To write the exponential equation of the form \( y = ab^x \), we can use the two points given: (-2, 1.28) and (1, 2.5).
Step 1: Set up the equations
Using the general form of the exponential equation \( y = ab^x \):
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For the point (-2, 1.28): \[ 1.28 = ab^{-2} \quad \text{or} \quad 1.28 = \frac{a}{b^2} \quad \text{(Equation 1)} \]
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For the point (1, 2.5): \[ 2.5 = ab^1 \quad \text{or} \quad 2.5 = ab \quad \text{(Equation 2)} \]
Step 2: Solve the equations
From Equation 2, we can express \( a \) in terms of \( b \): \[ a = \frac{2.5}{b} \quad \text{(substitute into Equation 1)} \]
Now substitute \( a \) in Equation 1: \[ 1.28 = \frac{2.5}{b} \cdot \frac{1}{b^2} \] \[ 1.28 = \frac{2.5}{b^3} \] Multiplying both sides by \( b^3 \) gives: \[ 1.28b^3 = 2.5 \] Now solve for \( b^3 \): \[ b^3 = \frac{2.5}{1.28} \approx 1.953125 \] Taking the cube root: \[ b \approx \sqrt[3]{1.953125} \approx 1.26 \]
Now substitute \( b \) back to find \( a \): \[ a = \frac{2.5}{1.26} \approx 1.98 \]
Step 3: Write the final equation
Now plug in the values of \( a \) and \( b \) into the exponential equation: \[ y \approx 1.98 \cdot (1.26)^x \]
Thus, the final exponential equation that represents the graph is approximately: \[ y \approx 1.98 \cdot (1.26)^x \]
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