Hmmm...
I think, f(x) is [tex]f(x)=\left(\dfrac{1}{3}\right)^x[/tex] not [tex]f(x)=\dfrac{1}{3}x[/tex].
The horizontal asymptote is y = -2.
Let [tex]g(x)=\left(\dfrac{1}{3}\right)^{x+a}+b[/tex].
We have two points on a graph ((-1, 1) and (0, -1). substitute:
[tex]1=\left(\dfrac{1}{3}\right)^{-1+a}+b\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\ \text{and}\ a^{-n}=\dfrac{1}{a^n}\\\\1=\left(\dfrac{1}{3}\right)^{-1}\cdot\left(\dfrac{1}{3}\right)^a+b\\\\1=3\cdot\left(\dfrac{1}{3}\right)^a+b\qquad\text{subtract}\ 3\cdot\left(\dfrac{1}{3}\right)^a\ \text{from both sides}\\\\b=1-3\cdot\left(\dfrac{1}{3}\right)^a\qquad(*)[/tex]
[tex]-1=\left(\dfrac{1}{3}\right)^{0+a}+b\\\\-1=\left(\dfrac{1}{3}\right)^a+b\qquad\text{subtract b from both sides}\\\\\left(\dfrac{1}{3}\right)^a=-1-b\\\\\text{substitute to}\ (*)\\\\b=1-3(-1-b)\qquad\text{use distributive property}\\\\b=1+3+3b\qquad\text{subtract 3b from both sides}\\\\-2b=4\qquad\text{divide both sides by (-2)}\\\\\boxed{b=-2}[/tex]
[tex]\text{Susbtitute the value of b to}\ (*):\\\\-2=1-3\left(\dfrac{1}{3}\right)^a\qquad\text{subtract 1 from both sides}\\\\-3=-3\left(\dfrac{1}{3}\right)^a\qquad\text{divide both sides by (-3)}\\\\1=\left(\dfrac{1}{3}\right)^a\\\\\left(\dfrac{1}{3}\right)^0=\left(\dfrac{1}{3}\right)^a\iff a=0\\\\\text{Therefore we have}\ g(x)=\left(\dfrac{1}{3}\right)^x-2\\\\Translate\ the\ graph\ of\ f(x)\ 2\ units\ down.[/tex]
