[tex]f(x)=b(a)^x\\\\\text{From the table}\\\\x=1\to y=-2\ and\ x=2\to y=-12.\\\text{Substitute:}\\\\\left\{\begin{array}{ccc}b(a)^1=-2\\b(a)^2=-12\end{array}\right\\\\\left\{\begin{array}{ccc}b(a)=-2\\b(a)(a)=-12\end{array}\right\\\\\text{substitute from the first equation to the second equation}\\\\-2(a)=-12\qquad\text{divide both sides by (-2)}\\\\a=6\\\\\text{Substitute the value of a to the first equation}\\\\b(6)=-2\qquad\text{divide both sides by 6}\\\\b=-\dfrac{2}{6}\to b=-\dfrac{1}{3}[/tex]
[tex]f(x)=-\dfrac{1}{3}(6)^x\\\\\text{check other values of x}\\\\for\ x=3\to f(3)=-\dfrac{1}{3}(6)^3=-\dfrac{1}{3}(216)=-72\qquad CORRECT\\\\for\ x=4\to f(4)=-\dfrac{1}{3}(6)^4=-\dfrac{1}{3}(1296)=-432\qquad CORRECT\\\\Answer:\ YES.\ It's\ an\ exponential\ function\ f(x)=-\dfrac{1}{3}(6)^x[/tex]
