To find the exponential function of the form \( y = a \cdot b^x \) that passes through the points \((-3, \frac{5}{64})\) and \((1, 20)\), we need to determine the constants \(a\) and \(b\). Here's a step-by-step process:
1. Set up the system of equations:
We have two points and we need to set up two equations using the exponential function form \(y = a \cdot b^x\).
For the point \((-3, \frac{5}{64}\)):
[tex]\[
\frac{5}{64} = a \cdot b^{-3}
\][/tex]
For the point \((1, 20)\):
[tex]\[
20 = a \cdot b^1
\][/tex]
2. Solve for \(a\) using the second equation:
[tex]\[
20 = a \cdot b
\][/tex]
[tex]\[
a = \frac{20}{b}
\][/tex]
3. Substitute \(a\) in the first equation:
[tex]\[
\frac{5}{64} = \left(\frac{20}{b}\right) \cdot b^{-3}
\][/tex]
Simplify the equation:
[tex]\[
\frac{5}{64} = 20 \cdot b^{-4}
\][/tex]
[tex]\[
\frac{5}{64} = \frac{20}{b^4}
\][/tex]
4. Solve for \(b\):
[tex]\[
\frac{5}{64} \cdot b^4 = 20
\][/tex]
[tex]\[
b^4 = 20 \cdot \frac{64}{5}
\][/tex]
[tex]\[
b^4 = 256
\][/tex]
[tex]\[
b = \sqrt[4]{256}
\][/tex]
[tex]\[
b = 4
\][/tex]
5. Solve for \(a\) using \(b = 4\):
Substitute \(b\) back into the equation \(a = \frac{20}{b}\):
[tex]\[
a = \frac{20}{4}
\][/tex]
[tex]\[
a = 5
\][/tex]
6. Write the exponential function:
Now that we have \(a = 5\) and \(b = 4\), the exponential function is:
[tex]\[
y = 5 \cdot 4^x
\][/tex]
So, the exponential function that passes through the points \((-3, \frac{5}{64})\) and \((1, 20)\) is:
[tex]\[
y = 5 \cdot 4^x
\][/tex]
