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k(-1,3)5-2 -14321(0, 1)y1.1, 3) (2, 3)1 2613VXAnalyze the graph of the exponential decay function.The initial value isThe base, or rate of change, isThe domain isD

k(-1,3)5-2 -14321(0, 1)y1.1, 3) (2, 3)1 2613VXAnalyze the graph of the exponential-example-1

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The general expression for an exponential function is the following:


y=ab^x

where b represents the base of the function.

In this case, notice that the graph of the function passes through the points (0,1) and (1,1/3), then, with these points we can find the values of a and b:


\begin{gathered} (0,1): \\ 1=ab⁰\Rightarrow a=1 \\ (1,(1)/(3)): \\ (1)/(3)=b¹\Rightarrow b=(1)/(3) \end{gathered}

therefore, we have that:

the initial value is a = 1

the base is b = 1/3

the domain of the function is all real numbers.

User Tim Barclay
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