To find the formula for the exponential function passing through the points [tex]\(\left(-3, \frac{2}{27}\right)\)[/tex] and [tex]\((3, 54)\)[/tex], we need to determine the parameters [tex]\(a\)[/tex] and [tex]\(b\)[/tex] in the general form of an exponential function:
[tex]\[ y = a \cdot b^x \][/tex]
1. Set up the equations using the given points:
Substitute the coordinates of the first point [tex]\((-3, \frac{2}{27})\)[/tex]:
[tex]\[ \frac{2}{27} = a \cdot b^{-3} \quad \text{(Equation 1)} \][/tex]
Substitute the coordinates of the second point [tex]\((3, 54)\)[/tex]:
[tex]\[ 54 = a \cdot b^3 \quad \text{(Equation 2)} \][/tex]
2. Eliminate [tex]\(a\)[/tex] by dividing Equation 2 by Equation 1:
Divide Equation 2 by Equation 1 to eliminate [tex]\(a\)[/tex]:
[tex]\[
\frac{54}{\frac{2}{27}} = \frac{a \cdot b^3}{a \cdot b^{-3}}
\][/tex]
Simplify the fraction:
[tex]\[
\frac{54 \cdot 27}{2} = b^6
\][/tex]
Calculate the left side of the equation:
[tex]\[
729 = b^6
\][/tex]
Solve for [tex]\(b\)[/tex]:
[tex]\[
b = 729^{1/6}
\][/tex]
We find:
[tex]\[
b = 3
\][/tex]
3. Solve for [tex]\(a\)[/tex] using [tex]\(b\)[/tex]:
Substitute [tex]\(b = 3\)[/tex] back into Equation 1:
[tex]\[
\frac{2}{27} = a \cdot 3^{-3}
\][/tex]
Simplify the exponent:
[tex]\[
\frac{2}{27} = a \cdot \frac{1}{27}
\][/tex]
Solve for [tex]\(a\)[/tex]:
[tex]\[
a = 2
\][/tex]
4. Write the final exponential function:
Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex] into the general form [tex]\(y = a \cdot b^x\)[/tex]:
[tex]\[
y = 2 \cdot 3^x
\][/tex]
Thus, the exponential function passing through the points [tex]\(\left(-3, \frac{2}{27}\right)\)[/tex] and [tex]\((3, 54)\)[/tex] is:
[tex]\[
y = 2 \cdot 3^x
\][/tex]
