Find a Sinusoidal Function for Each of the Graphs Below

Solution
The scaling along the yaxis is one unit for one large division and therefore the maximum value of y: y_{max} = 1 and the minimum value of y: y_{min} =  7. The scaling along the x axis is π for one large division and π/5 for one small division.
Points A and B mark the start and the end of one period P which is equal to 5π. These points are useful because they are maximum points with clear coordinates.
Since A and B are maximum points, it is easier to write an equation for the graph as y = a cos[ b(x  d) ] + c assuming that originally it is a cos(x), which starts with a maximum at x = 0, that is transformed by vertical and horizontal shifting (translation) and vertical and horizontal stretching/shringking.
Let us calculate a, and c.
a = (y_{max}  y_{min}) = (1  (7)) / 2 = 4. Which gives two possible values for a: a = 4 or a =  4
The graph between A and B has no reflection when compared to the period of cos(x) between 0 and 2π and we may therefore take a = 4.
c = (y_{max} + y_{min}) = (1 + (7)) / 2 =  3
Period: P = 2πb= 5π
solve the above for b to obtain: b = 2/5.
Again here we have two possible values for b: b = 2/5 and b = 2/5. We take b = 2/5 to make our calculations for d easier.
We now write the function for the graph as follows:
y = 4 cos[ (2/5)(x  d) ]  3
d indicates the shift. The shift is determined by comparing the graphs of y = 4 cos[ (2/5)(x) ]  3 ( note d = 0 ) and the given graph. We note that the shift (x coordinate of point A) d =  π/5 from the graph (one small division the the left). Hence the equation of the graph is:
y = 4 cos[ (2/5)(x  (π/5) ]  3 = 4 cos[ (2/5)(x + π/5) ]  3
We now check that the function found corresponds to the given graph by checking few points.
Point A: x = π/5; evaluate y at this value of x.
y(  π/5)= 4 cos[ (2/5)(  π/5 + π/5) ]  3 = 4 cos[ (2/5)( 0 ) ]  3 = 1 which corresponds to the value on the graph.
Point B: x = 4π + 4 π/5 = 24π/5 ( 4 small division after 4π)
y( 24π/5 )= 4 cos[ (2/5)( 24π/5 + π/5) ]  3 = 4 cos[ (2/5)( 25π/5) ]  3 = 4 cos[ (2/5)( 5π) ]  3 = 4 cos[ (2π) ]  3 = 1
which corresponds to the value on the graph.

Solution
Maximum value of y: y_{max} = 0.2 and the minimum value of y: y_{min} =  1.4 (one large division along the y axis is equal to 1 unit. A small division is 1/5 = 0.2). The scaling along the x axis is π for one large division and π/5 for one small division.
Points A and B mark the start and the end of one period P which is equal to 4π.
The coordinates of points A and B are: A(π/2 , 0.2 ) , B(9π/2 , 0.2).
The graph between A and B may be assumed that of a cos(x) that has been transformed. Hence a possible equation for the given graph is: y = a cos[ b(x  d) ] + c.
Let us calculate a, and c.
a = (y_{max}  y_{min}) = (0.2  (1.4)) / 2 = 0.8. Which gives two possible values for a: a = 0.8 or a =  0.8
The period between A and B has no reflection when compared to the period of cos(x) between 0 and 2π and we may therefore take a = 0.8.
c = (y_{max} + y_{min}) = (0.2 + (1.4)) / 2 =  0.6
Period: P = 2πb= 4π
solve for b to obtain: b = 1/2. Again here we have two possible values for b: b = 1/2 and b =  1/2. We take b = 1/2 to make our calculations for d easier.
We now write the function for the graph as follows:
y = 0.8 cos[ (1/2)(x  d) ]  0.6
d indicates the shift. The shift is determined by comparing the graphs of y = 0.8 cos[ (1/2)(x) ]  0.6 ( note d = 0 ) and the given graph. We note that the shift (x coordinate of point A) d = π/2 from the graph (one half a large division the the right). Hence the equation of the graph is:
y = 0.8 cos[ (1/2)(x  π/2) ]  0.6
We now check that the function found corresponds to the given graph by checking few point.
Point A: x = π/2; evaluate y at this value of x.
y(π/2)= 0.8 cos[ (1/2)(π/2  π/2) ]  0.6 = 0.8 cos (0)  0.6 = 0.2 , which corresponds to the value on the graph.
Point B: x = 4π + π/2 = 9π/2 ( half a large division after 4π)
y( 9π/2 )= 0.8 cos[ (1/2)(9π/2  π/2) ]  0.6 = 0.8 cos (2π)  0.6 = 0.2
which is equal to the value on the graph.

Solution
Maximum value of y: y_{max} = 0 and the minimum value of y: y_{min} =  2 (one large division along the y axis is equal to 1 unit). The scaling along the x axis is 1 unit for one large division and 1/5 = 0.2 for one small division.
Points A and B mark the start and the end of one period P which is calculated as follows: P = 2.6  0.6 = 2.
The coordinates of points A and B are: A(0.6 , 0 ) , B(2.6 , 0).
The graph between A and B may be assumed that of a sin(x) that has been transformed. Hence a possible equation for the given graph is: y = a sin[ b(x  d) ] + c.
Let us calculate a, and c.
a = (y_{max}  y_{min}) = (0  (2)) / 2 = 1. Which gives two possible values for a: a = 1 or a =  1
The period between A and B has no reflection when compared to the period of sin(x) between 0 and 2π and we may therefore take a = 1.
c = (y_{max} + y_{min}) = (0 + (2)) / 2 =  1
Period: P = 2πb= 2;
solve for b to obtain: b = π. Again here we have two possible values for b: b = π and b =  π. We take b = π to make our calculations for d simpler.
We now write the function for the graph as follows:
y = sin[ π(x  d) ]  1
d indicates the shift. The shift is determined by comparing the graphs of y = sin[ π(x) ]  1 ( note d = 0 ) and the given graph. We note the shift (x coordinate of A) d = 0.6 from the graph; shift to the right. Hence the equation of the graph is:
y = sin[ π (x  0.6) ]  1 = sin[ π(x  3/5) ]  1
We now check that the function found corresponds to the given graph by checking few point.
Point A: x = 0.6; evaluate y at this value of x.
y( 0.6)= sin[ π(0.6  3/5) ]  1 = sin[ π(0) ]  1 = 1 , which corresponds to the value on the graph.
Point B: x = 1.6
y( 1.6)= sin[ π(1.6  3/5) ]  1 = sin(π)  1 = 1 , which corresponds to the value on the graph.
Checking values at A and B is not enough because they would give same values if the function  sin[ πx  3π/5) ]  1 was used. We need to check a maximum or a minimum beside A and B. The first maximum point after point A is at x = 1 + (1/2)0.2 = 1.1
y( 1.1)= sin[ π(1.1  3/5) ]  1 = sin (0.5 π)  1 = 0 , which corresponds to the value on the graph.

Solution
Maximum value of y: y_{max} = 1 and the minimum value of y: y_{min} =  3 (one large division along the y axis is equal to 1 unit. A small division is 1/5 = 0.2). The scaling along the x axis is π/5 for one large division and π/25 for one small division.
Points A and B mark the start and the end of one period P which is equal to 8π/5  3π/5 = π.
The coordinates of points A and B are: A(3π/5 ,  1) , B(8π/5 ,  1).
The period between A and B may be considered as that of a cos(x) that has been transformed. Hence a possible equation for the given graph is: y = a cos[ b(x  d) ] + c.
Let us calculate a, and c.
a = (y_{max}  y_{min}) = (1  (3)) / 2 = 1. Which gives two possible values for a: a = 1 or a =  1 .
The period between A and B has no reflection when compared to the period of cos(x) between 0 and 2π and we may therefore take a = 1.
c = (y_{max} + y_{min}) = (1 + (3)) / 2 =  2
Period: P = 2πb= π
solve for b to obtain: b = 2. Two possible values for b: b = 2 and b =  2. We take b = 2 to make our calculations for d easier.
We now write the function for the graph as follows:
y = cos[ 2(x  d) ]  2
The x coordinate of point A indicates the shift d which is determined by comparing the graphs of y = cos[ 2(x) ]  2 ( note d = 0 ) and the given graph. We note that d = 3π/5 from the graph. Hence the equation of the graph is:
y = cos[ 2(x  3π/5) ]  2
We now check that the function found corresponds to the given graph by checking few points.
Point A: x = 3π/5 evaluate y at this value of x.
y( 3π/5 )= cos[ 2(3π/5  3π/5) ]  2 = cos(0)  2 =  1 , which corresponds to the value on the graph.
Point B: x = 8π/5
y( 8π/5 )= cos[ 2(8π/5  3π/5) ]  2 = cos (2π)  2 =  1
which is equal to the value on the graph.
