Question

Give a formula for the function illustrated using a vertical shift of an exponential function. The two points marked on the graph are
A=(−1,9) and B=(3,4). The red horizontal line is given by y=2, and is a horizontal asymptote of the function. Then y=??

Answer 1

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:
1. \(7 = \frac{a}{b}\) (Equation 1)
2. \(2 = a \cdot b^{3}\) (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.