To find the formula for the function with a vertical shift of an exponential function, we start with the standard form of an exponential function:
\[ y = a \cdot b^{x} + k \]
where:
- \(a\) is a scaling factor,
- \(b\) is the base of the exponential (greater than 0),
- \(k\) is the vertical shift (which corresponds to the horizontal asymptote).
Given that the horizontal asymptote is \(y = 2\), we have \(k = 2\). So, our function simplifies to:
\[ y = a \cdot b^{x} + 2 \]
Next, we use the two points provided, \(A = (-1, 9)\) and \(B = (3, 4)\), to create a system of equations to solve for \(a\) and \(b\).
Substituting point A into the equation:
\[ 9 = a \cdot b^{-1} + 2 \] \[ 9 - 2 = \frac{a}{b} \] \[ 7 = \frac{a}{b} \quad \text{(Equation 1)} \]
Now, substituting point B into the equation:
\[ 4 = a \cdot b^{3} + 2 \] \[ 4 - 2 = a \cdot b^{3} \] \[ 2 = a \cdot b^{3} \quad \text{(Equation 2)} \]
Now we have:
- \(7 = \frac{a}{b}\) (Equation 1)
- \(2 = a \cdot b^{3}\) (Equation 2)
From Equation 1, we can express \(a\) in terms of \(b\):
\[ a = 7b \]
Substituting \(a\) into Equation 2:
\[ 2 = (7b) \cdot b^{3} \] \[ 2 = 7b^{4} \] \[ b^{4} = \frac{2}{7} \] \[ b = \left(\frac{2}{7}\right)^{\frac{1}{4}} = \frac{\sqrt[4]{2}}{\sqrt[4]{7}} \]
Now substituting \(b\) back into Equation 1 to find \(a\):
\[ a = 7b = 7 \cdot \frac{\sqrt[4]{2}}{\sqrt[4]{7}} = \frac{7\sqrt[4]{2}}{\sqrt[4]{7}} \]
Now we can write the original function:
\[ y = \frac{7\sqrt[4]{2}}{\sqrt[4]{7}} \left(\frac{\sqrt[4]{2}}{\sqrt[4]{7}}\right)^{x} + 2 \]
Simplifying further, the final formula for the function is:
\[ y = \frac{7\sqrt[4]{2}}{\sqrt[4]{7}} \left(\frac{\sqrt[4]{2}}{\sqrt[4]{7}}\right)^{x} + 2 \]
This function will satisfy both points and the given asymptote.