1. The problem is to find the equation of an exponential function in the form $$f(x) = a b^x$$ given the values: $$f(0) = -3, f(1) = -4, f(2) = -\frac{16}{3}, f(3) = -\frac{64}{9}$$ 2. Recall that for an exponential function, when $$x=0$$, $$f(0) = a b^0 = a \cdot 1 = a$$. So, $$a = f(0) = -3$$. 3. Now use the value at $$x=1$$ to find $$b$$: $$f(1) = a b = -4$$ Substitute $$a = -3$$: $$-3 b = -4$$ Divide both sides by $$-3$$: $$b = \frac{-4}{-3} = \frac{4}{3}$$ 4. Verify with $$x=2$$: $$f(2) = a b^2 = -3 \left(\frac{4}{3}\right)^2 = -3 \cdot \frac{16}{9} = -\frac{48}{9} = -\frac{16}{3}$$ This matches the given value. 5. Verify with $$x=3$$: $$f(3) = a b^3 = -3 \left(\frac{4}{3}\right)^3 = -3 \cdot \frac{64}{27} = -\frac{192}{27} = -\frac{64}{9}$$ This also matches the given value. 6. Therefore, the equation is: $$f(x) = -3 \left(\frac{4}{3}\right)^x$$