Create an equation that results in at least on extraneous solution

(0, 1) and (-2, 9) are points on the graph.

Step-by-step explanation:

I am assuming that the x is an exponent 1/3^x.

Take each x values and plug it into  (1/3)^x

x = 0 , f(x) = (1/3)^0 = 1. Therefore (0, 1) is on the graph.

x = 3 , f(x) = 1/3^3 = 1/27 so (3,27) not on graph.

x = -2, f(x) = (1/3)^-2   1 / 1/9 = 9 so (-2, 9) is on graph.

x = -1, f(x) = (1/3)^-1 = 1 / 1/3 = 3 so (-1, -1/3) is not on the graph.