Because we don’t have the initial value, we substitute both points into an equation of the form
[tex]f(x)=ab^x[/tex]and then solve the system for a and b.
- Substituting (-3,4) gives:
[tex]4=ab^{-3}[/tex]- Substituting (2,3) gives:
[tex]3=ab^2[/tex]Use the first equation to solve for a in terms of b:
[tex]\begin{gathered} 4=ab^{-3} \\ \frac{4}{b^{-3}}=a \\ \frac{4}{\frac{1}{b^3}}=a \\ a=4b^3 \end{gathered}[/tex]Substitute a in the second equation, and solve for b:
[tex]\begin{gathered} 3=ab^2 \\ 3=4b^3b^2 \\ 3=4b^{3+2} \\ 3=4b^5 \\ \frac{3}{4}=b^5 \\ b=\sqrt[5]{\frac{3}{4}}=(\frac{3}{4})^{\frac{1}{5}}=0.944 \end{gathered}[/tex]Use the value of b in the first equation to solve for the value of a:
[tex]a=4b^3=4(0.944)^3=3.365[/tex]Thus, the equation is:
[tex]f(x)=3.365(0.944)^x[/tex]Next, we can graph the function:
Notice that the graph below passes through the initial points given in the problem (-3,4) and (2,3). So, the graph is an exponential decay function.
Answer:
- Equation
[tex]f(x)=3.365(0.944)^x[/tex]- This is an exponential decay function.
