Answer:[tex]g(x)\text{= (}\frac{1}{4})^{x+2}[/tex]Explanations:
An exponential function is given by the equation:
[tex]y=ab^x[/tex]
For the points (-2, 1) and (1, 1/64)
[tex]x_1=-2,y_1=1,x_2=1,y_2=\frac{1}{64}[/tex]
Substitute x₁, y₁, x₂, and y₂ into the functions to form two equations
[tex]\begin{gathered} y_1=ab^{x_1} \\ y_2=ab^{x_2} \end{gathered}[/tex][tex]\begin{gathered} \text{1 = }ab^{-2}\ldots\ldots(1) \\ \frac{1}{64}=ab^1\ldots\ldots.(2) \end{gathered}[/tex]
Divide equation (2) by equation (1)
[tex]\begin{gathered} \frac{1}{64}=\text{ }\frac{ab^1}{ab^{-2}} \\ \frac{1}{64}=b^3 \\ b\text{ = }\frac{1}{\sqrt[3]{64}} \\ \text{b = }\frac{1}{4} \end{gathered}[/tex]
Substitute the value b = 1/4 into the equation (2)
[tex]\begin{gathered} \frac{1}{64}=a\text{ }\frac{1}{4} \\ \text{a = }\frac{4}{64} \\ \text{a = }\frac{1}{16} \end{gathered}[/tex][tex]\begin{gathered} \text{Substituting the values of a and b into g(x) = ab}^x \\ g(x)\text{ = }\frac{1}{16}(\frac{1}{4})^x \\ \text{g(x)= (}\frac{1}{4})^2(\frac{1}{4})^x \\ g(x)\text{ = (}\frac{1}{4})^{x+2} \end{gathered}[/tex]
