Answer:
Last graph on the right.
Step-by-step explanation:
Try substituting some values for x and see which graph is valid.
when x = 0, y = f(0) = 3 [tex](2/3)^{0}[/tex] = 3 (1) = 3
When we compare this to the graphs, we immediately see that the first 2 are not correct because in those cases, when x=0, y = 6 (i.e not 3).
Next we try x = 1
when x = 1, y = f(1) = 3 [tex](2/3)^{1}[/tex] = 3 ([tex]\frac{2}{3}[/tex]) = 2
Comparing the graphs ones again, show that only the last graph has x=1 and y = 2.
