Find Exponential Functions from Their Graphs

Find the exponential function of the form \( y = b^x \) whose graph is shown below.

Solution:

• From the graph, when \( x = 1 \), we have \( y = 4 \).
• Substitute into \( y = b^x \): \( b^1 = 4 \implies b = 4 \).
• Hence, the function is: \( y = 4^x \).

Find the exponential function of the form \( y = b^{x-c} \) whose graph is shown below.

Solution:

• From the graph: at \( x = 2 \), \( y = 1 \); and at \( x = 3 \), \( y = 2 \).
• Substitute into \( y = b^{x-c} \): \[ b^{2 - c} = 1 \quad \text{(Equation 1)}, \quad b^{3 - c} = 2 \quad \text{(Equation 2)} \]
• Equation (1) yields \( b^{2-c} = b^0 \implies 2-c = 0 \implies c = 2 \).
• Substitute \( c = 2 \) into Equation (2): \( b^{3-2} = 2 \implies b = 2 \).
• Hence, the function is: \( y = 2^{x-2} \).

Find the exponential function of the form \( y = -b^x + d \) whose graph is shown below.

Solution:

• From the graph: at \( x = 0 \), \( y = 0 \); and at \( x = 1 \), \( y = -2 \).
• Substitute \( (0,0) \) into \( y = -b^x + d \): \[ -b^0 + d = 0 \implies -1 + d = 0 \implies d = 1 \]
• Substitute \( (1,-2) \) and \( d = 1 \): \[ -b^1 + 1 = -2 \implies -b = -3 \implies b = 3 \]
• Hence, the function is: \( y = -3^x + 1 \).

Find the exponential function of the form \( y = a \cdot b^x + d \) with horizontal asymptote \( y = 1 \).

Solution:

• Horizontal asymptote \( y = d \implies d = 1 \).
• From the graph: point \( (0, 4) \) gives: \[ a \cdot b^0 + 1 = 4 \implies a + 1 = 4 \implies a = 3 \]
• Point \( (1, 7) \) on the graph gives: \[ 3 \cdot b^1 + 1 = 7 \implies 3b = 6 \implies b = 2 \]
• Hence, the function is: \( y = 3 \cdot 2^x + 1 \).

Find the exponential function of the form \( y = a \cdot e^{x-1} + d \) with horizontal asymptote \( y = -2 \).

Solution:

• Horizontal asymptote: \( d = -2 \).
• From the graph, for \( x = 1 \), \( y = -4 \): \[ a \cdot e^{1-1} - 2 = -4 \implies a \cdot 1 - 2 = -4 \implies a = -2 \]
• Hence, the function is: \( y = -2 \cdot e^{x-1} - 2 \).

Find the exponential function of the form \( y = a \cdot b^{x-1} + d \) with horizontal asymptote \( y = -1 \).

Solution:

• Horizontal asymptote: \( d = -1 \).
• From the graph, using points \( (1,2) \) and \( (2,5) \): \[ a \cdot b^{1-1} - 1 = 2 \implies a \cdot 1 - 1 = 2 \implies a = 3 \]
• Substitute \( a = 3 \) and point \( (2,5) \): \[ 3 \cdot b^{2-1} - 1 = 5 \implies 3b = 6 \implies b = 2 \]
• Hence, the function is: \( y = 3 \cdot 2^{x-1} - 1 \).