To find the exponential function that represents the curve given the points in quadrant 1, we generally want to use the form of an exponential function, which is:
\[ f(x) = a \cdot b^x \]
where \( a \) is a constant and \( b \) is the base of the exponential function.
Since I cannot see the image you mentioned, I'll guide you on how to determine the exponential function:
Steps to Find the Exponential Function
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Identify Points: Look for the x and y coordinates of the plotted points on the graph. Let's assume the points are:
- Point 1: \((x_1, y_1)\)
- Point 2: \((x_2, y_2)\)
- Point 3: \((x_3, y_3)\)
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Set Up Equations: Using the points, set up a system of equations. For each point \((x_i, y_i)\):
- Equation 1: \( y_1 = a \cdot b^{x_1} \)
- Equation 2: \( y_2 = a \cdot b^{x_2} \)
- Equation 3: \( y_3 = a \cdot b^{x_3} \)
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Solve for \( a \) and \( b \): Choose any two equations and solve for \( a \) and \( b \) simultaneously. This might involve taking logarithms to isolate \( b \).
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Confirm with Third Point: After finding \( a \) and \( b \), plug in the coordinates of the third point to confirm that your function works.
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Evaluate \( f(5) \): Once you have the function, simply substitute \( x = 5 \) into the function to find \( f(5) \).
Example
For example, if the points were (1, 3), (2, 6), and (3, 12):
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Set up the equations:
- \( 3 = a \cdot b^1 \)
- \( 6 = a \cdot b^2 \)
- \( 12 = a \cdot b^3 \)
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Divide the second equation by the first: \[ \frac{6}{3} = \frac{a \cdot b^2}{a \cdot b^1} \implies 2 = b \]
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Plug \( b \) back into the first equation: \[ 3 = a \cdot 2^1 \implies a = \frac{3}{2} \]
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Therefore, our function is: \[ f(x) = \frac{3}{2} \cdot 2^x \]
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Now evaluate \( f(5) \): \[ f(5) = \frac{3}{2} \cdot 2^5 = \frac{3}{2} \cdot 32 = 48 \]
Answer
So, if you follow these steps with the actual points you have:
- \( f(x) = a \cdot b^x \)
- \( f(5) = \text{calculated value} \)
Just replace \( a \) and \( b \) with what you find from your points. Don’t forget to round \( f(5) \) to two decimal places as per your requirement.
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