Question

Find an exponential function that passes through the points (-2,6) and (2,1)

Answer 1

To find an exponential function of the form \( f(x) = ab^x \) that passes through the points (-2, 6) and (2, 1), we can set up the following equations using these points.

1. For the point (-2, 6):
\[
f(-2) = ab^{-2} = 6
\]
This simplifies to:
\[
\frac{a}{b^2} = 6 \quad \text{(Equation 1)}
\]

2. For the point (2, 1):
\[
f(2) = ab^2 = 1
\]
This simplifies to:
\[
ab^2 = 1 \quad \text{(Equation 2)}
\]

Now we have a system of two equations:
1. \(\frac{a}{b^2} = 6\)
2. \(ab^2 = 1\)

### Step 1: Solve for \(a\) in terms of \(b\)

From Equation 1:
\[
a = 6b^2
\]

### Step 2: Substitute \(a\) into Equation 2

Now substitute \(a\) into Equation 2:
\[
(6b^2)b^2 = 1
\]
Simplifying:
\[
6b^4 = 1
\]
Now solve for \(b^4\):
\[
b^4 = \frac{1}{6}
\]

Taking the fourth root gives:
\[
b = \left(\frac{1}{6}\right)^{1/4} = \frac{1}{\sqrt[4]{6}}
\]

### Step 3: Find \(a\)

Using \(b\) to find \(a\):
\[
a = 6b^2 = 6\left(\frac{1}{\sqrt[4]{6}}\right)^2 = 6 \cdot \frac{1}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}
\]

### Final Function
Now we have \(a\) and \(b\):
- \( a = \sqrt{6} \)
- \( b = \frac{1}{\sqrt[4]{6}} \)

Thus, the exponential function that passes through the points (-2, 6) and (2, 1) is:
\[
f(x) = \sqrt{6} \left(\frac{1}{\sqrt[4]{6}}\right)^x
\]

### Conclusion
You can express it as:
\[
f(x) = \sqrt{6} \cdot 6^{-x/4}
\]
This is the required exponential function.