see below
Step-by-step explanation:
we are given a exponential function
![\displaystyle y = \left(1/5\right) ^{x}](/tpl/images/1393/9515/a2225.png)
we want to figure out the missing values of y for x i.e (-2,-1, and 2) to do we can consider substituting so,
when x is -2 y should be
![\displaystyle y = \left(1/5\right) ^{ - 2}](/tpl/images/1393/9515/029f4.png)
rewrite ⅕ in exponent notation:
![\displaystyle y = \left((5) ^{ - 1} \right) ^{ - 2}](/tpl/images/1393/9515/a9599.png)
by law of exponent we obtain:
![\displaystyle y = (5) ^{ 2}](/tpl/images/1393/9515/fab90.png)
simplify square:
![\displaystyle y = \boxed{ 25}](/tpl/images/1393/9515/c1082.png)
when x is -1:
![\displaystyle y = \left(1/5\right) ^{ - 1}](/tpl/images/1393/9515/a3ec7.png)
rewrite ⅕ in exponent notation:
![\displaystyle y = \left((5) ^{ - 1} \right) ^{ - 1}](/tpl/images/1393/9515/dc332.png)
use law of exponent which yields:
![\displaystyle y = \boxed5](/tpl/images/1393/9515/43e3c.png)
when x is 3:
![\displaystyle y = \left(1/5\right) ^{ 3}](/tpl/images/1393/9515/35de7.png)
by law of exponent we obtain:
![\displaystyle y = 1/ {5}^{3}](/tpl/images/1393/9515/80491.png)
simplify square which yields:
![\displaystyle y = \frac{1}{ \boxed{125}}](/tpl/images/1393/9515/70af4.png)