Chapter 9 - Section 9.7 - Common Logarithms, Natural Logarithms, and Change of Base - Exercise Set - Page 586: 81

Please see the graph.

$f(x)=e^{-3x}$ $x=0$ $f(x)=e^{-3x}$ $f(0)=e^{-3*0}$ $f(0)=e^0$ $f(0)=1$ $x=1$ $f(x)=e^{-3x}$ $f(1)=e^{-3*1}$ $f(1)=e^{-3}$ $x=2$ $f(x)=e^{-3x}$ $f(2)=e^{-3*2}$ $f(2)=e^{-6}$ $(0,1)$, $(1,e^{-3})$, and $(2,e^{-6})$ are on the curve.