SOLUTION: The graph has an equation in the form of f(x) = a(1/2)^b(x+2)+k. What is the actual equation? Please view the graph at: https://docs.google.com/document/d/1Cl3HrcoR6jibNzxfD

Algebra ->  Coordinate-system -> SOLUTION: The graph has an equation in the form of f(x) = a(1/2)^b(x+2)+k. What is the actual equation? Please view the graph at: https://docs.google.com/document/d/1Cl3HrcoR6jibNzxfD      Log On


   

Question 1181009: The graph has an equation in the form of f(x) = a(1/2)^b(x+2)+k. What is the actual equation?

Please view the graph at:
https://docs.google.com/document/d/1Cl3HrcoR6jibNzxfDTIilS3imSplLY8nQUU0OpU9vLA/edit?usp=sharing
Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


Duplicate -- answer repeated below

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f(x) = a(1/2)^b(x+2)+k

NOTE: To make the form of the function absolutely clear, the whole exponent "b(x+2)" should be in parentheses: f(x) = a(1/2)^(b(x+2))+k

f%28x%29+=+a%281%2F2%29%5E%28b%28x%2B2%29%29%2Bk

Use the two given points and the horizontal asymptote to get three equations that can be solved to determine the constants a, b, and k.

(1) horizontal asymptote:
For large values of x, the decaying exponential goes to zero, making the function close to f(x)=k. Since the asymptote is y=4, we have k=4.

(2) f(-2) = 1:
f%28-2%29+=+a%281%2F2%29%5Eb%28-2%2B2%29%2B4
1+=+a%281%2F2%29%5E0%2B4
1+=+a%2B4
a=-3

(3) f(-3) = -8:
f%28-3%29+=+-3%281%2F2%29%5Eb%28-3%2B2%29%2B4
-8+=+-3%281%2F2%29%5Eb%28-1%29%2B4
-12+=+-3%282%5Eb%29
2%5Eb=4
b=2

a=-3; b=2; k=4

The function is

f%28x%29+=+-3%281%2F2%29%5E%282%28x%2B2%29%29%2B4

A graph:

graph%28400%2C400%2C-4%2C4%2C-6%2C6%2C-3%281%2F2%29%5E%282%28x%2B2%29%29%2B4%29