To find the exponential equation of the form \(y = ab^x\) that passes through the points (-1, 1/3) and (1, 3/4), we can set up a system of equations using the given points.
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From the point (-1, 1/3): \[ y = ab^{-1} = \frac{1}{3} \] This can be rewritten as: \[ \frac{a}{b} = \frac{1}{3} \quad \text{(Equation 1)} \]
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From the point (1, 3/4): \[ y = ab^1 = \frac{3}{4} \] This gives us: \[ ab = \frac{3}{4} \quad \text{(Equation 2)} \]
Now, we can express \(a\) in terms of \(b\) using Equation 1: \[ a = \frac{1}{3}b \]
Substituting \(a\) into Equation 2: \[ \left(\frac{1}{3}b\right)b = \frac{3}{4} \] \[ \frac{1}{3}b^2 = \frac{3}{4} \]
Next, we can multiply both sides by 3 to eliminate the fraction: \[ b^2 = \frac{3 \times 3}{4} = \frac{9}{4} \]
Now, taking the square root of both sides: \[ b = \frac{3}{2} \quad \text{(using the positive root since we are modeling growth)} \]
Now we can substitute \(b = \frac{3}{2}\) back into Equation 1 to find \(a\): \[ \frac{a}{\frac{3}{2}} = \frac{1}{3} \] Multiplying both sides by \(\frac{3}{2}\): \[ a = \frac{1}{3} \cdot \frac{3}{2} = \frac{1}{2} \]
Thus, we have found \(a\) and \(b\): \[ a = \frac{1}{2}, \quad b = \frac{3}{2} \]
Finally, the exponential equation is: \[ y = \frac{1}{2} \left(\frac{3}{2}\right)^x \]
This is the exponential equation that corresponds to the rising line in the graph given the specified points.
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