2. Exponential function:
[tex]f(x)=(\frac{1}{4})^x[/tex]
Substituting with x = 0, we get:
[tex]\begin{gathered} f(0)=(\frac{1}{4})^0 \\ f(0)=1 \end{gathered}[/tex]
Then, f(x) passes through the point (0 ,1)
Substituting with x = -1, we get:
[tex]\begin{gathered} f(-1)=(\frac{1}{4})^{-1} \\ f(-1)=4 \end{gathered}[/tex]
Then, f(x) passes through the point (-1 ,4)
Substituting with x = -2, we get:
[tex]\begin{gathered} f(-2)=(\frac{1}{4})^{-2} \\ f(-2)=4^2 \\ f(-2)=16 \end{gathered}[/tex]
Then, f(x) passes through the point (-2 ,16)
Substituting with x = 1, we get:
[tex]\begin{gathered} f(1)=(\frac{1}{4})^1 \\ f(1)=\frac{1}{4} \end{gathered}[/tex]
Then, f(x) passes through the point (1, 1/4)
Substituting with x = 2, we get:
[tex]\begin{gathered} f(2)=(\frac{1}{4})^2 \\ f(2)=\frac{1^2}{4^2}^{} \\ f(2)=\frac{1}{16}^{} \end{gathered}[/tex]
Then, f(x) passes through the point (2, 1/16)
f(x) has the form:
[tex]y=b^x^{}[/tex]
where b, the base, is between 0 and 1. This means that, when x tends to infinity, f(x) tends to zero, and when x tends to negative infinity, f(x) tends to
infinity.
Taking into account these characteristics and the points where f(x) passess, its graph is:
