F-LE Two Points Determine an Exponential Function I

The point \((0,2)\) is on the graph, so the equation should be satisfied by \(x=0\) and \(y=2\). This gives \(2=ab^0\). Because \(b^0=1\) we get \(2=a.\)

Now setting \(a= 2\) and substituting \(x=2\), \(y=\frac12\) we get \begin{align}\frac{1}{2} =& 2b^2 \text{ so} \ \frac{1}{4} =& b^2 \text{ so} \ \pm \frac{1}{2} =& b\end{align} The base \(b\) must be positive so \(f(x) = 2\left(\frac{1}{2}\right)^x\).

The value of the function decreases from 2 to \(\frac{1}{2}\) by multiplying 2 twice by \(b\).

$$\begin{align} 2 \times b \times b =& \frac{1}{2} \ 2b^2 =& \frac{1}{2} \ b^2 =& \frac{1}{4} \ b =& \pm \frac{1}{2}. \end{align}$$

The base, \(b\), must be positive so \(b = \frac{1}{2}\).

Substituting the point \((0, 2)\) yields

$$ \begin{align} 2=&a\left(\frac{1}{2}\right)^0 \text{ so} \ 2=&a \cdot 1 \text{ and} \ 2=&a. \end{align} $$

So \(f(x) = 2\left(\frac{1}{2}\right)^x\).