Both functions start up at the same y value when x = 0:
[tex]\begin{gathered} f(0)=50(\frac{1}{2})^0=50\cdot1=50 \\ g(0)=50(\frac{1}{3})^0=50\cdot1=50 \end{gathered}[/tex]
However, from there we will multiply by the exponential term.
Since the numbers 2 and 3 are on the denominator, they work on contrary, because when we divide by a higher number, we get a smaller result.
So, for x = 1, for example, since 3 > 2 and there are on the denominato, g(1) wil be less than f(1):
[tex]\begin{gathered} f(1)=50(\frac{1}{2})^1=50\cdot\frac{1}{2}=25 \\ g(1)=50(\frac{1}{3})^1=50\cdot\frac{1}{3}=16.66\ldots \end{gathered}[/tex]
So, We start by noticing that 3 > 2, but since they are on the denominator, the conclusion is the other way around: graph f lies above the graph g, so graph f is the graph 1 and graph g is the graph 2.
