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 November 11th, 2012, 12:36 PM #1 Newbie   Joined: Nov 2012 Posts: 5 Thanks: 0 write as an equation? The following points are part of a table of values for a function. (0, 2), (1, 4), (2, 10), (3, 2 Represent this function as: a) an equation
November 11th, 2012, 04:49 PM   #2
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Re: write as an equation?

Hello, vivalajuicy!

Quote:
 The following points are part of a table of values for a function: [color=beige]. . [/color]$(0,\,2),\;(1,\,4),\;(2,\,10),\;(3,\,28)$ Represent this function as: (a) an equation

I don't have the equation yet, but I have a recurrence.

[color=beige]. . [/color]$a_{n+1} \;=\;a_n\,+\,2\cdot3^{n-1}\;\;\;a_0\,=\,2$

 November 12th, 2012, 12:34 AM #3 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: write as an equation? y = 1+3^x
November 13th, 2012, 06:33 AM   #4
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Re: write as an equation?

Quote:
 Originally Posted by vivalajuicy The following points are part of a table of values for a function. (0, 2), (1, 4), (2, 10), (3, 2 Represent this function as: a) an equation
Four points determine a cubic function so we can write y= ax^3+ bx^2+ cx+ d. Using the values given,
d= 2,
a+ b+ c+ 2= 4 or a+ b+ c= 2
8a+ 4b+ 2c+ 2= 10 or 8a+ 4b+ 2c= 8, and, dividing by 2, 4a+ 2b+ c= 4.
27a+ 9b+ 3c+ 2= 28 or 27a+ 9b+ 3c= 26

If we subtract a+ b+ c= 2 from 4a+ 2b+ c= 4, we get 3a+ b= 2.
If we multiply a+ b+ c= 2 by 3 to get 3a+ 3b+ 3c= 6 and subtract that from 27a+ 9b+ 3c= 26 we get 24a+ 6b= 20 or 12a+ 3b= 10.

If we multiply 3a+ b= 2 by 3 we get 9a+ 3b= 6 and subtracting that from 12a+ 3b= 10, 3a= 4 so a= 4/3.
Now, you can put that back into 3a+ b= 2 to find b, and put those into one of the original equations to find c.

That is a completely different solution from Hoempa's but you have to understand that there can be an infinite number of functions whose graph will go through four given points.

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