It's easy to see that f(x) = -3^-x fits the bill, but how would you find the answer in general?
Suppose f(x) = ka^-x where k in RR and a in RR, a > 0. This will have the x axis as an asymptote, regardless of the values of the constants k and a.
Then if f(x) > 0:
log(f(x)) = log(ka^-x) = log(k) -x log(a)
or if f(x) < 0:
log (-f(x)) = log(-ka^-x) = log(-k) -x log(a)
In either case,
log(abs(f(x))) = log(abs(k))-x log(a)
=(-log(a))x+log(abs(k))
This is in the form of the equation of a line, with slope -log(a) and intercept log(abs(k)).
Let's try applying this to our given points:
(x_1, y_1) = (0, -1)
(x_2, y_2) = (-1, -3)
(x_3, y_3) = (-2, -9)
(x_1, log(abs(y_1))) = (0, log(1)) = (0, 0)
(x_2, log(abs(y_2))) = (-1, log(3))
(x_3, log(abs(y_3))) = (-2, log(9)) = (-2, 2 log(3))
These three points, lie along a line of slope -log(3) allowing us to deduce that a=3
The intercept, log(abs(k)) = 0 so k = +-1
Since f(x) < 0, k < 0 so k = -1 and f(x) = - 3^-x
