How do I find the function rule for an exponential function?

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July 17th 2007, 08:53 AM
BrettSTaylor

How do I find the function rule for an exponential function?

How do I find the function rule for an exponential function?

The function values are like this:

x- -2 -1 0 1 2

f(x)- 1 2 4 8 16

I know the answer is f(x)= 4(2)to the x power

I don't get how to come up with the function rule.

I'm teaching from the book Glencoe Algebra 1 1997.

The problem was on a supplemental review I printed out from http://www.mcps.k12.md.us/curriculum...B%20review.pdf
and it was problem #2 c.

The answer key was at http://www.mcps.k12.md.us/curriculum...view%20KEY.pdf

I can't see anyplace in the text that I have that explains how to determine the function rule other than to study the data table or look at the graph. I can't find any instruction or practice on this so any help or a link to someplace to learn and practice finding the function rule would be appreciated.

Brett Taylor
Teacher- Tomlinson Adult Education Center
I am certified in math grades 5-9 and mainly teach adult GED students but I have a few credit math students.

Thanks-
July 17th 2007, 09:09 AM
Plato

$f(x)=2^{x+2}$ works, does it not?
July 17th 2007, 09:13 AM
galactus

Just by looking at the data it appears to be something like $4\cdot{2^{x}}$

$4\cdot{2^{-2}}=1; \;\ 4\cdot{2^{-1}}=2, \;\ etc$

That appears to work. I just looked at it. Didn't used any particular method or formula.
July 17th 2007, 11:06 AM
earboth

Quote:

Originally Posted by BrettSTaylor

...
I can't see anyplace in the text that I have that explains how to determine the function rule other than to study the data table or look at the graph. ...

Hello,

maybe the authors of the test are refering to the question 1.1.2.3 where a (very simple) general equation of an exponential function is mentioned. The coordinates of the points must satisfy this equation that means if you plug in the x and y-values the general equation becomes true:

$(-2, 1): 1 = a \cdot b^{-2}$ [1]
$((-1, 2): 2 = a \cdot b^{-1}$ [2]

Divide [2] by [1]:
$\frac{2}{1} = \frac{a \cdot \frac{1}{b}}{a \cdot \frac{1}{b^2}}~ \Longrightarrow ~ b = 2$

Now plug in this value into [1]:

$1 = a \cdot 2^{-2} = \frac{a}{2^2}~ \Longleftrightarrow ~ a = 4$

The function is completely decribed by:

$y = 4 \cdot 2^x$

A more general form of an exponentila function could be:

$f(x) = a \cdot b^{k\cdot x + t} + c$ . That means you have 5 parameters. To determine these 5 values you need at least 5 points with their coordinates. The meaning of these parameters are:
a: A factor which describes the dilation(?) of the graph.
k: A factor which describes the dilation(?) of the graph.
t: Determines the horizontal translation.
c: Determines the vertical translation.