To find an exponential function in the form \( y = a(b)^x \) that passes through the points (-1, 10) and (1, 2/5), we first use the two points to create a system of equations.
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For the point (-1, 10): \[ 10 = a(b)^{-1} \quad \Rightarrow \quad 10 = \frac{a}{b} \quad \Rightarrow \quad a = 10b \]
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For the point (1, 2/5): \[ \frac{2}{5} = a(b)^{1} \quad \Rightarrow \quad \frac{2}{5} = ab \quad \Rightarrow \quad a = \frac{2}{5b} \]
Now we have two expressions for \( a \):
- \( a = 10b \)
- \( a = \frac{2}{5b} \)
Set these equal to each other: \[ 10b = \frac{2}{5b} \]
To solve for \( b \), first multiply both sides by \( 5b \) to eliminate the fraction: \[ 50b^2 = 2 \] \[ b^2 = \frac{2}{50} = \frac{1}{25} \quad \Rightarrow \quad b = \frac{1}{5} \quad (\text{since } b > 0) \]
Now, substitute \( b \) back into the equation for \( a \): \[ a = 10b = 10 \cdot \frac{1}{5} = 2 \]
Thus, we have \( a = 2 \) and \( b = \frac{1}{5} \).
Now we can write the equation of the exponential function: \[ y = 2\left(\frac{1}{5}\right)^x \]
This can also be rewritten as: \[ y = 2 \cdot 5^{-x} \]
Therefore, the equation of the exponential function is: \[ \boxed{y = 2\left(\frac{1}{5}\right)^x} \]