2. Exponential function:
![f(x)=((1)/(4))^x](https://img.qammunity.org/2023/formulas/mathematics/college/gv7joh3mfzvwo2em62ovzf2k09bt46e4l5.png)
Substituting with x = 0, we get:
![\begin{gathered} f(0)=((1)/(4))^0 \\ f(0)=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nri3ld0zh0lxjq4cesryhacg27aaxi9kfh.png)
Then, f(x) passes through the point (0 ,1)
Substituting with x = -1, we get:
![\begin{gathered} f(-1)=((1)/(4))^(-1) \\ f(-1)=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w76dp2fzmzit7af6dlav42nna6zhnx9mex.png)
Then, f(x) passes through the point (-1 ,4)
Substituting with x = -2, we get:
![\begin{gathered} f(-2)=((1)/(4))^(-2) \\ f(-2)=4^2 \\ f(-2)=16 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/feldjewo4m3saj70ewi56vym5wnqygowr4.png)
Then, f(x) passes through the point (-2 ,16)
Substituting with x = 1, we get:
![\begin{gathered} f(1)=((1)/(4))^1 \\ f(1)=(1)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eec6q59p7y45qz9dww268z3f6mdos360oo.png)
Then, f(x) passes through the point (1, 1/4)
Substituting with x = 2, we get:
![\begin{gathered} f(2)=((1)/(4))^2 \\ f(2)=(1^2)/(4^2)^{} \\ f(2)=(1)/(16)^{} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rp3b2fijtoz2vghp6rnbriov6h1n0zyq6l.png)
Then, f(x) passes through the point (2, 1/16)
f(x) has the form:
![y=b^x^{}](https://img.qammunity.org/2023/formulas/mathematics/college/qzotjtab372imvrxpl009bthv8bgta88ap.png)
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: