Practice problems to help you learn how to determine the equation of a sinusoidal function based on its graph. The general forms of sinusoidal functions are: \[ y = a \sin[ b ( x - d) ] + c \quad \text{or} \quad y = a \cos[ b ( x - d) ] + c \] Use these formulas to identify amplitude, period, phase shift, and vertical shift from the graph of the sine or cosine function.
The scaling along the y-axis is one unit for one large division and therefore the maximum value of y: \( y_{\text{max}} = 1 \) and the minimum value of y: \( y_{\text{min}} = -7 \).
The scaling along the x-axis is \( \pi \) for one large division and \( \dfrac{\pi}{5} \) for one small division.
Points A and B mark the start and the end of one period \( P \), which is equal to \( 5\pi \). 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 \[ y = a \cos\left[ b(x - d) \right] + c \] assuming that originally it is \( \cos(x) \), which starts with a maximum at \( x = 0 \), that is transformed by vertical and horizontal shifting (translation) and vertical and horizontal stretching/shrinking.
Let us calculate \( a \), and \( c \). \[ |a| = \dfrac{y_{\text{max}} - y_{\text{min}}}{2} = \dfrac{1 - (-7)}{2} = 4 \] 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 \( 2\pi \), and we may therefore take \( a = 4 \). \[ c = \dfrac{y_{\text{max}} + y_{\text{min}}}{2} = \dfrac{1 + (-7)}{2} = -3 \] \[ \text{Period: } P = \dfrac{2\pi}{|b|} = 5\pi \] Solve the above for \( |b| \) to obtain: \[ |b| = \dfrac{2}{5} \] Again here we have two possible values for \( b \): \( b = \dfrac{2}{5} \) and \( b = -\dfrac{2}{5} \). We take \( b = \dfrac{2}{5} \) to make our calculations for \( d \) easier.
We now write the function for the graph as follows: \[ y = 4 \cos\left[ \dfrac{2}{5}(x - d) \right] - 3 \] \( d \) indicates the shift. The shift is determined by comparing the graphs of \[ y = 4 \cos\left( \dfrac{2}{5}x \right) - 3 \quad \text{(note \( d = 0 \))} \] and the given graph. We note that the shift (x-coordinate of point A) \( d = -\dfrac{\pi}{5} \) from the graph (one small division to the left). Hence the equation of the graph is: \[ y = 4 \cos\left[ \dfrac{2}{5}\left(x - \left(-\dfrac{\pi}{5}\right)\right) \right] - 3 = 4 \cos\left[ \dfrac{2}{5}\left(x + \dfrac{\pi}{5} \right) \right] - 3 \]
We now check that the function found corresponds to the given graph by checking a few points.
Point A: \( x = -\dfrac{\pi}{5} \); evaluate \( y \) at this value of \( x \).
\[ y\left(-\dfrac{\pi}{5}\right) = 4 \cos\left[ \dfrac{2}{5} \left( -\dfrac{\pi}{5} + \dfrac{\pi}{5} \right) \right] - 3 = 4 \cos(0) - 3 = 1 \] which corresponds to the value on the graph.
Point B: \( x = 4\pi + \dfrac{4\pi}{5} = \dfrac{24\pi}{5} \) (4 small divisions after \( 4\pi \))
\[ y\left( \dfrac{24\pi}{5} \right) = 4 \cos\left[ \dfrac{2}{5} \left( \dfrac{24\pi}{5} + \dfrac{\pi}{5} \right) \right] - 3 = 4 \cos\left( \dfrac{2}{5} \cdot \dfrac{25\pi}{5} \right) - 3 = 4 \cos(2\pi) - 3 = 1 \] which corresponds to the value on the graph.
Maximum value of \( y \): \( y_{\text{max}} = 0.2 \) and the minimum value of \( y \): \( y_{\text{min}} = -1.4 \) (one large division along the \( y \)-axis is equal to 1 unit. A small division is \( \dfrac{1}{5} = 0.2 \)).
The scaling along the \( x \)-axis is \( \pi \) for one large division and \( \dfrac{\pi}{5} \) for one small division.
Points A and B mark the start and the end of one period \( P \), which is equal to \( 4\pi \).
The coordinates of points A and B are: \( A\left(\dfrac{\pi}{2}, 0.2\right) \), \( B\left(\dfrac{9\pi}{2}, 0.2\right) \).
The graph between A and B may be assumed to be that of a transformed \( \cos(x) \). Hence, a possible equation for the given graph is:
\[ y = a \cos\left[ b(x - d) \right] + c \]Let us calculate \( a \) and \( c \).
\[ |a| = \dfrac{y_{\text{max}} - y_{\text{min}}}{2} = \dfrac{0.2 - (-1.4)}{2} = 0.8 \] 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 \( 2\pi \), and we may therefore take \( a = 0.8 \).
\[ c = \dfrac{y_{\text{max}} + y_{\text{min}}}{2} = \dfrac{0.2 + (-1.4)}{2} = -0.6 \]Period: \( P = \dfrac{2\pi}{|b|} = 4\pi \)
Solve for \( |b| \) to obtain: \( |b| = \dfrac{1}{2} \). Again here we have two possible values for \( b \): \( b = \dfrac{1}{2} \) or \( b = -\dfrac{1}{2} \).
We take \( b = \dfrac{1}{2} \) to make our calculations for \( d \) easier.
We now write the function for the graph as follows:
\[ y = 0.8 \cos\left[ \dfrac{1}{2}(x - d) \right] - 0.6 \]\( d \) indicates the shift. The shift is determined by comparing the graphs of \( y = 0.8 \cos\left( \dfrac{1}{2}x \right) - 0.6 \) (note \( d = 0 \)) and the given graph. We note that the shift (x-coordinate of point A) \( d = \dfrac{\pi}{2} \) from the graph (one half a large division to the right). Hence, the equation of the graph is:
\[ y = 0.8 \cos\left[ \dfrac{1}{2}(x - \dfrac{\pi}{2}) \right] - 0.6 \]We now check that the function found corresponds to the given graph by checking a few points.
Point A: \( x = \dfrac{\pi}{2} \); evaluate \( y \) at this value of \( x \):
\[ y\left(\dfrac{\pi}{2}\right) = 0.8 \cos\left[ \dfrac{1}{2}\left(\dfrac{\pi}{2} - \dfrac{\pi}{2}\right) \right] - 0.6 = 0.8 \cos(0) - 0.6 = 0.2 \]which corresponds to the value on the graph.
Point B: \( x = 4\pi + \dfrac{\pi}{2} = \dfrac{9\pi}{2} \) (half a large division after \( 4\pi \))
\[ y\left(\dfrac{9\pi}{2}\right) = 0.8 \cos\left[ \dfrac{1}{2}\left(\dfrac{9\pi}{2} - \dfrac{\pi}{2}\right) \right] - 0.6 = 0.8 \cos(2\pi) - 0.6 = 0.2 \]which is equal to the value on the graph.
Maximum value of \( y \): \( y_{\text{max}} = 0 \) and the minimum value of \( y \): \( y_{\text{min}} = -2 \) (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 \( \dfrac{1}{5} = 0.2 \) 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 \( c \).
\[ |a| = \dfrac{y_{\text{max}} - y_{\text{min}}}{2} = \dfrac{0 - (-2)}{2} = 1 \] 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 \( 2\pi \), and we may therefore take \( a = 1 \).
\[ c = \dfrac{y_{\text{max}} + y_{\text{min}}}{2} = \dfrac{0 + (-2)}{2} = -1 \]Period: \[ P = \dfrac{2\pi}{|b|} = 2 \] Solve for \( |b| \) to obtain: \[ |b| = \pi \] Again here we have two possible values for \( b \): \( b = \pi \) and \( b = -\pi \). We take \( b = \pi \) to make our calculations for \( d \) simpler.
We now write the function for the graph as follows: \[ y = \sin[ \pi(x - d) ] - 1 \]
\( d \) indicates the shift. The shift is determined by comparing the graphs of \( y = \sin[ \pi x ] - 1 \) (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: \[ y = \sin[ \pi(x - 0.6) ] - 1 = \sin\left[ \pi\left(x - \dfrac{3}{5}\right) \right] - 1 \]
We now check that the function found corresponds to the given graph by checking a few points.
Point A: \( x = 0.6 \); evaluate \( y \) at this value of \( x \).
\[ y(0.6) = \sin\left[ \pi(0.6 - \dfrac{3}{5}) \right] - 1 = \sin(0) - 1 = -1 \] which corresponds to the value on the graph.
Point B: \( x = 2.6 \)
\[ y(2.6) = \sin\left[ \pi(2.6 - \dfrac{3}{5}) \right] - 1 = \sin(\pi \cdot 2) - 1 = -1 \] which corresponds to the value on the graph.
Checking values at A and B is not enough because they would give the same values if the function \[ - \sin\left( \pi x - \dfrac{3\pi}{5} \right) - 1 \] was used. We need to check a maximum or a minimum besides A and B.
The first maximum point after point A is at \[ x = 1 + \dfrac{1}{2} \cdot 0.2 = 1.1 \]
\[ y(1.1) = \sin\left[ \pi(1.1 - \dfrac{3}{5}) \right] - 1 = \sin\left( \dfrac{\pi}{2} \right) - 1 = 0 \] which corresponds to the value on the graph.
Maximum value of \( y \): \( y_{\text{max}} = -1 \) and the minimum value of \( y \): \( y_{\text{min}} = -3 \) (one large division along the y-axis is equal to 1 unit. A small division is \( \dfrac{1}{5} = 0.2 \)). The scaling along the x-axis is \( \dfrac{\pi}{5} \) for one large division and \( \dfrac{\pi}{25} \) for one small division.
Points A and B mark the start and the end of one period \( P \) which is equal to \[ P = \dfrac{8\pi}{5} - \dfrac{3\pi}{5} = \pi \].
The coordinates of points A and B are: \( A\left(\dfrac{3\pi}{5}, -1\right) \), \( B\left(\dfrac{8\pi}{5}, -1\right) \).
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: \[ y = a \cos\left[ b(x - d) \right] + c \]
Let us calculate \( a \), and \( c \). \[ |a| = \dfrac{y_{\text{max}} - y_{\text{min}}}{2} = \dfrac{-1 - (-3)}{2} = 1 \]
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 \( 2\pi \), and we may therefore take \( a = 1 \).
\[ c = \dfrac{y_{\text{max}} + y_{\text{min}}}{2} = \dfrac{-1 + (-3)}{2} = -2 \] \[ \text{Period: } P = \dfrac{2\pi}{|b|} = \pi \]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: \[ y = \cos\left[ 2(x - d) \right] - 2 \]
The x-coordinate of point A indicates the shift \( d \), which is determined by comparing the graphs of \( y = \cos\left[ 2x \right] - 2 \) (note \( d = 0 \)) and the given graph. We note that \( d = \dfrac{3\pi}{5} \) from the graph. Hence the equation of the graph is:
\[ y = \cos\left[ 2\left(x - \dfrac{3\pi}{5} \right) \right] - 2 \]
We now check that the function found corresponds to the given graph by checking few points.
Point A: \( x = \dfrac{3\pi}{5} \). Evaluate \( y \) at this value of \( x \): \[ y\left( \dfrac{3\pi}{5} \right) = \cos\left[ 2\left( \dfrac{3\pi}{5} - \dfrac{3\pi}{5} \right) \right] - 2 = \cos(0) - 2 = -1 \]
Which corresponds to the value on the graph.
Point B: \( x = \dfrac{8\pi}{5} \) \[ y\left( \dfrac{8\pi}{5} \right) = \cos\left[ 2\left( \dfrac{8\pi}{5} - \dfrac{3\pi}{5} \right) \right] - 2 = \cos(2\pi) - 2 = -1 \]
Which is equal to the value on the graph.