# Find a Sinusoidal Function Given its Graph

How to find the equation of a sinusoidal function of the form y = a sin( b ( x - d)) + c or y = a cos( b ( x - d)) + c given its graph.

 Find a Sinusoidal Function for Each of the Graphs Below Solution The scaling along the y-axis is one unit for one large division and therefore the maximum value of y: ymax = 1 and the minimum value of y: ymin = - 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| = (ymax - ymin) = (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 = (ymax + ymin) = (1 + (-7)) / 2 = - 3 Period: P = 2π|b|= 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: ymax = 0.2 and the minimum value of y: ymin = - 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| = (ymax - ymin) = (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 = (ymax + ymin) = (0.2 + (-1.4)) / 2 = - 0.6 Period: P = 2π|b|= 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: ymax = 0 and the minimum value of y: ymin = - 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| = (ymax - ymin) = (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 = (ymax + ymin) = (0 + (-2)) / 2 = - 1 Period: P = 2π|b|= 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: ymax = -1 and the minimum value of y: ymin = - 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| = (ymax - ymin) = (-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 = (ymax + ymin) = (-1 + (-3)) / 2 = - 2 Period: P = 2π|b|= 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.

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