6.1 Exponential functions  (Page 5/16)

Because we don’t have the initial value, we substitute both points into an equation of the form $f(x)=a{b}^x,$ and then solve the system for $a$ and $b.$

• Substituting $\left(-2,6\right)$ gives $6=a{b}^{-2}$
• Substituting $\left(2,1\right)$ gives $1=a{b}^2$

Use the first equation to solve for $a$ in terms of $b:$

Substitute $a$ in the second equation, and solve for $b:$

Use the value of $b$ in the first equation to solve for the value of $a:$

Thus, the equation is $f(x)=2.4492{(0.6389)}^x.$

We can graph our model to check our work. Notice that the graph in [link] passes through the initial points given in the problem, $\left(-2,\text{ 6}\right)$ and $\left(2,\text{ 1}\right).$ The graph is an example of an exponential decay function.

We can choose the y -intercept of the graph, $\left(0,3\right),$ as our first point. This gives us the initial value, $a=3.$ Next, choose a point on the curve some distance away from $\left(0,3\right)$ that has integer coordinates. One such point is $(2,12).$

Because we restrict ourselves to positive values of $b,$ we will use $b=2.$ Substitute $a$ and $b$ into the standard form to yield the equation $f(x)=3{(2)}^x.$