location of maximum and minimum values. If there is no vertical shift, they will also indicate x -intercepts. For example, suppose we want to graph the function   y = cos θ . We know that the period is 2 π , so we find the interval between key points as follows.

2 π 4 = π 2

Starting with   θ = 0 , we calculate the first y- value, add the length of the interval   π 2   to 0, and calculate the second y -value. We then add   π 2   repeatedly until the five key points are determined. The last value should equal the first value, as the calculations cover one full period. Making a table similar to [link] , we can see these key points clearly on the graph shown in [link] .

θ 0 π 2 π 3 π 2 2 π
y = cos θ 1 0 −1 0 1
Graph of y=cos(x) from -pi/2 to 5pi/2.

Graphing sinusoidal functions using key points

Graph the function   y = −4 cos ( π x )   using amplitude, period, and key points.

The amplitude is   | 4 | = 4.   The period is   2 π ω = 2 π π = 2.   (Recall that we sometimes refer to   B   as   ω . )   One cycle of the graph can be drawn over the interval   [ 0 , 2 ] .   To find the key points, we divide the period by 4. Make a table similar to [link] , starting with   x = 0   and then adding   1 2   successively to   x   and calculate   y .   See the graph in [link] .

x 0 1 2 1 3 2 2
y = −4 cos ( π x ) −4 0 4 0 −4
Graph of y=-4cos(pi*x) using the five key points: intervals of equal length representing 1/4 of the period. Here, the points are at 0, 1/2, 1, 3/2, and 2.
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Graph the function   y = 3 sin ( 3 x )   using the amplitude, period, and five key points.

x 3 sin ( 3 x )
0 0
π 6 3
π 3 0
π 2 −3
2 π 3 0
Graph of y=3sin(3x) using the five key points: intervals of equal length representing 1/4 of the period. Here, the points are at 0, pi/6, pi/3, pi/2, and 2pi/3.
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Modeling periodic behavior

We will now apply these ideas to problems involving periodic behavior.

Modeling an equation and sketching a sinusoidal graph to fit criteria

The average monthly temperatures for a small town in Oregon are given in [link] . Find a sinusoidal function of the form y = A sin ( B t C ) + D that fits the data (round to the nearest tenth) and sketch the graph.

Month Temperature, o F
January 42.5
February 44.5
March 48.5
April 52.5
May 58
June 63
July 68.5
August 69
September 64.5
October 55.5
November 46.5
December 43.5

Recall that amplitude is found using the formula

A = largest value  smallest value 2

Thus, the amplitude is

| A | = 69 42.5 2      = 13.25

The data covers a period of 12 months, so 2 π B = 12 which gives B = 2 π 12 = π 6 .

The vertical shift is found using the following equation.

D = highest value + lowest value 2

Thus, the vertical shift is

D = 69 + 42.5 2     = 55.8

So far, we have the equation y = 13.3 sin ( π 6 x C ) + 55.8.

To find the horizontal shift, we input the x and y values for the first month and solve for C .

    42.5 = 13.3 sin ( π 6 ( 1 ) C ) + 55.8 13.3 = 13.3 sin ( π 6 C )      1 = sin ( π 6 C ) sin θ = 1 θ = π 2 π 6 C = π 2 π 6 + π 2 = C            = 2 π 3

We have the equation y = 13.3 sin ( π 6 x 2 π 3 ) + 55.8. See the graph in [link] .

Graph of the equation y=13.3sin(pi/6 x - 2pi/3) + 55.8. The average value is a dotted horizontal line y=55.8, and the amplitude is 13.3
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Questions & Answers

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Conney Reply
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Abena Reply
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
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find the equation of the line if m=3, and b=-2
Ashley Reply
graph the following linear equation using intercepts method. 2x+y=4
ok, one moment
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it won't let me send an image?
also for the first one... y=mx+b so.... y=3x-2
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Please were did you get y=mx+b from
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
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0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
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"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Fiston Reply
x=-b+_Гb2-(4ac) ______________ 2a
Ahlicia Reply
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
Carlos Reply
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this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
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consider r(a+b) = ra + rb. The a and b are the trig identity.
How can you tell what type of parent function a graph is ?
Mary Reply
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
y=x will obviously be a straight line with a zero slope
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
yes, correction on my end, I meant slope of 1 instead of slope of 0
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Karim Reply
I don't understand
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
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unknown Reply
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is there any question in particular?
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Sure, are you in high school or college?
Hi, apologies for the delayed response. I'm in college.
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Ef Reply
Practice Key Terms 2

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