The table below displays the enrollment of freshmen taking Algebra 1. Geometry, or Algebra 2.
2 answers:
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Answer:
Recursive: f(1) = 1; f(n) = f(n-1)/3
Equation: f(x) = 5/3·(1/3)^(x-1)
Step-by-step explanation:
You know this is an exponential function because sequential lines in the table have a common ratio of f(x) values:
f(2)/f(1) = (5/9)/(5/3) = 3/9 = 1/3
f(3)/f(2) = (5/27)/(5/9) = 9/27 = 1/3
and it continues like that.
Besides telling you the function is exponential, it also tells you that each term is 1/3 of the previous term.
<h3>Recursive formula</h3>The value of f(1) is read from the table. For x = 1, that value is ...
f(1) = 5/3
As we noticed above, each term is 1/3 the previous term, so the recursive relation is ...
f(n) = (1/3)·f(n-1)
<h3>Equation</h3>The general form of the equation for a geometric (exponential) relation is ...
f(x) = f(1)·r^(x-1) . . . . . . where f(1) is the first term, and r is the common ratio
For this function, we have f(1) = 5/3 and r = 1/3, so the equation is ...
f(x) = 5/3·(1/3)^(x-1)
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<em>Additional comment</em>
When given values of x that count from 1, we often express the equation in terms of (x-1). That equation can be simplified so the exponent is x.
f(x) = 5·(1/3)^x
