One way the function can be written is as #y=f(x)=a*b^{x}# and the goal is to find #a# and #b#. Since #f(1)=1.5# and #f(-1)=6#, we get the following system of two equations and two unknowns: #1.5=ab#, #6=a/b#. One of the many ways to solve this system is to divide the second equation by the first to get #4=6/1.5=(a/b)/(ab)=b^{-2}# so that #b^{2}=\frac{1}{4}# an #b=\frac{1}{2}#. This then implies that #a=6b=6* 1/2=3# and the answer is #y=3*(1/2)^{x}=3*2^{-x}#.
Another way this type of problem is often solved is to write the function as #y=f(x)=a*e^{kx}# and the goal is to find #a# and #k#. The same points as above give the following system of equations: #1.5=ae^{k}#, #6=ae^{-k}#. Dividing the second equation by the first results in #4=6/1.5=(ae^{-k})/(ae^{k})=e^{-2k}# so that #-2k=ln(4)# and #k=-1/2*ln(4)=ln(4^{-1/2})=ln(0.5)\approx -0.693147#. Then #a=6e^{k}=6e^{ln(0.5)}=6*0.5=3# and the answer can be written in approximate form as #y=f(x)\approx 3e^{-0.693147x}#.
